From b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b Mon Sep 17 00:00:00 2001 From: priyanka Date: Wed, 24 Jun 2015 15:03:17 +0530 Subject: initial commit / add all books --- 2705/CH1/EX1.1/Ex1_1.sce | 17 +++++++++ 2705/CH1/EX1.10/Ex1_10.sce | 21 +++++++++++ 2705/CH1/EX1.11/Ex1_11.sce | 17 +++++++++ 2705/CH1/EX1.12/Ex1_12.sce | 25 +++++++++++++ 2705/CH1/EX1.13/Ex1_13.sce | 17 +++++++++ 2705/CH1/EX1.14/Ex1_14.sce | 24 ++++++++++++ 2705/CH1/EX1.2/Ex1_2.sce | 21 +++++++++++ 2705/CH1/EX1.3/Ex1_3.sce | 26 +++++++++++++ 2705/CH1/EX1.4/Ex1_4.sce | 26 +++++++++++++ 2705/CH1/EX1.5/Ex1_5.sce | 19 ++++++++++ 2705/CH1/EX1.6/Ex1_6.sce | 28 ++++++++++++++ 2705/CH1/EX1.7/Ex1_7.sce | 23 ++++++++++++ 2705/CH1/EX1.8/Ex1_8.sce | 22 +++++++++++ 2705/CH1/EX1.9/Ex1_9.sce | 27 ++++++++++++++ 2705/CH10/EX10.1/Ex10_1.sce | 31 +++++++++++++++ 2705/CH10/EX10.10/Ex10_10.sce | 64 +++++++++++++++++++++++++++++++ 2705/CH10/EX10.2/Ex10_2.sce | 47 +++++++++++++++++++++++ 2705/CH10/EX10.3/Ex10_3.sce | 44 ++++++++++++++++++++++ 2705/CH10/EX10.4/Ex10_4.sce | 46 +++++++++++++++++++++++ 2705/CH10/EX10.5/Ex10_5.sce | 31 +++++++++++++++ 2705/CH10/EX10.6/Ex10_6.sce | 49 ++++++++++++++++++++++++ 2705/CH10/EX10.7/Ex10_7.sce | 61 ++++++++++++++++++++++++++++++ 2705/CH10/EX10.8/Ex10_8.sce | 63 +++++++++++++++++++++++++++++++ 2705/CH10/EX10.9/Ex10_9.sce | 48 ++++++++++++++++++++++++ 2705/CH11/EX11.1/Ex11_1.sce | 45 ++++++++++++++++++++++ 2705/CH11/EX11.2/Ex11_2.sce | 51 +++++++++++++++++++++++++ 2705/CH11/EX11.3/Ex11_3.sce | 41 ++++++++++++++++++++ 2705/CH11/EX11.4/Ex11_4.sce | 20 ++++++++++ 2705/CH11/EX11.5/Ex11_5.sce | 52 ++++++++++++++++++++++++++ 2705/CH11/EX11.6/Ex11_6.sce | 45 ++++++++++++++++++++++ 2705/CH11/EX11.7/Ex11_7.sce | 37 ++++++++++++++++++ 2705/CH11/EX11.8/Ex11_8.sce | 54 +++++++++++++++++++++++++++ 2705/CH11/EX11.9/Ex11_9.sce | 48 ++++++++++++++++++++++++ 2705/CH12/EX12.1/Ex12_1.sce | 43 +++++++++++++++++++++ 2705/CH12/EX12.2/Ex12_2.sce | 33 ++++++++++++++++ 2705/CH12/EX12.3/Ex12_3.sce | 47 +++++++++++++++++++++++ 2705/CH12/EX12.4/Ex12_4.sce | 48 ++++++++++++++++++++++++ 2705/CH13/EX13.1/Ex13_1.sce | 24 ++++++++++++ 2705/CH13/EX13.2/Ex13_2.sce | 38 +++++++++++++++++++ 2705/CH13/EX13.3/Ex13_3.sce | 28 ++++++++++++++ 2705/CH13/EX13.4/Ex13_4.sce | 37 ++++++++++++++++++ 2705/CH13/EX13.5/Ex13_5.sce | 36 ++++++++++++++++++ 2705/CH14/EX14.1/Ex14_1.sce | 55 +++++++++++++++++++++++++++ 2705/CH14/EX14.2/Ex14_2.sce | 35 +++++++++++++++++ 2705/CH14/EX14.3/Ex14_3.sce | 59 +++++++++++++++++++++++++++++ 2705/CH14/EX14.4/Ex14_4.sce | 38 +++++++++++++++++++ 2705/CH14/EX14.5/Ex14_5.sce | 28 ++++++++++++++ 2705/CH14/EX14.6/Ex14_6.sce | 21 +++++++++++ 2705/CH14/EX14.7/Ex14_7.sce | 25 +++++++++++++ 2705/CH14/EX14.8/Ex14_8.sce | 36 ++++++++++++++++++ 2705/CH15/EX15.1/Ex15_1.sce | 17 +++++++++ 2705/CH15/EX15.10/Ex15_10.sce | 32 ++++++++++++++++ 2705/CH15/EX15.11/Ex15_11.sce | 37 ++++++++++++++++++ 2705/CH15/EX15.12/Ex15_12.sce | 48 ++++++++++++++++++++++++ 2705/CH15/EX15.13/Ex15_13.sce | 79 +++++++++++++++++++++++++++++++++++++++ 2705/CH15/EX15.14/Ex15_14.sce | 77 ++++++++++++++++++++++++++++++++++++++ 2705/CH15/EX15.15/Ex15_15.sce | 41 ++++++++++++++++++++ 2705/CH15/EX15.16/Ex15_16.sce | 48 ++++++++++++++++++++++++ 2705/CH15/EX15.17/Ex15_17.sce | 36 ++++++++++++++++++ 2705/CH15/EX15.18/Ex15_18.sce | 22 +++++++++++ 2705/CH15/EX15.2/Ex15_2.sce | 29 +++++++++++++++ 2705/CH15/EX15.3/Ex15_3.sce | 69 ++++++++++++++++++++++++++++++++++ 2705/CH15/EX15.4/Ex15_4.sce | 72 +++++++++++++++++++++++++++++++++++ 2705/CH15/EX15.5/Ex15_5.sce | 28 ++++++++++++++ 2705/CH15/EX15.6/Ex15_6.sce | 23 ++++++++++++ 2705/CH15/EX15.7/Ex15_7.sce | 68 +++++++++++++++++++++++++++++++++ 2705/CH15/EX15.8/Ex15_8.sce | 65 ++++++++++++++++++++++++++++++++ 2705/CH15/EX15.9/Ex15_9.sce | 72 +++++++++++++++++++++++++++++++++++ 2705/CH16/EX16.1/Ex16_1.sce | 45 ++++++++++++++++++++++ 2705/CH16/EX16.2/Ex16_2.sce | 47 +++++++++++++++++++++++ 2705/CH16/EX16.3/Ex16_3.sce | 62 ++++++++++++++++++++++++++++++ 2705/CH16/EX16.4/Ex16_4.sce | 68 +++++++++++++++++++++++++++++++++ 2705/CH16/EX16.5/Ex16_5.sce | 51 +++++++++++++++++++++++++ 2705/CH16/EX16.6/Ex16_6.sce | 65 ++++++++++++++++++++++++++++++++ 2705/CH17/EX17.1/Ex17_1.sce | 47 +++++++++++++++++++++++ 2705/CH17/EX17.2/Ex17_2.sce | 87 +++++++++++++++++++++++++++++++++++++++++++ 2705/CH17/EX17.3/Ex17_3.sce | 55 +++++++++++++++++++++++++++ 2705/CH17/EX17.4/Ex17_4.sce | 33 ++++++++++++++++ 2705/CH17/EX17.5/Ex17_5.sce | 61 ++++++++++++++++++++++++++++++ 2705/CH17/EX17.6/Ex17_6.sce | 33 ++++++++++++++++ 2705/CH18/EX18.1/Ex18_1.sce | 61 ++++++++++++++++++++++++++++++ 2705/CH18/EX18.2/Ex18_2.sce | 54 +++++++++++++++++++++++++++ 2705/CH19/EX19.1/Ex19_1.sce | 46 +++++++++++++++++++++++ 2705/CH19/EX19.2/Ex19_2.sce | 34 +++++++++++++++++ 2705/CH19/EX19.3/Ex19_3.sce | 52 ++++++++++++++++++++++++++ 2705/CH19/EX19.4/Ex19_4.sce | 73 ++++++++++++++++++++++++++++++++++++ 2705/CH2/EX2.1/Ex2_1.sce | 26 +++++++++++++ 2705/CH2/EX2.2/Ex2_2.sce | 22 +++++++++++ 2705/CH2/EX2.3/Ex2_3.sce | 23 ++++++++++++ 2705/CH2/EX2.4/Ex2_4.sce | 43 +++++++++++++++++++++ 2705/CH2/EX2.5/Ex2_5.sce | 27 ++++++++++++++ 2705/CH2/EX2.6/Ex2_6.sce | 38 +++++++++++++++++++ 2705/CH4/EX4.1/Ex4_1.sce | 23 ++++++++++++ 2705/CH4/EX4.10/Ex4_10.sce | 38 +++++++++++++++++++ 2705/CH4/EX4.11/Ex4_11.sce | 40 ++++++++++++++++++++ 2705/CH4/EX4.12/Ex4_12.sce | 29 +++++++++++++++ 2705/CH4/EX4.13/Ex4_13.sce | 65 ++++++++++++++++++++++++++++++++ 2705/CH4/EX4.14/Ex4_14.sce | 17 +++++++++ 2705/CH4/EX4.15/Ex4_15.sce | 51 +++++++++++++++++++++++++ 2705/CH4/EX4.16/Ex4_16.sce | 28 ++++++++++++++ 2705/CH4/EX4.17/Ex4_17.sce | 52 ++++++++++++++++++++++++++ 2705/CH4/EX4.18/Ex4_18.sce | 31 +++++++++++++++ 2705/CH4/EX4.19/Ex4_19.sce | 50 +++++++++++++++++++++++++ 2705/CH4/EX4.2/Ex4_2.sce | 42 +++++++++++++++++++++ 2705/CH4/EX4.20/Ex4_20.sce | 76 +++++++++++++++++++++++++++++++++++++ 2705/CH4/EX4.21/Ex4_21.sce | 42 +++++++++++++++++++++ 2705/CH4/EX4.3/Ex4_3.sce | 30 +++++++++++++++ 2705/CH4/EX4.4/Ex4_4.sce | 43 +++++++++++++++++++++ 2705/CH4/EX4.5/Ex4_5.sce | 25 +++++++++++++ 2705/CH4/EX4.8/Ex4_8.sce | 20 ++++++++++ 2705/CH4/EX4.9/Ex4_9.sce | 21 +++++++++++ 2705/CH5/EX5.1/Ex5_1.sce | 23 ++++++++++++ 2705/CH5/EX5.10/Ex5_10.sce | 27 ++++++++++++++ 2705/CH5/EX5.11/Ex5_11.sce | 40 ++++++++++++++++++++ 2705/CH5/EX5.12/Ex5_12.sce | 57 ++++++++++++++++++++++++++++ 2705/CH5/EX5.13/Ex5_13.sce | 35 +++++++++++++++++ 2705/CH5/EX5.14/Ex5_14.sce | 54 +++++++++++++++++++++++++++ 2705/CH5/EX5.15/Ex5_15.sce | 35 +++++++++++++++++ 2705/CH5/EX5.16/Ex5_16.sce | 40 ++++++++++++++++++++ 2705/CH5/EX5.17/Ex5_17.sce | 40 ++++++++++++++++++++ 2705/CH5/EX5.18/Ex5_18.sce | 41 ++++++++++++++++++++ 2705/CH5/EX5.19/Ex5_19.sce | 39 +++++++++++++++++++ 2705/CH5/EX5.2/Ex5_2.sce | 20 ++++++++++ 2705/CH5/EX5.20/Ex5_20.sce | 43 +++++++++++++++++++++ 2705/CH5/EX5.21/Ex5_21.sce | 43 +++++++++++++++++++++ 2705/CH5/EX5.22/Ex5_22.sce | 36 ++++++++++++++++++ 2705/CH5/EX5.23/Ex5_23.sce | 47 +++++++++++++++++++++++ 2705/CH5/EX5.24/Ex5_24.sce | 42 +++++++++++++++++++++ 2705/CH5/EX5.26/Ex5_26.sce | 45 ++++++++++++++++++++++ 2705/CH5/EX5.3/Ex5_3.sce | 20 ++++++++++ 2705/CH5/EX5.4/Ex5_4.sce | 20 ++++++++++ 2705/CH5/EX5.5/Ex5_5.sce | 25 +++++++++++++ 2705/CH5/EX5.6/Ex5_6.sce | 29 +++++++++++++++ 2705/CH5/EX5.7/Ex5_7.sce | 28 ++++++++++++++ 2705/CH5/EX5.8/Ex5_8.sce | 33 ++++++++++++++++ 2705/CH5/EX5.9/Ex5_9.sce | 22 +++++++++++ 2705/CH7/EX7.1/Ex7_1.sce | 25 +++++++++++++ 2705/CH7/EX7.2/Ex7_2.sce | 33 ++++++++++++++++ 2705/CH7/EX7.3/Ex7_3.sce | 30 +++++++++++++++ 2705/CH7/EX7.4/Ex7_4.sce | 30 +++++++++++++++ 2705/CH7/EX7.5/Ex7_5.sce | 53 ++++++++++++++++++++++++++ 2705/CH7/EX7.6/Ex7_6.sce | 67 +++++++++++++++++++++++++++++++++ 2705/CH7/EX7.7/Ex7_7.sce | 45 ++++++++++++++++++++++ 2705/CH7/EX7.8/Ex7_8.sce | 39 +++++++++++++++++++ 2705/CH7/EX7.9/Ex7_9.sce | 21 +++++++++++ 2705/CH8/EX8.1/Ex8_1.sce | 24 ++++++++++++ 2705/CH8/EX8.10/Ex8_10.sce | 44 ++++++++++++++++++++++ 2705/CH8/EX8.11/Ex8_11.sce | 33 ++++++++++++++++ 2705/CH8/EX8.12/Ex8_12.sce | 49 ++++++++++++++++++++++++ 2705/CH8/EX8.13/Ex8_13.sce | 46 +++++++++++++++++++++++ 2705/CH8/EX8.14/Ex8_14.sce | 40 ++++++++++++++++++++ 2705/CH8/EX8.15/Ex8_15.sce | 70 ++++++++++++++++++++++++++++++++++ 2705/CH8/EX8.16/Ex8_16.sce | 40 ++++++++++++++++++++ 2705/CH8/EX8.17/Ex8_17.sce | 48 ++++++++++++++++++++++++ 2705/CH8/EX8.18/Ex8_18.sce | 65 ++++++++++++++++++++++++++++++++ 2705/CH8/EX8.19/Ex8_19.sce | 51 +++++++++++++++++++++++++ 2705/CH8/EX8.20/Ex8_20.sce | 19 ++++++++++ 2705/CH8/EX8.22/Ex8_22.sce | 53 ++++++++++++++++++++++++++ 2705/CH8/EX8.23/Ex8_23.sce | 54 +++++++++++++++++++++++++++ 2705/CH8/EX8.24/Ex8_24.sce | 38 +++++++++++++++++++ 2705/CH8/EX8.3/Ex8_3.sce | 37 ++++++++++++++++++ 2705/CH8/EX8.4/Ex8_4.sce | 25 +++++++++++++ 2705/CH8/EX8.5/Ex8_5.sce | 21 +++++++++++ 2705/CH8/EX8.6/Ex8_6.sce | 37 ++++++++++++++++++ 2705/CH8/EX8.7/Ex8_7.sce | 25 +++++++++++++ 2705/CH8/EX8.8/Ex8_8.sce | 31 +++++++++++++++ 2705/CH8/EX8.9/Ex8_9.sce | 44 ++++++++++++++++++++++ 2705/CH9/EX9.1/Ex9_1.sce | 32 ++++++++++++++++ 2705/CH9/EX9.2/Ex9_2.sce | 35 +++++++++++++++++ 2705/CH9/EX9.3/Ex9_3.sce | 41 ++++++++++++++++++++ 2705/CH9/EX9.4/Ex9_4.sce | 22 +++++++++++ 2705/CH9/EX9.5/Ex9_5.sce | 34 +++++++++++++++++ 2705/CH9/EX9.6/Ex9_6.sce | 34 +++++++++++++++++ 2705/CH9/EX9.7/Ex9_7.sce | 44 ++++++++++++++++++++++ 174 files changed, 6923 insertions(+) create mode 100755 2705/CH1/EX1.1/Ex1_1.sce create mode 100755 2705/CH1/EX1.10/Ex1_10.sce create mode 100755 2705/CH1/EX1.11/Ex1_11.sce create mode 100755 2705/CH1/EX1.12/Ex1_12.sce create mode 100755 2705/CH1/EX1.13/Ex1_13.sce create mode 100755 2705/CH1/EX1.14/Ex1_14.sce create mode 100755 2705/CH1/EX1.2/Ex1_2.sce create mode 100755 2705/CH1/EX1.3/Ex1_3.sce create mode 100755 2705/CH1/EX1.4/Ex1_4.sce create mode 100755 2705/CH1/EX1.5/Ex1_5.sce create mode 100755 2705/CH1/EX1.6/Ex1_6.sce create mode 100755 2705/CH1/EX1.7/Ex1_7.sce create mode 100755 2705/CH1/EX1.8/Ex1_8.sce create mode 100755 2705/CH1/EX1.9/Ex1_9.sce create mode 100755 2705/CH10/EX10.1/Ex10_1.sce create mode 100755 2705/CH10/EX10.10/Ex10_10.sce create mode 100755 2705/CH10/EX10.2/Ex10_2.sce create mode 100755 2705/CH10/EX10.3/Ex10_3.sce create mode 100755 2705/CH10/EX10.4/Ex10_4.sce create mode 100755 2705/CH10/EX10.5/Ex10_5.sce create mode 100755 2705/CH10/EX10.6/Ex10_6.sce create mode 100755 2705/CH10/EX10.7/Ex10_7.sce create mode 100755 2705/CH10/EX10.8/Ex10_8.sce create mode 100755 2705/CH10/EX10.9/Ex10_9.sce create mode 100755 2705/CH11/EX11.1/Ex11_1.sce create mode 100755 2705/CH11/EX11.2/Ex11_2.sce create mode 100755 2705/CH11/EX11.3/Ex11_3.sce create mode 100755 2705/CH11/EX11.4/Ex11_4.sce create mode 100755 2705/CH11/EX11.5/Ex11_5.sce create mode 100755 2705/CH11/EX11.6/Ex11_6.sce create mode 100755 2705/CH11/EX11.7/Ex11_7.sce create mode 100755 2705/CH11/EX11.8/Ex11_8.sce create mode 100755 2705/CH11/EX11.9/Ex11_9.sce create mode 100755 2705/CH12/EX12.1/Ex12_1.sce create mode 100755 2705/CH12/EX12.2/Ex12_2.sce create mode 100755 2705/CH12/EX12.3/Ex12_3.sce create mode 100755 2705/CH12/EX12.4/Ex12_4.sce create mode 100755 2705/CH13/EX13.1/Ex13_1.sce create mode 100755 2705/CH13/EX13.2/Ex13_2.sce create mode 100755 2705/CH13/EX13.3/Ex13_3.sce create mode 100755 2705/CH13/EX13.4/Ex13_4.sce create mode 100755 2705/CH13/EX13.5/Ex13_5.sce create mode 100755 2705/CH14/EX14.1/Ex14_1.sce create mode 100755 2705/CH14/EX14.2/Ex14_2.sce create mode 100755 2705/CH14/EX14.3/Ex14_3.sce create mode 100755 2705/CH14/EX14.4/Ex14_4.sce create mode 100755 2705/CH14/EX14.5/Ex14_5.sce create mode 100755 2705/CH14/EX14.6/Ex14_6.sce create mode 100755 2705/CH14/EX14.7/Ex14_7.sce create mode 100755 2705/CH14/EX14.8/Ex14_8.sce create mode 100755 2705/CH15/EX15.1/Ex15_1.sce create mode 100755 2705/CH15/EX15.10/Ex15_10.sce create mode 100755 2705/CH15/EX15.11/Ex15_11.sce create mode 100755 2705/CH15/EX15.12/Ex15_12.sce create mode 100755 2705/CH15/EX15.13/Ex15_13.sce create mode 100755 2705/CH15/EX15.14/Ex15_14.sce create mode 100755 2705/CH15/EX15.15/Ex15_15.sce create mode 100755 2705/CH15/EX15.16/Ex15_16.sce create mode 100755 2705/CH15/EX15.17/Ex15_17.sce create mode 100755 2705/CH15/EX15.18/Ex15_18.sce create mode 100755 2705/CH15/EX15.2/Ex15_2.sce create mode 100755 2705/CH15/EX15.3/Ex15_3.sce create mode 100755 2705/CH15/EX15.4/Ex15_4.sce create mode 100755 2705/CH15/EX15.5/Ex15_5.sce create mode 100755 2705/CH15/EX15.6/Ex15_6.sce create mode 100755 2705/CH15/EX15.7/Ex15_7.sce create mode 100755 2705/CH15/EX15.8/Ex15_8.sce create mode 100755 2705/CH15/EX15.9/Ex15_9.sce create mode 100755 2705/CH16/EX16.1/Ex16_1.sce create mode 100755 2705/CH16/EX16.2/Ex16_2.sce create mode 100755 2705/CH16/EX16.3/Ex16_3.sce create mode 100755 2705/CH16/EX16.4/Ex16_4.sce create mode 100755 2705/CH16/EX16.5/Ex16_5.sce create mode 100755 2705/CH16/EX16.6/Ex16_6.sce create mode 100755 2705/CH17/EX17.1/Ex17_1.sce create mode 100755 2705/CH17/EX17.2/Ex17_2.sce create mode 100755 2705/CH17/EX17.3/Ex17_3.sce create mode 100755 2705/CH17/EX17.4/Ex17_4.sce create mode 100755 2705/CH17/EX17.5/Ex17_5.sce create mode 100755 2705/CH17/EX17.6/Ex17_6.sce create mode 100755 2705/CH18/EX18.1/Ex18_1.sce create mode 100755 2705/CH18/EX18.2/Ex18_2.sce create mode 100755 2705/CH19/EX19.1/Ex19_1.sce create mode 100755 2705/CH19/EX19.2/Ex19_2.sce create mode 100755 2705/CH19/EX19.3/Ex19_3.sce create mode 100755 2705/CH19/EX19.4/Ex19_4.sce create mode 100755 2705/CH2/EX2.1/Ex2_1.sce create mode 100755 2705/CH2/EX2.2/Ex2_2.sce create mode 100755 2705/CH2/EX2.3/Ex2_3.sce create mode 100755 2705/CH2/EX2.4/Ex2_4.sce create mode 100755 2705/CH2/EX2.5/Ex2_5.sce create mode 100755 2705/CH2/EX2.6/Ex2_6.sce create mode 100755 2705/CH4/EX4.1/Ex4_1.sce create mode 100755 2705/CH4/EX4.10/Ex4_10.sce create mode 100755 2705/CH4/EX4.11/Ex4_11.sce create mode 100755 2705/CH4/EX4.12/Ex4_12.sce create mode 100755 2705/CH4/EX4.13/Ex4_13.sce create mode 100755 2705/CH4/EX4.14/Ex4_14.sce create mode 100755 2705/CH4/EX4.15/Ex4_15.sce create mode 100755 2705/CH4/EX4.16/Ex4_16.sce create mode 100755 2705/CH4/EX4.17/Ex4_17.sce create mode 100755 2705/CH4/EX4.18/Ex4_18.sce create mode 100755 2705/CH4/EX4.19/Ex4_19.sce create mode 100755 2705/CH4/EX4.2/Ex4_2.sce create mode 100755 2705/CH4/EX4.20/Ex4_20.sce create mode 100755 2705/CH4/EX4.21/Ex4_21.sce create mode 100755 2705/CH4/EX4.3/Ex4_3.sce create mode 100755 2705/CH4/EX4.4/Ex4_4.sce create mode 100755 2705/CH4/EX4.5/Ex4_5.sce create mode 100755 2705/CH4/EX4.8/Ex4_8.sce create mode 100755 2705/CH4/EX4.9/Ex4_9.sce create mode 100755 2705/CH5/EX5.1/Ex5_1.sce create mode 100755 2705/CH5/EX5.10/Ex5_10.sce create mode 100755 2705/CH5/EX5.11/Ex5_11.sce create mode 100755 2705/CH5/EX5.12/Ex5_12.sce create mode 100755 2705/CH5/EX5.13/Ex5_13.sce create mode 100755 2705/CH5/EX5.14/Ex5_14.sce create mode 100755 2705/CH5/EX5.15/Ex5_15.sce create mode 100755 2705/CH5/EX5.16/Ex5_16.sce create mode 100755 2705/CH5/EX5.17/Ex5_17.sce create mode 100755 2705/CH5/EX5.18/Ex5_18.sce create mode 100755 2705/CH5/EX5.19/Ex5_19.sce create mode 100755 2705/CH5/EX5.2/Ex5_2.sce create mode 100755 2705/CH5/EX5.20/Ex5_20.sce create mode 100755 2705/CH5/EX5.21/Ex5_21.sce create mode 100755 2705/CH5/EX5.22/Ex5_22.sce create mode 100755 2705/CH5/EX5.23/Ex5_23.sce create mode 100755 2705/CH5/EX5.24/Ex5_24.sce create mode 100755 2705/CH5/EX5.26/Ex5_26.sce create mode 100755 2705/CH5/EX5.3/Ex5_3.sce create mode 100755 2705/CH5/EX5.4/Ex5_4.sce create mode 100755 2705/CH5/EX5.5/Ex5_5.sce create mode 100755 2705/CH5/EX5.6/Ex5_6.sce create mode 100755 2705/CH5/EX5.7/Ex5_7.sce create mode 100755 2705/CH5/EX5.8/Ex5_8.sce create mode 100755 2705/CH5/EX5.9/Ex5_9.sce create mode 100755 2705/CH7/EX7.1/Ex7_1.sce create mode 100755 2705/CH7/EX7.2/Ex7_2.sce create mode 100755 2705/CH7/EX7.3/Ex7_3.sce create mode 100755 2705/CH7/EX7.4/Ex7_4.sce create mode 100755 2705/CH7/EX7.5/Ex7_5.sce create mode 100755 2705/CH7/EX7.6/Ex7_6.sce create mode 100755 2705/CH7/EX7.7/Ex7_7.sce create mode 100755 2705/CH7/EX7.8/Ex7_8.sce create mode 100755 2705/CH7/EX7.9/Ex7_9.sce create mode 100755 2705/CH8/EX8.1/Ex8_1.sce create mode 100755 2705/CH8/EX8.10/Ex8_10.sce create mode 100755 2705/CH8/EX8.11/Ex8_11.sce create mode 100755 2705/CH8/EX8.12/Ex8_12.sce create mode 100755 2705/CH8/EX8.13/Ex8_13.sce create mode 100755 2705/CH8/EX8.14/Ex8_14.sce create mode 100755 2705/CH8/EX8.15/Ex8_15.sce create mode 100755 2705/CH8/EX8.16/Ex8_16.sce create mode 100755 2705/CH8/EX8.17/Ex8_17.sce create mode 100755 2705/CH8/EX8.18/Ex8_18.sce create mode 100755 2705/CH8/EX8.19/Ex8_19.sce create mode 100755 2705/CH8/EX8.20/Ex8_20.sce create mode 100755 2705/CH8/EX8.22/Ex8_22.sce create mode 100755 2705/CH8/EX8.23/Ex8_23.sce create mode 100755 2705/CH8/EX8.24/Ex8_24.sce create mode 100755 2705/CH8/EX8.3/Ex8_3.sce create mode 100755 2705/CH8/EX8.4/Ex8_4.sce create mode 100755 2705/CH8/EX8.5/Ex8_5.sce create mode 100755 2705/CH8/EX8.6/Ex8_6.sce create mode 100755 2705/CH8/EX8.7/Ex8_7.sce create mode 100755 2705/CH8/EX8.8/Ex8_8.sce create mode 100755 2705/CH8/EX8.9/Ex8_9.sce create mode 100755 2705/CH9/EX9.1/Ex9_1.sce create mode 100755 2705/CH9/EX9.2/Ex9_2.sce create mode 100755 2705/CH9/EX9.3/Ex9_3.sce create mode 100755 2705/CH9/EX9.4/Ex9_4.sce create mode 100755 2705/CH9/EX9.5/Ex9_5.sce create mode 100755 2705/CH9/EX9.6/Ex9_6.sce create mode 100755 2705/CH9/EX9.7/Ex9_7.sce (limited to '2705') diff --git a/2705/CH1/EX1.1/Ex1_1.sce b/2705/CH1/EX1.1/Ex1_1.sce new file mode 100755 index 000000000..e41b309e6 --- /dev/null +++ b/2705/CH1/EX1.1/Ex1_1.sce @@ -0,0 +1,17 @@ +clear ; +clc; +disp('Example 1.1'); + + + +// Given values +P = 700; //pressure,[kN/m^2] +V1 = .28; //initial volume,[m^3] +V2 = 1.68; //final volume,[m^3] + +//solution + +W = P*(V2-V1);// // Formula for work done at constant pressure is, [kJ] +mprintf('\n The Work done is = %f MJ\n',W*10^-3); + +//End diff --git a/2705/CH1/EX1.10/Ex1_10.sce b/2705/CH1/EX1.10/Ex1_10.sce new file mode 100755 index 000000000..801ac3a6f --- /dev/null +++ b/2705/CH1/EX1.10/Ex1_10.sce @@ -0,0 +1,21 @@ +clear; +clc; +disp('Example 1.10'); + + + +// Given values +m_dot = 3.045; // use of coal, [tonne/h] +c = 28; // calorific value of the coal, [MJ/kg] +P_out = 4.1; // output of turbine, [MW] + +// solution +m_dot = m_dot*10^3/3600; // [kg/s] + +P_in = m_dot*c; // power input by coal, [MW] + +n = P_out/P_in; // thermal efficiency formula + +mprintf('\n Thermal efficiency of the plant is = %f \n',n); + +//End diff --git a/2705/CH1/EX1.11/Ex1_11.sce b/2705/CH1/EX1.11/Ex1_11.sce new file mode 100755 index 000000000..de23f795c --- /dev/null +++ b/2705/CH1/EX1.11/Ex1_11.sce @@ -0,0 +1,17 @@ +clear; +clc; +disp('Example 1.11'); + + +// Given values +v = 50; // speed, [km/h] +F = 900; // Resistance to the motion of a car + +// solution +v = v*10^3/3600; // [m/s] + Power = F*v; // Power formula, [W] + +mprintf('\n The power output of the engine is = %f kW\n',Power*10^-3); + + // End + diff --git a/2705/CH1/EX1.12/Ex1_12.sce b/2705/CH1/EX1.12/Ex1_12.sce new file mode 100755 index 000000000..426f6c26d --- /dev/null +++ b/2705/CH1/EX1.12/Ex1_12.sce @@ -0,0 +1,25 @@ + +clear; +clc; +disp('Example 1.12'); + + + +// Given values +V = 230; // volatage, [volts] +I = 60; // current, [amps] +n_gen = .95; // efficiency of generator +n_eng = .92; // efficiency of engine + +// solution + +P_gen = V*I; // Power delivered by generator, [W] +P_gen=P_gen*10^-3; // [kW] + +P_in_eng=P_gen/n_gen;//Power input from engine,[kW] + +P_out_eng=P_in_eng/n_eng;//Power output from engine,[kW] + +mprintf('\n The power output from the engine is = %f kW\n',P_out_eng); + +// End diff --git a/2705/CH1/EX1.13/Ex1_13.sce b/2705/CH1/EX1.13/Ex1_13.sce new file mode 100755 index 000000000..c91fa71a0 --- /dev/null +++ b/2705/CH1/EX1.13/Ex1_13.sce @@ -0,0 +1,17 @@ +clear; +clc; +disp('Example 1.13'); + + + +// Given values +V = 230; // Voltage, [volts] +W = 4; // Power of heater, [kW] + +// solution + +// using equation P=VI +I = W/V; // current, [K amps] +mprintf('\n The current taken by heater is = %f amps \n',I*10^3); + +// End diff --git a/2705/CH1/EX1.14/Ex1_14.sce b/2705/CH1/EX1.14/Ex1_14.sce new file mode 100755 index 000000000..857faf8fc --- /dev/null +++ b/2705/CH1/EX1.14/Ex1_14.sce @@ -0,0 +1,24 @@ +clear; +clc; +disp('Example 1.14'); + + + +// Given values +P_out = 500; // output of power station, [MW] +c = 29.5; // calorific value of coal, [MJ/kg] +r=.28; + +// solution + +// since P represents only 28 percent of energy available from coal +P_coal = P_out/r; // [MW] + +m_coal = P_coal/c; // Mass of coal used, [kg/s] +m_coal = m_coal*3600; // [kg/h] + +//After one hour +m_coal = m_coal*1*10^-3; // [tonne] +mprintf('\n Mass of coal burnt by the power station in 1 hour is = %f tonne \n',m_coal); + +// End diff --git a/2705/CH1/EX1.2/Ex1_2.sce b/2705/CH1/EX1.2/Ex1_2.sce new file mode 100755 index 000000000..dab25d43a --- /dev/null +++ b/2705/CH1/EX1.2/Ex1_2.sce @@ -0,0 +1,21 @@ +clear; +clc; +disp('Example 1.2'); + + + +//Given values +P1 = 138; // initial pressure,[kN/m^2] +V1 = .112; //initial volume,[m^3] +P2 = 690; // final pressure,[kN/m^2] +Gama=1.4; // heat capacity ratio + +// solution + +// since gas is following, PV^1.4=constant,hence + +V2 =V1*(P1/P2)^(1/Gama); // final volume, [m^3] + +mprintf('\n The new volume of the gas is = %f m^3\n',V2) + +//End diff --git a/2705/CH1/EX1.3/Ex1_3.sce b/2705/CH1/EX1.3/Ex1_3.sce new file mode 100755 index 000000000..bddd96cd3 --- /dev/null +++ b/2705/CH1/EX1.3/Ex1_3.sce @@ -0,0 +1,26 @@ +clear; +clc; +disp('Example 1.3'); + + + +// Given values +P1 = 2070; // initial pressure, [kN/m^2] +V1 = .014; // initial volume, [m^3] +P2 = 207; // final pressure, [kN/m^2] +n=1.35; // polytropic index + +// solution + +// since gas is following PV^n=constant +// hence + +V2 = V1*(P1/P2)^(1/n); // final volume, [m^3] + +// calculation of workdone + +W=(P1*V1-P2*V2)/(1.35-1); // using work done formula for polytropic process, [kJ] + +mprintf('\n The Work done by gas during expansion is = %f kJ\n',W); + +//End diff --git a/2705/CH1/EX1.4/Ex1_4.sce b/2705/CH1/EX1.4/Ex1_4.sce new file mode 100755 index 000000000..6ae65d27c --- /dev/null +++ b/2705/CH1/EX1.4/Ex1_4.sce @@ -0,0 +1,26 @@ +clear; +clc; +disp('Example 1.4'); + + + +// Given values +P1 = 100; // initial pressure, [kN/m^2] +V1 = .056; // initial volume, [m^3] +V2 = .007; // final volume, [m^3] + +// To know P2 +// since process is hyperbolic so, PV=constant +// hence + +P2 = P1*V1/V2; // final pressure, [kN/m^2] + +mprintf('\n The final pressure is = %f kN/m^2\n',P2); + +// calculation of workdone + +W = P1*V1*log(V2/V1); // formula for work done in this process, [kJ] + +mprintf('\n Work done on the gas is = %f kJ\n',W); + +//End diff --git a/2705/CH1/EX1.5/Ex1_5.sce b/2705/CH1/EX1.5/Ex1_5.sce new file mode 100755 index 000000000..2e651d7e1 --- /dev/null +++ b/2705/CH1/EX1.5/Ex1_5.sce @@ -0,0 +1,19 @@ +clear; +clc; +disp('Example 1.5'); + + + +// Given values +m = 5; // mass, [kg] +t1 = 15; // inital temperature, [C] +t2 = 100; // final temperature, [C] +c = 450; // specific heat capacity, [J/kg K] + +// solution + +// using heat transfer equation,[1] +Q = m*c*(t2-t1); // [J] +mprintf('\n The heat required is = %f kJ\n',Q*10^-3); + +//End diff --git a/2705/CH1/EX1.6/Ex1_6.sce b/2705/CH1/EX1.6/Ex1_6.sce new file mode 100755 index 000000000..fe3ceb552 --- /dev/null +++ b/2705/CH1/EX1.6/Ex1_6.sce @@ -0,0 +1,28 @@ +clear; +clc; +disp('Example 1.6'); + + +// Given values +m_cop = 2; // mass of copper vessel, [kg] +m_wat = 6; // mass of water, [kg] +c_wat = 4.19; // specific heat capacity of water, [kJ/kg K] + +t1 = 20; // initial temperature, [C] +t2 = 90; // final temperature, [C] + +// From the table of average specific heat capacities +c_cop = .390; // specific heat capacity of copper,[kJ/kg k] + +// solution +Q_cop = m_cop*c_cop*(t2-t1); // heat required by copper vessel, [kJ] + +Q_wat = m_wat*c_wat*(t2-t1); // heat required by water, [kJ] + +// since there is no heat loss,so total heat transfer is sum of both +Q_total = Q_cop+Q_wat ; // [kJ] + +mprintf(' \n Required heat transfer to accomplish the change = %f kJ\n',Q_total); + +//End + diff --git a/2705/CH1/EX1.7/Ex1_7.sce b/2705/CH1/EX1.7/Ex1_7.sce new file mode 100755 index 000000000..5fb3d347d --- /dev/null +++ b/2705/CH1/EX1.7/Ex1_7.sce @@ -0,0 +1,23 @@ + +clear; +clc; +disp('Example 1.7'); + + +// Given values +m = 10; // mass of iron casting, [kg] +t1 = 200; // initial temperature, [C] +Q = -715.5; // [kJ], since heat is lost in this process + +// From the table of average specific heat capacities +c = .50; // specific heat capacity of casting iron, [kJ/kg K] + +// solution +// using heat equation +// Q = m*c*(t2-t1) + +t2 = t1+Q/(m*c); // [C] + +mprintf('\n The final temperature is t2 = %f C\n',t2); + +// End diff --git a/2705/CH1/EX1.8/Ex1_8.sce b/2705/CH1/EX1.8/Ex1_8.sce new file mode 100755 index 000000000..6febbea57 --- /dev/null +++ b/2705/CH1/EX1.8/Ex1_8.sce @@ -0,0 +1,22 @@ +clear; +clc; +disp('Example 1.8'); + + + +// Given values +m = 4; // mass of the liquid, [kg] +t1 = 15; // initial temperature, [C] +t2 = 100; // final temperature, [C] +Q = 714; // [kJ],required heat to accomplish this change + +// solution +// using heat equation +// Q=m*c*(t2-t1) + +// calculation of c +c=Q/(m*(t2-t1)); // heat capacity, [kJ/kg K] + +mprintf('\n The specific heat capacity of the liquid is c = %f kJ/kg K\n',c); + +//End diff --git a/2705/CH1/EX1.9/Ex1_9.sce b/2705/CH1/EX1.9/Ex1_9.sce new file mode 100755 index 000000000..a5cc15bad --- /dev/null +++ b/2705/CH1/EX1.9/Ex1_9.sce @@ -0,0 +1,27 @@ +clear; +clc; +disp('Example 1.9'); + + +// Given values +m_dot = 20.4; // mass flowrate of petrol, [kg/h] +c = 43; // calorific value of petrol, [MJ/kg] +n = .2; // Thermal efficiency of engine + +// solution +m_dot = 20.4/3600; // [kg/s] +c = 43*10^6; // [J/kg] + +// power output +P_out = n*m_dot*c; // [W] + +mprintf('\n The power output of the engine is = %f kJ\n',P_out*10^-3); + +// power rejected + +P_rej = m_dot*c*(1-n); // [W] +P_rej = P_rej*60*10^-6; // [MJ/min] + +mprintf('\n The energy rejected by the engine is = %f MJ/min \n',P_rej); + +//End diff --git a/2705/CH10/EX10.1/Ex10_1.sce b/2705/CH10/EX10.1/Ex10_1.sce new file mode 100755 index 000000000..563d4d450 --- /dev/null +++ b/2705/CH10/EX10.1/Ex10_1.sce @@ -0,0 +1,31 @@ +clear; +clc; +disp('Example 10.1'); + +// aim : To determine +// the equivalent evaporation + +// Given +P = 1.4;// [MN/m^2] +m = 8;// mass of water,[kg] +T1 = 39;// entering temperature,[C] +T2 = 100;// [C] +x = .95;//dryness fraction + +// solution +hf = 830.1;// [kJ/kg] +hfg = 1957.7;// [kJ/kg] +// steam is wet so specific enthalpy of steam is +h = hf+x*hfg;// [kJ/kg] + +// at 39 C +h1 = 163.4;// [kJ/kg] +// hence +q = h-h1;// [kJ/kg] +Q = m*q;// [kJ] + +evap = Q/2256.9;// equivalent evaporation[kg steam/(kg coal)] + +mprintf('\n The equivalent evaporation, from and at 100 C is = %f kg steam/kg coal\n ',evap); + +// End diff --git a/2705/CH10/EX10.10/Ex10_10.sce b/2705/CH10/EX10.10/Ex10_10.sce new file mode 100755 index 000000000..c0d346f7f --- /dev/null +++ b/2705/CH10/EX10.10/Ex10_10.sce @@ -0,0 +1,64 @@ +clear; +clc; +disp('Example 10.10'); + +// aim : To determine +// (a) the mass of steam bled to each feed heater in kg/kg of supply steam +// (b) the thermal efficiency of the arrangement + +// given values +P1 = 7;// steam initial pressure, [MN/m^2] +T1 = 273+500;// steam initil temperature, [K] +P2 = 2;// pressure at stage 1, [MN/m^2] +P3 = .5;// pressure at stage 2, [MN/m^2] +P4 = .05;// condenser pressure,[MN/m^2] +SE = .82;// stage efficiency of turbine + +// solution +// from the enthalpy-entropy chart(Fig10.23) values of specific enthalpies are +h1 = 3410;// [kJ/kg] +h2_prim = 3045;// [kJ/kg] +// h1-h2=SE*(h1-h2_prim), so +h2 = h1-SE*(h1-h2_prim);// [kJ/kg] + +h3_prim = 2790;// [kJ/kg] +// h2-h3=SE*(h2-h3_prim), so +h3 = h2-SE*(h2-h3_prim);// [kJ/kg] + +h4_prim = 2450;// [kJ/kg] +// h3-h4 = SE*(h3-h4_prim), so +h4 = h3-SE*(h3-h4_prim);// [kJ/kg] + +// from steam table +// @ 2 MN/m^2 +hf2 = 908.6;// [kJ/kg] +// @ .5 MN/m^2 +hf3 = 640.1;// [kJ/kg] +// @ .05 MN/m^2 +hf4 = 340.6;// [kJ/kg] + +// (a) +// for feed heater1 +m1 = (hf2-hf3)/(h2-hf3);// mass of bled steam, [kg/kg supplied steam] +// for feed heater2 +m2 = (1-m1)*(hf3-hf4)/(h3-hf4);// +mprintf('\n (a) The mass of steam bled in feed heater 1 is = %f kg/kg supply steam\n',m1); +mprintf('\n The mass of steam bled in feed heater 2 is = %f kg/kg supply steam\n',m2); + +// (b) +W = (h1-h2)+(1-m1)*(h2-h3)+(1-m1-m2)*(h3-h4);// theoretical work done, [kJ/kg] +Eb = h1-hf2;// energy input in the boiler, [kJ/kg] +TE1 = W/Eb;// thermal efficiency +mprintf('\n (b) The thermal efficiency of the arrangement is = %f percent\n',TE1*100); + +// If there is no feed heating +hf5 = hf4; +h5_prim = 2370;// [kJ/kg] +// h1-h5 = SE*(h1-h5_prim), so +h5 = h1-SE*(h1-h5_prim);// [kJ/kg] +Ei = h1-hf5;//energy input, [kJ/kg] +W = h1-h5;// theoretical work, [kJ/kg] +TE2 = W/Ei;// thermal efficiency +mprintf('\n The thermal efficiency if there is no feed heating is = %f percent\n',TE2*100); + +// End diff --git a/2705/CH10/EX10.2/Ex10_2.sce b/2705/CH10/EX10.2/Ex10_2.sce new file mode 100755 index 000000000..7993e77ac --- /dev/null +++ b/2705/CH10/EX10.2/Ex10_2.sce @@ -0,0 +1,47 @@ +clear; +clc; +disp('Example 10.2'); + +// aim : To determine +// the mass of oil used per hour and the fraction of enthalpy drop through the turbine +// heat transfer available per kilogram of exhaust steam + +// Given values +ms_dot = 5000;// generation of steam, [kg/h] +P1 = 1.8;// generated steam pressure, [MN/m^2] +T1 = 273+325;// generated steam temperature, [K] +Tf = 273+49.4;// feed temperature, [K] +neta = .8;// efficiency of boiler plant +c = 45500;// calorific value, [kJ/kg] +P = 500;// turbine generated power, [kW] +Pt = .18;// turbine exhaust pressure, [MN/m^2] +x = .98;// dryness farction of steam + +// solution +// using steam table at 1.8 MN/m^2 +hf1 = 3106;// [kJ/kg] +hg1 = 3080;// [kJ/kg] +// so +h1 = hf1-neta*(hf1-hg1);// [kJ/kg] +// again using steam table specific enthalpy of feed water is +hwf = 206.9;// [kJ/kg] +h_rais = ms_dot*(h1-hwf);// energy to raise steam, [kJ] + +h_fue = h_rais/neta;// energy from fuel per hour, [kJ] +m_oil = h_fue/c;// mass of fuel per hour, [kg] + +// from steam table at exhaust +hf = 490.7;// [kJ/kg] +hfg = 2210.8;// [kJ/kg] +// hence +h = hf+x*hfg;// [kJ/kg] +// now +h_drop = (h1-h)*ms_dot/3600;// specific enthalpy drop in turbine [kJ] +f = P/h_drop;// fraction ofenthalpy drop converted into work +// heat transfer available in exhaust is +Q = h-hwf;// [kJ/kg] +mprintf('\n The mass of oil used per hour is = %f kg\n',m_oil); +mprintf('\n The fraction of the enthalpy drop through the turbine that is converted into useful work is = %f\n',f); +mprintf('\n The heat transfer available in exhaust steam above 49.4 C is = %f kJ/kg\n',Q); + +// End diff --git a/2705/CH10/EX10.3/Ex10_3.sce b/2705/CH10/EX10.3/Ex10_3.sce new file mode 100755 index 000000000..5ea7c08db --- /dev/null +++ b/2705/CH10/EX10.3/Ex10_3.sce @@ -0,0 +1,44 @@ +clear; +clc; +disp('Example 10.3'); + +// aim : To determine +// (a) the thermal efficiency of the boiler +// (b) the equivalent evaporation of the boiler +// (c) the new coal consumption + +// given values +ms_dot = 5400;// steam feed rate, [kg/h] +P = 750;// steam pressure, [kN/m^2] +x = .98;// steam dryness fraction +Tf1 = 41.5;// feed water temperature, [C] +CV = 31000;// calorific value of coal used in the boiler, [kJ/kg] +mc1 = 670;// rate of burning of coal/h, [kg] +Tf2 = 100;// increased water temperature, [C] + +// solution +// (a) +SRC = ms_dot/mc1;// steam raised/kg coal, [kg] +hf = 709.3;// [kJ/kg] +hfg = 2055.5;// [kJ/kg] +h1 = hf+x*hfg;// specific enthalpy of steam raised, [kJ/kg] +// from steam table +hfw = 173.9;// specific enthalpy of feed water, [kJ/kg] +EOB = SRC*(h1-hfw)/CV;// efficiency of boiler +mprintf('\n (a) The thermal efficiency of the boiler is = %f percent\n',EOB*100); + +// (b) +he = 2256.9;// specific enthalpy of evaporation, [kJ/kg] +Ee = SRC*(h1-hfw)/he;// equivalent evaporation[kg/kg coal] +mprintf('\n (b) The equivalent evaporation of boiler is = %f kg/kg coal\n',Ee); + +// (c) +hw = 419.1;// specific enthalpy of feed water at 100 C, [kJ/kg] +Eos = ms_dot*(h1-hw);// energy of steam under new condition, [kJ/h] +neb = EOB+.05;// given condition new efficiency of boiler if 5%more than previous +Ec = Eos/neb;// energy from coal, [kJ/h] +mc2 = Ec/CV;// mass of coal used per hour in new condition, [kg] +mprintf('\n (c) Mass of coal used in new condition is = %f kg\n',mc2); +mprintf('\n The saving in coal per hour is = %f kg\n',mc1-mc2); + +// End diff --git a/2705/CH10/EX10.4/Ex10_4.sce b/2705/CH10/EX10.4/Ex10_4.sce new file mode 100755 index 000000000..c7d9ea745 --- /dev/null +++ b/2705/CH10/EX10.4/Ex10_4.sce @@ -0,0 +1,46 @@ +clear; +clc; +disp('Example 10.4'); + +// aim : To determine the +// (a) Heat transfer in the boiler +// (b) Heat transfer in the superheater +// (c) Gas used + +// given values +P = 100;// boiler operating pressure, [bar] +Tf = 256;// feed water temperature, [C] +x = .9;// steam dryness fraction. +Th = 450;// superheater exit temperature, [C] +m = 1200;// steam generation/h, [tonne] +TE = .92;// thermal efficiency +CV = 38;// calorific value of fuel, [MJ/m^3] + +// solution +// (a) +// from steam table +hw = 1115.4;// specific enthalpy of feed water, [kJ/kg] +// for wet steam +hf = 1408;// specific enthalpy, [kJ/kg] +hg = 2727.7;// specific enthalpy, [kJ/kg] +// so +h = hf+x*(hg-hf);// total specific enthalpy of wet steam, [kJ/kg] +// hence +Qb = m*(h-hw);// heat transfer/h for wet steam, [MJ] +mprintf('\n (a) The heat transfer/h in producing wet steam in the boiler is = %f MJ\n',Qb); + +// (b) +// again from steam table +// specific enthalpy of superheated stem at given condition is, +hs = 3244;// [kJ/kg] + +Qs = m*(hs-h);// heat transfer/h in superheater, [MJ] +mprintf('\n (b) The heat transfer/h in superheater is = %f MJ\n',Qs); + +// (c) +V = (Qb+Qs)/(TE*CV);// volume of gs used/h, [m^3] +mprintf('\n (c) The volume of gas used/h is = %f m^3\n',V); + +// There is calculation mistake in the book so our answer is not matching + +// End diff --git a/2705/CH10/EX10.5/Ex10_5.sce b/2705/CH10/EX10.5/Ex10_5.sce new file mode 100755 index 000000000..e7086049a --- /dev/null +++ b/2705/CH10/EX10.5/Ex10_5.sce @@ -0,0 +1,31 @@ +clear; +clc; +disp('Example 10.5'); + +//aim : To determine +// the flow rate of cooling water + +//Given values +P=24;//pressure, [kN/m^2] +ms_dot=1.8;//steam condense rate,[tonne/h] +x=.98;//dryness fraction +T1=21;//entrance temperature of cooling water,[C] +T2=57;//outlet temperature of cooling water,[C] + +//solution +//at 24 kN/m^2, for steam +hfg=2616.8;//[kJ/kg] +hf1=268.2;//[kJ/kg] +//hence +h1=hf1+x*(hfg-hf1);//[kJ/kg] + +//for cooling water +hf3=238.6;//[kJ/kg] +hf2=88.1;//[kJ/kg] + +//using equation [3] +//ms_dot*(hf3-hf2)=mw_dot*(h1-hf1),so +mw_dot=ms_dot*(h1-hf1)/(hf3-hf2);//[tonne/h] +disp('tonne/h',mw_dot,'The flow rate of the cooling water is =') + +//End diff --git a/2705/CH10/EX10.6/Ex10_6.sce b/2705/CH10/EX10.6/Ex10_6.sce new file mode 100755 index 000000000..b4d0e04fe --- /dev/null +++ b/2705/CH10/EX10.6/Ex10_6.sce @@ -0,0 +1,49 @@ +clear; +clc; +disp('Example 10.6'); + +// aim : To determine +// (a) the energy supplied in the boiler +// (b) the dryness fraction of the steam entering the condenser +// (c) the rankine efficiency + +// given values +P1 = 3.5;// steam entering pressure, [MN/m^2] +T1 = 273+350;// entering temperature, [K] +P2 = 10;//steam exhaust pressure, [kN/m^2] + +// solution +// (a) +// from steam table, at P1 is, +hf1 = 3139;// [kJ/kg] +hg1 = 3095;// [kJ/kg] +h1 = hf1-1.5/2*(hf1-hg1); +// at Point 3 +h3 = 191.8;// [kJ/kg] +Es = h1-h3;// energy supplied, [kJ/kg] +mprintf('\n (a) The energy supplied in boiler/kg steam is = %f kJ/kg\n',Es); + +// (b) +// at P1 +sf1 = 6.960;// [kJ/kg K] +sg1 = 6.587;// [kJ/kg K] +s1 = sf1-1.5/2*(sf1-sg1);// [kJ/kg K] +// at P2 +sf2 = .649;// [kJ/kg K] + sg2 = 8.151;// [kJ/kg K] + // s2=sf2+x2(sg2-sf2) + // theoretically expansion through turbine is isentropic so s1=s2 + // hence + s2 = s1; + x2 = (s2-sf2)/(sg2-sf2);// dryness fraction + mprintf('\n (b) The dryness fraction of steam entering the condenser is = %f \n',x2); + + // (c) + // at point 2 + hf2 = 191.8;// [kJ/kg] + hfg2 = 2392.9;// [kJ/kg] + h2 = hf2+x2*hfg2;// [kJ/kg] + Re = (h1-h2)/(h1-h3);// rankine efficiency + mprintf('\n (c) The Rankine efficiency is = %f percent\n',Re*100); + + // End diff --git a/2705/CH10/EX10.7/Ex10_7.sce b/2705/CH10/EX10.7/Ex10_7.sce new file mode 100755 index 000000000..b3e5be740 --- /dev/null +++ b/2705/CH10/EX10.7/Ex10_7.sce @@ -0,0 +1,61 @@ +clear; +clc; +disp('Example 10.7'); + +// aim : To determine +// the specific work done and compare this with that obtained when determining the rankine effficiency + +// given values +P1 = 1000;// steam entering pressure, [kN/m^2] +x1 = .97;// steam entering dryness fraction +P2 = 15;//steam exhaust pressure, [kN/m^2] +n = 1.135;// polytropic index + +// solution +// (a) +// from steam table, at P1 is +hf1 = 762.6;// [kJ/kg] +hfg1 = 2013.6;// [kJ/kg] +h1 = hf1+hfg1; // [kJ/kg] + +sf1 = 2.138;// [kJ/kg K] +sg1 = 6.583;// [kJ/kg K] +s1 = sf1+x1*(sg1-sf1);// [kJ/kg K] + +// at P2 +sf2 = .755;// [kJ/kg K] + sg2 = 8.009;// [kJ/kg K] +// s2 = sf2+x2(sg2-sf2) +// since expansion through turbine is isentropic so s1=s2 + // hence + s2 = s1; + x2 = (s2-sf2)/(sg2-sf2);// dryness fraction + + // at point 2 + hf2 = 226.0;// [kJ/kg] + hfg2 = 2373.2;// [kJ/kg] + h2 = hf2+x2*hfg2;// [kJ/kg] + +// at Point 3 +h3 = 226.0;// [kJ/kg] + +// (a) + Re = (h1-h2)/(h1-h3);// rankine efficiency + mprintf('\n (a) The Rankine efficiency is = %f percent\n',Re*100); + +// (b) +vg1 = .1943;// specific volume at P1, [m^3/kg] +vg2 = 10.02;// specific volume at P2, [m^3/kg] +V1 = x1*vg1;// [m^3/kg] +V2 = x2*vg2;// [m^3/kg] + +W1 = n/(n-1)*(P1*V1-P2*V2);// specific work done, [kJ/kg] + +// from rankine cycle +W2 = h1-h2;// [kJ/kg] +mprintf('\n (b) The specific work done is = %f kJ/kg\n',W1); +mprintf('\n The specific work done (from rankine) is = %f kJ/kg\n',W2); + +// there is calculation mistake in the book so our answer is not matching + +// End diff --git a/2705/CH10/EX10.8/Ex10_8.sce b/2705/CH10/EX10.8/Ex10_8.sce new file mode 100755 index 000000000..f65703173 --- /dev/null +++ b/2705/CH10/EX10.8/Ex10_8.sce @@ -0,0 +1,63 @@ +clear; +clc; +disp('Example 10.8'); + +// aim : To determine +// (a) the rankine fficiency +// (b) the specific steam consumption +// (c) the carnot efficiency of the cycle + +// given values +P1 = 1100;// steam entering pressure, [kN/m^2] +T1 = 273+250;// steam entering temperature, [K] +P2 = 280;// pressure at point 2, [kN/m^2] +P3 = 35;// pressure at point 3, [kN/m^2] + +// solution +// (a) +// from steam table, at P1 and T1 is +hf1 = 2943;// [kJ/kg] +hg1 = 2902;// [kJ/kg] +h1 = hf1-.1*(hf1-hg1); // [kJ/kg] + +sf1 = 6.926;// [kJ/kg K] +sg1 = 6.545;// [kJ/kg K] +s1 = sf1-.1*(sf1-sg1);// [kJ/kg K] + +// at P2 +sf2 = 1.647;// [kJ/kg K] + sg2 = 7.014;// [kJ/kg K] +// s2=sf2+x2(sg2-sf2) +// since expansion through turbine is isentropic so s1=s2 + // hence + s2 = s1; + x2 = (s2-sf2)/(sg2-sf2);// dryness fraction + + // at point 2 + hf2 = 551.4;// [kJ/kg] + hfg2 = 2170.1;// [kJ/kg] + h2 = hf2+x2*hfg2;// [kJ/kg] + vg2 = .646;// [m^3/kg] + v2 = x2*vg2;// [m^3/kg] + + // by Fig10.20. + A6125 = h1-h2;// area of 6125, [kJ/kg] + A5234 = v2*(P2-P3);// area 5234, [kJ/kg] + W = A6125+A5234;// work done + hf = 304.3;// specific enthalpy of water at condenser pressuer, [kJ/kg] + ER = h1-hf;// energy received, [kJ/kg] + Re = W/ER;// rankine efficiency + mprintf('\n (a) The rankine efficiency is = %f percent\n',Re*100); + + // (b) + kWh = 3600;// [kJ] + SSC = kWh/W;// specific steam consumption, [kJ/kWh] + mprintf('\n (b) The specific steam consumption is = %f kJ/kWh\n',SSC); + + // (c) + // from steam table +T3 = 273+72.7;// temperature at point 3 +CE = (T1-T3)/T1;// carnot efficiency +mprintf('\n (c) The carnot efficiency of the cycle is = %f percent\n',CE*100); + +// End diff --git a/2705/CH10/EX10.9/Ex10_9.sce b/2705/CH10/EX10.9/Ex10_9.sce new file mode 100755 index 000000000..cd4a60ca0 --- /dev/null +++ b/2705/CH10/EX10.9/Ex10_9.sce @@ -0,0 +1,48 @@ +clear; +clc; +disp('Example 10.9'); + +// aim : To determine +// (a) the theoretical power of steam passing through the turbine +// (b) the thermal efficiency of the cycle +// (c) the thermal efficiency of the cycle assuming there is no reheat + +// given values +P1 = 6;// initial pressure, [MN/m^2] +T1 = 450;// initial temperature, [C] +P2 = 1;// pressure at stage 1, [MN/m^2] +P3 = 1;// pressure at stage 2, [MN/m^2] +T3 = 370;// temperature, [C] +P4 = .02;// pressure at stage 3, [MN/m^2] +P5 = .02;// pressure at stage 4, [MN/m^2] +T5 = 320;// temperature, [C] +P6 = .02;// pressure at stage 5, [MN/m^2] +P7 = .02;// final pressure , [MN/m^2] + +// solution +// (a) +// using Fig 10.21 +h1 = 3305;// specific enthalpy, [kJ/kg] +h2 = 2850;// specific enthalpy, [kJ/kg] +h3 = 3202;// specific enthalpy, [kJ/kg] +h4 = 2810;// specific enthalpy, [kJ/kg] +h5 = 3115;// specific enthalpy, [kJ/kg] +h6 = 2630;// specific enthalpy, [kJ/kg] +h7 = 2215;// specific enthalpy, [kJ/kg] +W = (h1-h2)+(h3-h4)+(h5-h6);// specific work through the turbine, [kJ/kg] +mprintf('\n (a) The theoretical power/kg steam/s is = %f kW\n',W); + +// (b) +// from steam table +hf6 = 251.5;// [kJ/kg] + +TE1 = ((h1-h2)+(h3-h4)+(h5-h6))/((h1-hf6)+(h3-h2)+(h5-h4));// thermal efficiency +mprintf('\n (b) The thermal efficiency of the cycle is = %f percent\n',TE1*100); + +// (c) +// if there is no heat +hf7 = hf6; +TE2 = (h1-h7)/(h1-hf7);// thermal efficiency +mprintf('\n (c) The thermal efficiency of the cycle if there is no heat is = %f percent\n',TE2*100); + +// End diff --git a/2705/CH11/EX11.1/Ex11_1.sce b/2705/CH11/EX11.1/Ex11_1.sce new file mode 100755 index 000000000..973fd235e --- /dev/null +++ b/2705/CH11/EX11.1/Ex11_1.sce @@ -0,0 +1,45 @@ +clear; +clc; +disp('Example 11.1') + +// aim : To determine the +// (a) bore of the cylinder +// (b) piston stroke +// (c) speed of the engine + +// Given values +P_req = 60;// power required to develop, [kW] +P = 1.25;// boiler pressure, [MN/m^2] +Pb = .13;// back pressure, [MN/m^2] +cut_off = .3;// [stroke] +k = .82;// diagram factor +n = .78;// mechanical efficiency +LN = 3;// mean piston speed, [m/s] + +// solution +// (a) +r = 1/cut_off;// expansion ratio +Pm = P/r*(1+log(r))-Pb;// mean effective pressure, [MN/m^2] +P_ind = P_req/n;// Actual indicated power developed, [kW] +P_the = P_ind/k;// Theoretical indicated power developed, [kW] + +// using indicated_power=Pm*LN*A +// Hence +A = P_the/(Pm*LN)*10^-3;// piston area,[m^2] +d = sqrt(4*A/%pi)*10^3;// bore ,[mm] +mprintf('\n (a) The bore of the cylinder is = %f mm\n',d); + +// (b) +// given that stroke is 1.25 times bore +L = 1.25*d;// [mm] +mprintf('\n (b) The piston stroke is = %f mm\n',L); + +// (c) +// LN=mean piston speed, where L is stroke in meter and N is 2*rev/s,(since engine is double_acting) +// hence +rev_per_sec = LN/(2*L*10^-3);// [rev/s] + +rev_per_min = rev_per_sec*60;// [rev/min] +mprintf('\n (c) The speed of the engine is = %f rev/min\n',rev_per_min); + +// End diff --git a/2705/CH11/EX11.2/Ex11_2.sce b/2705/CH11/EX11.2/Ex11_2.sce new file mode 100755 index 000000000..e7f862fc2 --- /dev/null +++ b/2705/CH11/EX11.2/Ex11_2.sce @@ -0,0 +1,51 @@ +clear; +clc; +disp('Example 11.2') + +// aim : To determine the +// (a) the diameter of the cylinder +// (b) piston stroke +// (c) actual steam consumption and indicated thermal efficiency + +// Given values +P = 900;// inlet pressure, [kN/m^2] +Pb = 140;// exhaust pressure, [kN/m^2] +cut_off =.4;// [stroke] +k = .8;// diagram factor +rs = 1.2;// stroke to bore ratio +N = 4;// engine speed, [rev/s] +ip = 22.5;// power output from the engine, [kW] + +// solution +// (a) +r = 1/cut_off;// expansion ratio +Pm = P/r*(1+log(r))-Pb;// mean effective pressure, [kN/m^2] +Pm = Pm*k;// actual mean effective pressure, [kN/m^2] + +// using ip=Pm*L*A*N +// and L=r*d; where L is stroke and d is bore +d = (ip/(Pm*rs*%pi/4*2*N))^(1/3);// diameter of the cylinder, [m] + +mprintf('\n (a) The diameter of the cylinder is = %f mm\n',d*1000); + +// (b) +L = rs*d;// stroke, [m] +mprintf('\n (b) The piston stroke is = %f mm\n',L*1000); + +// (c) +SV = %pi/4*d^2*L;// stroke volume, [m^3] +V = SV*cut_off*2*240*60;// volume of steam consumed per hour, [m^3] +v = .2148;// specific volume at 900 kN/m^2, [m^3/kg] +SC = V/v;// steam consumed/h, [kg] +ASC = 1.5*SC;// actual steam consumption/h, [kg] +mprintf('\n (c) The actual steam consumption/h is = %f kg\n',ASC); + +m_dot = ASC/3600;// steam consumption,[kg/s] +// from steam table +hg = 2772.1;// specific enthalpy of inlet steam, [kJ/kg] +hfe = 458.4;// specific liquid enthalpy at exhaust pressure, [kJ/kg] + +ITE = ip/(m_dot*(hg-hfe));// indicated thermal efficiency +mprintf('\n The indicated thermal efficiency is = %f percent\n',ITE*100); + +// End diff --git a/2705/CH11/EX11.3/Ex11_3.sce b/2705/CH11/EX11.3/Ex11_3.sce new file mode 100755 index 000000000..14b2cf23b --- /dev/null +++ b/2705/CH11/EX11.3/Ex11_3.sce @@ -0,0 +1,41 @@ +clear; +clc; +disp('Example 11.3'); + +// aim : To determine +// (a) the diagram factor +// (b) the indicated thermal efficiency of the engine + +// given values +d = 250*10^-3;// cylinder diameter, [m] +L = 375*10^-3;// length of stroke, [m] +P = 1000;// steam pressure , [kPa] +x = .96;// dryness fraction of steam +Pb = 55;// exhaust pressure, [kPa] +r = 6;// expansion ratio +ip = 45;// indicated power developed, [kW] +N = 3.5;// speed of engine, [rev/s] +m = 460;// steam consumption, [kg/h] + +// solution +// (a) +Pm = P/r*(1+log(r))-Pb;// [kN/m^3] +A = %pi*(d)^2/4;// area, [m^2] +tip = Pm*L*A*N*2;// theoretical indicated power, [kW] +k = ip/tip;// diagram factor +mprintf('\n (a) The diagram factor is = %f\n',k); + +// (b) +// from steam table at 1 MN/m^2 +hf = 762.6;// [kJ/kg] +hfg = 2013.6;// [kJ/kg] +// so +h1 = hf+x*hfg;// specific enthalpy of steam at 1MN/m^2, [kJ/kg] +// minimum specific enthalpy in engine at 55 kPa +hf = 350.6;// [kJ/kg] +// maximum energy available in engine is +h = h1-hf;// [kJ/kg] +ITE = ip/(m*h/3600)*100;// indicated thermal efficiency +mprintf('\n (b) The indicated thermal efficiency is = %f percent\n ',ITE); + +// End diff --git a/2705/CH11/EX11.4/Ex11_4.sce b/2705/CH11/EX11.4/Ex11_4.sce new file mode 100755 index 000000000..4e480bf97 --- /dev/null +++ b/2705/CH11/EX11.4/Ex11_4.sce @@ -0,0 +1,20 @@ +clear; +clc; +disp('Example 11.4'); + +// aim : To determine +// steam consumption + +// given values +P1 = 11;// power, [kW] +m1 = 276;// steam use/h when developing power P1,[kW] +ip = 8;// indicated power output, [kW] +B = 45;// steam used/h at no load, [kg] + +// solution +// using graph of Fig.11.9 +A = (m1-B)/P1;// slop of line, [kg/kWh] +W = A*ip+B;// output, [kg/h] +mprintf('\n The steam consumption is = %f kg/h\n ',W); + +// End diff --git a/2705/CH11/EX11.5/Ex11_5.sce b/2705/CH11/EX11.5/Ex11_5.sce new file mode 100755 index 000000000..b4106e20d --- /dev/null +++ b/2705/CH11/EX11.5/Ex11_5.sce @@ -0,0 +1,52 @@ +clear; +clc; +disp('Example 11.5'); + +// aim : To determine +// (a) the intermediate pressure +// (b) the indicated power output +// (c) the steam consumption of the engine + +// given values +P1 = 1400;// initial pressure, [kN/m^2] +x = .9;// dryness fraction +P5 = 35;// exhaust pressure +k = .8;// diagram factor of low-pressure cylindaer +L = 350*10^-3;// stroke of both the cylinder, [m] +dhp = 200*10^-3;// diameter of high pressure cylinder, [m] +dlp = 300*10^-3;// diameter of low-pressure cylinder, [m] +N = 300;// engine speed, [rev/min] + +// solution +// taking reference Fig.11.13 +Ahp = %pi/4*dhp^2;// area of high-pressure cylinder, [m^2] +Alp = %pi/4*dlp^2;// area of low-pressure cylinder, [m^2] +// for equal initial piston loads +// (P1-P7)Ahp=(P7-P5)Alp +deff('[x]=f(P7)','x=(P1-P7)*Ahp-(P7-P5)*Alp'); +P7 = fsolve(0,f);// intermediate pressure, [kN/m^2] +mprintf('\n (a) The intermediate pressure is = %f kN/m^2\n ',P7); + +// (b) +V6 = Ahp*L;// volume of high-pressure cylinder, [m^3] +P2 = P1; +P6 = P7; +// using P2*V2=P6*V6 +V2 = P6*V6/P2; // [m^3] +V1 = Alp*L;// volume of low-pressure cylinder, [m^3] +R = V1/V2;// expansion ratio +Pm = P1/R*(1+log(R))-P5;// effective pressure of low-pressure cylinder, [kn/m^2] +Pm = k*Pm;// actual effective pressure, [kN/m^2] +ip = Pm*L*Alp*N*2/60;// indicated power, [kW] +mprintf('\n (b) The indicated power is = %f kW\n',ip); + +// (c) +COV = V1/ R;// cut-off volume in high-pressure cylinder, [m^3] +V = COV*N*2*60;// volume of steam admitted/h +// from steam table +vg = .1407;// [m^3/kg] +AV = x*vg;// specific volume of admission steam, [m^3/kg] +m = V/AV;// steam consumption, [kg/h] +mprintf('\n (c) The steam consumption of the engine is = %f kg/h\n',m); + +// End diff --git a/2705/CH11/EX11.6/Ex11_6.sce b/2705/CH11/EX11.6/Ex11_6.sce new file mode 100755 index 000000000..5cbaf958a --- /dev/null +++ b/2705/CH11/EX11.6/Ex11_6.sce @@ -0,0 +1,45 @@ +clear; +clc; +disp('Example 11.6'); + +// aim : To determine +// (a) the indicated power output +// (b) the diameter of high-pressure cylinder of the engine +// (c) the intermediate pressure + +// given values +P = 1100;// initial pressure, [kN/m^2] +Pb = 28;// exhaust pressure +k = .82;// diagram factor of low-pressure cylindaer +L = 600*10^-3;// stroke of both the cylinder, [m] +dlp = 600*10^-3;// diameter of low-pressure cylinder, [m] +N = 4;// engine speed, [rev/s] +R = 8;// expansion ratio + +// solution +// taking reference Fig.11.13 +// (a) +Pm = P/R*(1+log(R))-Pb;// effective pressure of low-pressure cylinder, [kn/m^2] +Pm = k*Pm;// actual effective pressure, [kN/m^2] +Alp = %pi/4*dlp^2;// area of low-pressure cylinder, [m^2] +ip = Pm*L*Alp*N*2;// indicated power, [kW] +mprintf('\n (a) The indicated power is = %f kW\n',ip); + +// (b) +// work done by both cylinder is same as area of diagram +w = Pm*Alp*L;// [kJ] +W = w/2;// work done/cylinder, [kJ] +V2 = Alp*L/8;// volume, [m63] +P2 = P;// [kN/m^2] +// using area A1267=P2*V2*log(V6/V2)=W +V6 = V2*exp(W/(P2*V2));// intermediate volume, [m^3] +// using Ahp*L=%pi/4*dhp^2*L=V6 +dhp = sqrt(V6*4/L/%pi);// diameter of high-pressure cylinder, [m] +mprintf('\n (b) The diameter of high-pressure cylinder is = %f mm\n',dhp*1000); + +// (c) +// using P2*V2=P6*V6 +P6 = P2*V2/V6; // intermediate pressure, [kN/m^2] +mprintf('\n (c) The intermediate opressure is = %f kN/m^2\n',P6); + +// End diff --git a/2705/CH11/EX11.7/Ex11_7.sce b/2705/CH11/EX11.7/Ex11_7.sce new file mode 100755 index 000000000..da5f5dd4e --- /dev/null +++ b/2705/CH11/EX11.7/Ex11_7.sce @@ -0,0 +1,37 @@ +clear; +clc; +disp('Example 11.7'); + +// aim : To determine +// (a) The speed of the engine +// (b) the diameter of the high pressure cylinder + +// given values +ip = 230;// indicated power, [kW] +P = 1400;// admission pressure, [kN/m^2] +Pb = 35;// exhaust pressure, [kN/m^2] +R = 12.5;// expansion ratio +d1 = 400*10^-3;// diameter of low pressure cylinder, [m] +L = 500*10^-3;// stroke of both the cylinder, [m] +k = .78;// diagram factor +rv = 2.5;// expansion ratio of high pressure cylinder + +// solution +// (a) +Pm = P/R*(1+log(R))-Pb;// mean effective pressure in low pressure cylinder, [kN/m^2] +ipt = ip/k;// theoretical indicated power, [kw] +// using ip=Pm*L*A*N +A = %pi/4*d1^2;// area , [m^2] +N = ipt/(Pm*L*A*2);// speed, [rev/s] +mprintf('\n (a) The engine speed is = %f rev/s\n',N); + +// (b) +Vl = A*L;// volume of low pressure cylinder, [m^3] +COV = Vl/R;// cutt off volume of hp cylinder, [m^3] +V = COV*rv;// total volume, [m^3] + +// V = %pi/4*d^2*L, so +d = sqrt(4*V/%pi/L);// diameter of high pressure cylinder, [m] +mprintf('\n (b) The diameter of the high pressure cylinder is = %f mm\n',d*1000); + +// End diff --git a/2705/CH11/EX11.8/Ex11_8.sce b/2705/CH11/EX11.8/Ex11_8.sce new file mode 100755 index 000000000..4189dbf58 --- /dev/null +++ b/2705/CH11/EX11.8/Ex11_8.sce @@ -0,0 +1,54 @@ +clear; +clc; +disp('Example 11.8'); + +// aim : To determine +// (a) the actual and hypothetical mean effective pressures referred to the low-pressure cylinder +// (b) the overall diagram factor +// (c) the indicated power + +// given values +P = 1100;// steam supply pressure, [kN/m^2] +Pb = 32;// back pressure, [kN/m^2] +d1 = 300*10^-3;// cylinder1 diameter, [m] +d2 = 600*10^-3;// cylinder2 diameter, [m] +L = 400*10^-3;// common stroke of both cylinder, [m] + +A1 = 12.5;// average area of indicated diagram for HP, [cm^2] +A2 = 11.4;// average area of indicated diagram for LP, [cm^2] + +P1 = 270;// indicator calibration, [kN/m^2/ cm] +P2 = 80;// spring calibration, [kN/m^2/ cm] +N = 2.7;// engine speed, [rev/s] +l = .75;// length of both diagram, [m] + +// solution +// (a) +// for HP cylinder +Pmh = P1*A1/7.5;// [kN/m^2] +F = Pmh*%pi/4*d1^2;// force on HP, [kN] +PmH = Pmh*(d1/d2)^2;// pressure referred to LP cylinder, [kN/m^2] +PmL = P2*A2/7.5;// pressure for LP cylinder, [kN/m^2] +PmA = PmH+PmL;// actual effective pressure referred to LP cylinder, [kN/m^2] + +Ah = %pi/4*d1^2;// area of HP cylinder, [m^2] +Vh = Ah*L;// volume of HP cylinder, [m^3] +CVh = Vh/3;// cut-off volume of HP cylinder, [m^3] +Al = %pi/4*d2^2;// area of LP cylinder, [m^2] +Vl = Al*L;// volume of LP cylinder, [m^3] + +R = Vl/CVh;// expansion ratio +Pm = P/R*(1+log(R))-Pb;// hypothetical mean effective pressure referred to LP cylinder, [kN/m^2] + +mprintf('\n (a) The actual mean effective pressure referred to LP cylinder is = %f kN/m^2\n',PmA); +mprintf('\n The hypothetical mean effective pressure referred to LP cylinder is = %f kN/m^2\n',Pm); + +// (a) +ko = PmA/Pm;// overall diagram factor +mprintf('\n (b) The overall diagram factor is = %f\n',ko); + +// (c) +ip = PmA*L*Al*N*2;// indicated power, [kW] +mprintf('\n (c) The indicated power is = %f kW\n',ip); + +// End diff --git a/2705/CH11/EX11.9/Ex11_9.sce b/2705/CH11/EX11.9/Ex11_9.sce new file mode 100755 index 000000000..e020aaa3e --- /dev/null +++ b/2705/CH11/EX11.9/Ex11_9.sce @@ -0,0 +1,48 @@ +clear; +clc; +disp('Example 11.9'); + +// aim : To determine +// (a) the actual and hypothetical mean effective pressures referred to the low-pressure cylinder +// (b) the overall diagram factor +// (c) the pecentage of the total indicated power developed in each cylinder + +// given values +P = 1400;// steam supply pressure, [kN/m^2] +Pb = 20;// back pressure, [kN/m^2] +Chp = .6;// cut-off in HP cylinder, [stroke] +dh = 300*10^-3;// HP diameter, [m] +di = 500*10^-3;// IP diameter, [m] +dl = 900*10^-3;// LP diameter, [m] + +Pm1 = 590;// actual pressure of HP cylinder, [kN/m^2] +Pm2 = 214;// actual pressure of IP cylinder, [kN/m^2] +Pm3 = 88;// actual pressure of LP cylinder, [kN/m^2] + +// solution +// (a) +// for HP cylinder +PmH = Pm1*(dh/dl)^2;// PmH referred to LP cylinder, [kN/m^2] +// for IP cylinder +PmI = Pm2*(di/dl)^2;// PmI referred to LP cylinder, [kN/m^2] +PmA = PmH+PmI+Pm3;// actual mean effective pressure referred to LP cylinder, [kN/m^2] + +R = dl^2/(dh^2*Chp);// expansion ratio +Pm = P/R*(1+log(R))-Pb;// hypothetical mean effective pressure referred to LP cylinder, [kN/m^2] + +mprintf('\n (a) The actual mean effective pressure referred to LP cylinder is = %f kN/m^2\n',PmA); +mprintf('\n The hypothetical mean effective pressure referred to LP cylinder is = %f kN/m^2\n',Pm); + +// (b) +ko = PmA/Pm;// overall diagram factor +mprintf('\n (b) The overall diagram factor is = %f\n',ko); + +// (c) +HP = PmH/PmA*100;// %age of indicated power developed in HP +IP = PmI/PmA*100; // %age of indicated power developed in IP +LP = Pm3/PmA*100; // %age of indicated power developed in LP +mprintf('\n (c) The pecentage of the total indicated power developed in HP cylinder is = %f percent\n',HP); +mprintf('\n The pecentage of the total indicated power developed in IP cylinder is = %f percent\n',IP); +mprintf('\n The pecentage of the total indicated power developed in LP cylinder is = %f percent\n',LP); + +// End diff --git a/2705/CH12/EX12.1/Ex12_1.sce b/2705/CH12/EX12.1/Ex12_1.sce new file mode 100755 index 000000000..6acd59e30 --- /dev/null +++ b/2705/CH12/EX12.1/Ex12_1.sce @@ -0,0 +1,43 @@ +clear; +clc; +disp('Example 12.1'); + +// aim : To determine the +// (a) throat area +// (b) exit area +// (c) Mach number at exit + +// Given values +P1 = 3.5;// inlet pressure of air, [MN/m^2] +T1 = 273+500;// inlet temperature of air, [MN/m^2] +P2 = .7;// exit pressure, [MN/m^2] +m_dot = 1.3;// flow rate of air, [kg/s] +Gamma = 1.4;// heat capacity ratio +R = .287;// [kJ/kg K] + +// solution +// given expansion may be considered to be adiabatic and to follow the law PV^Gamma=constant +// using ideal gas law +v1 = R*T1/P1*10^-3;// [m^3/kg] +Pt = P1*(2/(Gamma+1))^(Gamma/(Gamma-1));// critical pressure, [MN/m^2] + +// velocity at throat is +Ct = sqrt(2*Gamma/(Gamma-1)*P1*10^6*v1*(1-(Pt/P1)^(((Gamma-1)/Gamma))));// [m/s] +vt = v1*(P1/Pt)^(1/Gamma);// [m^3/kg] +// using m_dot/At=Ct/vt +At = m_dot*vt/Ct*10^6;// throat area, [mm^2] +mprintf('\n (a) The throat area is = %f mm^2\n',At); + +// (b) +// at exit +C2 = sqrt(2*Gamma/(Gamma-1)*P1*10^6*v1*(1-(P2/P1)^(((Gamma-1)/Gamma))));// [m/s] +v2 = v1*(P1/P2)^(1/Gamma);// [m^3/kg] +A2 = m_dot*v2/C2*10^6;// exit area, [mm^2] + +mprintf('\n (b) The exit area is = %f mm^2\n',A2); + +// (c) +M = C2/Ct; +mprintf('\n (c) The Mach number at exit is = %f\n',M); + +// End diff --git a/2705/CH12/EX12.2/Ex12_2.sce b/2705/CH12/EX12.2/Ex12_2.sce new file mode 100755 index 000000000..fd3cd191b --- /dev/null +++ b/2705/CH12/EX12.2/Ex12_2.sce @@ -0,0 +1,33 @@ +clear; +clc; +disp('Example 12.2'); + +// aim : To determine the increases in pressure, temperature and internal energy per kg of air + +// Given values +T1 = 273;// [K] +P1 = 140;// [kN/m^2] +C1 = 900;// [m/s] +C2 = 300;// [m/s] +cp = 1.006;// [kJ/kg K] +cv =.717;// [kJ/kg K] + +// solution +R = cp-cv;// [kJ/kg K] +Gamma = cp/cv;// heat capacity ratio +// for frictionless adiabatic flow, (C2^2-C1^2)/2=Gamma/(Gamma-1)*R*(T1-T2) + +T2 =T1-((C2^2-C1^2)*(Gamma-1)/(2*Gamma*R))*10^-3; // [K] +T_inc = T2-T1;// increase in temperature [K] + +P2 = P1*(T2/T1)^(Gamma/(Gamma-1));// [MN/m^2] +P_inc = (P2-P1)*10^-3;// increase in pressure,[MN/m^2] + +U_inc = cv*(T2-T1);// Increase in internal energy per kg,[kJ/kg] +mprintf('\n The increase in pressure is = %f MN/m^2\n',P_inc); +mprintf('\n Increase in temperature is = %f K\n',T_inc); +mprintf('\n Increase in internal energy is = %f kJ/kg\n',U_inc); + +// there is minor variation in result + +// End diff --git a/2705/CH12/EX12.3/Ex12_3.sce b/2705/CH12/EX12.3/Ex12_3.sce new file mode 100755 index 000000000..37084b47a --- /dev/null +++ b/2705/CH12/EX12.3/Ex12_3.sce @@ -0,0 +1,47 @@ +clear; +clc; +disp('Example 12.3'); + +// aim : To determine the +// (a) throat and exit areas +// (b) degree of undercooling at exit +// Given values +P1 = 2;// inlet pressure of air, [MN/m^2] +T1 = 273+325;// inlet temperature of air, [MN/m^2] +P2 = .36;// exit pressure, [MN/m^2] +m_dot = 7.5;// flow rate of air, [kg/s] +n = 1.3;// polytropic index + +// solution +// (a) +// using steam table +v1 = .132;// [m^3/kg] +// given expansion following law PV^n=constant + +Pt = P1*(2/(n+1))^(n/(n-1));// critical pressure, [MN/m^2] + +//velocity at throat is +Ct = sqrt(2*n/(n-1)*P1*10^6*v1*(1-(Pt/P1)^(((n-1)/n))));// [m/s] +vt = v1*(P1/Pt)^(1/n);// [m^3/kg] +// using m_dot/At=Ct/vt +At = m_dot*vt/Ct*10^6;// throat area, [mm^2] +mprintf('\n (a) The throat area is = %f mm^2\n',At); + +// at exit +C2 = sqrt(2*n/(n-1)*P1*10^6*v1*(1-(P2/P1)^(((n-1)/n))));// [m/s] +v2 = v1*(P1/P2)^(1/n);// [m^3/kg] +A2 = m_dot*v2/C2*10^6;// exit area, [mm^2] + +mprintf('\n The exit area is = %f mm^2\n',A2); + +// (b) +T2 = T1*(P2/P1)^((n-1)/n);//outlet temperature, [K] +t2 = T2-273;//[C] +// at exit pressure saturation temperature is +ts = 139.9;// saturation temperature,[C] +Doc = ts-t2;// Degree of undercooling,[C] +mprintf('\n (b) The Degree of undercooling at exit is = %f C\n',Doc); + +// There is some calculation mistake in the book so answer is not matching + +// End diff --git a/2705/CH12/EX12.4/Ex12_4.sce b/2705/CH12/EX12.4/Ex12_4.sce new file mode 100755 index 000000000..0bc59c65e --- /dev/null +++ b/2705/CH12/EX12.4/Ex12_4.sce @@ -0,0 +1,48 @@ +clear; +clc; +disp('Example 12.4'); + +// aim : To determine the +// (a) throat and exit velocities +// (b) throat and exit areas + +// Given values +P1 = 2.2;// inlet pressure, [MN/m^2] +T1 = 273+260;// inlet temperature, [K] +P2 = .4;// exit pressure,[MN/m^2] +eff = .85;// efficiency of the nozzle after throat +m_dot = 11;// steam flow rate in the nozzle, [kg/s] + +// solution +// (a) +// assuming steam is following same law as previous question 12.3 +Pt = .546*P1;// critical pressure,[MN/m^2] +// from Fig. 12.6 +h1 = 2940;// [kJ/kg] +ht = 2790;// [kJ/kg] + +Ct = sqrt(2*(h1-ht)*10^3);// [m/s] + +// again from Fig. 12.6 +h2_prime = 2590;// [kJ/kg] +// using eff = (ht-h2)/(ht-h2_prime) + +h2 = ht-eff*(ht-h2_prime); // [kJ/kg] + +C2 = sqrt(2*(h1-h2)*10^3);// [m/s] + +// (b) +// from chart +vt = .16;// [m^3/kg] +v2 = .44;// [m^3/kg] +// using m_dot*v=A*C +At = m_dot*vt/Ct*10^6;// throat area, [mm^2] + +A2 = m_dot*v2/C2*10^6;// throat area, [mm^2] + +mprintf('\n (a) The throat velocity is = %f m/s\n',Ct); +mprintf('\n The exit velocity is = %f m/s\n',C2); +mprintf('\n (b) The throat area is = %f mm^2\n',At); +mprintf('\n The throat area is = %f mm^2\n',A2); + +// End diff --git a/2705/CH13/EX13.1/Ex13_1.sce b/2705/CH13/EX13.1/Ex13_1.sce new file mode 100755 index 000000000..55203b728 --- /dev/null +++ b/2705/CH13/EX13.1/Ex13_1.sce @@ -0,0 +1,24 @@ +clear; +clc; +disp('Example 13.1'); + +// aim : To determine +// the power developed for a steam flow of 1 kg/s at the blades and the kinetic energy of the steam finally leaving the wheel + +// Given values +alfa = 20;// blade angle, [degree] +Cai = 375;// steam exit velocity in the nozzle,[m/s] +U = 165;// blade speed, [m/s] +loss = .15;// loss of velocity due to friction + +// solution +// using Fig13.12, +Cvw = 320;// change in velocity of whirl, [m/s] +cae = 132.5;// absolute velocity at exit, [m/s] +Pds = U*Cvw*10^-3;// Power developed for steam flow of 1 kg/s, [kW] +Kes = cae^2/2*10^-3;// Kinetic energy change of steam, [kW/kg] + +mprintf('\n The power developed for a steam flow of 1 kg/s is = %f kW\n',Pds) +mprintf('\n The energy of steam finally leaving the wheel is = %f kW/kg\n',Kes); + +// End diff --git a/2705/CH13/EX13.2/Ex13_2.sce b/2705/CH13/EX13.2/Ex13_2.sce new file mode 100755 index 000000000..601a28f65 --- /dev/null +++ b/2705/CH13/EX13.2/Ex13_2.sce @@ -0,0 +1,38 @@ +clear; +clc; +disp('Example 13.2'); + +// aim : To determine +// (a) the entry angle of the blades +// (b) the work done per kilogram of steam per second +// (c) the diagram efficiency +// (d) the end-thrust per kilogram of steam per second + +// given values +Cai = 600;// steam velocity, [m/s] +sia = 25;// steam inlet angle with blade, [degree] +U = 255;// mean blade speed, [m/s] +sea = 30;// steam exit angle with blade,[degree] + +// solution +// (a) +// using Fig.13.13(diagram for example 13.2) +eab = 41.5;// entry angle of blades, [degree] +mprintf('\n (a) The angle of blades is = %f degree\n',eab); + +// (b) +Cwi_plus_Cwe = 590;// velocity of whirl, [m/s] +W = U*(Cwi_plus_Cwe);// work done on the blade,[W/kg] +mprintf('\n (b) The work done on the blade is = %f kW/kg\n',W*10^-3); + +// (c) +De = 2*U*(Cwi_plus_Cwe)/Cai^2;// diagram efficiency +mprintf('\n (c) The diagram efficiency is = %f percent\n',De*100); + +// (d) +// again from the diagram +Cfe_minus_Cfi = -90;// change invelocity of flow, [m/s] +Eth = Cfe_minus_Cfi;// end-thrust, [N/kg s] +mprintf('\n (d) The End-thrust is = %f N/kg',Eth); + +// End diff --git a/2705/CH13/EX13.3/Ex13_3.sce b/2705/CH13/EX13.3/Ex13_3.sce new file mode 100755 index 000000000..511e4e183 --- /dev/null +++ b/2705/CH13/EX13.3/Ex13_3.sce @@ -0,0 +1,28 @@ +clear; +clc; +disp('Example 13.3'); + +// aim : To determine +// (a) the power output of the turbine +// (b) the diagram efficiency + +// given values +U = 150;// mean blade speed, [m/s] +Cai1 = 675;// nozzle speed, [m/s] +na = 20;// nozzle angle, [degree] +m_dot = 4.5;// steam flow rate, [kg/s] + +// solution +// from Fig. 13.15(diagram 13.3) +Cw1 = 915;// [m/s] +Cw2 = 280;// [m/s] + +// (a) +P = m_dot*U*(Cw1+Cw2);// power of turbine,[W] +mprintf('\n (a) The power of turbine is = %f kW\n',P*10^-3); + +// (b) +De = 2*U*(Cw1+Cw2)/Cai1^2;// diagram efficiency +mprintf('\n (b) The diagram efficiency is = %f percent\n',De*100); + +// End diff --git a/2705/CH13/EX13.4/Ex13_4.sce b/2705/CH13/EX13.4/Ex13_4.sce new file mode 100755 index 000000000..2f9f9c42a --- /dev/null +++ b/2705/CH13/EX13.4/Ex13_4.sce @@ -0,0 +1,37 @@ +clear; +clc; +disp('Example 13.4'); + +// aim : To determine +// (a) the power output of the stage +// (b) the specific enthalpy drop in the stage +// (c) the percentage increase in relative velocity in the moving blades due to expansion in the bladse + +// given values +N = 50;// speed, [m/s] +d = 1;// blade ring diameter, [m] +nai = 50;// nozzle inlet angle, [degree] +nae = 30;// nozzle exit angle, [degree] +m_dot = 600000;// steam flow rate, [kg/h] +se = .85;// stage efficiency + +// solution +// (a) +U = %pi*d*N;// mean blade speed, [m/s] +// from Fig. 13.17(diagram 13.4) +Cwi_plus_Cwe = 444;// change in whirl speed, [m/s] +P = m_dot*U*Cwi_plus_Cwe/3600;// power output of the stage, [W] +mprintf('\n (a) The power output of the stage is = %f MW\n',P*10^-6); + +// (b) +h = U*Cwi_plus_Cwe/se;// specific enthalpy,[J/kg] +mprintf('\n (b) The specific enthalpy drop in the stage is = %f kJ/kg\n ',h*10^-3); + +// (c) +// again from diagram +Cri = 224;// [m/s] +Cre = 341;// [m/s] +Iir = (Cre-Cri)/Cri;// increase in relative velocity +mprintf('\n (c) The increase in relative velocity is = %f percent\n',Iir*100); + +// End diff --git a/2705/CH13/EX13.5/Ex13_5.sce b/2705/CH13/EX13.5/Ex13_5.sce new file mode 100755 index 000000000..b2c3c6672 --- /dev/null +++ b/2705/CH13/EX13.5/Ex13_5.sce @@ -0,0 +1,36 @@ +clear; +clc; +disp('Example 13.5'); + +// aim : To determine +// (a) the blade height of the stage +// (b) the power developed in the stage +// (c) the specific enthalpy drop at the stage + +// given values +U = 60;// mean blade speed, [m/s] +P = 350;// steam pressure, [kN/m^2] +T = 175;// steam temperature, [C] +nai = 30;// stage inlet angle, [degree] +nae = 20;// stage exit angle, [degree] + +// solution +// (a) +m_dot = 13.5;// steam flow rate, [kg/s] +// at given T and P +v = .589;// specific volume, [m^3/kg] +// given H=d/10, so +H = sqrt(m_dot*v/(%pi*10*60));// blade height, [m] +mprintf('\n (a) The blade height at this stage is = %f mm\n',H*10^3); + +// (b) +Cwi_plus_Cwe = 270;// change in whirl speed, [m/s] +P = m_dot*U*(Cwi_plus_Cwe);// power developed, [W] +mprintf('\n (b) The power developed is = %f kW\n',P*10^-3); + +// (c) +s = .85;// stage efficiency +h = U*Cwi_plus_Cwe/s;// specific enthalpy,[J/kg] +mprintf('\n (a) The specific enthalpy drop in the stage is = %f kJ/kg',h*10^-3); + +// End diff --git a/2705/CH14/EX14.1/Ex14_1.sce b/2705/CH14/EX14.1/Ex14_1.sce new file mode 100755 index 000000000..404a8d8d4 --- /dev/null +++ b/2705/CH14/EX14.1/Ex14_1.sce @@ -0,0 +1,55 @@ +clear; +clc; +disp(' Example 14.1'); + +// aim : To determine +// (a) the free air delivered +// (b) the volumetric efficiency +// (c) the air delivery temperature +// (d) the cycle power +// (e) the isothermal efficiency + +// given values +d = 200*10^-3;// bore, [m] +L = 300*10^-3;// stroke, [m] +N = 500;// speed, [rev/min] +n = 1.3;// polytropic index +P1 = 97;// intake pressure, [kN/m^2] +T1 = 273+20;// intake temperature, [K] +P3 = 550;// compression pressure, [kN/m^2] + +// solution +// (a) +P4 = P1; +P2 = P3; +Pf = 101.325;// free air pressure, [kN/m^2] +Tf = 273+15;// free air temperature, [K] +SV = %pi/4*d^2*L;// swept volume, [m^3] +V3 = .05*SV;// [m^3] +V1 = SV+V3;// [m^3] +V4 = V3*(P3/P4)^(1/n);// [m^3] +ESV = (V1-V4)*N;// effective swept volume/min, [m^3] +// using PV/T=constant +Vf = P1*ESV*Tf/(Pf*T1);// free air delivered, [m^3/min] +mprintf('\n (a) The free air delivered is = %f m^3/min\n',Vf); + +// (b) +VE = Vf/(N*(V1-V3));// volumetric efficiency +mprintf('\n (b) The volumetric efficiency is = %f percent\n',VE*100); + +// (c) +T2 = T1*(P2/P1)^((n-1)/n);// free air temperature, [K] +mprintf('\n (c) The air delivery temperature is = %f C\n',T2-273); + +// (d) +CP = n/(n-1)*P1*(V1-V4)*((P2/P1)^((n-1)/n)-1)*N/60;// cycle power, [kW] + mprintf('\n (d) The cycle power is = %f kW\n',CP); + +// (e) +// neglecting clearence +W = n/(n-1)*P1*V1*((P2/P1)^((n-1)/n)-1) +Wi = P1*V1*log(P2/P1);// isothermal efficiency +IE = Wi/W;// isothermal efficiency +mprintf('\n (e) The isothermal efficiency neglecting clearence is = %f percent\n',IE*100); + +// End diff --git a/2705/CH14/EX14.2/Ex14_2.sce b/2705/CH14/EX14.2/Ex14_2.sce new file mode 100755 index 000000000..ec512aa18 --- /dev/null +++ b/2705/CH14/EX14.2/Ex14_2.sce @@ -0,0 +1,35 @@ +clear; +clc; +disp(' Example 14.2'); + +// aim : To determine +// (a) the intermediate pressure +// (b) the total volume of each cylinder +// (c) the cycle power + +// given values +v1 = .2;// air intake, [m^3/s] +P1 = .1;// intake pressure, [MN/m^2] +T1 = 273+16;// intake temperature, [K] +P3 = .7;// final pressure, [MN/m^2] +n = 1.25;// compression index +N = 10;// speed, [rev/s] + +// solution +// (a) +P2 = sqrt(P1*P3);// intermediate pressure, [MN/m^2] +mprintf('\n (a) The intermediate pressure is = %f MN/m^2\n',P2); + +// (b) +V1 = v1/N;// total volume,[m^3] +// since intercooling is perfect so 2 lie on the isothermal through1, P1*V1=P2*V2 +V2 = P1*V1/P2;// volume, [m^3] +mprintf('\n (b) The total volume of the HP cylinder is = %f litres\n',V2*10^3); + + // (c) + CP = 2*n/(n-1)*P1*v1*((P2/P1)^((n-1)/n)-1);// cycle power, [MW] + mprintf('\n (c) The cycle power is = %f MW\n',CP*10^3); + + // there is calculation mistake in the book so answer is not matching + + // End diff --git a/2705/CH14/EX14.3/Ex14_3.sce b/2705/CH14/EX14.3/Ex14_3.sce new file mode 100755 index 000000000..7b899feb5 --- /dev/null +++ b/2705/CH14/EX14.3/Ex14_3.sce @@ -0,0 +1,59 @@ +clear; +clc; +disp(' Example 14.3'); + +// aim : To determine +// (a) the intermediate pressures +// (b) the effective swept volume of the LP cylinder +// (c) the temperature and the volume of air delivered per stroke at 15 bar +// (d) the work done per kilogram of air + +// given values +d = 450*10^-3;// bore , [m] +L = 300*10^-3;// stroke, [m] +cl = .05;// clearence +P1 = 1; // intake pressure, [bar] +T1 = 273+18;// intake temperature, [K] +P4 = 15;// final delivery pressure, [bar] +n = 1.3;// compression and expansion index +R = .29;// gas constant, [kJ/kg K] + +// solution +// (a) +k=(P4/P1)^(1/3); +// hence +P2 = k*P1;// intermediare pressure, [bar] +P3 = k*P2;// intermediate pressure, [bar] + +mprintf('\n (a) The intermediate pressure is P2 = %f bar\n',P2); +mprintf('\n The intermediate pressure is P3= %f bar\n',P3); + +// (b) +SV = %pi*d^2/4*L;// swept volume of LP cylinder, [m^3] +// hence +V7 = cl*SV;// volume, [m^3] +V1 = SV+V7;// volume, [m^3] +// also +P7 = P2; +P8 = P1; +V8 = V7*(P7/P8)^(1/n);// volume, [m^3] +ESV = V1-V8;// effective swept volume of LP cylinder, [m^3] + +mprintf('\n (b) The effective swept volume of the LP cylinder is = %f litres\n',ESV*10^3); + +// (c) +T9 = T1; +P9 = P3; +T4 = T9*(P4/P9)^((n-1)/n);// delivery temperature, [K] +// now using P4*(V4-V5)/T4=P1*(V1-V8)/T1 +V4_minus_V5 = P1*T4*(V1-V8)/(P4*T1);// delivery volume, [m^3] + +mprintf('\n (c) The delivery temperature is = %f C\n',T4-273); +mprintf('\n The delivery volume is = %f litres\n',V4_minus_V5*10^3); + +// (d) + +W = 3*n*R*T1*((P2/P1)^((n-1)/n)-1)/(n-1);// work done/kg ,[kJ] +mprintf('\n (d) The work done per kilogram of air is = %f kJ\n',W); + +// End diff --git a/2705/CH14/EX14.4/Ex14_4.sce b/2705/CH14/EX14.4/Ex14_4.sce new file mode 100755 index 000000000..8cfa8e686 --- /dev/null +++ b/2705/CH14/EX14.4/Ex14_4.sce @@ -0,0 +1,38 @@ +clear; +clc; +disp(' Example 14.4'); + +// aim : To determine +// (a) the final pressure and temperature +// (b) the energy required to drive the compressor + +// given values +rv = 5;// pressure compression ratio +m_dot = 10;// air flow rate, [kg/s] +P1 = 100;// initial pressure, [kN/m^2] +T1 = 273+20;// initial temperature, [K] +n_com = .85;// isentropic efficiency of compressor +Gama = 1.4;// heat capacity ratio +cp = 1.005;// specific heat capacity, [kJ/kg K] + +// solution +// (a) +T2_prim = T1*(rv)^((Gama-1)/Gama);// temperature after compression, [K] +// using isentropic efficiency=(T2_prim-T1)/(T2-T1) +T2 = T1+(T2_prim-T1)/n_com;// final temperature, [K] +P2 = rv*P1;// final pressure, [kN/m^2] +mprintf('\n (a) The final temperature is = %f C\n',T2-273); +mprintf('\n (b) The final pressure is = %f kN/m^2\n',P2); + +// (b) +E = m_dot*cp*(T1-T2);// energy required, [kW] +mprintf('\n (b) The energy required to drive the compressor is = %f kW',E); +if(E<0) + disp('The negative sign indicates energy input'); +else + disp('The positive sign indicates energy output'); +end + + // End + + diff --git a/2705/CH14/EX14.5/Ex14_5.sce b/2705/CH14/EX14.5/Ex14_5.sce new file mode 100755 index 000000000..c9d23cd92 --- /dev/null +++ b/2705/CH14/EX14.5/Ex14_5.sce @@ -0,0 +1,28 @@ +clear; +clc; +disp(' Example 14.5'); + +// aim : To determine +// the power absorbed in driving the compressor + +// given values +FC = .68;// fuel consumption rate, [kg/min] +P1 = 93;// initial pressure, [kN/m^2] +P2 = 200;// final pressure, [kN/m^2] +T1 = 273+15;// initial temperature, [K] +d = 1.3;// density of mixture, [kg/m^3] +n_com = .82;// isentropic efficiency of compressor +Gama = 1.38;// heat capacity ratio + +// solution +R = P1/(d*T1);// gas constant, [kJ/kg K] +// for mixture +cp = Gama*R/(Gama-1);// heat capacity, [kJ/kg K] +T2_prim = T1*(P2/P1)^((Gama-1)/Gama);// temperature after compression, [K] +// using isentropic efficiency=(T2_prim-T1)/(T2-T1) +T2 = T1+(T2_prim-T1)/n_com;// final temperature, [K] +m_dot = FC*15/60;// given condition, [kg/s] +P = m_dot*cp*(T2-T1);// power absorbed by compressor, [kW] +mprintf('\n The power absorbed by compressor is = %f kW\n',P); + +// End diff --git a/2705/CH14/EX14.6/Ex14_6.sce b/2705/CH14/EX14.6/Ex14_6.sce new file mode 100755 index 000000000..7e971e4b6 --- /dev/null +++ b/2705/CH14/EX14.6/Ex14_6.sce @@ -0,0 +1,21 @@ +clear; +clc; +disp(' Example 14.6'); + +// aim : To determine +// the power required to drive the blower + +// given values +m_dot = 1;// air capacity, [kg/s] +rp = 2;// pressure ratio +P1 = 1*10^5;// intake pressure, [N/m^2] +T1 = 273+70;// intake temperature, [K] +R = .29;// gas constant, [kJ/kg k] + +// solution +V1_dot = m_dot*R*T1/P1*10^3;// [m^3/s] +P2 = rp*P1;// final pressure, [n/m^2] +P = V1_dot*(P2-P1);// power required, [W] +mprintf('\n The power required to drive the blower is = %f kW\n',P*10^-3); + +// End diff --git a/2705/CH14/EX14.7/Ex14_7.sce b/2705/CH14/EX14.7/Ex14_7.sce new file mode 100755 index 000000000..a367f2e9e --- /dev/null +++ b/2705/CH14/EX14.7/Ex14_7.sce @@ -0,0 +1,25 @@ +clear; +clc; +disp(' Example 14.7'); + +// aim : To determine +// the power required to drive the vane pump + +// given values +m_dot = 1;// air capacity, [kg/s] +rp = 2;// pressure ratio +P1 = 1*10^5;// intake pressure, [N/m^2] +T1 = 273+70;// intake temperature, [K] +Gama = 1.4;// heat capacity ratio +rv = .7;// volume ratio + +// solution +V1 = .995;// intake pressure(as given previous question),[m^3/s] +// using P1*V1^Gama=P2*V2^Gama, so +P2 = P1*(1/rv)^Gama;// pressure, [N/m^2] +V2 = rv*V1;// volume,[m^3/s] +P3 = rp*P1;// final pressure, [N/m^2] +P = Gama/(Gama-1)*P1*V1*((P2/P1)^((Gama-1)/Gama)-1)+V2*(P3-P2);// power required,[W] +mprintf('\n The power required to drive the vane pump is = %f kW\n',P*10^-3); + +// End diff --git a/2705/CH14/EX14.8/Ex14_8.sce b/2705/CH14/EX14.8/Ex14_8.sce new file mode 100755 index 000000000..e851d6a3b --- /dev/null +++ b/2705/CH14/EX14.8/Ex14_8.sce @@ -0,0 +1,36 @@ +clear; +clc; +disp(' Example 14.8'); + +// aim : To determine +// the total temperature and pressure of the mixture + +// given values +rp = 2.5;// static pressure ratio +FC = .04;// fuel consumption rate, [kg/min] +P1 = 60;// inilet pressure, [kN/m^2] +T1 = 273+5;// inilet temperature, [K] +n_com = .84;// isentropic efficiency of compressor +Gama = 1.39;// heat capacity ratio +C2 = 120;//exit velocity from compressor, [m/s] +rm = 13;// air-fuel ratio +cp = 1.005;// heat capacity ratio + +// solution +P2 = rp*P1;// given condition, [kN/m^2] +T2_prim = T1*(P2/P1)^((Gama-1)/Gama);// temperature after compression, [K] +// using isentropic efficiency=(T2_prim-T1)/(T2-T1) +T2 = T1+(T2_prim-T1)/n_com;// final temperature, [K] +m_dot = FC*(rm+1);// mass of air-fuel mixture, [kg/s] +P = m_dot*cp*(T2-T1);// power to drive compressor, [kW] +mprintf('\n The power required to drive compressor is = %f kW\n',P); + +Tt2 = T2+C2^2/(2*cp*10^3);// total temperature,[K] +Pt2 = P2*(Tt2/T2)^(Gama/(Gama-1));// total pressure, [kN/m^2] +mprintf('\n The temperature in the engine is = %f C\n',Tt2-273); +mprintf('\n The pressure in the engine cylinder is = %f kN/m^2\n',Pt2); + +// There is calculation mistake in the book + + +// End diff --git a/2705/CH15/EX15.1/Ex15_1.sce b/2705/CH15/EX15.1/Ex15_1.sce new file mode 100755 index 000000000..153a3eea6 --- /dev/null +++ b/2705/CH15/EX15.1/Ex15_1.sce @@ -0,0 +1,17 @@ +clear; +clc; +disp('Example 15.1'); + +// aim : To determine +// the thermal efficiency of the cycle + +// given values +T1 = 273+400;// temperature limit, [K] +T3 = 273+70;// temperature limit, [K] + +// solution +// using equation [15] of section 15.3 +n_the = (T1-T3)/T1*100;// thermal efficiency +mprintf('\n The thermal efficiency of the cycle is = %f percent\n',n_the); + +// End diff --git a/2705/CH15/EX15.10/Ex15_10.sce b/2705/CH15/EX15.10/Ex15_10.sce new file mode 100755 index 000000000..5748bff66 --- /dev/null +++ b/2705/CH15/EX15.10/Ex15_10.sce @@ -0,0 +1,32 @@ +clear; +clc; +disp('Example 10'); + +// aim : To determine +// (a) the maximum temperature attained during the cycle +// (b) the thermal efficiency of the cycle + +// given value +rva =7.5;// volume ratio of adiabatic expansion +rvc =15;// volume ratio of compression +P1 = 98;// initial pressure, [kn/m^2] +T1 = 273+44;// initial temperature, [K] +P4 = 258;// pressure at the end of the adiabatic expansion, [kN/m^2] +Gama = 1.4;// heat capacity ratio + +// solution +// by seeing diagram +// for process 4-1, P4/T4=P1/T1 +T4 = T1*(P4/P1);// [K] +// for process 3-4 +T3 = T4*(rva)^(Gama-1); +mprintf('\n (a) The maximum temperature during the cycle is = %f C\n',T3-273); + +// (b) + +// for process 1-2, +T2 = T1*(rvc)^(Gama-1);// [K] +n_the = 1-(T4-T1)/((Gama)*(T3-T2));// thermal efficiency +mprintf('\n (b) The thermal efficiency of the cycle is = %f percent\n',n_the*100); + +// End diff --git a/2705/CH15/EX15.11/Ex15_11.sce b/2705/CH15/EX15.11/Ex15_11.sce new file mode 100755 index 000000000..9045f1cd3 --- /dev/null +++ b/2705/CH15/EX15.11/Ex15_11.sce @@ -0,0 +1,37 @@ +clear; +clc; +disp('Example 15.11'); + +// aim : To determine +// (a) the thermal efficiency of the cycle +// (b) the indicared power of the cycle + +// given values +// taking basis one second +rv = 11;// volume ratio +P1 = 96;// initial pressure , [kN/m^2] +T1 = 273+18;// initial temperature, [K] +Gama = 1.4;// heat capacity ratio + +// solution +// taking reference Fig. 15.24 +// (a) +Beta = 2;// ratio of V3 and V2 +TE = 1-(Beta^(Gama)-1)/((rv^(Gama-1))*Gama*(Beta-1));// thermal efficiency +mprintf('\n (a) the thermal efficiency of the cycle is = %f percent\n ',TE*100); + +// (b) +// let V1-V2=.05, so +V2 = .05*.1;// [m^3] +// from this +V1 = rv*V2;// [m^3] +V3 = Beta*V2;// [m^3] +V4 = V1;// [m^3] +P2 = P1*(V1/V2)^(Gama);// [kN/m^2] +P3 = P2;// [kn/m^2] +P4=P3*(V3/V4)^(Gama);// [kN/m^2] +// indicated power +W = P2*(V3-V2)+((P3*V3-P4*V4)-(P2*V2-P1*V1))/(Gama-1);// indicated power, [kW] +mprintf('\n (c) The indicated power of the cycle is = %f kW\n',W); + +// End diff --git a/2705/CH15/EX15.12/Ex15_12.sce b/2705/CH15/EX15.12/Ex15_12.sce new file mode 100755 index 000000000..43bddcdf8 --- /dev/null +++ b/2705/CH15/EX15.12/Ex15_12.sce @@ -0,0 +1,48 @@ +clear; +clc; +disp('Example 15.12'); + +// aim : To determine +// (a) the pressure and temperature at the end of compression +// (b) the pressure and temperature at the end of the constant volume process +// (c) the temperature at the end of constant pressure process + +// given values +P1 = 103;// initial pressure, [kN/m^2] +T1 = 273+22;// initial temperature, [K] +rv = 16;// volume ratio of the compression +Q = 244;//heat added, [kJ/kg] +Gama = 1.4;// heat capacity ratio +cv = .717;// heat capacity, [kJ/kg k] + +// solution +// taking reference as Fig.15.26 +// (a) +// for compression +// rv = V1/V2 +P2 = P1*(rv)^Gama;// pressure at end of compression, [kN/m^2] +T2 = T1*(rv)^(Gama-1);// temperature at end of compression, [K] +mprintf('\n (a) The pressure at the end of compression is = %f MN/m^2\n',P2*10^-3); +mprintf('\n The temperature at the end of compression is = %f C\n',T2-273); + +// (b) +// for constant volume process, +// Q = cv*(T3-T2), so +T3 = T2+Q/cv;// temperature at the end of constant volume, [K] + +// so for constant volume, P/T=constant, hence +P3 = P2*(T3/T2);// pressure at the end of constant volume process, [kN/m^2] +mprintf('\n (b) The pressure at the end of constant volume process is = %f MN/m^2\n ',P3*10^-3); +mprintf('\n The temperature at the end of constant volume process is = %f C\n',T3-273); + +// (c) +S = rv-1;// stroke +// assuming +V3 = 1;// [volume] +//so +V4 = V3+S*.03;// [volume] +// also for constant process V/T=constant, hence +T4 = T3*(V4/V3);// temperature at the end of constant presure process, [k] +mprintf('\n (c) The temperature at the end of constant pressure process is = %f C\n',T4-273); + +// End diff --git a/2705/CH15/EX15.13/Ex15_13.sce b/2705/CH15/EX15.13/Ex15_13.sce new file mode 100755 index 000000000..50608fe00 --- /dev/null +++ b/2705/CH15/EX15.13/Ex15_13.sce @@ -0,0 +1,79 @@ +clear; +clc; +disp('Example 15.13'); + +// aim : To determine +// (a) the pressure, volume and temperature at cycle process change points +// (b) the net work done +// (c) the thermal efficiency +// (d) the heat received +// (e) the work ratio +// (f) the mean effective pressure +// (g) the carnot efficiency + + +// given values +rv = 15;// volume ratio +P1 = 97*10^-3;// initial pressure , [MN/m^2] +V1 = .084;// initial volume, [m^3] +T1 = 273+28;// initial temperature, [K] +T4 = 273+1320;// maximum temperature, [K] +P3 = 6.2;// maximum pressure, [MN/m^2] +cp = 1.005;// specific heat capacity at constant pressure, [kJ/kg K] +cv = .717;// specific heat capacity at constant volume, [kJ/kg K] + +// solution +// taking reference Fig. 15.27 +// (a) +R = cp-cv;// gas constant, [kJ/kg K] +Gama = cp/cv;// heat capacity ratio +// for process 1-2 +V2 = V1/rv;// volume at stage2, [m^3] +// using PV^(Gama)=constant for process 1-2 +P2 = P1*(V1/V2)^(Gama);// pressure at stage2,. [MN/m^2] +T2 = T1*(V1/V2)^(Gama-1);// temperature at stage 2, [K] + +// for process 2-3 +// since volumee is constant in process 2-3 , so using P/T=constant, so +T3 = T2*(P3/P2);// volume at stage 3, [K] +V3 = V2;// volume at stage 3, [MN/m^2] + +// for process 3-4 +P4 = P3;// pressure at stage 4, [m^3] +// since in stage 3-4 P is constant, so V/T=constant, +V4 = V3*(T4/T3);// temperature at stage 4,[K] + +// for process 4-5 +V5 = V1;// volume at stage 5, [m^3] +P5 = P4*(V4/V5)^(Gama);// pressure at stage5,. [MN/m^2] +T5 = T4*(V4/V5)^(Gama-1);// temperature at stage 5, [K] + +mprintf('\n (a) P1 = %f kN/m^2, V1 = %f m^3, t1 = %f C,\n P2 = %f MN/m^2, V2 = %f m^3, t2 = %f C,\n P3 = %f MN/m^2, V3 = %f m^3, t3 = %f C,\n P4 = %f MN/m^2, V4 = %f m^3, t4 = %f C,\n P5 = %fkN/m^2, V5 = %fm^3, t5 = %fC\n',P1*10^3,V1,T1-273,P2,V2,T2-273,P3,V3,T3-273,P4,V4,T4-273,P5*10^3,V5,T5-273); + + +// (b) +W = (P3*(V4-V3)+((P4*V4-P5*V5)-(P2*V2-P1*V1))/(Gama-1))*10^3;// work done, [kJ] +mprintf('\n (b) The net work done is = %f kJ\n',W); + +// (c) +TE = 1-(T5-T1)/((T3-T2)+Gama*(T4-T3));// thermal efficiency +mprintf('\n (c) The thermal efficiency is = %f percent\n',TE*100); + +// (d) +Q = W/TE;// heat received, [kJ] +mprintf('\n (d) The heat received is = %f kJ\n',Q); + +// (e) +PW = P3*(V4-V3)+(P4*V4-P5*V5)/(Gama-1) +WR = W*10^-3/PW;// work ratio +mprintf('\n (f) The work ratio is = %f\n',WR); + +// (e) +Pm = W/(V1-V2);// mean effective pressure, [kN/m^2] +mprintf('\n (e) The mean effective pressure is = %f kN/m^2\n',Pm); + +// (f) +CE = (T4-T1)/T4;// carnot efficiency +mprintf('\n (f) The carnot efficiency is = %f percent\n',CE*100); + +// End diff --git a/2705/CH15/EX15.14/Ex15_14.sce b/2705/CH15/EX15.14/Ex15_14.sce new file mode 100755 index 000000000..ebf067b5b --- /dev/null +++ b/2705/CH15/EX15.14/Ex15_14.sce @@ -0,0 +1,77 @@ +clear; +clc; +disp('Example 15.14'); + +// aim : To determine +// (a) the thermal efficiency +// (b) the heat received +// (c) the heat rejected +// (d) the net work +// (e) the work ratio +// (f) the mean effective pressure +// (g) the carnot efficiency + + +// given values +P1 = 101;// initial pressure , [kN/m^2] +V1 = 14*10^-3;// initial volume, [m^3] +T1 = 273+15;// initial temperature, [K] +P3 = 1850;// maximum pressure, [kN/m^2] +V2 = 2.8*10^-3;// compressed volume, [m^3] +Gama = 1.4;// heat capacity +R = .29;// gas constant, [kJ/kg k] + +// solution +// taking reference Fig. 15.29 +// (a) +// for process 1-2 +// using PV^(Gama)=constant for process 1-2 +P2 = P1*(V1/V2)^(Gama);// pressure at stage2,. [MN/m^2] +T2 = T1*(V1/V2)^(Gama-1);// temperature at stage 2, [K] + +// for process 2-3 +// since volumee is constant in process 2-3 , so using P/T=constant, so +T3 = T2*(P3/P2);// volume at stage 3, [K] + +// for process 3-4 +P4 = P1; +T4 = T3*(P4/P3)^((Gama-1)/Gama);// temperature + +TE = 1-Gama*(T4-T1)/(T3-T2);// thermal efficiency +mprintf('\n (a) The thermal efficiency is = %f percent\n',TE*100); + +// (b) +cv = R/(Gama-1);// heat capacity at copnstant volume, [kJ/kg k] +m = P1*V1/(R*T1);// mass of gas, [kg] +Q1 = m*cv*(T3-T2);// heat received, [kJ/cycle] +mprintf('\n (b) The heat received is = %f kJ/cycle\n',Q1); + +// (c) +cp = Gama*cv;// heat capacity at constant at constant pressure, [kJ/kg K] +Q2 = m*cp*(T4-T1);// heat rejected, [kJ/cycle] +mprintf('\n (c) The heat rejected is = %f kJ/cycle\n',Q2); + +// (d) +W = Q1-Q2;// net work , [kJ/cycle] +mprintf('\n (d) The net work is = %f kJ/cycle\n',W); + +// (e) +// pressure is constant for process 1-4, so V/T=constant +V4 = V1*(T4/T1);// volume, [m^3] +V3 = V2;// for process 2-3 +P4 = P1;// for process 1-4 +PW = (P3*V3-P1*V1)/(Gama-1);// positive work done, [kJ/cycle] +WR = W/PW;// work ratio +mprintf('\n (e) The work ratio is = %f\n',WR); + +// (f) +Pm = W/(V4-V2);// mean effective pressure, [kN/m^2] +mprintf('\n (f) The mean effefctive pressure is = %f kN/m^2\n',Pm); + +// (g) +CE = (T3-T1)/T3;// carnot efficiency +mprintf('\n (g) The carnot efficiency is = %f percent\n',CE*100); + +// there is minor variation in answer reported in the book + +// End diff --git a/2705/CH15/EX15.15/Ex15_15.sce b/2705/CH15/EX15.15/Ex15_15.sce new file mode 100755 index 000000000..d6b24e902 --- /dev/null +++ b/2705/CH15/EX15.15/Ex15_15.sce @@ -0,0 +1,41 @@ +clear; +clc; +disp('Example 15.15'); + +// aim : To determine +// (a) the net work done +// (b) the ideal thermal efficiency +// (c) the thermal efficiency if the process of generation is not included + +// given values +P1 = 110;// initial pressure, [kN/m^2) +T1 = 273+30;// initial temperature, [K] +V1 = .05;// initial volume, [m^3] +V2 = .005;// volume, [m^3] +T3 = 273+700;// temperature, [m^3] +R = .289;// gas constant, [kJ/kg K] +cv = .718;// heat capacity, [kJ/kg K] + +// solution +// (a) +m = P1*V1/(R*T1);// mass , [kg] +W = m*R*(T3-T1)*log(V1/V2);// work done, [kJ] +mprintf('\n (a) The net work done is = %f kJ\n',W); + +// (b) +n_the = (T3-T1)/T3;// ideal thermal efficiency +mprintf('\n (b) The ideal thermal efficiency is = %f percent\n',n_the*100); + +// (c) +V4 = V1; +V3 = V2; +T4 = T3; +T2 = T1; + +Q_rej = m*cv*(T4-T1)+m*R*T1*log(V1/V2);// heat rejected +Q_rec = m*cv*(T3-T2)+m*R*T3*log(V4/V3);// heat received + +n_th = (1-Q_rej/Q_rec);// thermal efficiency +mprintf('\n (c) the thermal efficiency if the process of regeneration is not included is = %f percent\n',n_th*100); + +// End diff --git a/2705/CH15/EX15.16/Ex15_16.sce b/2705/CH15/EX15.16/Ex15_16.sce new file mode 100755 index 000000000..5417fa898 --- /dev/null +++ b/2705/CH15/EX15.16/Ex15_16.sce @@ -0,0 +1,48 @@ +clear; +clc; +disp('Example 15.16'); + +// aim : To determine +// (a) the maximum temperature +// (b) the net work done +// (c) the ideal thermal efficiency +// (d) the thermal efficiency if the process of regeneration is not included + +// given values +P1 = 100;// initial pressure, [kN/m^2) +T1 = 273+20;// initial temperature, [K] +V1 = .08;// initial volume, [m^3] +rv = 5;// volume ratio +R = .287;// gas constant, [kJ/kg K] +cp = 1.006;// heat capacity, [kJ/kg K] +V3_by_V2 = 2; + +// solution +// (a) +// using Fig.15.33 +// process 1-2 is isothermal +T2 = T1; +// since process 2-3 isisobaric, so V/T=constant +T3 = T2*(V3_by_V2);// maximumtemperature, [K] +mprintf('\n (a) The maximum temperature is = %f C\n',T3-273); + +// (b) +m = P1*V1/(R*T1);// mass , [kg] +W = m*R*(T3-T1)*log(rv);// work done, [kJ] +mprintf('\n (b) The net work done is = %f kJ\n',W); + +// (c) +TE = (T3-T1)/T3;// ideal thermal efficiency +mprintf('\n (c) The ideal thermal efficiency is = %f percent\n',TE*100); + +// (d) +T4 = T3; +T2 = T1; + +Q_rej = m*cp*(T4-T1)+m*R*T1*log(rv);// heat rejected +Q_rec = m*cp*(T3-T2)+m*R*T3*log(rv);// heat received + +n_th = (1-Q_rej/Q_rec);// thermal efficiency +mprintf('\n (d) the thermal efficiency if the process of regeneration is not included is = %f percent\n',n_th*100); + +// End diff --git a/2705/CH15/EX15.17/Ex15_17.sce b/2705/CH15/EX15.17/Ex15_17.sce new file mode 100755 index 000000000..5182839fe --- /dev/null +++ b/2705/CH15/EX15.17/Ex15_17.sce @@ -0,0 +1,36 @@ +clear; +clc; +disp('Example 15.17'); + +// aim : To determine +// (a) the net work done +// (b) thethermal efficiency + +// given values +m = 1;// mass of air, [kg] +T1 = 273+230;// initial temperature, [K] +P1 = 3450;// initial pressure, [kN/m^2] +P2 = 2000;// pressure, [kN/m^2] +P3 = 140;// pressure, [kN/m^2] +P4 = P3; +Gama = 1.4; // heat capacity ratio +cp = 1.006;// heat capacity, [kJ/kg k] + +// solution +T2 =T1;// isothermal process 1-2 +// process 2-3 and 1-4 are adiabatic so +T3 = T2*(P3/P2)^((Gama-1)/Gama);// temperature, [K] +T4 = T1*(P4/P1)^((Gama-1)/Gama);// [K] +R = cp*(Gama-1)/Gama;// gas constant, [kJ/kg K] +Q1 = m*R*T1*log(P1/P2);// heat received, [kJ] +Q2 = m*cp*(T3-T4);// heat rejected + +//hence +W = Q1-Q2;// work done +mprintf('\n (a) The net work done is = %f kJ\n',W); + +// (b) +TE = 1-Q2/Q1;// thermal efficiency +mprintf('\n (b) The thermal efficiency is = %f percent\n',TE*100); + +// End diff --git a/2705/CH15/EX15.18/Ex15_18.sce b/2705/CH15/EX15.18/Ex15_18.sce new file mode 100755 index 000000000..2932ac9d0 --- /dev/null +++ b/2705/CH15/EX15.18/Ex15_18.sce @@ -0,0 +1,22 @@ +clear; +clc; +disp('Example 15.18'); + +// aim : To determine +// thermal eficiency +// carnot efficiency + +// given values +rv = 5;// volume ratio +Gama = 1.4;// heat capacity ratio + +// solution +// under given condition + +TE = 1-(1/Gama*(2-1/rv^(Gama-1)))/(1+2*((Gama-1)/Gama)*log(rv/2));// thermal efficiency +mprintf('\n The thermal efficiency is = %f percent\n',TE*100); + +CE = 1-1/(2*rv^(Gama-1));// carnot efficiency +mprintf('\n The carnot efficiency is = %f \n',CE*100); + +// End diff --git a/2705/CH15/EX15.2/Ex15_2.sce b/2705/CH15/EX15.2/Ex15_2.sce new file mode 100755 index 000000000..2d6440345 --- /dev/null +++ b/2705/CH15/EX15.2/Ex15_2.sce @@ -0,0 +1,29 @@ +clear; +clc; +disp('Example 15.2'); + +// aim : To determine +// (a) the volume ratios of the isothermal and adiabatic processes +// (b) the thermal efficiency of the cycle + +// given values +T1 = 273+260;// temperature, [K] +T3 = 273+21;// temperature, [K] +er = 15;// expansion ratio +Gama = 1.4;// heat capacity ratio + +// solution +// (a) +T2 = T1; +T4 = T3; +// for adiabatic process +rva = (T1/T4)^(1/(Gama-1));// volume ratio of adiabatic +rvi = er/rva;// volume ratio of isothermal +mprintf('\n (a) The volume ratio of the adiabatic process is = %f\n',rva); +mprintf('\n The volume ratio of the isothermal process is = %f\n',rvi); + +// (b) +n_the = (T1-T4)/T1*100;// thermal efficiency +mprintf('\n (b) The thermal efficiency of the cycle is = %f percent\n',n_the); + +// End diff --git a/2705/CH15/EX15.3/Ex15_3.sce b/2705/CH15/EX15.3/Ex15_3.sce new file mode 100755 index 000000000..8cab979fa --- /dev/null +++ b/2705/CH15/EX15.3/Ex15_3.sce @@ -0,0 +1,69 @@ +clear; +clc; +disp('Example 15.3'); + +// aim : To determine +// (a) the pressure, volume and temperature at each corner of the cycle +// (b) the thermal efficiency of the cycle +// (c) the work done per cycle +// (d) the work ratio + +// given values +m = 1;// mass of air, [kg] +P1 = 1730;// initial pressure of carnot engine, [kN/m^2] +T1 = 273+300;// initial temperature, [K] +R = .29;// [kJ/kg K] +Gama = 1.4;// heat capacity ratio + +// solution +// taking reference Fig. 15.15 +// (a) +// for the isothermal process 1-2 +// using ideal gas law +V1 = m*R*T1/P1;// initial volume, [m^3] +T2 = T1; +V2 = 3*V1;// given condition +// for isothermal process, P1*V1=P2*V2, so +P2 = P1*(V1/V2);// [MN/m^2] +// for the adiabatic process 2-3 +V3 = 6*V1;// given condition +T3 = T2*(V2/V3)^(Gama-1); +// also for adiabatic process, P2*V2^Gama=P3*V3^Gama, so +P3 = P2*(V2/V3)^Gama; +// for the isothermal process 3-4 +T4 = T3; +// for both adiabatic processes, the temperataure ratio is same, +// T1/T4 = T2/T3=(V4/V1)^(Gama-1)=(V3/V2)^(Gama-1), so +V4 = 2*V1; +// for isothermal process, 3-4, P3*V3=P4*V4, so +P4 = P3*(V3/V4); +disp('(a) At line 1'); +mprintf('\n V1 = %f m^3, t1 = %f C, P1 = %f kN/m^2\n',V1,T1-273,P1); + +disp('At line 2'); +mprintf('\n V2 = %f m^3, t2 = %f C, P2 = %f kN/m^2\n',V2,T2-273,P2); + +disp('At line 3'); +mprintf('\n V3 = %f m^3, t3 = %f C, P3 = %f kN/m^2\n',V3,T3-273,P3); + + +disp('At line 4'); +mprintf('\n V4 = %f m^3, t4 = %f C, P4 = %f kN/m^2\n',V4,T4-273,P4); + + +// (b) +n_the = (T1-T3)/T1;// thermal efficiency +mprintf('\n (b) The thermal efficiency of the cycle is = %f percent\n',n_the*100); + +// (c) +W = m*R*T1*log(V2/V1)*n_the;// work done, [J] +mprintf('\n (c) The work done per cycle is = %f kJ\n',W); + +// (d) +wr = (T1-T3)*log(V2/V1)/(T1*log(V2/V1)+(T1-T3)/(Gama-1));// work ratio +mprintf('\n (d) The work ratio is = %f\n',wr); + +// there is calculation mistake in the book so answer is not matching + +// End + diff --git a/2705/CH15/EX15.4/Ex15_4.sce b/2705/CH15/EX15.4/Ex15_4.sce new file mode 100755 index 000000000..e13c274f1 --- /dev/null +++ b/2705/CH15/EX15.4/Ex15_4.sce @@ -0,0 +1,72 @@ +clear; +clc; +disp('Example 15.4'); + +// aim : To determine +// (a) the pressure, volume and temperature at cycle state points +// (b) the heat received +// (c) the work done +// (d) the thermal efficiency +// (e) the carnot efficiency +// (f) the work ration +// (g) the mean effective pressure + +// given values +ro = 8;// overall volume ratio; +rv = 6;// volume ratio of adiabatic compression +P1 = 100;// initial pressure , [kN/m^2] +V1 = .084;// initial volume, [m^3] +T1 = 273+28;// initial temperature, [K] +Gama = 1.4;// heat capacity ratio +cp = 1.006;// specific heat capacity, [kJ/kg K] + +// solution +// taking reference Fig. 15.18 +// (a) +V2 = V1/rv;// volume at stage2, [m^3] +V4 = ro*V2;// volume at stage 4;[m^3] +// using PV^(Gama)=constant for process 1-2 +P2 = P1*(V1/V2)^(Gama);// pressure at stage2,. [kN/m^2] +T2 = T1*(V1/V2)^(Gama-1);// [K] + +P3 = P2;// pressure at stage 3, [kN/m^2] +V3 = V4/rv;// volume at stage 3, [m^3] +// since pressure is constant in process 2-3 , so using V/T=constant, so +T3 = T2*(V3/V2);// temperature at stage 3, [K] + +// for process 1-4 +T4 = T1*(V4/V1);// temperature at stage4, [K +P4 = P1;// pressure at stage4, [kN/m^2] + +mprintf('\n (a) P1 = %f kN/m^2, V1 = %f m^3, t1 = %f C,\n P2 = %f kN/m^2, V2 = %f m^3, t2 = %f C,\n P3 = %f kN/m^2, V3 = %f m^3, t3 = %f C,\n P4 = %f kN/m^2, V4 = %f m^3, t4 = %f C\n',P1,V1,T1-273,P2,V2,T2-273,P3,V3,T3-273,P4,V4,T4-273); + +// (b) +R = cp*(Gama-1)/Gama;// gas constant, [kJ/kg K] +m = P1*V1/(R*T1);// mass of gas, [kg] +Q = m*cp*(T3-T2);// heat received, [kJ] +mprintf('\n (b) The heat received is = %f kJ\n',Q); + +// (c) +W = P2*(V3-V2)-P1*(V4-V1)+((P3*V3-P4*V4)-(P2*V2-P1*V1))/(Gama-1);// work done, [kJ] +mprintf('\n (c) The work done is = %f kJ\n',W); + +// (d) +TE = 1-T1/T2;// thermal efficiency +mprintf('\n (d) The thermal efficiency is = %f percent\n',TE*100); + +// (e) +CE = (T3-T1)/T3;// carnot efficiency +mprintf('\n (e) The carnot efficiency is = %f percent\n',CE*100); + +// (f) +PW = P2*(V3-V2)+(P3*V3-P4*V4)/(Gama-1);// positive work done, [kj] +WR = W/PW;// work ratio +mprintf('\n (f) The work ratio is = %f\n',WR); + +// (g) +Pm = W/(V4-V2);// mean effective pressure, [kN/m^2] +mprintf('\n (g) The mean effective pressure is = %f kN/m^2\n',Pm); + +// there is minor variation in answer reported in the book + +// End diff --git a/2705/CH15/EX15.5/Ex15_5.sce b/2705/CH15/EX15.5/Ex15_5.sce new file mode 100755 index 000000000..7de39d424 --- /dev/null +++ b/2705/CH15/EX15.5/Ex15_5.sce @@ -0,0 +1,28 @@ +clear; +clc; +disp('Example 15.5'); + +// aim : To determine +// (a) the actual thermal efficiency of the turbine +// (b) the specific fuel consumption of the turbine in kg/kWh + +// given values +P2_by_P1 = 8; +n_tur = .6;// ideal turbine thermal efficiency +c = 43*10^3;// calorific value of fuel, [kJ/kg] +Gama = 1.4;// heat capacity ratio + +// solution +// (a) +rv = P2_by_P1; +n_tur_ide = 1-1/(P2_by_P1)^((Gama-1)/Gama);// ideal thermal efficiency +ate = n_tur_ide*n_tur;// actual thermal efficiency +mprintf('\n (a) The actual thermal efficiency of the turbine is = %f percent\n',ate*100); + +// (b) +ewf = c*ate;// energy to work fuel, [kJ/kg] +kWh = 3600;// energy equivalent ,[kJ] +sfc = kWh/ewf;// specific fuel consumption, [kg/kWh] +mprintf('\n (b) The specific fuel consumption of the turbine is = %f kg/kWh',sfc); + +// End diff --git a/2705/CH15/EX15.6/Ex15_6.sce b/2705/CH15/EX15.6/Ex15_6.sce new file mode 100755 index 000000000..d0bad01c2 --- /dev/null +++ b/2705/CH15/EX15.6/Ex15_6.sce @@ -0,0 +1,23 @@ +clear; +clc; +disp('Example 15.6'); + +// aim : To determine +// the relative efficiency of the engine + +// given values +d = 80;// bore, [mm] +l = 85;// stroke, [mm] +V1 = .06*10^6;// clearence volume, [mm^3] +ate = .22;// actual thermal efficiency of the engine +Gama = 1.4;// heat capacity ratio + +// solution +sv = %pi*d^2/4*l;// stroke volume, [mm^3] +V2 = sv+V1;// [mm^3] +rv = V2/V1; +ite = 1-(1/rv)^(Gama-1);// ideal thermal efficiency +re = ate/ite;// relative thermal efficiency +mprintf('\n The relative efficiency of the engine is = %f percent\n',re*100); + +// End diff --git a/2705/CH15/EX15.7/Ex15_7.sce b/2705/CH15/EX15.7/Ex15_7.sce new file mode 100755 index 000000000..0f950939f --- /dev/null +++ b/2705/CH15/EX15.7/Ex15_7.sce @@ -0,0 +1,68 @@ +clear; +clc; +disp('Example 15.7'); + +// aim : To determine +// (a) the pressure, volume and temperature at each cycle process change points +// (b) the heat transferred to air +// (c) the heat rejected by the air +// (d) the ideal thermal efficiency +// (e) the work done +// (f) the mean effective pressure + +// given values +m = 1;// mass of air, [kg] +rv = 6;// volume ratio of adiabatic compression +P1 = 103;// initial pressure , [kN/m^2] +T1 = 273+100;// initial temperature, [K] +P3 = 3450;// maximum pressure, [kN/m^2] +Gama = 1.4;// heat capacity ratio +R = .287;// gas constant, [kJ/kg K] + +// solution +// taking reference Fig. 15.20 +// (a) +// for point 1 +V1 = m*R*T1/P1;// initial volume, [m^3] + +// for point 2 +V2 = V1/rv;// volume at point 2, [m^3] +// using PV^(Gama)=constant for process 1-2 +P2 = P1*(V1/V2)^(Gama);// pressure at point 2,. [kN/m^2] +T2 = T1*(V1/V2)^(Gama-1);// temperature at point 2,[K] + +// for point 3 +V3 = V2;// volume at point 3, [m^3] +// since volume is constant in process 2-3 , so using P/T=constant, so +T3 = T2*(P3/P2);// temperature at stage 3, [K] + +// for point 4 +V4 = V1;// volume at point 4, [m^3] +P4 = P3*(V3/V4)^Gama;// pressure at point 4, [kN/m^2] +// again since volume is constant in process 4-1 , so using P/T=constant, so +T4 = T1*(P4/P1);// temperature at point 4, [K] + +mprintf('\n (a) P1 = %f kN/m^2, V1 = %f m^3, t1 = %f C,\n P2 = %f kN/m^2, V2 = %f m^3, t2 = %f C,\n P3 = %f kN/m^2, V3 = %f m^3, t3 = %f C,\n P4 = %f kN/m^2, V4 = %f m^3, t4 = %f C\n',P1,V1,T1-273,P2,V2,T2-273,P3,V3,T3-273,P4,V4,T4-273); + +// (b) +cv = R/(Gama-1);// specific heat capacity, [kJ/kg K] +Q23 = m*cv*(T3-T2);// heat transferred, [kJ] +mprintf('\n (b) The heat transferred to the air is = %f kJ\n',Q23); + +// (c) +Q34 = m*cv*(T4-T1);// heat rejected by air, [kJ] +mprintf('\n (c) The heat rejected by the air is = %f kJ\n',Q34); + +// (d) +TE = 1-Q34/Q23;// ideal thermal efficiency +mprintf('\n (d) The ideal thermal efficiency is = %f percent\n',TE*100); + +// (e) +W = Q23-Q34;// work done ,[kJ] +mprintf('\n (e) The work done is = %f kJ\n',W); + +// (f) +Pm = W/(V1-V2);// mean effective pressure, [kN/m^2] +mprintf('\n (f) The mean effefctive pressure is = %f kN/m^2\n',Pm); + +// End diff --git a/2705/CH15/EX15.8/Ex15_8.sce b/2705/CH15/EX15.8/Ex15_8.sce new file mode 100755 index 000000000..a37208dcc --- /dev/null +++ b/2705/CH15/EX15.8/Ex15_8.sce @@ -0,0 +1,65 @@ +clear; +clc; +disp('Example 15.8'); + +// aim : To determine +// (a) the pressure, volume and temperature at cycle state points +// (b) the thermal efficiency +// (c) the theoretical output +// (d) the mean effective pressure +// (e) the carnot efficiency + +// given values +rv = 9;// volume ratio +P1 = 101;// initial pressure , [kN/m^2] +V1 = .003;// initial volume, [m^3] +T1 = 273+18;// initial temperature, [K] +P3 = 4500;// maximum pressure, [kN/m^2] +N = 3000; +cp = 1.006;// specific heat capacity at constant pressure, [kJ/kg K] +cv = .716;// specific heat capacity at constant volume, [kJ/kg K] + +// solution +// taking reference Fig. 15.20 +// (a) +// for process 1-2 +Gama = cp/cv;// heat capacity ratio +R = cp-cv;// gas constant, [kJ/kg K] +V2 = V1/rv;// volume at stage2, [m^3] +// using PV^(Gama)=constant for process 1-2 +P2 = P1*(V1/V2)^(Gama);// pressure at stage2,. [kN/m^2] +T2 = T1*(V1/V2)^(Gama-1);// [K] + +// for process 2-3 +V3 = V2;// volume at stage 3, [m^3] +// since volume is constant in process 2-3 , so using P/T=constant, so +T3 = T2*(P3/P2);// temperature at stage 3, [K] + +// for process 3-4 +V4 = V1;// volume at stage 4 +// using PV^(Gama)=constant for process 3-4 +P4 = P3*(V3/V4)^(Gama);// pressure at stage2,. [kN/m^2] +T4 = T3*(V3/V4)^(Gama-1);// temperature at stage 4,[K] + +mprintf('\n (a) P1 = %f kN/m^2, V1 = %f m^3, t1 = %f C,\n P2 = %f kN/m^2, V2 = %f m^3, t2 = %f C,\n P3 = %f kN/m^2, V3 = %f m^3, t3 = %f C,\n P4 = %f kN/m^2, V4 = %f m^3, t4 = %f C\n',P1,V1,T1-273,P2,V2,T2-273,P3,V3,T3-273,P4,V4,T4-273); + +// (b) +TE = 1-(T4-T1)/(T3-T2);// thermal efficiency +mprintf('\n (b) The thermal efficiency is = %f percent\n',TE*100); + +// (c) +m = P1*V1/(R*T1);// mass os gas, [kg] +W = m*cv*((T3-T2)-(T4-T1));// work done, [kJ] +Wt = W*N/60;// workdone per minute, [kW] +mprintf('\n (c) The theoretical output is = %f kW\n',Wt); + +// (d) +Pm = W/(V1-V2);// mean effective pressure, [kN/m^2] +mprintf('\n (g) The mean effefctive pressure is = %f kN/m^2\n',Pm); + +// (e) +CE = (T3-T1)/T3;// carnot efficiency +mprintf('\n (e) The carnot efficiency is = %f percent\n',CE*100); + + +// End diff --git a/2705/CH15/EX15.9/Ex15_9.sce b/2705/CH15/EX15.9/Ex15_9.sce new file mode 100755 index 000000000..7482cbe06 --- /dev/null +++ b/2705/CH15/EX15.9/Ex15_9.sce @@ -0,0 +1,72 @@ +clear; +clc; +disp('Example 15.9'); + +// aim : To determine +// (a) the pressure and temperature at cycle process change points +// (b) the work done +// (c) the thermal efficiency +// (d) the work ratio +// (e) the mean effective pressure +// (f) the carnot efficiency + + +// given values +rv = 16;// volume ratio of compression +P1 = 90;// initial pressure , [kN/m^2] +T1 = 273+40;// initial temperature, [K] +T3 = 273+1400;// maximum temperature, [K] +cp = 1.004;// specific heat capacity at constant pressure, [kJ/kg K] +Gama = 1.4;// heat capacoty ratio + +// solution +cv = cp/Gama;// specific heat capacity at constant volume, [kJ/kg K] +R = cp-cv;// gas constant, [kJ/kg K] +// for one kg of gas +V1 = R*T1/P1;// initial volume, [m^3] +// taking reference Fig. 15.22 +// (a) +// for process 1-2 +// using PV^(Gama)=constant for process 1-2 +// also rv = V1/V2 +P2 = P1*(rv)^(Gama);// pressure at stage2,. [kN/m^2] +T2 = T1*(rv)^(Gama-1);// temperature at stage 2, [K] + +// for process 2-3 +P3 = P2;// pressure at stage 3, [kN/m^2] +V2 = V1/rv;//[m^3] +// since pressure is constant in process 2-3 , so using V/T=constant, so +V3 = V2*(T3/T2);// volume at stage 3, [m^3] + +// for process 1-4 +V4 = V1;// [m^3] +P4 = P3*(V3/V4)^(Gama) +// since in stage 1-4 volume is constant, so P/T=constant, +T4 = T1*(P4/P1);// temperature at stage 4,[K] + +mprintf('\n (a) P1 = %f kN/m^2, t1 = %f C,\n P2 = %f kN/m^2, t2 = %f C,\n P3 = %f kN/m^2, t3 = %f C,\n P4 = %f kN/m^2, t4 = %f C\n',P1,T1-273,P2,T2-273,P3,T3-273,P4,T4-273); + +// (b) +W = cp*(T3-T2)-cv*(T4-T1);// work done, [kJ] +mprintf('\n (b) The work done is = %f kJ\n',W); + +// (c) +TE = 1-(T4-T1)/((T3-T2)*Gama);// thermal efficiency +mprintf('\n (c) The thermal efficiency is = %f percent\n',TE*100); + +// (d) +PW = cp*(T3-T2)+R*(T3-T4)/(Gama-1);// positive work done +WR = W/PW;// work ratio +mprintf('\n (d) The work ratio is = %f\n',WR); + +// (e) +Pm = W/(V1-V2);// mean effective pressure, [kN/m^2] +mprintf('\n (e) The mean effefctive pressure is = %f kN/m^2\n',Pm); + +// (f) +CE = (T3-T1)/T3;// carnot efficiency +mprintf('\n (f) The carnot efficiency is = %f percent\n',CE*100); + +// value of t2 printed in the book is incorrect + +// End diff --git a/2705/CH16/EX16.1/Ex16_1.sce b/2705/CH16/EX16.1/Ex16_1.sce new file mode 100755 index 000000000..b3251fa8a --- /dev/null +++ b/2705/CH16/EX16.1/Ex16_1.sce @@ -0,0 +1,45 @@ +clear; +clc; +disp('Example 16.1'); + +// aim : To determine +// (a) the net power output of the turbine plant if the turbine is coupled to the compresser +// (b) the thermal efficiency of the plant +// (c) the work ratio + +// Given values +P1 = 100;// inlet pressure of compressor, [kN/m^2] +T1 = 273+18;// inlet temperature, [K] +P2 = 8*P1;// outlet pressure of compressor, [kN/m^2] +n_com = .85;// isentropic efficiency of compressor +T3 = 273+1000;//inlet temperature of turbine, [K] +P3 = P2;// inlet pressure of turbine, [kN/m^2] +P4 = 100;// outlet pressure of turbine, [kN/m^2] +n_tur = .88;// isentropic efficiency of turbine +m_dot = 4.5;// air mass flow rate, [kg/s] +cp = 1.006;// [kJ/kg K] +Gamma = 1.4;// heat capacity ratio + +// (a) +// For the compressor +T2_prime = T1*(P2/P1)^((Gamma-1)/Gamma);// [K] +T2 = T1+(T2_prime-T1)/n_com;// exit pressure of compressor, [K] + +// for turbine +T4_prime = T3*(P4/P3)^((Gamma-1)/Gamma);// [K] +T4 = T3-(T3-T4_prime)*n_tur;// exit temperature of turbine, [K] + +P_output = m_dot*cp*((T3-T4)-(T2-T1));// [kW] +mprintf('\n (a) The net power output is = %f kW\n',P_output); + +// (b) +n_the = ((T3-T4)-(T2-T1))/(T3-T2)*100;// thermal efficiency +mprintf('\n (b) The thermal efficiency of the plant is = %f percent\n',n_the); + +// (c) +P_pos = m_dot*cp*(T3-T4);// Positive cycle work, [kW] + +W_ratio = P_output/P_pos;// work ratio +mprintf('\n (c) The work ratio is = %f\n',W_ratio) + +// End diff --git a/2705/CH16/EX16.2/Ex16_2.sce b/2705/CH16/EX16.2/Ex16_2.sce new file mode 100755 index 000000000..e76ab804c --- /dev/null +++ b/2705/CH16/EX16.2/Ex16_2.sce @@ -0,0 +1,47 @@ +clear; +clc; +disp('Example 16.2'); + +// aim : To determine +// (a) the pressure ratiowhich will give the maximum net work output +// (b) the maximum net specific work output +// (c) the thermal efficiency at maximum work output +// (d) the work ratio at maximum work output +// (e) the carnot efficiency within the cycle temperature limits + +// Given values +// taking the refrence as Fig.16.35 +T3 = 273+1080;// [K] +T1 = 273+10;// [K] +cp = 1.007;// [kJ/kg K] +Gamma = 1.41;// heat capacity ratio + +// (a) +r_pmax = (T3/T1)^((Gamma)/(Gamma-1));// maximum pressure ratio +// for maximum net work output +r_p = sqrt(r_pmax); +mprintf('\n (a) The pressure ratio which give the maximum network output is = %f\n',r_p); + +// (b) +T2 = T1*(r_p)^((Gamma-1)/Gamma);// [K] +// From equation [23] +T4 = T2; +W_max = cp*((T3-T4)-(T2-T1));// Maximum net specific work output, [kJ/kg] + +mprintf('\n (b) The maximum net specific work output is = %f kJ/kg\n',W_max); + +// (c) +W = cp*(T3-T2); +n_the = W_max/W;// thermal efficiency +mprintf('\n (c) The thermal efficiency at maximum work output is = %f percent\n ',n_the*100); + +// (d) +// From the equation [26] +W_ratio = n_the;// Work ratio +mprintf('\n (d) The work ratio at maximum work output is = %f\n',W_ratio); + +// (e) +n_carnot = (T3-T1)/T3*100;// carnot efficiency +mprintf('\n (e) The carnot efficiency within the cycle temperature limits is = %f percent\n',n_carnot); + +// End diff --git a/2705/CH16/EX16.3/Ex16_3.sce b/2705/CH16/EX16.3/Ex16_3.sce new file mode 100755 index 000000000..de93273ac --- /dev/null +++ b/2705/CH16/EX16.3/Ex16_3.sce @@ -0,0 +1,62 @@ +clc; +disp('Example 16.3'); + +// aim : To determine +// (a) the net power output of the plant +// (b) the exhaust temperature from the heat exchanger +// (c) the thermal efficiency of the plant +// (d) the thermal efficiency of the plant if there were no heat exchanger +// (e) the work ratio + +// Given values +T1 = 273+15;// temperature, [K] +P1 = 101;// pressure, [kN/m^2] +P2 = 6*P1; // [kN/m^2] +eff = .65;// effectiveness of the heat exchanger, +T3 = 273+870;// temperature, [K] +P4 = 101;// [kN/m^2] +n_com = .85;// efficiency of compressor, +n_tur = .80;// efficiency of turbine +m_dot = 4;// mass flow rate, [kg/s] +Gama = 1.4;// heat capacity ratio +cp = 1.005;// [kJ/kg K] + +// solution +// (a) +// For compressor +T2_prim = T1*(P2/P1)^((Gama-1)/Gama);// [K] + +// using n_com = (T2_prim-T1)/(T2-T1)') + +T2 = T1+(T2_prim-T1)/n_com +// For turbine +P3 = P2; +T4_prim = T3*(P4/P3)^((Gama-1)/Gama);// [K] + +T4=T3-n_tur*(T3-T4_prim); // [K] +P_out = m_dot*cp*((T3-T4)-(T2-T1));// net power output, [kW] +mprintf('\n (a) The net power output of the plant is = %f kW\n',P_out); + +// (b) +mtd = T4-T2;// maximum temperature drop for heat transfer, [K] +atd = eff*mtd;// actual temperature, [K] +et = T4-atd;// Exhaust temperature from heat exchanger, [K] +t6 = et-273;// [C] +mprintf('\n (b) The exhaust temperature from the heat exchanger is = %f C\n',t6); + +// (c) +T5 = T2+atd;// [K] +n_the = ((T3-T4)-(T2-T1))/(T3-T5)*100;// thermal effficiency +mprintf('\n (c) The thermal efficiency of the plant is = %f percent\n',n_the); + +// (d) +// with no heat exchanger +n_the = ((T3-T4)-(T2-T1))/(T3-T2)*100;// thermal efficiency without heat exchanger +mprintf('\n (d) The thermal efficiency of the plant if there wereno heat exchanger is = %f percent\n',n_the); + +// (e) +P_pos = m_dot*cp*(T3-T4);// positive cycle work;// [kW] +w_rat = P_out/P_pos;// work ratio +mprintf('\n (e) The work ratio is = %f\n',w_rat) + +// End diff --git a/2705/CH16/EX16.4/Ex16_4.sce b/2705/CH16/EX16.4/Ex16_4.sce new file mode 100755 index 000000000..01cd7211b --- /dev/null +++ b/2705/CH16/EX16.4/Ex16_4.sce @@ -0,0 +1,68 @@ +clear; +clc; +disp('Example 16.4'); + +// aim : To determine +// (a) the pressure and temperature as the air leaves the compressor turbine +// (b) the power output from the free power turbine +// (c) the thermal efficiency of the plant +// (d) the work ratio +// (e) the carnot efficiency within the cycle temperature limits + +// Given values +T1 = 273+19;// temperature, [K] +P1 = 100;// pressure, [kN/m^2] +P2 = 8*P1; // [kN/m^2] +P3 = P2;// [kN/m^2] +T3 = 273+980;// temperature, [K] +n_com = .85;// efficiency of rotary compressor +P5 = 100;// [kN/m^2] +n_cum = .88;// isentropic efficiency of combustion chamber compressor, +n_tur = .86;// isentropic efficiency of turbine +m_dot = 7;// mass flow rate of air, [kg/s] +Gama = 1.4;// heat capacity ratio +cp = 1.006;// [kJ/kg K] + +// solution +// (a) +// For compressor +T2_prim = T1*(P2/P1)^((Gama-1)/Gama);// [K] + +T2 = T1+(T2_prim-T1)/n_com;// temperature, [K] + +// for compressor turbine +// T3-T4 = T2-T1,because compressor turbine power=compressor power so +T4 = T3-(T2-T1);//turbine exit temperature, [K] +T4_prim = T3-(T3-T4)/n_cum;// [K] + +// For turbine +// T4_prim = T3*(P4/P3)^((Gama-1)/Gama) +P4 = P3*(T4_prim/T3)^(Gama/(Gama-1));// exit air pressure of air, [kN/m^2] + +mprintf('\n (a) The temperature as the air leaves the compressor turbine is = %f C\n',T4-273); +mprintf('\n The pressure as the air leaves the compressor turbine is = %f kN/m^2\n',P4); + +// (b) +T5_prim = T4*(P5/P4)^((Gama-1)/Gama);// [K] + + +T5 = T4-n_tur*(T4-T5_prim);// temperature, [K] + +PO = m_dot*cp*(T4-T5);// power output +mprintf('\n (b) The power output from the free power turbine is = %f kW\n',PO); + +// (c) + +n_the = (T4-T5)/(T3-T2)*100;// thermal effficiency +mprintf('\n (c) The thermal efficiency of the plant is = %f percent\n',n_the); + +// (d) + +WR = (T4-T5)/(T3-T5);// work ratio +mprintf('\n (d) The work ratio is = %f\n',WR); + +// (e) +CE = (T3-T1)/T3;// carnot efficiency +mprintf('\n (e) The carnot efficiency is = %f percent\n',CE*100); + +// End diff --git a/2705/CH16/EX16.5/Ex16_5.sce b/2705/CH16/EX16.5/Ex16_5.sce new file mode 100755 index 000000000..3b2cd40b4 --- /dev/null +++ b/2705/CH16/EX16.5/Ex16_5.sce @@ -0,0 +1,51 @@ +clear; +clc; +disp('Example 16.5'); + +// aim : To determine +// (a) the pressure and temperature of the air compression +// (b) the power developed by the gas turbine +// (c) the temperature and pressure of the airentering the exhaust jet as it leaves the gas turbine + +// Given values +T1 = 273-22.4;// temperature, [K] +P1 = 470;// pressure, [bar] +P2 = 30*P1; // [kN/m^2] +P3 = P2;// [kN/m^2] +T3 = 273+960;// temperature, [K] +r = 1.25;// ratio of turbine power to compressor power +n_tur = .86;// isentropic efficiency of turbine +m_dot = 80;// mass flow rate of air, [kg/s] +Gama = 1.41;// heat capacity ratio +cp = 1.05;// [kJ/kg K] + +// solution +// (a) +// For compressor +T2_prim = T1*(P2/P1)^((Gama-1)/Gama);// [K] +// using n_tur=(T2_prim-T1)/(T2-T1) +T2 = T1+(T2_prim-T1)/n_tur;// temperature, [K] + +mprintf('\n (a) The pressure of the air after compression is = %f bar\n',P2); + +mprintf('\n The temperature of the air after compression is = %f C\n',T2-273); + +// (b) +Td = r*(T2-T1);// temperature drop in turbine, [K] +PO = m_dot*cp*Td;// power output, [kW] +mprintf('\n (b) The power developed by the gas turbine is = %f MW\n',PO*10^-3); + +// (c) +t3 = T3-273;// [C] +t4 = t3-Td;// temeprerature of air leaving turbine,[K] +Tdi = Td/n_tur;// isentropic temperature drop, [K] +T4_prim = t3-Tdi+273;// temperature, [K] +// using T4_prim=T3*(P4/P3)^((Gama-1)/Gama) +P4 = P3*(T4_prim/T3)^(Gama/(Gama-1));// exit air pressure of air, [kN/m^2] + +mprintf('\n (c) The air pressure as it leaves the gas turbine is = %f bar\n',P4); + +// Result in the book is not matching because they have taken pressure in mbar but in in question it is given in bar + +// End + diff --git a/2705/CH16/EX16.6/Ex16_6.sce b/2705/CH16/EX16.6/Ex16_6.sce new file mode 100755 index 000000000..5a9318ec8 --- /dev/null +++ b/2705/CH16/EX16.6/Ex16_6.sce @@ -0,0 +1,65 @@ +clc; +disp('Example 16.6'); + +// aim : To determine +// (a) the mass of fuel oil used by the gas turbine +// (b) the mass flow of steam from the boiler +// (c) the theoretical output from the steam turbine +// (d) the overall theoretical thermal efficiency of the plant + +// given values +Po = 150;// generating plant output, [MW] +n_the1 = .35;// thermal efficiency +CV = 43;// calorific value of fuel, [MJ] +me = 400;// flow rate of exhaust gas, [kg/s] +T = 90;// boiler exit temperature, [C] +T1 = 550;// exhaust gas temperature, [C] +P2 = 10;// steam generation pressure, [MN/m^2] +T2 = 450;// boiler exit temperature, [C] +Tf = 140;// feed water temperature, [C] +n_tur = .86;// turbine efficiency +P3 = .5;// exhaust temperature, [MN/m^2] +n_boi = .92;// boiler thermal efficiency +cp = 1.1;// heat capacity, [kJ/kg] + + +// solution +// (a) +ER = Po*3600/n_the1;// energy requirement from the fuel, [MJ/h] +mf = ER/CV*10^-3;// fuel required, [tonne/h] +mprintf('\n (a) The mass of fuel oil used by the gas is = %f tonne/h\n',mf); + +// (b) + +ET = me*cp*(T1-T)*3600*n_boi;// energy transferred to steam,[kJ/h] +// from steam table +h1 = 3244;// specific enthalpy, [kJ/kg] +hf = 588.5;// specific enthalpy, [kJ/kg] +ERR = h1-hf;// energy required to raise steam, [kJ/kg] +ms = ET/ERR*10^-3;// mass flow of steam, [tonne/h] +mprintf('\n (b) The mass flow rate of steam from the boiler is = %f tonne/h\n',ms); + +// again from steam table +s1 = 6.424;// specific entropy, [kJ/kg K] +sf2 = 1.86;// specific entropy, [kJ/kg K +sg2 = 6.819;// specific entropy, [kJ/kg K] + +hf2 = 640.1;// specific enthalpy,[kJ/kg] +hg2 = 2747.5;// specific enthalpy, [kJ/kg] +// for ths process s1=s2=sf2+x2*(sg2-sf2) +s2 = s1; +// hence +x2 = (s2-sf2)/(sg2-sf2);// dryness fraction + +h2_prim = hf2+x2*(hg2-hf2);// specific enthalpy of steam, [kJ/kg] + +TO = n_tur*(h1-h2_prim);//theoretical steam turbine output, [kJ/kg] +TOt = TO*ms/3600;// total theoretical steam turbine output, [MW] + +mprintf('\n (c) The theoretical output from the steam turbine is = %f MW\n',TOt); + +// (d) +n_tho = (Po+TOt)*n_the1/Po;// overall theoretical thermal efficiency +mprintf('\n (d) The overall thermal efficiency is = %f percent\n',n_tho*100); + +// End diff --git a/2705/CH17/EX17.1/Ex17_1.sce b/2705/CH17/EX17.1/Ex17_1.sce new file mode 100755 index 000000000..1222d2ae4 --- /dev/null +++ b/2705/CH17/EX17.1/Ex17_1.sce @@ -0,0 +1,47 @@ +clear; +clc; +disp('Example 17.1'); + +// aim : To determine +// the indicated and brake output and the mechanicl efficiency +// draw up an overall energy balance and as % age + +// given values +h = 21;// height of indicator diagram, [mm] +ic = 27;// indicator calibration, [kN/m^2 per mm] +sv = 14*10^-3;// swept volume of the cylinder;,[m^3] +N = 6.6;// speed of engine, [rev/s] +ebl = 77;// effective brake load, [kg] +ebr = .7;// effective brake radious, [m] +fc = .002;// fuel consumption, [kg/s] +CV = 44000;// calorific value of fuel, [kJ/kg] +cwc = .15;// cooling water circulation, [kg/s] +Ti = 38;// cooling water inlet temperature, [C] +To = 71;// cooling water outlet temperature, [C] +c = 4.18;// specific heat capacity of water, [kJ/kg] +eeg = 33.6;// energy to exhaust gases, [kJ/s] +g = 9.81;// gravitational acceleration, [m/s^2] + +// solution +PM = ic*h;// mean effective pressure, [kN/m^2] +LA = sv;// swept volume of the cylinder, [m^3] +ip = PM*LA*N/2;// indicated power,[kW] +T = ebl*g*ebr;// torque, [N*m] +bp = 2*%pi*N*T;// brake power, [W] +n_mech = bp/ip*10^-3;// mechanical efficiency +mprintf('\n The Indicated power is = %f kW\n',ip); +mprintf('\n The Brake power is = %f kW\n',bp*10^-3); +mprintf('\n The mechanical efficiency is = %f percent\n',n_mech); + +ef = CV*fc;// energy from fuel, [kJ/s] +eb = bp*10^-3;// energy to brake power,[kJ/s] +ec = cwc*c*(To-Ti);// energy to coolant,[kJ/s] +es = ef-(eb+ec+eeg);// energy to surrounding,[kJ/s] + +disp('Energy can be tabulated as :-'); +disp('----------------------------------------------------------------------------------------------------'); +disp(' kJ/s Percentage ') +disp('----------------------------------------------------------------------------------------------------'); +mprintf('\n Energy from fuel %f %f\n Energy to brake power %f %f\n Energy to coolant %f %f\n Energy to exhaust %f %f\n Energy to suroundings,etc. %f %f\n',ef,ef/ef*100,eb,eb/ef*100,ec,ec/ef*100,eeg,eeg/ef*100,es,es/ef*100); + +// End diff --git a/2705/CH17/EX17.2/Ex17_2.sce b/2705/CH17/EX17.2/Ex17_2.sce new file mode 100755 index 000000000..16f568583 --- /dev/null +++ b/2705/CH17/EX17.2/Ex17_2.sce @@ -0,0 +1,87 @@ +clear; +clc; +disp('Example 17.2'); + +// aim : To determine +// (a) bp +// (b) ip +// (c) mechanical efficiency +// (d) indicated thermal efficiency +// (e) brake specific steam consumption +// (f) draw up complete energy account for the test one-minute basis taking 0 C as datum + +// given values +d = 200*10^-3;// cylinder diameter, [mm] +L = 250*10^-3;// stroke, [mm] +N = 5;// speed, [rev/s] +r = .75/2;// effective radious of brake wheel, [m] +Ps = 800;// stop valve pressure, [kN/m^2] +x = .97;// dryness fraction of steam +BL = 136;// brake load, [kg] +SL = 90;// spring balance load, [N] +PM = 232;// mean effective pressure, [kN/m^2] +Pc = 10;// condenser pressure, [kN/m^2] +m_dot = 3.36;// steam consumption, [kg/min] +CC = 113;// condenser cooling water, [kg/min] +Tr = 11;// temperature rise of condenser cooling water, [K] +Tc = 38;// condensate temperature, [C] +C = 4.18;// heat capacity of water, [kJ/kg K] +g = 9.81;// gravitational acceleration, [m/s^2] + +// solution +// from steam table +// at 800 kN/m^2 +tf1 = 170.4;// saturation temperature, [C] +hf1 = 720.9;// [kJ/kg] +hfg1 = 2046.5;// [kJ/kg] +hg1 = 2767.5;// [kJ/kg] +vg1 = .2403;// [m^3/kg] + +// at 10 kN/m^2 +tf2 = 45.8;// saturation temperature, [C] +hf2 = 191.8;// [kJ/kg] +hfg2 = 2392.9;// [kJ/kg] +hg2 = 2584.8;// [kJ/kg] +vg2 = 14.67;// [m^3/kg] + +// (a) +T = (BL*g-SL)*r;// torque, [Nm] +bp = 2*%pi*N*T*10^-3;// brake power,[W] +mprintf('\n (a) The brake power is = %f kW\n',bp); + +// (b) +A = %pi*d^2/4;// area, [m^2] +ip = PM*L*A*N*2;// double-acting so*2, [kW] +mprintf('\n (b) The indicated power is = %f kW\n',ip); + +// (c) +n_mec = bp/ip;// mechanical efficiency +mprintf('\n (c) The mechanical efficiency is = %f percent\n',n_mec*100); + +// (d) +h = hf1+x*hfg1;// [kJ/kg] +hf = hf2; +ITE = ip/((m_dot/60)*(h-hf));// indicated thermal efficiency +mprintf('\n (d) The indicated thermal efficiency is = %f percent\n',ITE*100); +// (e) +Bsc=m_dot*60/bp;// brake specific steam consumption, [kg/kWh] +mprintf('\n (e) The brake steam consumption is = %f kg/kWh\n',Bsc); + +// (f) +// energy balanvce reckoned from 0 C +Es = m_dot*h;// energy supplied, [kJ] +Eb = bp*60;// energy to bp, [kJ] +Ecc = CC*C*Tr;// energy to condensate cooling water, [kJ] +Ec = m_dot*C*Tc;// energy to condensate, [kJ] +Ese = Es-Eb-Ecc-Ec;// energy to surrounding,etc, [kJ] + +mprintf('\n (f) Energy supplied/min is = %f kJ\n',Es); + +mprintf('\n Energy to bp/min is = %f kJ\n',Eb); +mprintf('\n Energy to condenser cooling water/min is = %f kJ\n',Ecc); +mprintf('\n Energy to condensate/min is = %f kJ\n',Ec); +mprintf('\n Energy to surrounding, etc/min is = %f kJ\n',Ese); + +// answer in the book is misprinted + +// End diff --git a/2705/CH17/EX17.3/Ex17_3.sce b/2705/CH17/EX17.3/Ex17_3.sce new file mode 100755 index 000000000..baddc9cc9 --- /dev/null +++ b/2705/CH17/EX17.3/Ex17_3.sce @@ -0,0 +1,55 @@ +clear; +clc; +disp('Example 17.3'); + +// aim : To determine +// (a) the brake power +// (b) the brake specific fuel consumption +// (c) the indicated thermal efficiency +// (d) the energy balance, expressing the various items + +// given values +t = 30;// duration of trial, [min] +N = 1750;// speed of engine, [rev/min] +T = 330;// brake torque, [Nm] +mf = 9.35;// fuel consumption, [kg] +CV = 42300;// calorific value of fuel, [kJ/kg] +cwc = 483;// jacket cooling water circulation, [kg] +Ti = 17;// inlet temperature, [C] +To = 77;// outlet temperature, [C] +ma = 182;// air consumption, [kg] +Te = 486;// exhaust temperature, [C] +Ta = 17;// atmospheric temperature, [C] +n_mec = .83;// mechanical efficiency +c = 1.25;// mean specific heat capacity of exhaust gas, [kJ/kg K] +C = 4.18;// specific heat capacity, [kJ/kg K] + +// solution +// (a) +bp = 2*%pi*N*T/60*10^-3;// brake power, [kW] +mprintf('\n (a) The Brake power is = %f kW\n',bp); + +// (b) +bsf = mf*2/bp;//brake specific fuel consumption, [kg/kWh] +mprintf('\n (b) The brake specific fuel consumption is = %f kg/kWh\n',bsf); + +// (c) +ip = bp/n_mec;// indicated power, [kW] +ITE = ip/(2*mf*CV/3600);// indicated thermal efficiency +mprintf('\n (c) The indicated thermal efficiency is = %f percent\n',ITE*100); + +// (d) +// taking basis one minute +ef = CV*mf/30;// energy from fuel, [kJ] +eb = bp*60;// energy to brake power,[kJ] +ec = cwc/30*C*(To-Ti);// energy to cooling water,[kJ] +ee = (ma+mf)/30*c*(Te-Ta);// energy to exhaust, [kJ] +es = ef-(eb+ec+ee);// energy to surrounding,etc,[kJ] + +mprintf('\n (d) Energy from fuel is = %f kJ\n',ef); +mprintf('\n Energy to brake power is = %f kJ\n',eb); +mprintf('\n Energy to cooling water is = %f kJ\n',ec); +mprintf('\n Energy to exhaust is = %f kJ\n',ee); +mprintf('\n Energy to surrounding, etc is = %f kJ\n',es); + +// End diff --git a/2705/CH17/EX17.4/Ex17_4.sce b/2705/CH17/EX17.4/Ex17_4.sce new file mode 100755 index 000000000..e50619a8c --- /dev/null +++ b/2705/CH17/EX17.4/Ex17_4.sce @@ -0,0 +1,33 @@ +clear; +clc; +disp('Example 17.4'); + +// aim : To determine +// (a) the indicated power of the engine +// (b) the mechanical efficiency of the engine + +// given values +bp = 52;// brake power output, [kW] +bp1 = 40.5;// brake power of cylinder cut1, [kW] +bp2 = 40.2;// brake power of cylinder cut2, [kW] +bp3 = 40.1;// brake power of cylinder cut3, [kW] +bp4 = 40.6;// brake power of cylinder cut4, [kW] +bp5 = 40.7;// brake power of cylinder cut5, [kW] +bp6 = 40.0;// brake power of cylinder cut6, [kW] + +// sollution +ip1 = bp-bp1;// indicated power of cylinder cut1, [kW] +ip2 = bp-bp2;// indicated power of cylinder cut2, [kW] +ip3 = bp-bp3;// indicated power of cylinder cut3, [kW] +ip4 = bp-bp4;// indicated power of cylinder cut4, [kW] +ip5 = bp-bp5;// indicated power of cylinder cut5, [kW] +ip6 = bp-bp6;// indicated power of cylinder cut6, [kW] + +ip = ip1+ip2+ip3+ip4+ip5+ip6;// indicated power of engine,[kW] +mprintf('\n (a) The indicated power of the engine is = %f kW\n',ip); + +// (b) +n_mec = bp/ip;// mechanical efficiency +mprintf('\n (b) The mechanical efficiency of the engine is = %f percent\n',n_mec*100); + +// End diff --git a/2705/CH17/EX17.5/Ex17_5.sce b/2705/CH17/EX17.5/Ex17_5.sce new file mode 100755 index 000000000..23733b131 --- /dev/null +++ b/2705/CH17/EX17.5/Ex17_5.sce @@ -0,0 +1,61 @@ +clear; +clc; +disp('Example 17.5'); + +// aim : To determine +// the brake power,indicated power and mechanicl efficiency +// draw up an energy balance and as % age of the energy supplied + +// given values +N = 50;// speed, [rev/s] +BL = 267;// break load.,[N] +BL1 = 178;// break load of cylinder cut1, [N] +BL2 = 187;// break load of cylinder cut2, [N] +BL3 = 182;// break load of cylinder cut3, [N] +BL4 = 182;// break load of cylinder cut4, [N] + +FC = .568/130;// fuel consumption, [L/s] +s = .72;// specific gravity of fuel +CV = 43000;// calorific value of fuel, [kJ/kg] + +Te = 760;// exhaust temperature, [C] +c = 1.015;// specific heat capacity of exhaust gas, [kJ/kg K] +Ti = 18;// cooling water inlet temperature, [C] +To = 56;// cooling water outlet temperature, [C] +mw = .28;// cooling water flow rate, [kg/s] +Ta = 21;// ambient tempearture, [C] +C = 4.18;// specific heat capacity of cooling water, [kJ/kg K] + +// solution +bp = BL*N/455;// brake power of engine, [kW] +bp1 = BL1*N/455;// brake power of cylinder cut1, [kW] +i1 = bp-bp1;// indicated power of cylinder cut1, [kW] +bp2 = BL2*N/455;// brake power of cylinder cut2, [kW] +i2 = bp-bp2;// indicated power of cylinder cut2, [kW] +bp3 = BL3*N/455;// brake power of cylinder cut3, [kW] +i3 = bp-bp3;// indicated power of cylinder cut3, [kW] +bp4 = BL4*N/455;// brake power of cylinder cut4, [kW] +i4 = bp-bp4;// indicated power of cylinder cut4, [kW] + +ip = i1+i2+i3+i4;// indicated power of engine, [kW] +n_mec = bp/ip;// mechanical efficiency + +mprintf('\n The Brake power is = %f kW\n',bp); +mprintf('\n The Indicated power is = %f kW\n',ip); +mprintf('\n The mechanical efficiency is = %f percent\n',n_mec*100); + +mf = FC*s;// mass of fuel/s, [kg] +ef = CV*mf;// energy from fuel/s, [kJ] +me = 15*mf;// mass of exhaust/s,[kg],(given in condition) +ee = me*c*(Te-Ta);// energy to exhaust/s,[kJ] +ec = mw*C*(To-Ti);// energy to cooling water/s,[kJ] +es = ef-(ee+ec+bp);// energy to surrounding,etc/s,[kJ] + +disp('Energy can be tabulated as :-'); +disp('----------------------------------------------------------------------------------------------------'); +disp(' kJ/s Percentage ') +disp('----------------------------------------------------------------------------------------------------'); +mprintf('\n Energy from fuel %f %f\n Energy to brake power %f %f\n Energy to exhaust %f %f\n Energy to coolant %f %f\n Energy to suroundings,etc. %f %f\n',ef,ef/ef*100,bp,bp/ef*100,ee,ee/ef*100,ec,ec/ef*100,es,es/ef*100); + +// there is minor variation in the result reported in the book +// End diff --git a/2705/CH17/EX17.6/Ex17_6.sce b/2705/CH17/EX17.6/Ex17_6.sce new file mode 100755 index 000000000..7f34d5492 --- /dev/null +++ b/2705/CH17/EX17.6/Ex17_6.sce @@ -0,0 +1,33 @@ +clear; +clc; +disp('Example 17.6'); + +// aim : To determine +// (a) the break power of engine +// (b) the fuel consumption of the engine +// (c) the brake thermal efficiency of the engine + +// given values +d = 850*10^-3;// bore , [m] +L = 2200*10^-3;// stroke, [m] +PMb = 15;// BMEP of cylinder, [bar] +N = 95/60;// speed of engine, [rev/s] +sfc = .2;// specific fuel oil consumption, [kg/kWh] +CV = 43000;// calorific value of the fuel oil, [kJ/kg] + +// solution +// (a) +A = %pi*d^2/4;// area, [m^2] +bp = PMb*L*A*N*8/10;// brake power,[MW] +mprintf('\n (a) The brake power is = %f MW\n',bp); + + // (b) + FC = bp*sfc;// fuel consumption, [kg/h] + mprintf('\n (b) The fuel consumption is = %f tonne/h\n',FC); + + // (c) + mf = FC/3600;// fuel used, [kg/s] + n_the = bp/(mf*CV);// brake thermal efficiency + mprintf('\n (c) The brake thermal efficiency is = %f percent\n',n_the*100); + + // End diff --git a/2705/CH18/EX18.1/Ex18_1.sce b/2705/CH18/EX18.1/Ex18_1.sce new file mode 100755 index 000000000..85da4efdc --- /dev/null +++ b/2705/CH18/EX18.1/Ex18_1.sce @@ -0,0 +1,61 @@ +clear; +clc; +disp('Example 18.1'); + +// aim : To determine +// (a) the coefficient of performance +// (b) the mass flow of the refrigerant +// (c) the cooling water required by the condenser + +// given values +P1 = 462.47;// pressure limit, [kN/m^2] +P3 = 1785.90;// pressure limit, [kN/m^2] +T2 = 273+59;// entering saturation temperature, [K] +T5 = 273+32;// exit temperature of condenser, [K] +d = 75*10^-3;// bore, [m] +L = d;// stroke, [m] +N = 8;// engine speed, [rev/s] +VE = .8;// olumetric efficiency +cpL = 1.32;// heat capacity of liquid, [kJ/kg K] +c = 4.187;// heat capacity of water, [kj/kg K] + +// solution +// from given table +// at P1 +h1 = 231.4;// specific enthalpy, [kJ/kg] +s1 = .8614;// specific entropy,[ kJ/kg K +v1 = .04573;// specific volume, [m^3/kg] + +// at P3 +h3 = 246.4;// specific enthalpy, [kJ/kg] +s3 = .8093;// specific entropy,[ kJ/kg K +v3 = .04573;// specific volume, [m^3/kg] +T3= 273+40;// saturation temperature, [K] +h4 = 99.27;// specific enthalpy, [kJ/kg] +// (a) +s2 = s1;// specific entropy, [kJ/kg k] +// using s2=s3+cpv*log(T2/T3) +cpv = (s2-s3)/log(T2/T3);// heat capacity, [kj/kg k] + +// from Fig.18.8 +T4 = T3; +h2 = h3+cpv*(T2-T3);// specific enthalpy, [kJ/kg] +h5 = h4-cpL*(T4-T5);// specific enthalpy, [kJ/kg] +h6 = h5; +COP = (h1-h6)/(h2-h1);// coefficient of performance +mprintf('\n (a) The coefficient of performance of the refrigerator is = %f\n',COP); + +// (b) +SV = %pi/4*d^2*L;// swept volume of compressor/rev, [m^3] +ESV = SV*VE*N*3600;// effective swept volume/h, [m^3] +m = ESV/v1;// mass flow of refrigerant/h,[kg] +mprintf('\n (b) The mass flow of refrigerant/h is = %f kg\n',m); + +// (c) +dT = 12;// temperature limit, [C] +Q = m*(h2-h5);// heat transfer in condenser/h, [kJ] +// using Q=m_dot*c*dT, so +m_dot = Q/(c*dT);// mass flow of water required, [kg/h] +mprintf('\n (c) The mass flow of water required is = %f kg/h\n',m_dot); + +// End diff --git a/2705/CH18/EX18.2/Ex18_2.sce b/2705/CH18/EX18.2/Ex18_2.sce new file mode 100755 index 000000000..7d917dfd9 --- /dev/null +++ b/2705/CH18/EX18.2/Ex18_2.sce @@ -0,0 +1,54 @@ +clear; +clc; +disp('Example 18.2'); + +// aim : To determine +// (a) the mass flow of R401 +// (b) the dryness fraction of R401 at the entry to the evaporator +// (c) the power of driving motor +// (d) the ratio of heat transferred from condenser to the power required to the motor + +// given values +P1 = 411.2;// pressure limit, [kN/m^2] +P3 = 1118.9;// pressure limit, [kN/m^2] +Q = 100*10^3;// heat transfer from the condenser,[kJ/h] +T2 = 273+60;// entering saturation temperature, [K] + +// given +// from given table +// at P1 +h1 = 409.3;// specific enthalpy, [kJ/kg] +s1 = 1.7431;// specific entropy,[ kJ/kg K + +// at P3 +h3 = 426.4;// specific enthalpy, [kJ/kg] +s3 = 1.7192;// specific entropy,[ kJ/kg K +T3 = 273+50;// saturation temperature, [K] +h4 = 265.5;// specific enthalpy, [kJ/kg] +// (a) +s2 = s1;// specific entropy, [kJ/kg k] +// using s2=s3+cpv*log(T2/T3) +cpv = (s2-s3)/log(T2/T3);// heat capacity, [kj/kg k] + +// from Fig.18.8 +h2 = h3+cpv*(T2-T3);// specific enthalpy, [kJ/kg] +Qc = h2-h4;// heat transfer from condenser, [kJ/kg] +mR401 = Q/Qc;// mass flow of R401, [kg] + mprintf('\n (a) The mass flow of R401 is = %f kg/h\n',mR401); + +// (b) +hf1 = 219;// specific enthalpy, [kJ/kg] +h5 = h4; +// using h5=hf1+s5*(h1-hf1),so +x5 = (h5-hf1)/(h1-hf1);// dryness fraction +mprintf('\n (b) The dryness fraction of R401 at the entry to the evaporator is = %f\n',x5); + +// (c) +P = mR401*(h2-h1)/3600/.7;// power to driving motor, [kW] + mprintf('\n (c) The power to driving motor is = %f kW\n',P); + +// (d) +r = Q/3600/P;// ratio +mprintf('\n (d) The ratio of heat transferred from condenser to the power required to the motor is = %f : 1\n',r); + +// End diff --git a/2705/CH19/EX19.1/Ex19_1.sce b/2705/CH19/EX19.1/Ex19_1.sce new file mode 100755 index 000000000..feed53150 --- /dev/null +++ b/2705/CH19/EX19.1/Ex19_1.sce @@ -0,0 +1,46 @@ +clear; +clc; +disp('Example 19.1'); + +// aim : To compare the moisture content and the true specific volumes of atmosphere air +// (a) temperature is 12 C and the air is saturaded +// (b) temperature is 31 C and air is .75 saturated + +// Given values +P_atm = 101.4;// atmospheric pressure, [kN/m^2] +R = .287;// [kJ/kg K] + +// solution +// (a) +T = 273+12;// air temperature, [K] +// From steam table at 12 C +p = 1.4;// [kN/m^2] +vg = 93.9;// [m^3/kg] +pa = P_atm-p;// partial pressure of the dry air, [kN/m^2] +va = R*T/pa;// [m^3/kg] + +mw = va/vg;// mass of water vapor in the air,[kg] +v = va/(1+mw);// specific volume of humid air, [m^3/kg] + +mprintf('\n (a) The mass of water vapor in the humid air is = %f kg\n',mw); +mprintf('\n The specific volume of humid air is = %f m^3/kg\n',v); + +// (b) +x = .75;// dryness fraction +T = 273+31;// air temperature, [K] +// From steam table +p = 4.5;// [kN/m^2] +vg = 31.1;// [m^3/kg] +pa = P_atm-p;// [kN/m^2] +va = R*T/pa;// [m^3/kg] +mw1= va/vg;// mass of water vapor in the air, [kg] +mw_actual = mw1*x;// actual mass of vapor, [kg] +v = va/(1+mw_actual);// true specific volume of humid air,[m^3/kg] + +mprintf('\n (b) The mass of water vapor in the humid air is = %f kg\n',mw1); +mprintf('\n The specific volume of humid air is = %f m^3/kg\n',v); + +ewv = mw_actual/mw ; +mprintf('\n On the warm day the air conteains %f times the mass of water vapor as on the cool day \n',ewv); + +// End diff --git a/2705/CH19/EX19.2/Ex19_2.sce b/2705/CH19/EX19.2/Ex19_2.sce new file mode 100755 index 000000000..99dc20724 --- /dev/null +++ b/2705/CH19/EX19.2/Ex19_2.sce @@ -0,0 +1,34 @@ +clear; +clc; +disp('Example 19.2'); + +// aim : To determine +// (a) the partial pressures of the vapor and the dry air +// (b) the specific humidity of the mixture +// (c) the composition of the mixture + +// Given values +phi = .65;// Relative humidity +T = 2733+20;// temperature, [K] +p = 100;// barometric pressure, [kN/m^2] + +// solution +// (a) +// From the steam table at 20 C +pg = 2.34;// [kN/m^2] +ps = phi*pg;// partial pressure of vapor, [kN/m^2] +pa = p-ps;// partial pressure of dry air, [kN/m^2] +mprintf('\n (a) The partial pressure of vapor is = %f kN/m^2\n',ps); +mprintf('\n The partial pressure of dry air is = %f kN/m^2\n',pa); + +// (b) +// from equation [15] +omega = .622*ps/(p-ps);// specific humidity of the mixture +mprintf('\n (b) The specific humidity of the mixture is = %f kg/kg dry air\n',omega); + +// (c) +// using eqn [1] from section 19.2 +y = 1/(1+omega);// composition of the mixture +mprintf('\n (c) The composition of the mixture is = %f\n',y); + +// End diff --git a/2705/CH19/EX19.3/Ex19_3.sce b/2705/CH19/EX19.3/Ex19_3.sce new file mode 100755 index 000000000..54f4d0111 --- /dev/null +++ b/2705/CH19/EX19.3/Ex19_3.sce @@ -0,0 +1,52 @@ +clear; +clc; +disp('Example 19.3'); + +// aim : To determine +// (a) the specific humidity +// (b) the dew point +// (c) the degree of superheat of the superheated vapor +// (d) the mass of condensate formed per kg of dry air if the moist air is cooled to 12 C + +// Given values +t = 25;// C +T = 273+25;// moist air temperature, [K] +phi = .6;// relative humidity +p = 101.3;// barometric pressure, [kN/m^2] +R = .287;// [kJ/kg K] + +// solution +// (a) +// From steam table at 25 C +pg = 3.17;// [kN/m^2] +ps = phi*pg;// partial pressure of the vapor, [kN/m^2] +omega = .622*ps/(p-ps);// the specific humidity of air + +mprintf('\n (a) The specific humidity is = %f kg/kg air\n',omega); + +// (b) +// Dew point is saturated temperature at ps is, +t_dew = 16+2*(1.092-1.817)/(2.062-1.817);// [C] +mprintf('\n (b) The dew point is = %f C\n',t_dew); + +// (c) +Dos = t-t_dew;// degree of superheat, [C] +mprintf('\n (c) The degree of superheat is = %f C\n',Dos); + +// (d) +// at 25 C +pa = p-ps;// [kN/m^2] +va = R*T/pa;// [m^3/kg] +// at 16.69 C +vg = 73.4-(73.4-65.1)*.69/2;// [m^3/kg] +ms1= va/vg; +// at 12 C +vg = 93.8;// [m^3/kg] +ms2 = va/vg; + +m = ms1-ms2;// mas of condensate +mprintf('\n (d) The mass of condensate is = %f kg/kg dry air\n',m); + +// there is calculation mistake in the book so answer is no matching + +// End diff --git a/2705/CH19/EX19.4/Ex19_4.sce b/2705/CH19/EX19.4/Ex19_4.sce new file mode 100755 index 000000000..0cec4f91c --- /dev/null +++ b/2705/CH19/EX19.4/Ex19_4.sce @@ -0,0 +1,73 @@ +clear; +clc; +disp(' Example 19.4'); + +// aim : To determine +// (a) the volume of external saturated air +// (b) the mass of air +// (c) the heat transfer +// (d) the heat transfer required by the combind water vapour + +// given values +Vb = 56000;// volume of building, [m^3] +T2 = 273+20;// temperature of air in thebuilding, [K] +phi = .6;// relative humidity +T1 = 8+273;// external air saturated temperature, [K] +p0 = 101.3;// atmospheric pressure, [kN/m^2] +cp = 2.093;// heat capacity of saturated steam, [kJ/kg K] +R = .287;// gas constant, [kJ/kg K] + +// solution +// from steam table at 20 C saturation pressure of steam is, +pg = 2.34;// [kN/m^2] + +// (a) +pvap = phi*pg;// partial pressure of vapor, [kN/m^2] +P = p0-pvap;// partial pressure of air, [kN/m^2] +V = 2*Vb;// air required, [m^3] +// at 8 C saturation pressure ia +pvap = 1.072;// [kN/m^2] +P2 = p0-pvap;// partial pressure of entry at 8 C, [kN/m^2] + +// using P1*V1/T1=P2*V2/T2; +V2 = P*V*T1/(T2*P2);// air required at 8 C, [m^3/h] +mprintf('\n (a) The volume of air required is = %f m^3/h\n',V2); + +// (b) +// assuming +pg = 1.401;// pressure, [kN/m^2] +Tg = 273+12;// [K] +vg = 93.8;// [m^3/kg] +// at constant pressure +v = vg*T2/Tg;// volume[m^3/kg] +mv = V/v;// mass of vapor in building at 20 C, [kg/h] +// from steam table at 8 C +vg2 = 121;// [m^3/kg] +mve = V2/vg2;// mass of vapor supplied with saturated entry air, [kg/h] +mw = mv-mve;// mass of water added, [kg/h] +mprintf('\n (b) The mass of water added is = %f kg/h\n ',mw); + +// (c) +// for perfect gas +m = P2*V2/(R*T1);// [kg/h] +Cp = .287;// heat capacity, [kJ/kg K] +Q = m*Cp*(T2-T1);// heat transfer by dry air,[kJ/h] +mprintf('\n (c) The heat transfer required by dry air is = %f MJ/h\n',Q*10^-3); + +// (d) +// from steam table +h1 = 2516.2;// specific enthalpy of saturated vapor at 8 C,[kJ/kg] +hs = 2523.6;// specific enthalpy of saturated vapor at 20 C, [kJ/kg] +h2 = hs+cp*(T2-T1);// specific enthalpy of vapor at 20 c, [kJ/kg] +Q1 = mve*(h2-h1);// heat transfer required for vapor, [kJ] + +// again from steam table +hf1 = 33.6;// [kJ/kg] +hg3 = 2538.2;// [kJ/kg] +Q2 = mw*(hg3-hf1);// heat transfer required for water, [kJ/h] +Qt = Q1+Q2;// total heat transfer, [kJ/h] +mprintf('\n (d) The heat transferred required for vapor+supply water is = %f MJ/h\n',Qt*10^-3); + +// there is minor variation in the answer reported in the book + +// End diff --git a/2705/CH2/EX2.1/Ex2_1.sce b/2705/CH2/EX2.1/Ex2_1.sce new file mode 100755 index 000000000..1d5f07b56 --- /dev/null +++ b/2705/CH2/EX2.1/Ex2_1.sce @@ -0,0 +1,26 @@ +clear; +clc; +disp('Example 2.1'); + + + +// Given values +Q = 2500; // Heat transferred into the system, [kJ] +W = 1400; // Work transferred from the system, [kJ] + +// solution + +// since process carried out on a closed system, so using equation [4] +del_E = Q-W; // Change in total energy, [kJ] + +mprintf('\n The Change in total energy is, del_E = %f kJ\n',del_E); + +if(del_E>0) + disp('Since del_E is positive, so there is an increase in total enery') +else + disp('Since del_E is negative, so there is an decrease in total enery') +end + +// There is mistake in the book's results unit + +// End diff --git a/2705/CH2/EX2.2/Ex2_2.sce b/2705/CH2/EX2.2/Ex2_2.sce new file mode 100755 index 000000000..095f933f2 --- /dev/null +++ b/2705/CH2/EX2.2/Ex2_2.sce @@ -0,0 +1,22 @@ +clear; +clc; +disp('Example 2.2'); + + +// Given values +del_E = 3500; // Increase in total energy of the system, [kJ] +W = -4200; // Work transfer into the system, [kJ] + +// solution +// since process carried out on a closed system, so using equation [3] +Q = del_E+W;// [kJ] + +mprintf('\n The Heat transfer is, Q = %f kJ \n',Q); + +if(Q>0) + disp('Since Q>0, so heat is transferred into the system') +else + disp('Since Q<0, so heat is transferred from the system') +end + +// End diff --git a/2705/CH2/EX2.3/Ex2_3.sce b/2705/CH2/EX2.3/Ex2_3.sce new file mode 100755 index 000000000..0f6f799d4 --- /dev/null +++ b/2705/CH2/EX2.3/Ex2_3.sce @@ -0,0 +1,23 @@ +clear; +clc; +disp('Example 2.3'); + + + +// Given values +Q = -150; // Heat transferred out of the system, [kJ/kg] +del_u = -400; // Internal energy decreased ,[kJ/kg] + +// solution +// using equation [3],the non flow energy equation +// Q=del_u+W +W = Q-del_u; // [kJ/kg] +mprintf('\n The Work done is, W = %f kJ/kg \n',W); + +if(W>0) + disp('Since W>0, so Work done by the engine per kilogram of working substance') +else + disp('Since <0, so Work done on the engine per kilogram of working substance') +end + +// End diff --git a/2705/CH2/EX2.4/Ex2_4.sce b/2705/CH2/EX2.4/Ex2_4.sce new file mode 100755 index 000000000..9102fc8b3 --- /dev/null +++ b/2705/CH2/EX2.4/Ex2_4.sce @@ -0,0 +1,43 @@ +clear; +clc; +disp('Example 2.4'); + + + +// Given values +m_dot = 4; // fluid flow rate, [kg/s] +Q = -40; // Heat loss to the surrounding, [kJ/kg] + +// At inlet +P1 = 600; // pressure ,[kn/m^2] +C1 = 220; // velocity ,[m/s] +u1 = 2200; // internal energy, [kJ/kg] +v1 = .42; // specific volume, [m^3/kg] + +// At outlet +P2 = 150; // pressure, [kN/m^2] +C2 = 145; // velocity, [m/s] +u2 = 1650; // internal energy, [kJ/kg] +v2 = 1.5; // specific volume, [m^3/kg] + +// solution +// for steady flow energy equation for the open system is given by +// u1+P1*v1+C1^2/2+Q=u2+P2*v2+C2^2/2+W +// hence + +W = (u1-u2)+(P1*v1-P2*v2)+(C1^2/2-C2^2/2)*10^-3+Q; // [kJ/kg] + +mprintf('\n workdone is, W = %f kJ/kg ',W); + +if(W>0) + disp('Since W>0, so Power is output from the system') +else + disp('Since <0, so Power is input to the system') +end + +// Hence + +P_out = W*m_dot; // power out put from the system, [kW] +mprintf('\n The power output from the system is = %f kW \n',P_out); + +// End diff --git a/2705/CH2/EX2.5/Ex2_5.sce b/2705/CH2/EX2.5/Ex2_5.sce new file mode 100755 index 000000000..f43f37c08 --- /dev/null +++ b/2705/CH2/EX2.5/Ex2_5.sce @@ -0,0 +1,27 @@ +clear; +clc; +disp('Example 2.5'); + + + +// Given values +del_P = 154.45; // pressure difference across the die, [MN/m^2] +rho = 11360; // Density of the lead, [kg/m^3] +c = 130; // specific heat capacity of the lead, [J/kg*K] + +// solution +// since there is no cooling and no externel work is done, so energy balane becomes +// P1*V1+U1=P2*V2+U2 ,so +// del_U=U2-U1=P1*V1-P2*V2 + +// also, for temperature rise, del_U=m*c*t, where, m is mass; c is specific heat capacity; and t is temperature rise + +// Also given that lead is incompressible, so V1=V2=V and assuming one m^3 of lead + +// using above equations +t = del_P/(rho*c)*10^6 ;// temperature rise [C] + +mprintf('\n The temperature rise of the lead is = %f C\n',t); + +// End + diff --git a/2705/CH2/EX2.6/Ex2_6.sce b/2705/CH2/EX2.6/Ex2_6.sce new file mode 100755 index 000000000..186ca15e2 --- /dev/null +++ b/2705/CH2/EX2.6/Ex2_6.sce @@ -0,0 +1,38 @@ +clear; +clc; +disp('Example 2.6'); + + +// Given values +m_dot = 4.5; // mass flow rate of air, [kg/s] +Q = -40; // Heat transfer loss, [kJ/kg] +del_h = -200; // specific enthalpy reduce, [kJ/kg] + +C1 = 90; // inlet velocity, [m/s] +v1 = .85; // inlet specific volume, [m^3/kg] + +v2 = 1.45; // exit specific volume, [m^3/kg] +A2 = .038; // exit area of turbine, [m^2] + +// solution + +// part (a) +// At inlet, by equation[4], m_dot=A1*C1/v1 +A1 = m_dot*v1/C1;//inlet area, [m^2] +mprintf('\n (a) The inlet area is, A1 = %f m^2 \n',A1); + +// part (b), +// At outlet, since mass flow rate is same, so m_dot=A2*C2/v2, hence +C2 = m_dot*v2/A2; // Exit velocity,[m/s] +mprintf('\n (b) The exit velocity is, C2 = %f m/s \n',C2); + +// part (c) +// using steady flow equation, h1+C1^2/2+Q=h2+C2^2/2+W +W = -del_h+(C1^2/2-C2^2/2)*10^-3+Q; // [kJ/kg] + +// Hence power developed is +P = W*m_dot;// [kW] +mprintf('\n (c) The power developed by the turbine system is = %f kW \n',P); + +// End + diff --git a/2705/CH4/EX4.1/Ex4_1.sce b/2705/CH4/EX4.1/Ex4_1.sce new file mode 100755 index 000000000..afa29662b --- /dev/null +++ b/2705/CH4/EX4.1/Ex4_1.sce @@ -0,0 +1,23 @@ +clear; +clc; +disp('Example 4.1'); + +// aim : To determine +// the enthalpy + +// Given values +P = .50;// Pressure, [MN/m^2] + +// solution + +// From steam tables, at given pressure +hf = 640.1;// specific liquid enthalpy ,[kJ/kg] +hfg = 2107.4;// specific enthalpy of evaporation ,[kJ/kg] +hg = 2747.5; // specific enthalpy of dry saturated steam ,[kJ/kg] +tf = 151.8; // saturation temperature,[C] + +mprintf('\n The specific liquid enthalpy is = %f kJ/kg \n',hf); +mprintf('\n The specific enthalpy of evaporation is = %f kJ/kg \n',hfg); +mprintf('\n The specific enthalpy of dry saturated steam is = %f kJ/kg \n',hg); + +// End diff --git a/2705/CH4/EX4.10/Ex4_10.sce b/2705/CH4/EX4.10/Ex4_10.sce new file mode 100755 index 000000000..972a1eed3 --- /dev/null +++ b/2705/CH4/EX4.10/Ex4_10.sce @@ -0,0 +1,38 @@ +clear; +clc; +disp('Example .10'); + +// aim : To determine +// (a) the mass of steam entering the heater +// (b) the mass of water entering the heater + +// Given values +x = .95;// Dryness fraction +P = .7;// pressure,[MN/m^2] +d = 25;// internal diameter of heater,[mm] +C = 12; // steam velocity in the pipe,[m/s] + +// solution +// from steam table at .7 MN/m^2 pressure +hf = 697.1;// [kJ/kg] +hfg = 2064.9;// [kJ/kg] +hg = 2762.0; // [kJ/kg] +vg = .273; // [m^3/kg] + +// (a) +v = x*vg; // [m^3/kg] +ms_dot = %pi*(d*10^-3)^2*C*3600/(4*v);// mass of steam entering, [kg/h] +mprintf('\n (a) The mass of steam entering the heater is = %f kg/h \n',ms_dot); + +// (b) +h = hf+x*hfg;// specific enthalpy of steam entering heater,[kJ/kg] +// again from steam tables +hf1 = 376.8;// [kJ/kg] at 90 C +hf2 = 79.8;// [kJ/kg] at 19 C + +// using energy balance,mw_dot*(hf1-hf2)=ms_dot*(h-hf1) +mw_dot = ms_dot*(h-hf1)/(hf1-hf2);// mass of water entering to heater,[kg/h] + +mprintf('\n (b) The mass of water entering the heater is = %f kg/h \n',mw_dot); + +// End diff --git a/2705/CH4/EX4.11/Ex4_11.sce b/2705/CH4/EX4.11/Ex4_11.sce new file mode 100755 index 000000000..89819556c --- /dev/null +++ b/2705/CH4/EX4.11/Ex4_11.sce @@ -0,0 +1,40 @@ +clear; +clc; +disp('Example 4.11'); + +// aim: To determine +// the change of internal energy + +// Given values +m = 1.5;// mass of steam,[kg] +P1 = 1;// initial pressure, [MN/m^2] +t = 225;// temperature, [C] +P2 = .28;// final pressure, [MN/m^2] +x = .9;// dryness fraction of steam at P2 + +// solution + +// from steam table at P1 +h1 = 2886;// [kJ/kg] +v1 = .2198; // [m^3/kg] +// hence +u1 = h1-P1*v1*10^3;// internal energy [kJ/kg] + +// at P2 +hf2 = 551.4;// [kJ/kg] +hfg2 = 2170.1;// [kJ/kg] +vg2 = .646; // [m^3/kg] +// so +h2 = hf2+x*hfg2;// [kj/kg] +v2 = x*vg2;// [m^3/kg] + +// now +u2 = h2-P2*v2*10^3;// [kJ/kg] + +// hence change in specific internal energy is +del_u = u2-u1;// [kJ/kg] + +del_u = m*del_u;// [kJ]; +mprintf('\n The change in internal energy is = %f kJ \n',del_u); + +// End diff --git a/2705/CH4/EX4.12/Ex4_12.sce b/2705/CH4/EX4.12/Ex4_12.sce new file mode 100755 index 000000000..2ef1bbed3 --- /dev/null +++ b/2705/CH4/EX4.12/Ex4_12.sce @@ -0,0 +1,29 @@ +clear; +clc; +disp('Example 4.12'); + +// aim : To determine +// the dryness fraction of steam after throttling + +// given values +P1 = 1.4;// pressure before throttling, [MN/m^2] +x1 = .7;// dryness fraction before throttling +P2 = .11;// pressure after throttling, [MN/m^2] + +// solution +// from steam table +hf1 = 830.1;// [kJ/kg] +hfg1 = 1957.7;// [kJ/kg] +h1 = hf1 + x1*hfg1; // [kJ/kg] + +hf2 = 428.8;// [kJ/kg] +hfg2 = 2250.8;// [kJ/kg] + +// now for throttling, +// hf1+x1*hfg1=hf2+x2*hfg2; where x2 is dryness fraction after throttling + +x2=(h1-hf2)/hfg2; // final dryness fraction + +mprintf('\n Dryness fraction of steam after throttling is = %f \n',x2); + +// End diff --git a/2705/CH4/EX4.13/Ex4_13.sce b/2705/CH4/EX4.13/Ex4_13.sce new file mode 100755 index 000000000..108099539 --- /dev/null +++ b/2705/CH4/EX4.13/Ex4_13.sce @@ -0,0 +1,65 @@ +clear; +clc; +disp('Example 4.13'); + +// aim : To determine +// the dryness fraction of steam +// and the internal diameter of the pipe + +// Given values + +// steam1 +P1 = 2;// pressure before throttling, [MN/m^2] +t = 300;// temperature,[C] +ms1_dot = 2;// steam flow rate, [kg/s] +P2 = 800;// pressure after throttling, [kN/m^2] + +// steam2 +P = 800;// pressure, [N/m^2] +x2 = .9;// dryness fraction +ms2_dot = 5; // [kg/s] + +// solution +// (a) +// from steam table specific enthalpy of steam1 before throttling is +hf1 = 3025;// [kJ/kg] +// for throttling process specific enthalpy will same so final specific enthalpy of steam1 is +hf2 = hf1; +// hence +h1 = ms1_dot*hf2;// [kJ/s] + +// calculation of specific enthalpy of steam2 +hf2 = 720.9;// [kJ/kg] +hfg2 = 2046.5;// [kJ/kg] +// hence +h2 = hf2+x2*hfg2;// specific enthalpy, [kJ/kg] +h2 = ms2_dot*h2;// total enthalpy, [kJ/s] + +// after mixing +m_dot = ms1_dot+ms2_dot;// total mass of mixture,[kg/s] +h = h1+h2;// Total enthalpy of the mixture,[kJ/s] +h = h/7;// [kJ/kg] + +// At pressure 800 N/m^2 +hf = 720.9;// [kJ/kg] +hfg = 2046.5;// [kJ/kg] +// so total enthalpy is,hf+x*hfg, where x is dryness fraction of mixture and which is equal to h +// hence +x = (h-hf)/hfg;// dryness fraction after mixing +mprintf('\n (a) The condition of the resulting mixture is dry with dryness fraction = %f \n',x); + +// (b) +// Given +C = 15;// velocity, [m/s] +// from steam table +v = .1255;// [m^/kg] +A = ms1_dot*v/C;// area, [m^2] +// using ms1_dot = A*C/v, where A is cross section area in m^2 and +// A = %pi*d^2/4, where d is diameter of the pipe + +// calculation of d +d = sqrt(4*A/%pi); // diameter, [m] + +mprintf('\n (b) The internal diameter of the pipe is = %f mm \n',d*1000); + +// End diff --git a/2705/CH4/EX4.14/Ex4_14.sce b/2705/CH4/EX4.14/Ex4_14.sce new file mode 100755 index 000000000..caec80859 --- /dev/null +++ b/2705/CH4/EX4.14/Ex4_14.sce @@ -0,0 +1,17 @@ +clear; +clc; +disp('Example 4.14'); + +// aim : To estimate +// the dryness fraction + +// Given values +M = 1.8;// mass of condensate, [kg] +m = .2;// water collected, [kg] + +// solution +x = M/(M+m);// formula for calculation of dryness fraction using seprating calorimeter + +mprintf(' \n The dryness fraction of the steam entering seprating calorimeter is = %f \n',x); + +// End diff --git a/2705/CH4/EX4.15/Ex4_15.sce b/2705/CH4/EX4.15/Ex4_15.sce new file mode 100755 index 000000000..e28726c11 --- /dev/null +++ b/2705/CH4/EX4.15/Ex4_15.sce @@ -0,0 +1,51 @@ +clear; +clc; +disp('Example 4.15'); + +// aim : To determine +// the dryness fraction of the steam at 2.2 MN/m^2 + +// Given values +P1 = 2.2;// [MN/m^2] +P2 = .13;// [MN/m^2] +t2 = 112;// [C] +tf2 = 150;// temperature, [C] + +// solution +// from steam table, at 2.2 MN/m^2 +// saturated steam at 2 MN/m^2 Pressure +hf1 = 931;// [kJ/kg] +hfg1 = 1870;// [kJ/kg] +hg1 = 2801;// [kJ/kg] + +// for superheated steam +// at .1 MN/m^2 +hg2 = 2675;// [kJ/kg] +hg2_150 = 2777;// specific enthalpy at 150 C, [kJ/kg] +tf2 = 99.6;// saturation temperature, [C] + +// at .5 MN/m^2 +hg3 = 2693;// [kJ/kg] +hg3_150 = 2773;// specific enthalpy at 150 C, [kJ/kg] +tf3 = 111.4;// saturation temperature, [C] + +Table_P_h1 = [[.1,.5];[hg2,hg3]];// where, P in MN/m^2 and h in [kJ/kg] +hg = interpln(Table_P_h1,.13);// specific entahlpy at .13 MN/m^2, [kJ/kg] + +Table_P_h2 = [[.1,.5];[hg2_150,hg3_150]];// where, P in MN/m^2 and h in [kJ/kg] +hg_150 = interpln(Table_P_h2,.13);// specific entahlpy at .13 MN/m^2 and 150 C, [kJ/kg] + +Table_P_tf = [[.1,.5];[tf2,tf3]];// where, P in MN/m^2 and h in [kJ/kg] +tf = interpln(Table_P_tf,.13);// saturation temperature, [C] + +// hence +h2 = hg+(hg_150-hg)/(t2-tf)/(tf2-tf);// specific enthalpy at .13 MN/m^2 and 112 C, [kJ/kg] + +// now since process is throttling so h2=h1 +// and h1 = hf1+x1*hfg1, so +x1 = (h2-hf1)/hfg1;// dryness fraction +mprintf(' \n The dryness fraction of steam is = %f \n',x1); + +// There is a calculation mistake in book so answer is not matching + +// End diff --git a/2705/CH4/EX4.16/Ex4_16.sce b/2705/CH4/EX4.16/Ex4_16.sce new file mode 100755 index 000000000..895917d3a --- /dev/null +++ b/2705/CH4/EX4.16/Ex4_16.sce @@ -0,0 +1,28 @@ +clear; +clc; +disp('Example 4.16'); + +// aim : To determine +// the minimum dryness fraction of steam + +// Given values +P1 = 1.8;// testing pressure,[MN/m^2] +P2 = .11;// pressure after throttling,[MN/m^2] + +// solution +// from steam table +// at .11 MN/m^2 steam is completely dry and specific enthalpy is +hg = 2680;// [kJ/kg] + +// before throttling steam is wet, so specific enthalpy is=hf+x*hfg, where x is dryness fraction +// from steam table +hf = 885;// [kJ/kg] +hfg = 1912;// [kJ/kg] + +// now for throttling process,specific enthalpy will same before and after +// hence +x = (hg-hf)/hfg; +mprintf('\n The minimum dryness fraction of steam is x = %f \n',x); + +// End + diff --git a/2705/CH4/EX4.17/Ex4_17.sce b/2705/CH4/EX4.17/Ex4_17.sce new file mode 100755 index 000000000..b85b38646 --- /dev/null +++ b/2705/CH4/EX4.17/Ex4_17.sce @@ -0,0 +1,52 @@ +clear; +clc; +disp('Example 4.17'); + +// aim : To determine the +// (a) mass of steam in the vessel +// (b) final dryness of the steam +// (c) amount of heat transferrred during the cooling process + +// Given values +V1 = .8;// [m^3] +P1 = 360;// [kN/m^2] +P2 = 200;// [kN/m^2] + +// solution + +// (a) +// at 360 kN/m^2 +vg1 = .510;// [m^3] +m = V1/vg1;// mass of steam,[kg] +mprintf('\n (a) The mass of steam in the vessel is = %f kg\n',m); + +// (b) +// at 200 kN/m^2 +vg2 = .885;// [m^3/kg] +// the volume remains constant so +x = vg1/vg2;// final dryness fraction +mprintf('\n (b) The final dryness fraction of the steam is = %f \n',x); + +// (c) +// at 360 kN/m^2 +h1 = 2732.9;// [kJ/kg] +// hence +u1 = h1-P1*vg1;// [kJ/kg] + +// at 200 kN/m^2 +hf = 504.7;// [kJ/kg] +hfg=2201.6;//[kJ/kg] +// hence +h2 = hf+x*hfg;// [kJ/kg] +// now +u2 = h2-P2*vg1;// [kJ/kg] +// so +del_u = u2-u1;// [kJ/kg] +// from the first law of thermodynamics del_U+W=Q, +W = 0;// because volume is constant +del_U = m*del_u;// [kJ] +// hence +Q = del_U;// [kJ] +mprintf('\n (c) The amount of heat transferred during cooling process is = %f kJ \n',Q); + +// End diff --git a/2705/CH4/EX4.18/Ex4_18.sce b/2705/CH4/EX4.18/Ex4_18.sce new file mode 100755 index 000000000..61a3ef637 --- /dev/null +++ b/2705/CH4/EX4.18/Ex4_18.sce @@ -0,0 +1,31 @@ +clear; +clc; +disp('Example 4.18'); + +// aim : To determine +// the heat received by the steam per kilogram + +// Given values +// initial +P1 = 4;// pressure, [MN/m^2] +x1 = .95; // dryness fraction + +// final +t2 = 350;// temperature,[C] + +// solution + +// from steam table, at 4 MN/m^2 and x1=.95 +hf = 1087.4;// [kJ/kg] +hfg = 1712.9;// [kJ/kg] +// hence +h1 = hf+x1*hfg;// [kJ/kg] + +// since pressure is kept constant ant temperature is raised so at this condition +h2 = 3095;// [kJ/kg] + +// so by energy balance +Q = h2-h1;// Heat received,[kJ/kg] +mprintf('\n The heat received by the steam is = %f kJ/kg \n',Q); + +// End diff --git a/2705/CH4/EX4.19/Ex4_19.sce b/2705/CH4/EX4.19/Ex4_19.sce new file mode 100755 index 000000000..edc6f9d44 --- /dev/null +++ b/2705/CH4/EX4.19/Ex4_19.sce @@ -0,0 +1,50 @@ +clear; +clc; +disp('Example 4.19'); + +// aim : To determine the condition of the steam after +// (a) isothermal compression to half its initial volume,heat rejected +// (b) hyperbolic compression to half its initial volume + +// Given values +V1 = .3951;// initial volume,[m^3] +P1 = 1.5;// initial pressure,[MN/m^2] + +// solution + +// (a) +// from steam table, at 1.5 MN/m^2 +hf1 = 844.7;// [kJ/kg] +hfg1 = 1945.2;// [kJ/kg] +hg1 = 2789.9;// [kJ/kg] +vg1 = .1317;// [m^3/kg] + +// calculation +m = V1/vg1;// mass of steam,[kg] +vg2b = vg1/2;// given,[m^3/kg](vg2b is actual specific volume before compression) +x1 = vg2b/vg1;// dryness fraction +h1 = m*(hf1+x1*hfg1);// [kJ] +Q = m*x1*hfg1;// heat loss,[kJ] +mprintf('\n (a) The Quantity of steam present is = %f kg \n',m); +mprintf('\n Dryness fraction is = %f \n',x1); +mprintf('\n The enthalpy is = %f kJ \n',h1); +mprintf('\n The heat loss is = %f kJ \n',Q); + +// (b) +V2 = V1/2; +// Given compression is according to the law PV=Constant,so +P2 = P1*V1/V2;// [MN/m^2] +// from steam table at P2 +hf2 = 1008.4;// [kJ/kg] +hfg2 = 1793.9;// [kJ/kg] +hg2 = 2802.3;// [kJ/kg] +vg2 = .0666;// [m^3/kg] + +// calculation +x2 = vg2b/vg2;// dryness fraction +h2 = m*(hf2+x2*hfg2);// [kJ] + +mprintf('\n (b) The dryness fraction is = %f \n',x2); +mprintf('\n The enthalpy is = %f kJ\n',h2); + +// End diff --git a/2705/CH4/EX4.2/Ex4_2.sce b/2705/CH4/EX4.2/Ex4_2.sce new file mode 100755 index 000000000..d3c8e64dd --- /dev/null +++ b/2705/CH4/EX4.2/Ex4_2.sce @@ -0,0 +1,42 @@ +clear; +clc; +disp('Example 4.2'); + +// aim : To determine +// saturation temperature and enthalpy + +// Given values +P = 2.04;// pressure, [MN/m^2] + +// solution +// since in the steam table values of enthalpy and saturation temperature at 2 and 2.1 MN?m^2 are given, so for knowing required values at given pressure,there is need to do interpolation + +// calculation of saturation temperature +// from steam table +Table_P_tf = [[2.1,2.0];[214.9,212.4]]; // P in [MN/m^2] and tf in [C] +// using interpolation +tf = interpln(Table_P_tf,2.04);// saturation temperature at given condition +mprintf('\n The Saturation temperature is = %f C \n',tf); + +// calculation of specific liquid enthalpy +// from steam table +Table_P_hf = [[2.1,2.0];[920.0,908.6]];// P in [MN/m^2] and hf in [kJ/kg] +// using interpolation +hf = interpln(Table_P_hf,2.04); // enthalpy at given condition, [kJ/kg] +mprintf('\n The Specific liquid enthalpy is = %f kJ/kg \n',hf); + +// calculation of specific enthalpy of evaporation +// from steam table +Table_P_hfg = [[2.1,2.0];[1878.2,1888.6]];// P in [MN/m^2] and hfg in [kJ/kg] +// using interpolation +hfg = interpln(Table_P_hfg,2.04); // enthalpy at given condition, [kJ/kg] +mprintf('\n The Specific enthalpy of evaporation is = %f kJ/kg \n',hfg); + +// calculation of specific enthalpy of dry saturated steam +// from steam table +Table_P_hg = [[2.1,2.0];[2798.2,2797.2]];//P in [MN/m^2] and hg in [kJ/kg] +// using interpolation +hg = interpln(Table_P_hg,2.04); // enthalpy at given condition, [kJ/kg] +mprintf('\n The Specific enthalpy of dry saturated steam is = %f kJ/kg \n',hg); + +// End diff --git a/2705/CH4/EX4.20/Ex4_20.sce b/2705/CH4/EX4.20/Ex4_20.sce new file mode 100755 index 000000000..47eddb9ac --- /dev/null +++ b/2705/CH4/EX4.20/Ex4_20.sce @@ -0,0 +1,76 @@ +clear; +clc; +disp('Example 4.20'); + +// aim : To determine the +// (a) mass of steam +// (b) work transfer +// (c) change of internal energy +// (d) heat exchange b/w the steam and surroundings + +// Given values +P1 = 2.1;// initial pressure,[MN/m^2] +x1 = .9;// dryness fraction +V1 = .427;// initial volume,[m^3] +P2 = .7;// final pressure,[MN/m^2] +// Given process is polytropic with +n = 1.25; // polytropic index + +// solution +// from steam table + +// at 2.1 MN/m^2 +hf1 = 920.0;// [kJ/kg] +hfg1=1878.2;// [kJ/kg] +hg1=2798.2;// [kJ/kg] +vg1 = .0949;// [m^3/kg] + +// and at .7 MN/m^2 +hf2 = 697.1;// [kJ/kg] +hfg2 = 2064.9;// [kJ/kg] +hg2 = 2762.0;// [kJ/kg] +vg2 = .273;// [m^3/kg] + +// (a) +v1 = x1*vg1;// [m^3/kg] +m = V1/v1;// [kg] +mprintf('\n (a) The mass of steam present is = %f kg\n',m); + +// (b) +// for polytropic process +v2 = v1*(P1/P2)^(1/n);// [m^3/kg] + +x2 = v2/vg2;// final dryness fraction +// work transfer +W = m*(P1*v1-P2*v2)*10^3/(n-1);// [kJ] +mprintf('\n (b) The work transfer is = %f kJ\n',W); + +// (c) +// initial +h1 = hf1+x1*hfg1;// [kJ/kg] +u1 = h1-P1*v1*10^3;// [kJ/kg] + +// final +h2 = hf2+x2*hfg2;// [kJ/kg] +u2 = h2-P2*v2*10^3;// [kJ/kg] + +del_U = m*(u2-u1);// [kJ] +mprintf('\n (c) The change in internal energy is = %f kJ',del_U); +if(del_U<0) + disp('since del_U<0,so this is loss of internal energy') +else + disp('since del_U>0,so this is gain in internal energy') +end + +// (d) +Q = del_U+W;// [kJ] +mprintf('\n (d) The heat exchange between the steam and surrounding is = %f kJ',Q); +if(Q<0) + disp('since Q<0,so this is loss of heat energy to surrounding') +else + disp('since Q>0,so this is gain in heat energy to the steam') +end + +// there are minor vairations in the values reported in the book + +// End diff --git a/2705/CH4/EX4.21/Ex4_21.sce b/2705/CH4/EX4.21/Ex4_21.sce new file mode 100755 index 000000000..6aafe47b0 --- /dev/null +++ b/2705/CH4/EX4.21/Ex4_21.sce @@ -0,0 +1,42 @@ +clear; +clc; +disp('Example 4.21'); + +// aim : To determine the +// (a) volume occupied by steam +// (b)(1) final dryness fraction of steam +// (2) Change of internal energy during expansion + +// (a) +// Given values +P1 = .85;// [mN/m^2] +x1 = .97; + +// solution +// from steam table, at .85 MN/m^2, +vg1 = .2268;// [m^3/kg] +// hence +v1 = x1*vg1;// [m^3/kg] +mprintf('\n (a) The volume occupied by 1 kg of steam is = %f m^3/kg\n',v1); + +// (b)(1) +P2 = .17;// [MN/m^2] +// since process is polytropic process with +n = 1.13; // polytropic index +// hence +v2 = v1*(P1/P2)^(1/n);// [m^3/kg] + +// from steam table at .17 MN/m^2 +vg2 = 1.031;// [m^3/kg] +// steam is wet so +x2 = v2/vg2;// final dryness fraction +mprintf('\n (b)(1) The final dryness fraction of the steam is = %f \n',x2); + +// (2) +W = (P1*v1-P2*v2)*10^3/(n-1);// [kJ/kg] +// since process is adiabatic, so +del_u = -W;// [kJ/kg] +mprintf('\n (2) The change in internal energy of the steam during expansion is = %f kJ/kg (This is a loss of internal energy)\n',del_u); +// There are minor variation in the answer + +// End diff --git a/2705/CH4/EX4.3/Ex4_3.sce b/2705/CH4/EX4.3/Ex4_3.sce new file mode 100755 index 000000000..7b6b2dd19 --- /dev/null +++ b/2705/CH4/EX4.3/Ex4_3.sce @@ -0,0 +1,30 @@ +clear; +clc; +disp('Example 4.3'); + +// aim : To determine +// the specific enthalpy + +// given values +P = 2; // pressure ,[MN/m^2] +t = 250; // Temperature, [C] +cp = 2.0934; // average value of specific heat capacity, [kJ/kg K] + +// solution + +// looking up steam table it shows that at given pressure saturation temperature is 212.4 C,so +tf = 212.4; // [C] +// hence, +Degree_of_superheat = t-tf;// [C] +// from table at given temperature 250 C +h = 2902; // specific enthalpy of steam at 250 C ,[kJ/kg] +mprintf('\nThe specific enthalpy of steam at 2 MN/m^2 with temperature 250 C is = %f kJ/kg \n',h); + +// Also from steam table enthalpy at saturation temperature is +hf = 2797.2 ;// [kJ/kg] +// so enthalpy at given temperature is +h = hf+cp*(t-tf);// [kJ/kg] + +mprintf('\n The specific enthalpy at given T and P by alternative path is = %f kJ/kg \n',h); + +// End diff --git a/2705/CH4/EX4.4/Ex4_4.sce b/2705/CH4/EX4.4/Ex4_4.sce new file mode 100755 index 000000000..2af9a5594 --- /dev/null +++ b/2705/CH4/EX4.4/Ex4_4.sce @@ -0,0 +1,43 @@ +clear; +clc; +disp('Example 4.4'); + +// aim : To determine +// the specific enthalpy of steam + +// Given values +P = 2.5;// pressure, [MN/m^2] +t = 320; // temperature, [C] + +// solution +// from steam table at given condition the saturation temperature of steam is 223.9 C, therefore steam is superheated +tf = 223.9;// [C] + +// first let's calculate estimated enthalpy +// again from steam table + +hg = 2800.9;// enthalpy at saturation temp, [kJ/kg] +cp =2.0934;// specific heat capacity of steam,[kJ/kg K] + +// so enthalpy at given condition is +h = hg+cp*(t-tf);// [kJ/kg] +mprintf('\n The estimated specific enthalpy is = %f kJ/kg \n',h); + +// calculation of accurate specific enthalpy +// we need double interpolation for this + +// first interpolation w.r.t. to temperature +// At 2 MN/m^2 +Table_t_h = [[325,300];[3083,3025]];// where, t in [C] and h in [kJ/kg] +h1 = interpln(Table_t_h,320); // [kJ/kg] + +// at 4 MN/m^2 +Table_t_h = [[325,300];[3031,2962]]; // t in [C] and h in [kJ/kg] +h2 = interpln(Table_t_h,320); // [kJ/kg] + +// now interpolation w.r.t. pressure +Table_P_h = [[4,2];[h2,h1]]; // where P in NM/m^2 and h1,h2 in kJ/kg +h = interpln(Table_P_h,2.5);// [kJ/kg] +mprintf('\n The accurate specific enthalpy of steam at pressure of 2.5 MN/m^2 and with a temperature 320 C is = %f kJ/kg \n',h); + +// End diff --git a/2705/CH4/EX4.5/Ex4_5.sce b/2705/CH4/EX4.5/Ex4_5.sce new file mode 100755 index 000000000..c2c030046 --- /dev/null +++ b/2705/CH4/EX4.5/Ex4_5.sce @@ -0,0 +1,25 @@ +clear; +clc; +disp('Example 4.5'); + +// aim : To determine +// the specific enthalpy + +// Given values +P = 70; // pressure, [kn/m^2] +x = .85; // Dryness fraction + +// solution + +// from steam table, at given pressure +hf = 376.8;// [kJ/kg] +hfg = 2283.3;// [kJ/kg] + +// now using equation [2] +h = hf+x*hfg;// specific enthalpy of wet steam,[kJ/kg] + +mprintf('\n The specific enthalpy of wet steam is = %f kJ/kg \n',h); + +// There is minor variation in the book's answer + +// End diff --git a/2705/CH4/EX4.8/Ex4_8.sce b/2705/CH4/EX4.8/Ex4_8.sce new file mode 100755 index 000000000..7cfe99980 --- /dev/null +++ b/2705/CH4/EX4.8/Ex4_8.sce @@ -0,0 +1,20 @@ +clear; +clc; +disp('Example 4.8'); + +// aim : To determine +// the specific volume of wet steam + +// Given values +P = 1.25; // pressure, [MN/m^2] +x = .9; // dry fraction + +// solution +// from steam table at given pressure +vg = .1569;// [m^3/kg] +// hence +v = x*vg; // [m^3/kg] + +mprintf('\nThe specific volume of wet steam is = %f m^3/kg \n',v); + +// End diff --git a/2705/CH4/EX4.9/Ex4_9.sce b/2705/CH4/EX4.9/Ex4_9.sce new file mode 100755 index 000000000..d9417bfea --- /dev/null +++ b/2705/CH4/EX4.9/Ex4_9.sce @@ -0,0 +1,21 @@ +clear; +clc; +disp('Example 4.9'); + +// aim : To determine +// the specific volume + +// Given values +t = 325; // temperature, [C] +P = 2; // pressure, [MN/m^2] + +// solution +// from steam table at given t and P +vf = .1321; // [m^3/kg] +tf = 212.4; // saturation temperature, [C] + +mprintf('\n The specific volume of steam at pressure of 2 MN/m^2 and with temperature 325 C is = %f m^3/kg \n',vf); +doh= t-tf; // degree of superheat, [C] +mprintf('\n The degree of superheat is = %f C\n',doh); + +// End diff --git a/2705/CH5/EX5.1/Ex5_1.sce b/2705/CH5/EX5.1/Ex5_1.sce new file mode 100755 index 000000000..dd8a53487 --- /dev/null +++ b/2705/CH5/EX5.1/Ex5_1.sce @@ -0,0 +1,23 @@ +clear; +clc; +disp('Example 5.1'); + +// aim : To determine +// new pressure exerted on the air and the difference in two mercury column level + +// Given values +P1 = 765;// atmospheric pressure, [mmHg] +V1 = 20000;// [mm^3] +V2 = 17000;// [mm^3] + +// solution + +// using boyle's law P*V=constant +// hence +P2 = P1*V1/V2;// [mmHg] +mprintf('\n The new pressure exerted on the air is = %f mmHg \n',P2); + +del_h = P2-P1;// difference in Height of mercury column level +mprintf('\n The difference in the two mercury column level is = %f mm\n',del_h); + +// End diff --git a/2705/CH5/EX5.10/Ex5_10.sce b/2705/CH5/EX5.10/Ex5_10.sce new file mode 100755 index 000000000..b5d45cd32 --- /dev/null +++ b/2705/CH5/EX5.10/Ex5_10.sce @@ -0,0 +1,27 @@ +clear; +clc; +disp('Example 5.10'); + +// aim : To determine +// the new temperature of the gas + +// Given values +V1 = .015;// original volume,[m^3] +T1 = 273+285;// original temperature,[K] +V2 = .09;// final volume,[m^3] + +// solution +// Given gas is following the law,P*V^1.35=constant +// so process is polytropic with +n = 1.35; // polytropic index + +// hence +T2 = T1*(V1/V2)^(n-1);// final temperature, [K] + +t2 = T2-273;// [C] + +mprintf('\n The new temperature of the gas is = %f C \n',t2); + +// there is minor error in book's answer + +// End diff --git a/2705/CH5/EX5.11/Ex5_11.sce b/2705/CH5/EX5.11/Ex5_11.sce new file mode 100755 index 000000000..42acb405c --- /dev/null +++ b/2705/CH5/EX5.11/Ex5_11.sce @@ -0,0 +1,40 @@ +clear; +clc; +disp('Example 5.11'); + +// aim : To determine the +// (a) original and final volume of the gas +// (b) final pressure of the gas +// (c) final temperature of the gas + +// Given values +m = .675;// mass of the gas,[kg] +P1 = 1.4;// original pressure,[MN/m^2] +T1 = 273+280;// original temperature,[K] +R = .287;//gas constant,[kJ/kg K] + +// solution + +// (a) +// using characteristic equation, P1*V1=m*R*T1 +V1 = m*R*T1*10^-3/P1;// [m^3] +// also Given +V2 = 4*V1;// [m^3] +mprintf('\n (a) The original volume of the gas is = %f m^3\n',V1); +mprintf('\n and The final volume of the gas is = %f m^3\n',V2); + +// (b) +// Given that gas is following the law P*V^1.3=constant +// hence process is polytropic with +n = 1.3; // polytropic index +P2 = P1*(V1/V2)^n;// formula for polytropic process,[MN/m^2] +mprintf('\n (b) The final pressure of the gas is = %f kN/m^2\n',P2*10^3); + +// (c) +// since mass is constant so,using P*V/T=constant +// hence +T2 = P2*V2*T1/(P1*V1);// [K] +t2 = T2-273;// [C] +mprintf('\n (c) The final temperature of the gas is = %f C\n',t2); + +// End diff --git a/2705/CH5/EX5.12/Ex5_12.sce b/2705/CH5/EX5.12/Ex5_12.sce new file mode 100755 index 000000000..e522db8a2 --- /dev/null +++ b/2705/CH5/EX5.12/Ex5_12.sce @@ -0,0 +1,57 @@ +clear; +clc; +disp('Example 5.12'); + +// aim : T0 determine +// (a) change in internal nergy of the air +// (b) work done +// (c) heat transfer + +// Given values +m = .25;// mass, [kg] +P1 = 140;// initial pressure, [kN/m^2] +V1 = .15;// initial volume, [m^3] +P2 = 1400;// final volume, [m^3] +cp = 1.005;// [kJ/kg K] +cv = .718;// [kJ/kg K] + +// solution + +// (a) +// assuming ideal gas +R = cp-cv;// [kJ/kg K] +// also, P1*V1=m*R*T1,hence +T1 = P1*V1/(m*R);// [K] + +// given that process is polytropic with +n = 1.25; // polytropic index +T2 = T1*(P2/P1)^((n-1)/n);// [K] + +// Hence, change in internal energy is, +del_U = m*cv*(T2-T1);// [kJ] +mprintf('\n (a) The change in internal energy of the air is del_U = %f kJ',del_U); +if(del_U>0) + disp('since del_U>0, so it is gain of internal energy to the air') +else + disp('since del_U<0, so it is gain of internal energy to the surrounding') +end + +// (b) +W = m*R*(T1-T2)/(n-1);// formula of work done for polytropic process,[kJ] +mprintf('\n (b) The work done is W = %f kJ',W); +if(W>0) + disp('since W>0, so the work is done by the air') +else + disp('since W<0, so the work is done on the air') +end + +// (c) +Q = del_U+W;// using 1st law of thermodynamics,[kJ] +mprintf('\n (c) The heat transfer is Q = %f kJ',Q); +if(Q>0) + disp('since Q>0, so the heat is received by the air') +else + disp('since Q<0, so the heat is rejected by the air') +end + +// End diff --git a/2705/CH5/EX5.13/Ex5_13.sce b/2705/CH5/EX5.13/Ex5_13.sce new file mode 100755 index 000000000..e2fc9c1b2 --- /dev/null +++ b/2705/CH5/EX5.13/Ex5_13.sce @@ -0,0 +1,35 @@ +clear; +clc; +disp('Example 5.13'); + +// aim : To determine the +// final volume, work done and the change in internal energy + +// Given values +P1 = 700;// initial pressure,[kN/m^2] +V1 = .015;// initial volume, [m^3] +P2 = 140;// final pressure, [kN/m^2] +cp = 1.046;// [kJ/kg K] +cv = .752; // [kJ/kg K] + +// solution + +Gamma = cp/cv; +// for adiabatic expansion, P*V^gamma=constant, so +V2 = V1*(P1/P2)^(1/Gamma);// final volume, [m^3] +mprintf('\n The final volume of the gas is V2 = %f m^3\n',V2); + +// work done +W = (P1*V1-P2*V2)/(Gamma-1);// [kJ] +mprintf('\n The work done by the gas is = %f kJ\n',W); + +// for adiabatic process +del_U = -W;// [kJ] +mprintf('\n The change of internal energy is = %f kJ',del_U); +if(del_U>0) + disp('since del_U>0, so the the gain in internal energy of the gas ') +else + disp('since del_U<0, so this is a loss of internal energy from the gas') +end + +// End diff --git a/2705/CH5/EX5.14/Ex5_14.sce b/2705/CH5/EX5.14/Ex5_14.sce new file mode 100755 index 000000000..25771ce9f --- /dev/null +++ b/2705/CH5/EX5.14/Ex5_14.sce @@ -0,0 +1,54 @@ +clear; +clc; +disp('Example 5.14'); + +// aim : To determine the +// (a)heat transfer +// (b)change of internal energy +// (c)mass of gas + +// Given values +V1 = .4;// initial volume, [m^3] +P1 = 100;// initial pressure, [kN/m^2] +T1 = 273+20;// temperature, [K] +P2 = 450;// final pressure,[kN/m^2] +cp = 1.0;// [kJ/kg K] +Gamma = 1.4; // heat capacity ratio + +// solution + +// (a) +// for the isothermal compression,P*V=constant,so +V2 = V1*P1/P2;// [m^3] +W = P1*V1*log(P1/P2);// formula of workdone for isothermal process,[kJ] + +// for isothermal process, del_U=0;so +Q = W; +mprintf('\n (a) The heat transferred during compression is Q = %f kJ\n',Q); + + +// (b) +V3 = V1; +// for adiabatic expansion +// also + +P3 = P2*(V2/V3)^Gamma;// [kN/m^2] +W = -(P3*V3-P2*V2)/(Gamma-1);// work done formula for adiabatic process,[kJ] +// also, Q=0,so using Q=del_U+W +del_U = -W;// [kJ] +mprintf('\n (b) The change of the internal energy during the expansion is,del_U = %f kJ\n',del_U); + +// (c) +// for ideal gas +// cp-cv=R, and cp/cv=gamma, hence +R = cp*(1-1/Gamma);// [kj/kg K] + +// now using ideal gas equation +m = P1*V1/(R*T1);// mass of the gas,[kg] +mprintf('\n (c) The mass of the gas is,m = %f kg\n',m); + +// There is calculation mistake in the book + + +// End + diff --git a/2705/CH5/EX5.15/Ex5_15.sce b/2705/CH5/EX5.15/Ex5_15.sce new file mode 100755 index 000000000..e5ae9aa6a --- /dev/null +++ b/2705/CH5/EX5.15/Ex5_15.sce @@ -0,0 +1,35 @@ +clear; +clc; +disp('Example 5.15'); + +// aim : To determine +// the heat transferred and polytropic specific heat capacity + +// Given values +P1 = 1;// initial pressure, [MN/m^2] +V1 = .003;// initial volume, [m^3] +P2 = .1;// final pressure,[MN/m^2] +cv = .718;// [kJ/kg*K] +Gamma=1.4;// heat capacity ratio + +// solution +// Given process is polytropic with +n = 1.3;// polytropic index +// hence +V2 = V1*(P1/P2)^(1/n);// final volume,[m^3] +W = (P1*V1-P2*V2)*10^3/(n-1);// work done,[kJ] +// so +Q = (Gamma-n)*W/(Gamma-1);// heat transferred,[kJ] + +mprintf('\n The heat received or rejected by the gas during this process is Q = %f kJ',Q); +if(Q>0) + disp('since Q>0, so heat is received by the gas') +else + disp('since Q<0, so heat is rejected by the gas') +end + +// now +cn = cv*(Gamma-n)/(n-1);// polytropic specific heat capacity,[kJ/kg K] +mprintf('\n The polytropic specific heat capacity is cn = %f kJ/kg K\n',cn); + +// End diff --git a/2705/CH5/EX5.16/Ex5_16.sce b/2705/CH5/EX5.16/Ex5_16.sce new file mode 100755 index 000000000..37712f1ac --- /dev/null +++ b/2705/CH5/EX5.16/Ex5_16.sce @@ -0,0 +1,40 @@ +clear; +clc; +disp('Example 5.16'); + +// aim : To determine the +// (a) initial partial pressure of the steam and air +// (b) final partial pressure of the steam and air +// (c) total pressure in the container after heating + +// Given values +T1 = 273+39;// initial temperature,[K] +P1 = 100;// pressure, [MN/m^2] +T2 = 273+120.2;// final temperature,[K] + +// solution + +// (a) +// from the steam tables, the pressure of wet steam at 39 C is +Pw1 = 7;// partial pressure of wet steam,[kN/m^2] +// and by Dalton's law +Pa1 = P1-Pw1;// initial pressure of air, [kN/m^2] + +mprintf('\n (a) The initial partial pressure of the steam is = %f kN/m^2',Pw1); +mprintf('\n The initial partial pressure of the air is = %f kN/m^2\n',Pa1); + +// (b) +// again from steam table, at 120.2 C the pressure of wet steam is +Pw2 = 200;// [kN/m^2] + +// now since volume is constant so assuming air to be ideal gas so for air P/T=contant, hence +Pa2 = Pa1*T2/T1 ;// [kN/m^2] + +mprintf(' \n(b) The final partial pressure of the steam is = %f kN/m^2',Pw2); +mprintf('\n The final partial pressure of the air is = %f kN/m^2\n',Pa2); + +// (c) +Pt = Pa2+Pw2;// using dalton's law, total pressure,[kN/m^2] +mprintf('\n (c) The total pressure after heating is = %f kN/m^2\n',Pt); + +// End diff --git a/2705/CH5/EX5.17/Ex5_17.sce b/2705/CH5/EX5.17/Ex5_17.sce new file mode 100755 index 000000000..187f6250a --- /dev/null +++ b/2705/CH5/EX5.17/Ex5_17.sce @@ -0,0 +1,40 @@ +clear; +clc; +disp('Example 5.17'); + +// aim : To determine +// the partial pressure of the air and steam, and the mass of the air + +// Given values +P1 = 660;// vaccum gauge pressure on condenser [mmHg] +P = 765;// atmospheric pressure, [mmHg] +x = .8;// dryness fraction +T = 273+41.5;// temperature,[K] +ms_dot = 1500;// condense rate of steam,[kg/h] +R = .29;// [kJ/kg] + +// solution +Pa = (P-P1)*.1334;// absolute pressure,[kN/m^2] +// from steam table, at 41.5 C partial pressure of steam is +Ps = 8;// [kN/m^2] +// by dalton's law, partial pressure of air is +Pg = Pa-Ps;// [kN/m^2] + +mprintf('\n The partial pressure of the air in the condenser is = %f kN/m^2\n',Pg); +mprintf('\n The partial pressure of the steam in the condenser is = %f kN/m^2\n',Ps); + +// also +vg = 18.1;// [m^3/kg] +// so +V = x*vg;// [m^3/kg] +// The air associated with 1 kg of the steam will occupiy this same volume +// for air, Pg*V=m*R*T,so +m = Pg*V/(R*T);// [kg/kg steam] +// hence +ma = m*ms_dot;// [kg/h] + +mprintf('\n The mass of air which will associated with this steam is = %f kg\n',ma); + +// There is misprint in book + +// End diff --git a/2705/CH5/EX5.18/Ex5_18.sce b/2705/CH5/EX5.18/Ex5_18.sce new file mode 100755 index 000000000..3266c108e --- /dev/null +++ b/2705/CH5/EX5.18/Ex5_18.sce @@ -0,0 +1,41 @@ +clear; +clc; +disp('Example 5.18'); + +// aim : To determine the +// (a) final pressure +// (b) final dryness fraction of the steam + +// Given values +P1 = 130;// initial pressure, [kN/m^2] +T1 = 273+75.9;// initial temperature, [K] +x1 = .92;// initial dryness fraction +T2 = 273+120.2;// final temperature, [K] + +// solution + +// (a) +// from steam table, at 75.9 C +Pws = 40;// partial pressure of wet steam[kN/m^2] +Pa = P1-Pws;// partial pressure of air, [kN/m^2] +vg = 3.99// specific volume of the wet steam, [m^3/kg] +// hence +V1 = x1*vg;// [m^3/kg] +V2 = V1/5;// [m^3/kg] +// for air, mass is constant so, Pa*V1/T1=P2*V2/T2,also given ,V1/V2=5,so +P2 = Pa*V1*T2/(V2*T1);// final pressure,[kN/m^2] + +// now for steam at 120.2 C +Ps = 200;// final partial pressure of steam,[kN/m^2] +// so by dalton's law total pressure in cylindert is +Pt = P2+Ps;// [kN/m^2] +mprintf('\n (a) The final pressure in the cylinder is = %f kN/m^2\n',Pt); + +// (b) +// from steam table at 200 kN/m^2 +vg = .885;// [m^3/kg] +// hence +x2 = V2/vg;// final dryness fraction of the steam +mprintf('\n (b) The final dryness fraction of the steam is = %f\n ',x2); + +// End diff --git a/2705/CH5/EX5.19/Ex5_19.sce b/2705/CH5/EX5.19/Ex5_19.sce new file mode 100755 index 000000000..830fd787f --- /dev/null +++ b/2705/CH5/EX5.19/Ex5_19.sce @@ -0,0 +1,39 @@ +clear; +clc; +disp('Example 5.19') + +// aim : To determine the +// (a) Gamma, +// (b) del_U + +// Given Values +P1 = 1400;// [kN/m^2] +P2 = 100;// [kN/m^2] +P3 = 220;// [kN/m^2] +T1 = 273+360;// [K] +m = .23;// [kg] +cp = 1.005;// [kJ/kg*K] + +// Solution +T3 = T1;// since process 1-3 is isothermal + +// (a) +// for process 1-3, P1*V1=P3*V3,so +V3_by_V1 = P1/P3; +// also process 1-2 is adiabatic,so P1*V1^(Gamma)=P2*V2^(Gamma),hence +// and process process 2-3 is iso-choric so,V3=V2 and +V2_by_V1 = V3_by_V1; +// hence, +Gamma = log(P1/P2)/log(P1/P3); // heat capacity ratio + +mprintf('\n (a) The value of adiabatic index Gamma is = %f\n',Gamma); + +// (b) +cv = cp/Gamma;// [kJ/kg K] +// for process 2-3,P3/T3=P2/T2,so +T2 = P2*T3/P3;// [K] + +// now +del_U = m*cv*(T2-T1);// [kJ] +mprintf('\n (b) The change in internal energy during the adiabatic expansion is U2-U1 = %f kJ (This is loss of internal energy)\n',del_U); +// End diff --git a/2705/CH5/EX5.2/Ex5_2.sce b/2705/CH5/EX5.2/Ex5_2.sce new file mode 100755 index 000000000..3e80ae408 --- /dev/null +++ b/2705/CH5/EX5.2/Ex5_2.sce @@ -0,0 +1,20 @@ +clear; +clc; +disp('Example 5.2'); + +// aim : To determine +// the new volume + +// Given values +P1 = 300;// original pressure,[kN/m^2] +V1 = .14;// original volume,[m^3] + +P2 = 60;// new pressure after expansion,[kn/m^2] + +// solution +// since temperature is constant so using boyle's law P*V=constant +V2 = V1*P1/P2;// [m^3] + +mprintf('\n The new volume after expansion is = %f m^3\n',V2); + +// End diff --git a/2705/CH5/EX5.20/Ex5_20.sce b/2705/CH5/EX5.20/Ex5_20.sce new file mode 100755 index 000000000..daf3343a7 --- /dev/null +++ b/2705/CH5/EX5.20/Ex5_20.sce @@ -0,0 +1,43 @@ +clear; +clc; +disp('Example 5.20'); + +// aim : To determine +// the mass of oxygen and heat transferred + +// Given values +V1 = 300;// [L] +P1 = 3.1;// [MN/m^2] +T1 = 273+18;// [K] +P2 = 1.7;// [MN/m^2] +T2 = 273+15;// [K] +Gamma = 1.4; // heat capacity ratio +// density condition +P = .101325;// [MN/m^2] +T = 273;// [K] +V = 1;// [m^3] +m = 1.429;// [kg] + +// hence +R = P*V*10^3/(m*T);// [kJ/kg*K] +// since volume is constant +V2 = V1;// [L] +// for the initial conditions in the cylinder,P1*V1=m1*R*T1 +m1 = P1*V1/(R*T1);// [kg] + +// after some of the gas is used +m2 = P2*V2/(R*T2);// [kg] +// The mass of oxygen remaining in cylinder is m2 kg,so +// Mass of oxygen used is +m_used = m1-m2;// [kg] +mprintf('\n The mass of oxygen used = %f kg\n',m_used); + +// for non-flow process,Q=del_U+W +// volume is constant so no external work is done so,Q=del_U +cv = R/(Gamma-1);// [kJ/kg*K] + +// heat transfer is +Q = m2*cv*(T1-T2);// (kJ) +mprintf('\n The amount of heat transferred through the cylinder wall is = %f kJ\n',Q); + +// End diff --git a/2705/CH5/EX5.21/Ex5_21.sce b/2705/CH5/EX5.21/Ex5_21.sce new file mode 100755 index 000000000..a5cdcb432 --- /dev/null +++ b/2705/CH5/EX5.21/Ex5_21.sce @@ -0,0 +1,43 @@ +clear; +clc; +disp('Example 5.21'); + +// aim : To determine the +// (a) work transferred during the compression +// (b) change in internal energy +// (c) heat transferred during the compression + +// Given values +V1 = .1;// initial volume, [m^3] +P1 = 120;// initial pressure, [kN/m^2] +P2 = 1200; // final pressure, [kN/m^2] +T1 = 273+25;// initial temperature, [K] +cv = .72;// [kJ/kg*K] +R = .285;// [kJ/kg*K] + +// solution + +// (a) +// given process is polytropic with +n = 1.2; // polytropic index +// hence +V2 = V1*(P1/P2)^(1/n);// [m^3] +W = (P1*V1-P2*V2)/(n-1);// workdone formula, [kJ] +mprintf('\n (a) The work transferred during the compression is = %f kJ\n',W); + +// (b) +// now mass is constant so, +T2 = P2*V2*T1/(P1*V1);// [K] +// using, P*V=m*R*T +m = P1*V1/(R*T1);// [kg] + +// change in internal energy is +del_U = m*cv*(T2-T1);// [kJ] +mprintf('\n (b) The change in internal energy is = %f kJ\n',del_U); + +// (c) +Q = del_U+W;// [kJ] +mprintf('\n (c) The heat transferred during the compression is = %f kJ\n',Q); + +// End + diff --git a/2705/CH5/EX5.22/Ex5_22.sce b/2705/CH5/EX5.22/Ex5_22.sce new file mode 100755 index 000000000..756233f57 --- /dev/null +++ b/2705/CH5/EX5.22/Ex5_22.sce @@ -0,0 +1,36 @@ +clear; +clc; +disp('Example 5.22'); + +// aim : To determine the +// (a) new pressure of the air in the receiver +// (b) specific enthalpy of air at 15 C + +// Given values +V1 = .85;// [m^3] +T1 = 15+273;// [K] +P1 = 275;// pressure,[kN/m^2] +m = 1.7;// [kg] +cp = 1.005;// [kJ/kg*K] +cv = .715;// [kJ/kg*K] + +// solution + +// (a) + +R = cp-cv;// [kJ/kg*K] +// assuming m1 is original mass of the air, using P*V=m*R*T +m1 = P1*V1/(R*T1);// [kg] +m2 = m1+m;// [kg] +// again using P*V=m*R*T +// P2/P1=(m2*R*T2/V2)/(m1*R*T1/V1); and T1=T2,V1=V2,so +P2 = P1*m2/m1;// [kN/m^2] +mprintf('\n (a) The new pressure of the air in the receiver is = %f kN/m^2\n',P2); + +// (b) +// for 1 kg of air, h2-h1=cp*(T1-T0) +// and if 0 is chosen as the zero enthalpy, then +h = cp*(T1-273);// [kJ/kg] +mprintf('\n (b) The specific enthalpy of the air at 15 C is = %f kJ/kg\n',h); + +// End diff --git a/2705/CH5/EX5.23/Ex5_23.sce b/2705/CH5/EX5.23/Ex5_23.sce new file mode 100755 index 000000000..69c2910b0 --- /dev/null +++ b/2705/CH5/EX5.23/Ex5_23.sce @@ -0,0 +1,47 @@ +clear; +clc; +disp('Example 5.23'); + +// aim : T determine the +// (a) characteristic gas constant of the gas +// (b) cp, +// (c) cv, +// (d) del_u +// (e) work transfer + +// Given values +P = 1;// [bar] +T1 = 273+15;// [K] +m = .9;// [kg] +T2 = 273+250;// [K] +Q = 175;// heat transfer,[kJ] + +// solution + +// (a) +// using, P*V=m*R*T, given, +m_by_V = 1.875; +// hence +R = P*100/(T1*m_by_V);// [kJ/kg*K] +mprintf('\n (a) The characteristic gas constant of the gas is R = %f kJ/kg K\n',R); + +// (b) +// using, Q=m*cp*(T2-T1) +cp = Q/(m*(T2-T1));// [kJ/kg K] +mprintf('\n (b) The specific heat capacity of the gas at constant pressure cp = %f kJ/kg K\n',cp); + +// (c) +// we have, cp-cv=R,so +cv = cp-R;// [kJ/kg*K] +mprintf('\n (c) The specific heat capacity of the gas at constant volume cv = %f kJ/kg K\n',cv); + +// (d) +del_U = m*cv*(T2-T1);// [kJ] +mprintf('\n (d) The change in internal energy is = %f kJ\n',del_U); + +// (e) +// using, Q=del_U+W +W = Q-del_U;// [kJ] +mprintf('\n (e) The work transfer is W = %f kJ\n',W); + +// End diff --git a/2705/CH5/EX5.24/Ex5_24.sce b/2705/CH5/EX5.24/Ex5_24.sce new file mode 100755 index 000000000..70324999b --- /dev/null +++ b/2705/CH5/EX5.24/Ex5_24.sce @@ -0,0 +1,42 @@ +clear; +clc; +disp('Example 5.24'); + +// aim : To determine the +// (a) work transfer, +// (b)del_U and, +// (c)heat transfer + +// Given values +V1 = .15;// [m^3] +P1 = 1200;// [kN/m^2] +T1 = 273+120;// [K] +P2 = 200;// [kN/m^2] +cp = 1.006;//[kJ/kg K] +cv = .717;// [kJ/kg K] + +// solution + +// (a) +// Given, PV^1.32=constant, so it is polytropic process with +n = 1.32;// polytropic index +// hence +V2 = V1*(P1/P2)^(1/n);// [m^3] +// now, W +W = (P1*V1-P2*V2)/(n-1);// [kJ] +mprintf('\n (a) The work transfer is W = %f kJ\n',W); + +// (b) +R = cp-cv;// [kJ/kg K] +m = P1*V1/(R*T1);// gas law,[kg] +// also for polytropic process +T2 = T1*(P2/P1)^((n-1)/n);// [K] +// now for gas, +del_U = m*cv*(T2-T1);// [kJ] +mprintf('\n (b) The change of internal energy is del_U = %f kJ\n',del_U); + +// (c) +Q = del_U+W;// first law of thermodynamics,[kJ] +mprintf('\n (c) The heat transfer Q = %f kJ\n',Q); + +// End diff --git a/2705/CH5/EX5.26/Ex5_26.sce b/2705/CH5/EX5.26/Ex5_26.sce new file mode 100755 index 000000000..83f0cc59c --- /dev/null +++ b/2705/CH5/EX5.26/Ex5_26.sce @@ -0,0 +1,45 @@ +clear; +clc; +disp('Example 5.26'); + +// aim : To determine +// the volume of the pressure vessel and the volume of the gas before transfer + +// Given values + +P1 = 1400;// initial pressure,[kN/m^2] +T1 = 273+85;// initial temperature,[K] + +P2 = 700;// final pressure,[kN/m^2] +T2 = 273+60;// final temperature,[K] + +m = 2.7;// mass of the gas passes,[kg] +cp = .88;// [kJ/kg] +cv = .67;// [kJ/kg] + +// solution + +// steady flow equation is, u1+P1*V1+C1^2/2+Q=u2+P2*V2+C2^2/2+W [1], +// given, there is no kinetic energy change and neglecting potential energy term +W = 0;// no external work done +// so final equation is,u1+P1*v1+Q=u2 [2] +// also u2-u1=cv*(T2-T1) +// hence Q=cv*(T2-T1)-P1*v1 [3] +// and for unit mass P1*v1=R*T1=(cp-cv)*T1 [4] +// so finally +Q = cv*(T2-T1)-(cp-cv)*T1;// [kJ/kg] +// so total heat transferred is +Q = m*Q;// [kJ] + +// using eqn [4] +v1 = (cp-cv)*T1/P1;// [m^3/kg] +// Total volume is +V1 = m*v1;// [m^3] + +// using ideal gas equation P1*V1/T1=P2*V2/T2 +V2 = P1*T2*V1/(P2*T1);// final volume,[m^3] + +mprintf('\n The volume of gas before transfer is = %f m^3\n',V1); +mprintf('\n The volume of pressure vessel is = %f m^3\n',V2); + +// End diff --git a/2705/CH5/EX5.3/Ex5_3.sce b/2705/CH5/EX5.3/Ex5_3.sce new file mode 100755 index 000000000..71188b43b --- /dev/null +++ b/2705/CH5/EX5.3/Ex5_3.sce @@ -0,0 +1,20 @@ +clear; +clc; +disp('Example 5.3'); + +// aim : To determine +// the new volume of the gas + +// Given values +V1 = 10000;// [mm^3] +T1 = 273+18;// [K] +T2 = 273+85;// [K] + +// solution +// since pressure exerted on the apparatus is constant so using charle's law V/T=constant +// hence +V2 = V1*T2/T1;// [mm^3] + +mprintf('\n The new volume of the gas trapped in the apparatus is = %f mm^3\n',V2); + +// End diff --git a/2705/CH5/EX5.4/Ex5_4.sce b/2705/CH5/EX5.4/Ex5_4.sce new file mode 100755 index 000000000..954ced5f4 --- /dev/null +++ b/2705/CH5/EX5.4/Ex5_4.sce @@ -0,0 +1,20 @@ +clear; +clc; +disp('Example 5.4'); + +// aim : To determine +// the final temperature + +// Given values +V1 = .2;// original volume,[m^3] +T1 = 273+303;// original temperature, [K] +V2 = .1;// final volume, [m^3] + +// solution +// since pressure is constant, so using charle's law V/T=constant +// hence +T2 = T1*V2/V1;// [K] +t2 = T2-273;// [C] +mprintf('\n The final temperature of the gas is = %f C\n',t2); + +// End diff --git a/2705/CH5/EX5.5/Ex5_5.sce b/2705/CH5/EX5.5/Ex5_5.sce new file mode 100755 index 000000000..894496057 --- /dev/null +++ b/2705/CH5/EX5.5/Ex5_5.sce @@ -0,0 +1,25 @@ +clear; +clc; +disp('Example 5.5'); + +// aim : To determine +// the new volume of the gas + +// Given values + +// initial codition +P1 = 140;// [kN/m^2] +V1 = .1;// [m^3] +T1 = 273+25;// [K] + +// final condition +P2 = 700;// [kN/m^2] +T2 = 273+60;// [K] + +// by charasteristic equation, P1*V1/T1=P2*V2/T2 + +V2=P1*V1*T2/(T1*P2);// final volume, [m^3] + +mprintf('\nThe new volume of the gas is = %f m^3\n',V2); + +// End diff --git a/2705/CH5/EX5.6/Ex5_6.sce b/2705/CH5/EX5.6/Ex5_6.sce new file mode 100755 index 000000000..40f5fc559 --- /dev/null +++ b/2705/CH5/EX5.6/Ex5_6.sce @@ -0,0 +1,29 @@ +clear; +clc; +disp('Example 5.6'); + +// aim : To determine +// the mas of the gas and new temperature + +// Given values +P1 = 350;// [kN/m^2] +V1 = .03;// [m^3] +T1 = 273+35;// [K] +R = .29;// Gas constant,[kJ/kg K] + +// solution +// using charasteristic equation, P*V=m*R*T +m = P1*V1/(R*T1);// [Kg] +mprintf('\n The mass of the gas present is = %f kg\n',m); + +// Now the gas is compressed +P2 = 1050;// [kN/m^2] +V2 = V1; +// since mass of the gas is constant so using, P*V/T=constant +// hence +T2 = T1*P2/P1// [K] +t2 = T2-273;// [C] + +mprintf('\n The new temperature of the gas is = %f C\n',t2); + +// End diff --git a/2705/CH5/EX5.7/Ex5_7.sce b/2705/CH5/EX5.7/Ex5_7.sce new file mode 100755 index 000000000..85cf1c2f9 --- /dev/null +++ b/2705/CH5/EX5.7/Ex5_7.sce @@ -0,0 +1,28 @@ +clear; +clc; +disp('Example 5.7'); + +// aim : To determine +// the heat transferred to the gas and its final pressure + +// Given values +m = 2;// masss of the gas, [kg] +V1 = .7;// volume,[m^3] +T1 = 273+15;// original temperature,[K] +T2 = 273+135;// final temperature,[K] +cv = .72;// specific heat capacity at constant volume,[kJ/kg K] +R = .29;// gas law constant,[kJ/kg K] + +// solution +Q = m*cv*(T2-T1);// Heat transferred at constant volume,[kJ] +mprintf('\n The heat transferred to the gas is = %f kJ\n',Q); + +// Now,using P1*V1=m*R*T1 +P1 = m*R*T1/V1;// [kN/m^2] + +// since volume of the system is constant, so P1/T1=P2/T2 +// hence +P2 = P1*T2/T1;// final pressure,[kN/m^2] +mprintf('\n The final pressure of the gas is = %f kN/m^2 \n',P2); + +// End diff --git a/2705/CH5/EX5.8/Ex5_8.sce b/2705/CH5/EX5.8/Ex5_8.sce new file mode 100755 index 000000000..b99a3fdae --- /dev/null +++ b/2705/CH5/EX5.8/Ex5_8.sce @@ -0,0 +1,33 @@ +clear; +clc; +disp('Example 5.8'); + +// aim : To determine +// the heat transferred from the gas and the work done on the gas + +// Given values +P1 = 275;// pressure, [kN/m^2] +V1 = .09;// volume,[m^3] +T1 = 273+185;// initial temperature,[K] +T2 = 273+15;// final temperature,[K] +cp = 1.005;// specific heat capacity at constant pressure,[kJ/kg K] +R = .29;// gas law constant,[kJ/kg K] + +// solution +// using P1*V1=m*R*T1 +m = P1*V1/(R*T1);// mass of the gas + +// calculation of heat transfer +Q = m*cp*(T2-T1);// Heat transferred at constant pressure,[kJ] +mprintf('\n The heat transferred to the gas is = %f kJ\n',Q); + +// calculation of work done +// Now,since pressure is constant so, V/T=constant +// hence +V2 = V1*T2/T1;// [m^3] + +W = P1*(V2-V1);// formula for work done at constant pressure,[kJ] +mprintf('\n Work done on the gas during the process is = %f kJ\n',W); + +// End + diff --git a/2705/CH5/EX5.9/Ex5_9.sce b/2705/CH5/EX5.9/Ex5_9.sce new file mode 100755 index 000000000..016e3d2ee --- /dev/null +++ b/2705/CH5/EX5.9/Ex5_9.sce @@ -0,0 +1,22 @@ +clear; +clc; +disp('Example 5.9'); + +// aim : To determine +// the new pressure of the gas + +// Given values +P1 = 300;// original pressure,[kN/m^2] +T1 = 273+25;// original temperature,[K] +T2 = 273+180;// final temperature,[K] + +// solution +// since gas compressing according to the law,P*V^1.4=constant +// so,for polytropic process,T1/T2=(P1/P2)^((n-1)/n),here n=1.4 + +// hence +P2 = P1*(T2/T1)^((1.4)/(1.4-1));// [kN/m^2] + +mprintf('\n The new pressure of the gas is = %f kN/m^2\n',P2); + +// End diff --git a/2705/CH7/EX7.1/Ex7_1.sce b/2705/CH7/EX7.1/Ex7_1.sce new file mode 100755 index 000000000..3a5463218 --- /dev/null +++ b/2705/CH7/EX7.1/Ex7_1.sce @@ -0,0 +1,25 @@ +clear; +clc; +disp('Example 7.1'); + +// aim : To determine +// the specific enthalpy of water + +// Given values +Tf = 273+100;// Temperature,[K] + +// solution +// from steam table +cpl = 4.187;// [kJ/kg K] +// using equation [8] +sf = cpl*log(Tf/273.16);// [kJ/kg*K] +mprintf('\n The specific entropy of water is = %f kJ/kg K\n',sf); + +// using steam table +sf = 1.307;// [kJ/kg K] +mprintf('\n From table The accurate value of sf in this case is = %f kJ/kg K\n',sf); + +// There is small error in book's final value of sf + + +// End diff --git a/2705/CH7/EX7.2/Ex7_2.sce b/2705/CH7/EX7.2/Ex7_2.sce new file mode 100755 index 000000000..38d974b14 --- /dev/null +++ b/2705/CH7/EX7.2/Ex7_2.sce @@ -0,0 +1,33 @@ + +clear; +clc; +disp('Example 7.2'); + +// aim : To determine +// the specific entropy + +// Given values +P = 2;// pressure,[MN/m^2] +x = .8;// dryness fraction + +// solution +// from steam table at given pressure +Tf = 485.4;// [K] +cpl = 4.187;// [kJ/kg K] +hfg = 1888.6;// [kJ/kg] + +// (a) finding entropy by calculation +s = cpl*log(Tf/273.16)+x*hfg/Tf;// formula for entropy calculation + +mprintf('\n (a) The specific entropy of wet steam is = %f kJ/kg K\n',s); + +// (b) calculation of entropy using steam table +// from steam table at given pressure +sf = 2.447;// [kJ/kg K] +sfg = 3.89;// [kJ/kg K] +// hence +s = sf+x*sfg;// [kJ/kg K] + +mprintf('\n (b) The specific entropy using steam table is = %f kJ/kg K\n',s); + +// End diff --git a/2705/CH7/EX7.3/Ex7_3.sce b/2705/CH7/EX7.3/Ex7_3.sce new file mode 100755 index 000000000..21e0a7087 --- /dev/null +++ b/2705/CH7/EX7.3/Ex7_3.sce @@ -0,0 +1,30 @@ +clear; +clc; +disp('Example 7.3'); + +// aim : To determine +// the specific entropy of steam + +// Given values +P = 1.5;//pressure,[MN/m^2] +T = 273+300;//temperature,[K] + +// solution + +// (a) +// from steam table +cpl = 4.187;// [kJ/kg K] +Tf = 471.3;// [K] +hfg = 1946;// [kJ/kg] +cpv = 2.093;// [kJ/kg K] + +// usung equation [2] +s = cpl*log(Tf/273.15)+hfg/Tf+cpv*log(T/Tf);// [kJ/kg K] +mprintf('\n (a) The specific entropy of steam is = %f kJ/kg K\n',s); + +// (b) +// from steam tables +s = 6.919;// [kJ/kg K] +mprintf('\n (b) The accurate value of specific entropy from steam table is = %f kJ/kg K\n',s); + +// End diff --git a/2705/CH7/EX7.4/Ex7_4.sce b/2705/CH7/EX7.4/Ex7_4.sce new file mode 100755 index 000000000..c908c859e --- /dev/null +++ b/2705/CH7/EX7.4/Ex7_4.sce @@ -0,0 +1,30 @@ +clear; +clc; +disp('Example 7.4'); + +// aim : To determine +// the dryness fraction of steam + +// Given values +P1 = 2;// initial pressure, [MN/m^2] +t = 350;// temperature, [C] +P2 = .28;// final pressure, [MN/m^2] + +// solution +// at 2 MN/m^2 and 350 C,steam is superheated because the saturation temperature is 212.4 C +// From steam table +s1 = 6.957;// [kJ/kg K] + +// for isentropic process +s2 = s1; +// also +sf2 = 1.647;// [kJ/kg K] +sfg2 = 5.368;// [kJ/kg K] + +// using +// s2 = sf2+x2*sfg2, where x2 is dryness fraction of steam +// hence +x2 = (s2-sf2)/sfg2; +mprintf('\n The final dryness fraction of steam is x2 = %f\n',x2); + +// End diff --git a/2705/CH7/EX7.5/Ex7_5.sce b/2705/CH7/EX7.5/Ex7_5.sce new file mode 100755 index 000000000..53cec3d7a --- /dev/null +++ b/2705/CH7/EX7.5/Ex7_5.sce @@ -0,0 +1,53 @@ +clear; +clc; +disp('Example 7.5'); + +// aim : To determine +// the final condition of steam... +// the change in specific entropy during hyperbolic process + +// Given values +P1 = 2;// pressure, [MN/m^2] +t = 250;// temperature, [C] +P2 = .36;// pressure, [MN/m^2] +P3 = .06;// pressure, [MN/m^2] + +// solution + +// (a) +// from steam table +s1 = 6.545;// [kJ/kg K] +// at .36 MN/m^2 +sg = 6.930;// [kJ/kg*K] + +sf2 = 1.738;// [kJ/kg K] +sfg2 = 5.192;// [kJ/kg K] +vg2 = .510;// [m^3] + +// so after isentropic expansion, steam is wet +// hence, s2=sf2+x2*sfg2, where x2 is dryness fraction +// also +s2 = s1; +// so +x2 = (s2-sf2)/sfg2; +// and +v2 = x2*vg2;// [m^3] + +// for hyperbolic process +// P2*v2=P3*v3 +// hence +v3 = P2*v2/P3;// [m^3] + +mprintf('\n (a) From steam table at .06 MN/m^2 steam is superheated and has temperature of 100 C with specific volume is = %f m^3/kg\n',v3); + +// (b) +// at this condition +s3 = 7.609;// [kJ/kg*K] +// hence +change_s23 = s3-sg;// change in specific entropy during the hyperblic process[kJ/kg*K] +mprintf('\n (b) The change in specific entropy during the hyperbolic process is = %f kJ/kg K\n',change_s23); + +// In the book they have taken sg instead of s2 for part (b), so answer is not matching + +// End + diff --git a/2705/CH7/EX7.6/Ex7_6.sce b/2705/CH7/EX7.6/Ex7_6.sce new file mode 100755 index 000000000..b9e9a3beb --- /dev/null +++ b/2705/CH7/EX7.6/Ex7_6.sce @@ -0,0 +1,67 @@ + +clear; +clc; +disp('Example 7.6'); + +// aim : To determine the +// (a) heat transfer during the expansion and +// (b) work done durind the expansion + +// given values +m = 4.5; // mass of steam,[kg] +P1 = 3; // initial pressure,[MN/m^2] +T1 = 300+273; // initial temperature,[K] + +P2 = .1; // final pressure,[MN/m^2] +x2 = .96; // dryness fraction at final stage + +// solution +// for state point 1,using steam table +s1 = 6.541;// [kJ/kg/K] +u1 = 2751;// [kJ/kg] + + // for state point 2 + sf2 = 1.303;// [kJ/kg/K] + sfg2 = 6.056;// [kJ/kg/k] + T2 = 273+99.6;// [K] + hf2 = 417;// [kJ/kg] + hfg2 = 2258;// [kJ/kg] + vg2 = 1.694;// [m^3/kg] + + // hence + s2 = sf2+x2*sfg2;// [kJ/kg/k] + h2 = hf2+x2*hfg2;// [kJ/kg] + u2 = h2-P2*x2*vg2*10^3;// [kJ/kg] + + // Diagram of example 7.6 + x = [s1 s2]; + y = [T1 T2]; +plot2d(x,y); + xtitle('Diagram for example 7.6(T vs s)'); + xlabel('Entropy (kJ/kg K)'); + ylabel('Temperature (K)'); + +x = [s1,s1]; +y = [0,T1]; +plot2d(x,y); + +x = [s2,s2]; +y = [0,T2]; +plot2d(x,y); + + // (a) + // Q_rev is area of T-s diagram + Q_rev = (T1+T2)/2*(s2-s1);// [kJ/kg] + // so total heat transfer is + Q_rev = m*Q_rev;// [kJ] + + // (b) + del_u = u2-u1;// change in internal energy, [kJ/kg] + // using 1st law of thermodynamics + W = Q_rev-m*del_u;// [kJ] + +mprintf('\n (a) The heat transfer during the expansion is = %f kJ (received)\n',Q_rev); + + mprintf('\n (b) The work done during the expansion is = %f kJ\n',W); + + // End diff --git a/2705/CH7/EX7.7/Ex7_7.sce b/2705/CH7/EX7.7/Ex7_7.sce new file mode 100755 index 000000000..9a68eddbd --- /dev/null +++ b/2705/CH7/EX7.7/Ex7_7.sce @@ -0,0 +1,45 @@ + +clear; +clc; +disp('Example 7.7'); + +// aim : To determine the +// (a) change of entropy +// (b) The approximate change of entropy obtained by dividing the heat transferred by the gas by the mean absolute temperature during the compression + +// Given values +P1 = 140;// initial pressure,[kN/m^2] +V1 = .14;// initial volume, [m^3] +T1 = 273+25;// initial temperature,[K] + P2 = 1400;// final pressure [kN/m^2] + n = 1.25; // polytropic index + cp = 1.041;// [kJ/kg K] + cv = .743;// [kJ/kg K] + + // solution + // (a) + R = cp-cv;// [kJ/kg/K] + // using ideal gas equation + m = P1*V1/(R*T1);// mass of gas,[kg] + // since gas is following law P*V^n=constant ,so + V2 = V1*(P1/P2)^(1/n);// [m^3] + + // using eqn [9] + del_s = m*(cp*log(V2/V1)+cv*log(P2/P1));// [kJ/K] + mprintf('\n (a) The change of entropy is = %f kJ/K\n',del_s); + + // (b) + W = (P1*V1-P2*V2)/(n-1);// polytropic work,[kJ] + Gamma = cp/cv;// heat capacity ratio + Q = (Gamma-n)/(Gamma-1)*W;// heat transferred,[kJ] + + // Again using polytropic law + T2 = T1*(V1/V2)^(n-1);// final temperature, [K] + T_avg = (T1+T2)/2;// mean absolute temperature, [K] + + // so approximate change in entropy is + del_s = Q/T_avg;// [kJ/K] + + mprintf('\n (b) The approximate change of entropy obtained by dividing the heat transferred by the gas by the mean absolute temperature during the compression = %f kJ/K\n',del_s); + + // End diff --git a/2705/CH7/EX7.8/Ex7_8.sce b/2705/CH7/EX7.8/Ex7_8.sce new file mode 100755 index 000000000..711505b0c --- /dev/null +++ b/2705/CH7/EX7.8/Ex7_8.sce @@ -0,0 +1,39 @@ +clear; +clc; +disp('Example 7.8'); + +// aim : To determine +// the change of entropy + +// Given values +m = .3;// [kg] +P1 = 350;// [kN/m^2] +T1 = 273+35;// [K] +P2 = 700;// [kN/m^2] +V3 = .2289;// [m^3] +cp = 1.006;// [kJ/kg K] +cv = .717;// [kJ/kg K] + +// solution +// for constant volume process +R = cp-cv;// [kJ/kg K] +// using PV=mRT +V1 = m*R*T1/P1;// [m^3] + +// for constant volume process P/T=constant,so +T2 = T1*P2/P1;// [K] +s21 = m*cv*log(P2/P1);// formula for entropy change for constant volume process +mprintf('\n The change of entropy in constant volume process is = %f kJ/kg K\n',s21); + +// 'For the above part result given in the book is wrong + +V2 = V1; +// for constant pressure process +T3 = T2*V3/V2;// [K] +s32 = m*cp*log(V3/V2);// [kJ/kg K] + +mprintf('\n The change of entropy in constant pressure process is = %f kJ/kg K\n',s32); + +// there is misprint in the book's result + +// End diff --git a/2705/CH7/EX7.9/Ex7_9.sce b/2705/CH7/EX7.9/Ex7_9.sce new file mode 100755 index 000000000..05f58353b --- /dev/null +++ b/2705/CH7/EX7.9/Ex7_9.sce @@ -0,0 +1,21 @@ +clear; +clc; +disp('Example 7.9'); + +// aim : To determine +// the change of entropy + +// Given values +P1 = 700;// initial pressure, [kN/m^2] +T1 = 273+150;// Temperature ,[K] +V1 = .014;// initial volume, [m^3] +V2 = .084;// final volume, [m^3] + +// solution +// since process is isothermal so +T2 = T1; +// and using fig.7.10 +del_s = P1*V1*log(V2/V1)/T1 ;// [kJ/K] +mprintf('\n The change of entropy is = %f kJ/kg K\n',del_s); + +// End diff --git a/2705/CH8/EX8.1/Ex8_1.sce b/2705/CH8/EX8.1/Ex8_1.sce new file mode 100755 index 000000000..dc5881db5 --- /dev/null +++ b/2705/CH8/EX8.1/Ex8_1.sce @@ -0,0 +1,24 @@ + +clear; +clc; +disp('Example 8.1'); + +// aim : To determine +// the stoichiometric mass of air required to burn 1 kg the fuel + +// Given values +C = .72;// mass fraction of C; [kg/kg] +H2 = .20;// mass fraction of H2;, [kg/kg] +O2 = .08;// mass fraction of O2, [kg/kg] +aO2=.232;// composition of oxygen in air + +// solution +// for 1kg of fuel +mO2 = 8/3*C+8*H2-O2;// mass of O2, [kg] + +// hence stoichiometric mass of O2 required is +msO2 = mO2/aO2;// [kg] + +mprintf('\n The stoichiometric mass of air required to burn 1 kg the fuel should be = %f kg\n',msO2); + +// End diff --git a/2705/CH8/EX8.10/Ex8_10.sce b/2705/CH8/EX8.10/Ex8_10.sce new file mode 100755 index 000000000..079878869 --- /dev/null +++ b/2705/CH8/EX8.10/Ex8_10.sce @@ -0,0 +1,44 @@ +clear; +clc; +disp('Example 8.10'); + +// aim : To determine +// volumetric composition of the products of combustion + +// given values +C = .86;// mass composition of carbon +H = .14;// mass composition of hydrogen +Ea = .20;// excess air for combustion +O2 = .23;// mass composition of O2 in air + +MCO2 = 44;// moleculer mass of CO2 +MH2O = 18;// moleculer mass of H2O +MO2 = 32;// moleculer mass of O2 +MN2 = 28;// moleculer mass of N2, + + +// solution +sO2 = (8/3*C+8*H);// stoichiometric O2 required, [kg/kg petrol] +sair = sO2/O2;// stoichiometric air required, [kg/kg petrol] +// for one kg petrol +mCO2 = 11/3*C;// mass of CO2,[kg] +mH2O = 9*H;// mass of H2O, [kg] +mO2 = Ea*sO2;// mass of O2, [kg] +mN2 = 14.84*(1+Ea)*(1-O2);// mass of N2, [kg] + +mt = mCO2+mH2O+mO2+mN2;// total mass, [kg] +// percentage mass composition +x1 = mCO2/mt*100;// mass composition of CO2 +x2 = mH2O/mt*100;// mass composition of H2O +x3 = mO2/mt*100;// mass composition of O2 +x4 = mN2/mt*100;// mass composition of N2 + +vt = x1/MCO2+x2/MH2O+x3/MO2+x4/MN2;// total volume of petrol +v1 = x1/MCO2/vt*100;// %age composition of CO2 by volume +v2 = x2/MH2O/vt*100;// %age composition of H2O by volume +v3 = x3/MO2/vt*100;// %age composition of O2 by volume +v4 = x4/MN2/vt*100;// %age composition of N2 by volume + +mprintf('\nThe percentage composition of CO2 by volume is = %f\n,\nThe percentage composition of H2O by volume is = %f\n,\nThe percentage composition of O2 by volume is = %f\n,\nThe percentage composition of N2 by volume is = %f\n',v1,v2,v3,v4); + +// End diff --git a/2705/CH8/EX8.11/Ex8_11.sce b/2705/CH8/EX8.11/Ex8_11.sce new file mode 100755 index 000000000..f9ee92977 --- /dev/null +++ b/2705/CH8/EX8.11/Ex8_11.sce @@ -0,0 +1,33 @@ +clear; +clc; +disp('Example 8.11'); + +// aim : To determine +// the energy carried away by the dry flue gas/kg of fuel burned + +// given values +C = .78;// mass composition of carbon +H2 = .06;// mass composition of hydrogen +O2 = .09;// mass composition of oxygen +Ash = .07;// mass composition of ash +Ea = .50;// excess air for combustion +aO2 = .23;// mass composition of O2 in air +Tb = 273+20;// boiler house temperature, [K] +Tf = 273+320;// flue gas temperature, [K] +c = 1.006;// specific heat capacity of dry flue gas, [kJ/kg K] + +// solution +// for one kg of fuel +sO2 = (8/3*C+8*H2);// stoichiometric O2 required, [kg/kg fuel] +sO2a = sO2-O2;// stoichiometric O2 required from air, [kg/kg fuel] +sair = sO2a/aO2;// stoichiometric air required, [kg/kg fuel] +ma = sair*(1+Ea);// actual air supplied/kg of fuel, [kg] +// total mass of flue gas/kg fuel is +mf = ma+1;// [kg] +mH2 = 9*H2;//H2 produced, [kg] +// hence, mass of dry flue gas/kg coall is +m = mf-mH2;// [kg] +Q = m*c*(Tf-Tb);// energy carried away by flue gas, [kJ] +mprintf('\n The energy carried away by the dry flue gas/kg is = %f kg\n',Q); + +// End diff --git a/2705/CH8/EX8.12/Ex8_12.sce b/2705/CH8/EX8.12/Ex8_12.sce new file mode 100755 index 000000000..ec6dd7d76 --- /dev/null +++ b/2705/CH8/EX8.12/Ex8_12.sce @@ -0,0 +1,49 @@ +clear; +clc; +disp('Example 8.12'); + +// aim : To determine +// (a) the stoichiometric volume of air for the complete combustion of 1 m^3 +// (b) the percentage volumetric analysis of the products of combustion + +// given values +N2 = .018;// volumetric composition of N2 +CH4 = .94;// volumetric composition of CH4 +C2H6 = .035;// volumetric composition of C2H6 +C3H8 = .007;// volumetric composition of C3H8 +aO2 = .21;// O2 composition in air + +// solution +// (a) +// for CH4 +// CH4 +2 O2= CO2 + 2 H2O +sva1 = 2/aO2;// stoichiometric volume of air, [m^3/m^3 CH4] +svn1 = sva1*(1-aO2);// stoichiometric volume of nitrogen in the air, [m^3/m^3 CH4] + +// for C2H6 +// 2 C2H6 +7 O2= 4 CO2 + 6 H2O +sva2 = 7/2/aO2;// stoichiometric volume of air, [m^3/m^3 C2H6] +svn2 = sva2*(1-aO2);// stoichiometric volume of nitrogen in the air, [m^3/m^3 C2H6] + +// for C3H8 +// C3H8 +5 O2=3 CO2 + 4 H2O +sva3 = 5/aO2;// stoichiometric volume of air, [m^3/m^3 C3H8] +svn3 = sva3*(1-aO2);// stoichiometric volume of nitrogen in the air, [m^3/m^3 C3H8] + +Sva = CH4*sva1+C2H6*sva2+C3H8*sva3;// stoichiometric volume of air required, [m^3/m^3 gas] +mprintf('\n (a) The stoichiometric volume of air for the complete combustion = %f m^3m^3 gas\n',Sva); + +// (b) +// for one m^3 of natural gas +vCO2 = CH4*1+C2H6*2+C3H8*3;// volume of CO2 produced, [m^3] +vH2O = CH4*2+C2H6*3+C3H8*4;// volume of H2O produced, [m^3] +vN2 = CH4*svn1+C2H6*svn2+C3H8*svn3+N2;// volume of N2 produced, [m^3] + +vg = vCO2+vH2O+vN2;// total volume of gas, [m^3] +x1 = vCO2/vg*100;// volume percentage of CO2 produced +x2 = vH2O/vg*100;// volume percentage of H2O produced +x3 = vN2/vg*100;// volume percentage of N2 produced + +mprintf('\n (b) The percentage volumetric composition of CO2 in produced is = %f\n,\n The percentage volumetric composition of H2O in produced is = %f\n,\n The percentage volumetric composition of N2 in produced is = %f\n',x1,x2,x3); + +// End diff --git a/2705/CH8/EX8.13/Ex8_13.sce b/2705/CH8/EX8.13/Ex8_13.sce new file mode 100755 index 000000000..87f8573ec --- /dev/null +++ b/2705/CH8/EX8.13/Ex8_13.sce @@ -0,0 +1,46 @@ +clear; +clc; +disp('Example 8.13'); + +// aim : To determine +// (a) the volume of air taken by the fan +// (b) the percentage composition of dry flue gas + +// gien values +C = .82;// mass composition of carbon +H = .08;// mass composition of hydrogen + O = .03;// mass composition of oxygen + A = .07;// mass composition of ash +mc = .19;// coal uses, [kg/s] + ea = .3;// percentage excess air of oxygen in the air required for combustion +Oa = .23;// percentage of oxygen by mass in the air + + // solution + // (a) + P = 100;// air pressure, [kN/m^2] + T = 18+273;// air temperature, [K] + R = .287;// [kJ/kg K] + // basis one kg coal + sO2 = 8/3*C+8*H;// stoichiometric O2 required, [kg] + aO2 = sO2-.03;// actual O2 required, [kg] +tO2 = aO2/Oa;// theoretical O2 required, [kg] +Aa = tO2*(1+ea);// actual air supplied, [kg] +m = Aa*mc;// Air supplied, [kg/s] + +// now using P*V=m*R*T +V = m*R*T/P;// volume of air taken ,[m^3/s] +mprintf('\n (a) Volume of air taken by fan is = %f m^3/s\n',V); + +// (b) +mCO2 = 11/3*C;// mass of CO2 produced, [kg] +mO2 = aO2*.3;// mass of O2 produces, [kg] +mN2 = Aa*.77;// mass of N2 produced, [[kg] +mt = mCO2+mO2+mN2;// total mass, [kg] + +mprintf('\n (b) Percentage mass composition of CO2 is = %f percent \n',mCO2/mt*100); +mprintf('\n Percentage mass composition of O2 is = %f percent\n',mO2/mt*100) +mprintf('\n Percentage mass composition of N2 is = %f percent \n',mN2/mt*100) + + + +// End diff --git a/2705/CH8/EX8.14/Ex8_14.sce b/2705/CH8/EX8.14/Ex8_14.sce new file mode 100755 index 000000000..a93839efd --- /dev/null +++ b/2705/CH8/EX8.14/Ex8_14.sce @@ -0,0 +1,40 @@ +clear; +clc; +disp('Example 8.14'); + +// aim : To determine +// (a) the mass of fuel used per cycle +// (b) the actual mass of air taken in per cycle +// (c) the volume of air taken in per cycle + +// given values +W = 15;// work done, [kJ/s] +N = 5;// speed, [rev/s] +C = .84;// mass composition of carbon +H = .16;// mass composition of hydrogen +ea = 1;// percentage excess air supplied +CV = 45000;// calorificvalue of fuel, [kJ/kg] +n_the = .3;// thermal efficiency +P = 100;// pressuer, [kN/m^2] +T = 273+15;// temperature, [K] +R = .29;// gas constant, [kJ/kg K] + +// solution +// (a) +E = W*2/N/n_the;// energy supplied, [kJ/cycle] +mf = E/CV;// mass of fuell used, [kg] +mprintf('\n (a) Mass of fuel used per cycle is = %f g\n',mf*10^3); + +// (b) +// basis 1 kg fuel +mO2 = C*8/3+8*H;// mass of O2 requirea, [kg] +smO2 = mO2/.23;// stoichiometric mass of air, [kg] +ma = smO2*(1+ea);// actual mass of air supplied, [kg] +m = ma*mf;// mass of air supplied, [kg/cycle] +mprintf('\n (b) The mass of air supplied per cycle is = %f kg\n',m); + +// (c) +V = m*R*T/P;// volume of air, [m^3] +mprintf('\n (c) The volume of air taken in per cycle is = %f m^3\n',V); + +// End diff --git a/2705/CH8/EX8.15/Ex8_15.sce b/2705/CH8/EX8.15/Ex8_15.sce new file mode 100755 index 000000000..a4adab52d --- /dev/null +++ b/2705/CH8/EX8.15/Ex8_15.sce @@ -0,0 +1,70 @@ +clear; +clc; +disp('Example 8.15'); + +// aim : To determine +// (a) the mass of coal used per hour +// (b) the mass of air used per hour +// (c) the percentage analysis of the flue gases by mass + +// given values +m = 900;// mass of steam boiler generate/h, [kg] +x = .96;// steam dryness fraction +P = 1400;// steam pressure, [kN/m^2] +Tf = 52;// feed water temperature, [C] +BE = .71;// boiler efficiency +CV = 33000;// calorific value of coal, [kJkg[ +ea = .22;// excess air supply +aO2 = .23;// oxygen composition in air +c = 4.187;// specific heat capacity of water, [kJ/kg K] + +// coal composition +C = .83;// mass composition of carbon +H2 = .05;// mass composition of hydrogen +O2 = .03;// mass composition of oxygen +ash = .09;// mass composition of ash + +// solution +// from steam table at pressure P +hf = 830.1;// specific enthalpy, [kJ/kg] +hfg = 1957.1;// specific enthalpy, [kJ/kg] +hg = 2728.8;// specific enthalpy, [kJ/kg] + +// (a) +h = hf+x*hfg;// specific enthalpy of steam generated by boiler, [kJ/kg] +hfw = c*Tf;// specific enthalpy of feed water, [kJ/kg] +Q = m*(h-hfw);// energy to steam/h, [kJ] +Qf = Q/BE;// energy required from fuel/h, [kJ] +mc = Qf/CV;// mass of coal/h,[kg] +mprintf('\n (a) The mass of coal used per hour is = %f kg\n',mc); + +// (b) +// for one kg coal +mO2 = 8/3*C+8*H2+-O2;// actual mass of O2 required, [kg] +mta = mO2/aO2;// theoretical mass of air, [kg] +ma = mta*(1+ea);// mass of air supplied, [kg] +mas = ma*mc;// mass of air supplied/h, [kg] +mprintf('\n (b) The mass of air supplied per hour is = %f kg\n',mas); + + +// (c) +// for one kg coal +mCO2 = 11/3*C;// mass of CO2 produced, [kg] +mH2O = 9*H2;// mass of H2O produced, [kg] +mO2 = mO2*ea;// mass of excess O2 in flue gas, [kg] +mN2 = ma*(1-aO2);// mass of N2 in flue gas, [kg] + +mt = mCO2+mH2O+mO2+mN2;// total mass of gas +x1 = mCO2/mt*100;// mass percentage composition of CO2 +x2 = mH2O/mt*100;// mass percentage composition of H2O +x3 = mO2/mt*100;// mass percentage composition of O2 +x4 = mN2/mt*100;// mass percentage composition of N2 + +mprintf('\n (c) The mass percentage composition of CO2 = %f,\n The mass percentage composition of H2O = %f,\n The mass percentage composition of O2 = %f,\n The mass percentage composition of N2 = %f',x1,x2,x3,x4); + +// mass of coal taken in part (b) is wrong so answer is not matching + +// End + + + diff --git a/2705/CH8/EX8.16/Ex8_16.sce b/2705/CH8/EX8.16/Ex8_16.sce new file mode 100755 index 000000000..5e1808b5d --- /dev/null +++ b/2705/CH8/EX8.16/Ex8_16.sce @@ -0,0 +1,40 @@ +clear; +clc; +disp('Example 8.16'); + +// aim : To determine +// (a) volume of gas +// (b) (1) the average molecular mass of air +// (2) the value of R +// (3) the mass of 1 m^3 of air at STP + +// given values +n = 1;// moles of gas, [kmol] +P = 101.32;// standard pressure, [kN/m^2] +T = 273;// gas tempearture, [K] + +O2 = 21;// percentage volume composition of oxygen in air +N2 = 79;// percentage volume composition of nitrogen in air +R = 8.3143;// molar gas constant, [kJ/kg K] +mO2 = 32;// moleculer mass of O2 +mN2 = 28;// moleculer mass of N2 + +// solution +// (a) +V = n*R*T/P;// volume of gas, [m^3] +mprintf('\n (a) The volume of the gas is = %f m^3\n',V); + +// (b) +//(1) +Mav = (O2*mO2+N2*mN2)/(O2+N2);// average moleculer mass of air +mprintf('\n (b)(1) The average moleculer mass of air is = %f g/mol\n',Mav); + +// (2) +Rav = R/Mav;// characteristic gas constant, [kJ/kg k] +mprintf('\n (2) The value of R is = %f kJ/kg K\n',Rav); + +// (3) +rho = Mav/V;// density of air, [kg/m^3] +mprintf('\n (3) The mass of one cubic metre of air at STP is = %f kg/m^3\n',rho); + +// End diff --git a/2705/CH8/EX8.17/Ex8_17.sce b/2705/CH8/EX8.17/Ex8_17.sce new file mode 100755 index 000000000..d1099ee95 --- /dev/null +++ b/2705/CH8/EX8.17/Ex8_17.sce @@ -0,0 +1,48 @@ +clear; +clc; +disp('Example 8.17'); + +// aim : To determine +// (a) the partial pressure of each gas in the vessel +// (b) the volume of the vessel +// (c) the total pressure in the gas when temperature is raised to228 C + +// given values +MO2 = 8;// mass of O2, [kg] +MN2 = 7;// mass of N2, [kg] +MCO2 = 22;// mass of CO2, [kg] + +P = 416;// total pressure in the vessel, [kN/m^2] +T = 273+60;// vessel temperature, [K] +R = 8.3143;// gas constant, [kJ/kmol K] + +mO2 = 32;// molculer mass of O2 +mN2 = 28;// molculer mass of N2 +mCO2 = 44;// molculer mass of CO2 + +// solution +// (a) +n1 = MO2/mO2;// moles of O2, [kmol] +n2 = MN2/mN2;// moles of N2, [kmol] +n3 = MCO2/mCO2;// moles of CO2, [kmol] + +n = n1+n2+n3;// total moles in the vessel, [kmol] +// since,Partial pressure is proportinal, so +P1 = n1*P/n;// partial pressure of O2, [kN/m^2] +P2 = n2*P/n;// partial pressure of N2, [kN/m^2] +P3 = n3*P/n;// partial pressure of CO2, [kN/m^2] + +mprintf('\n (a)The partial pressure of O2 is = %f kN/m^2,\n, The partial pressure of N2 is = %f kN/m^2,\n The partial pressure of CO2 is = %f kN/m^2,\n',P1,P2,P3); + +// (b) +// assuming ideal gas +V = n*R*T/P;// volume of the container, [m^3] +mprintf('\n (b) The volume of the container is = %f m^3\n',V); + +// (c) +T2 = 273+228;// raised vessel temperature, [K] +// so volume of vessel will constant , P/T=constant +P2 = P*T2/T;// new pressure in the vessel , [kn/m62] +mprintf('\n (c) The new total pressure in the vessel is = %f kN/m^2\n',P2); + +// End diff --git a/2705/CH8/EX8.18/Ex8_18.sce b/2705/CH8/EX8.18/Ex8_18.sce new file mode 100755 index 000000000..c0961970a --- /dev/null +++ b/2705/CH8/EX8.18/Ex8_18.sce @@ -0,0 +1,65 @@ +clear; +clc; +disp('Example 8.18'); + +// aim : To determine +// the actual mass of air supplied/kg coal +// the velocity of flue gas + +// given values +mc = 635;// mass of coal burn/h, [kg] +ea = .25;// excess air required +C = .84;// mass composition of carbon +H2 = .04;// mass composition of hydrogen +O2 = .05;// mass composition of oxygen +ash = 1-(C+H2+O2);// mass composition of ash + +P1 = 101.3;// pressure, [kJn/m^2] +T1 = 273;// temperature, [K] +V1 = 22.4;// volume, [m^3] + +T2 = 273+344;// gas temperature, [K] +P2 = 100;// gas pressure, [kN/m^2] +A = 1.1;// cross section area, [m^2] +aO2 = .23;// composition of O2 in air + +mCO2 = 44;// moleculer mass of carbon +mH2O = 18;// molecular mass of hydrogen +mO2 = 32;// moleculer mas of oxygen +mN2 = 28;// moleculer mass of nitrogen + +// solution +mtO2 = 8/3*C+8*H2-O2;// theoretical O2 required/kg coal, [kg] +msa= mtO2/aO2;// stoichiometric mass of air supplied/kg coal, [kg] +mas = msa*(1+ea);// actual mass of air supplied/kg coal, [kg] + +m1 = 11/3*C;// mass of CO2/kg coal produced, [kg] +m2 = 9*H2;// mass of H2/kg coal produced, [kg] +m3 = mtO2*ea;// mass of O2/kg coal produced, [kg] +m4 = mas*(1-aO2);// mass of N2/kg coal produced, [kg] + +mt = m1+m2+m3+m4;// total mass, [kg] +x1 = m1/mt*100;// %age mass composition of CO2 produced +x2 = m2/mt*100;// %age mass composition of H2O produced +x3 = m3/mt*100;// %age mass composition of O2 produced +x4 = m4/mt*100;// %age mass composition of N2 produced + +vt = x1/mCO2+x2/mH2O+x3/mO2+x4/mN2;// total volume +v1 = x1/mCO2/vt*100;// %age volume composition of CO2 +v2 = x2/mH2O/vt*100;// %age volume composition of H2O +v3 = x3/mO2/vt*100;// %age volume composition of O2 +v4 = x4/mN2/vt*100;// %age volume composition of N2 + +Mav = (v1*mCO2+v2*mH2O+v3*mO2+v4*mN2)/(v1+v2+v3+v4);// average moleculer mass, [kg/kmol] +// since no of moles is constant so PV/T=constant +V2 = P1*V1*T2/(P2*T1);//volume, [m^3] + +mp = mt*mc/3600;// mass of product of combustion/s, [kg] + +V = V2*mp/Mav;// volume of flowing gas /s,[m^3] + +v = V/A;// velocity of flue gas, [m/s] +mprintf('\n The actual mass of air supplied is = %f kg/kg coal\n',mas); +mprintf('\n The velocity of flue gas is = %f m/s\n',v); + +// End diff --git a/2705/CH8/EX8.19/Ex8_19.sce b/2705/CH8/EX8.19/Ex8_19.sce new file mode 100755 index 000000000..30bfc9211 --- /dev/null +++ b/2705/CH8/EX8.19/Ex8_19.sce @@ -0,0 +1,51 @@ +clear; +clc; +disp('Example 8.19'); + +// aim : To determine +// (a) the temperature of the gas after compression +// (b) the density of the air-gas mixture + +// given values +CO = 26;// %age volume composition of CO +H2 = 16;// %age volume composition of H2 +CH4 = 7;// %age volume composition of CH4 +N2 = 51;// %age volume composition of N2 + +P1 = 103;// gas pressure, [kN/m^2] +T1 = 273+21;// gas temperature, [K] +rv = 7;// volume ratio + +aO2 = 21;// %age volume composition of O2 in the air +c = 21;// specific heat capacity of diatomic gas, [kJ/kg K] +cCH4 = 36;// specific heat capacity of CH4, [kJ/kg K] +R = 8.3143;// gas constant, [kJ/kg K] + +mCO = 28;// moleculer mass of carbon +mH2 = 2;// molecular mass of hydrogen +mCH4 = 16;// moleculer mas of methane +mN2 = 28;// moleculer mass of nitrogen +mO2 = 32;// moleculer mass of oxygen + +// solution +// (a) +Cav = (CO*c+H2*c+CH4*cCH4+N2*c+100*2*c)/(100+200);// heat capacity, [kJ/kg K] + +Gama = (Cav+R)/Cav;// heat capacity ratio +// rv = V1/V2 +// process is polytropic, so +T2 = T1*(rv)^(Gama-1);// final tempearture, [K] +mprintf('\n (a) The temperature of the gas after compression is = %f C\n',T2-273); + +// (b) + +Mav = (CO*mCO+H2*mH2+CH4*mCH4+N2*mN2+42*mO2+158*mN2)/(100+200) + +// for 1 kmol of gas +V = R*T1/P1;// volume of one kmol of gas, [m^3] +// hence +rho = Mav/V;// density of gas, [kg/m^3] + +mprintf('\n (b) The density of air-gas mixture is = %f kg/m^3\n',rho); + +// End diff --git a/2705/CH8/EX8.20/Ex8_20.sce b/2705/CH8/EX8.20/Ex8_20.sce new file mode 100755 index 000000000..51e9a6f72 --- /dev/null +++ b/2705/CH8/EX8.20/Ex8_20.sce @@ -0,0 +1,19 @@ +clear; +clc; +disp('Example 8.20'); + +// aim : to determine +// stoichiometric equation for combustion of hydrogen + +// solution +// equation with algebric coefficient is +// H2+aO2+79/21*aN2=bH2O+79/21*aN2 +// by equating coefficients +b = 1; +a = b/2; +// so equation becomes +// 2 H2+ O2+3.76 N2=2 H2O+3.76 N2 +disp('The required stoichiometric equation is = '); +disp('2 H2+ O2+3.76 N2 = 2 H2O+3.76 N2'); + +// End diff --git a/2705/CH8/EX8.22/Ex8_22.sce b/2705/CH8/EX8.22/Ex8_22.sce new file mode 100755 index 000000000..9e5bddbd5 --- /dev/null +++ b/2705/CH8/EX8.22/Ex8_22.sce @@ -0,0 +1,53 @@ +clear; +clc; +disp('Example 8.22'); + +// aim : To determine +// the percentage gravimetric analysis of the total products of combustion + +// given values +CO = 12;// %age volume composition of CO +H2 = 41;// %age volume composition of H2 +CH4 = 27;// %age volume composition of CH4 +O2 = 2;// %age volume composition of O2 +CO2 = 3;// %age volume composition of CO2 +N2 = 15;// %age volume composition of N2 + +mCO2 = 44;// moleculer mass of CO2,[kg/kmol] +mH2O = 18;// moleculer mass of H2O, [kg/kmol] +mO2 = 32;// moleculer mass of O2, [kg/kmol] +mN2 = 28;// moleculer mass of N2, [kg/kmol] + +ea = 15;// %age excess air required +aO2 = 21;// %age air composition in the air + +// solution +// combustion equation by no. of moles +// 12CO + 41H2 + 27CH4 + 2O2 + 3CO2 + 15N2 + aO2+79/21*aN2 = bCO2 + dH2O + eO2 + 15N2 +79/21*aN2 +// equating C coefficient +b = 12+27+3;// [mol] +// equatimg H2 coefficient +d = 41+2*27;// [mol] +// O2 required is 15 % extra,so +// e/(e-a)=.15 so e=.13a +// equating O2 coefficient +// 2+3+a=b+d/2 +e + +a = (b+d/2-5)/(1-.13); +e = .13*a;// [mol] + +// gravimetric analysis of product +v1 = b*mCO2;// gravimetric volume of CO2 +v2 = d*mH2O ;// gravimetric volume of H2O +v3 = e*mO2;// gravimetric volume of O2 +v4 = 15*mN2 +79/21*a*mN2;// gravimetric volume of N2 + +vt = v1+v2+v3+v4;// total +x1 = v1/vt*100;// percentage gravimetric of CO2 +x2 = v2/vt*100;// percentage gravimetric of H2O +x3 = v3/vt*100;// percentage gravimetric of O2 +x4 = v4/vt*100;// percentage gravimetric of N2 + +mprintf('\n Percentage gravimetric composition of CO2 = %f\n ,\n Percentage gravimetric composition of H2O = %f\n\n Percentage gravimetric composition of O2 = %f\n\n Percentage gravimetric composition of N2 = %f\n',x1,x2,x3,x4); + +// End diff --git a/2705/CH8/EX8.23/Ex8_23.sce b/2705/CH8/EX8.23/Ex8_23.sce new file mode 100755 index 000000000..6a5a20ce2 --- /dev/null +++ b/2705/CH8/EX8.23/Ex8_23.sce @@ -0,0 +1,54 @@ +clear; +clc; +disp('Example 8.23'); + +// aim : To determine +// (a) the actual quantity of air supplied/kg of fuel +// (b) the volumetric efficiency of the engine + +// given values +d = 300*10^-3;// bore,[m] +L = 460*10^-3;// stroke,[m] +N = 200;// engine speed, [rev/min] + +C = 87;// %age mass composition of Carbon in the fuel +H2 = 13;// %age mass composition of H2 in the fuel + +mc = 6.75;// fuel consumption, [kg/h] + +CO2 = 7;// %age composition of CO2 by volume +O2 = 10.5;// %age composition of O2 by volume +N2 = 7;// %age composition of N2 by volume + +mC = 12;// moleculer mass of CO2,[kg/kmol] +mH2 = 2;// moleculer mass of H2, [kg/kmol] +mO2 = 32;// moleculer mass of O2, [kg/kmol] +mN2 = 28;// moleculer mass of N2, [kg/kmol] + +T = 273+17;// atmospheric temperature, [K] +P = 100;// atmospheric pressure, [kn/m^2] +R =.287;// gas constant, [kJ/kg k] + +// solution +// (a) +// combustion equation by no. of moles +// 87/12 C + 13/2 H2 + a O2+79/21*a N2 = b CO2 + d H2O + eO2 + f N2 +// equating coefficient +b = 87/12;// [mol] +a = 22.7;// [mol] +e = 10.875;// [mol] +f = 11.8*b;// [mol] +// so fuel side combustion equation is +// 87/12 C + 13/2 H2 +22.7 O2 +85.5 N2 +mair = ( 22.7*mO2 +85.5*mN2)/100;// mass of air/kg fuel, [kg] +mprintf('\n (a) The mass of actual air supplied per kg of fuel is = %f kg\n',mair); + +// (b) +m = mair*mc/60;// mass of air/min, [kg] +V = m*R*T/P;// volumetric flow of air/min, [m^3] +SV = %pi/4*d^2*L*N/2;// swept volume/min, [m^3] + +VE = V/SV;// volumetric efficiency +mprintf('\n (b) The volumetric efficiency of the engine is = %fpercent\n',VE*100); + +// End diff --git a/2705/CH8/EX8.24/Ex8_24.sce b/2705/CH8/EX8.24/Ex8_24.sce new file mode 100755 index 000000000..42ac180f3 --- /dev/null +++ b/2705/CH8/EX8.24/Ex8_24.sce @@ -0,0 +1,38 @@ +clear; +clc; +disp('Example 8.24'); + +// aim : To determine +// the mass of air supplied/kg of fuel burnt + +// given values +// gas composition in the fuel +C = 84;// %age mass composition of Carbon in the fuel +H2 = 14;// %age mass composition of H2 in the fuel +O2f = 2;// %age mass composition of O2 in the fuel + +// exhaust gas composition +CO2 = 8.85;// %age composition of CO2 by volume +CO = 1.2// %age composition of CO by volume +O2 = 6.8;// %age composition of O2 by volume +N2 = 83.15;// %age composition of N2 by volume + +mC = 12;// moleculer mass of CO2,[kg/kmol] +mH2 = 2;// moleculer mass of H2, [kg/kmol] +mO2 = 32;// moleculer mass of O2, [kg/kmol] +mN2 = 28;// moleculer mass of N2, [kg/kmol] + +// solution +// combustion equation by no. of moles +// 84/12 C + 14/2 H2 +2/32 O2 + a O2+79.3/20.7*a N2 = b CO2 + d CO2+ eO2 + f N2 +g H2 +// equating coefficient and given condition +b = 6.16;// [mol] +a = 15.14;// [mol] +d = .836;// [mol] +f = 69.3*d;// [mol] +// so fuel side combustion equation is +// 84/12 C + 14/2 H2 +2/32 O2 + 15.14 O2 +85.5 N2 +mair = ( a*mO2 +f*mN2)/100;// mass of air/kg fuel, [kg] +mprintf('\n The mass of air supplied per kg of fuel is = %f kg\n',mair); + +// End diff --git a/2705/CH8/EX8.3/Ex8_3.sce b/2705/CH8/EX8.3/Ex8_3.sce new file mode 100755 index 000000000..6050727e8 --- /dev/null +++ b/2705/CH8/EX8.3/Ex8_3.sce @@ -0,0 +1,37 @@ +clear; +clc; +disp('Example 8.3'); + +// aim : To determine +// the stoichiometric mass of air +// the products of combustion both by mass and as percentage + +// Given values +C = .82;// mass composition C +H2 = .12;// mass composition of H2 +O2 = .02;// mass composition of O2 +S = .01;// mass composition of S +N2 = .03;// mass composition of N2 + + // solution +// for 1kg fuel +mo2 = 8/3*C+8*H2-O2+S*1;// total mass of O2 required, [kg] +sa = mo2/.232;// stoichimetric air, [kg] +mprintf('\n The stoichiometric mass of air is = %f kg/kg fuel\n',sa); + +// for one kg fuel +mCO2 = C*11/3;// mass of CO2 produced, [kg] +mH2O = H2*9;// mass of H2O produced, [kg] +mSO2 = S*2;// mass of SO2 produce, [kg] +mN2 = C*8.84+H2*26.5-O2*.768/.232+S*3.3+N2;// mass of N2 produced, [kg] + +mt = mCO2+mH2O+mSO2+mN2;// total mass of product, [kg] + +x1 = mCO2/mt*100;// %age mass composition of CO2 produced +x2 = mH2O/mt*100;// %age mass composition of H2O produced +x3 = mSO2/mt*100;// %age mass composition of SO2 produced +x4 = mN2/mt*100;// %age mass composition of N2 produced + +mprintf('\n CO2 produced = %f kg/kg fuel, percentage composition = %f,\n H2O produced = %f kg/kg fuel, percentage composition = %f,\n SO2 produced = %f kg/kg fuel, percentage composition = %f,\n N2 produced = %f kg/kg fuel, percentage composition = %f',mCO2,x1,mH2O,x2,mSO2,x3,mN2,x4); + +// End diff --git a/2705/CH8/EX8.4/Ex8_4.sce b/2705/CH8/EX8.4/Ex8_4.sce new file mode 100755 index 000000000..6938b2413 --- /dev/null +++ b/2705/CH8/EX8.4/Ex8_4.sce @@ -0,0 +1,25 @@ +clear; +clc; +disp('Example 8.4'); + +// aim : To determine +// the stoichiometric volume of air required for complete combution of 1 m^3 of the gas + +// Given values +H2 = .14;// volume fraction of H2 +CH4 = .02;// volume fraction of CH4 +CO = .22;// volume fraction of CO +CO2 = .05;// volume fraction of CO2 +O2 = .02;// volume fraction of O2 +N2 = .55;// volume fraction of N2 + +// solution +// for 1 m^3 of fuel +Va = .5*H2+2*CH4+.5*CO-O2;// [m^3] + +// stoichiometric air required is +Vsa = Va/.21;// [m^3] + +mprintf('\n The stoichiometric volume of air required for complete combustion is = %f m^3/m^3 fuel\n',Vsa); + +// End diff --git a/2705/CH8/EX8.5/Ex8_5.sce b/2705/CH8/EX8.5/Ex8_5.sce new file mode 100755 index 000000000..a69209cb1 --- /dev/null +++ b/2705/CH8/EX8.5/Ex8_5.sce @@ -0,0 +1,21 @@ +clear; +clc; +disp('Example 8.5'); + +// aim : To determine +// the volume of the air required + +// Given values +H2 = .45;// volume fraction of H2 +CO = .40;// volume fraction of CO +CH4 = .15;// volume fraction of CH4 + +// solution +V = 2.38*(H2+CO)+9.52*CH4;// stoichimetric volume of air, [m^3] + +mprintf('\n The volume of air required is = %f m^3/m^3 fuel\n',V); + +// Result in the book is misprinted + +// End + diff --git a/2705/CH8/EX8.6/Ex8_6.sce b/2705/CH8/EX8.6/Ex8_6.sce new file mode 100755 index 000000000..fe97bb2fb --- /dev/null +++ b/2705/CH8/EX8.6/Ex8_6.sce @@ -0,0 +1,37 @@ +clear; +clc; +disp('Example 8.6'); + +// aim : To determine +// the stoichiometric volume of air for the complete combustion +// the products of combustion + +// given values +CH4 = .142;// volumetric composition of CH4 +CO2 = .059;// volumetric composition of CO2 +CO = .360;// volumetric composition of CO +H2 = .405;// volumetric composition of H2 +O2 = .005;// volumetric composition of O2 +N2 = .029;// volumetric composition of N2 + +aO2 = .21;// O2 composition into air by volume + +// solution +svO2 = CH4*2+CO*.5+H2*.5-O2;// stroichiometric volume of O2 required, [m^3/m^3 fuel] +svair = svO2/aO2;// stroichiometric volume of air required, [m^3/m^3 fuel] +mprintf('\n Stoichiometric volume of air required is = %f m^3/m^3 fuel\n',svair); + +// for one m^3 fuel +vN2 = CH4*7.52+CO*1.88+H2*1.88-O2*.79/.21+N2;// volume of N2 produced, [m^3] +vCO2 = CH4*1+CO2+CO*1;// volume of CO2 produced, [m^3] +vH2O = CH4*2+H2*1;// volume of H2O produced, [m^3] + +vt = vN2+vCO2+vH2O;// total volume of product, [m^3] + +x1 = vN2/vt*100;// %age composition of N2 in product, +x2 = vCO2/vt*100;// %age composition of CO2 in product +x3 = vH2O/vt*100;// %age composition of H2O in product + +mprintf('\n N2 in products = %fm^3/m^3 fuel, percentage composition = %f,\n CO2 in products = %f m^3/m^3 fuel, percentage composition = %f,\n H2O in products = %fm^3/m^3 fuel, percentage composition = %f',vN2,x1,vCO2,x2,vH2O,x3); + +// End diff --git a/2705/CH8/EX8.7/Ex8_7.sce b/2705/CH8/EX8.7/Ex8_7.sce new file mode 100755 index 000000000..534f4f01e --- /dev/null +++ b/2705/CH8/EX8.7/Ex8_7.sce @@ -0,0 +1,25 @@ +clear; +clc; +disp('Example 8.7'); + +// aim : To determine +// the percentage analysis of the gas by mass + +// Given values +CO2 = 20;// percentage volumetric composition of CO2 +N2 = 70;// percentage volumetric composition of N2 +O2 = 10;// percentage volumetric composition of O2 + +mCO2 = 44;// moleculer mas of CO2 +mN2 = 28;// moleculer mass of N2 +mO2 = 32;// moleculer mass of O2 + +// solution +mgas = CO2*mCO2+N2*mN2+O2*mO2;// moleculer mass of gas +m1 = CO2*mCO2/mgas*100;// percentage composition of CO2 by mass +m2 = N2*mN2/mgas*100;// percentage composition of N2 by mass +m3 = O2*mO2/mgas*100;// percentage composition of O2 by mass + +mprintf('\n Mass percentage of CO2 is = %f\n\n Mass percentage of N2 is = %f\n\n Mass percentage of O2 is = %f\n',m1,m2,m3 ) + +// End diff --git a/2705/CH8/EX8.8/Ex8_8.sce b/2705/CH8/EX8.8/Ex8_8.sce new file mode 100755 index 000000000..347ec4c01 --- /dev/null +++ b/2705/CH8/EX8.8/Ex8_8.sce @@ -0,0 +1,31 @@ +clear; +clc; +disp('Example 8.8'); + +// aim : To determine +// the percentage composition of the gas by volume + +// given values +CO = 30;// %age mass composition of CO +N2 = 20;// %age mass composition of N2 +CH4 = 15;// %age mass composition of CH4 +H2 = 25;// %age mass composition of H2 +O2 = 10;// %age mass composition of O2 + +mCO = 28;// molculer mass of CO +mN2 = 28;// molculer mass of N2 +mCH4 = 16;// molculer mass of CH4 +mH2 = 2;// molculer mass of H2 +mO2 = 32;// molculer mass of O2 + +// solution +vg = CO/mCO+N2/mN2+CH4/mCH4+H2/mH2+O2/mO2; +v1 = CO/mCO/vg*100;// %age volume composition of CO +v2 = N2/mN2/vg*100;// %age volume composition of N2 +v3 = CH4/mCH4/vg*100;// %age volume composition of CH4 +v4 = H2/mH2/vg*100;// %age volume composition of H2 +v5 = O2/mO2/vg*100;// %age volume composition of O2 + +mprintf('\n The percentage composition of CO by volume is = %f\n,\nThe percentage composition of N2 by volume is = %f\n\nThe percentage composition of CH4 by volume is = %f\n\nThe percentage composition of H2 by volume is = %f\n\nThe percentage composition of O2by volume is=%f',v1,v2,v3,v4,v5); + +// End diff --git a/2705/CH8/EX8.9/Ex8_9.sce b/2705/CH8/EX8.9/Ex8_9.sce new file mode 100755 index 000000000..0528a2575 --- /dev/null +++ b/2705/CH8/EX8.9/Ex8_9.sce @@ -0,0 +1,44 @@ +clear; +clc; +disp('Example 8.9'); + +// aim : To determine +// the mass of air supplied per kilogram of fuel burnt + +// given values +CO2 = 8.85;// volume composition of CO2 +CO = 1.2;// volume composition of CO +O2 = 6.8;// volume composition of O2 +N2 = 83.15;// volume composition of N2 + +// composition of gases in the fuel oil +C = .84;// mass composition of carbon +H = .14;// mass composition of hydrogen +o2 = .02;// mass composition of oxygen + +mC = 12;// moleculer mass of Carbon +mCO2 = 44;// molculer mass of CO2 +mCO = 28;// molculer mass of CO +mN2 = 28;// molculer mass of N2 +mO2 = 32;// molculer mass of O2 +aO2 = .23;// mass composition of O2 in air + +// solution +ma = (8/3*C+8*H-o2)/aO2;// theoretical mass of air/kg fuel, [kg] + +mgas = CO2*mCO2+CO*mCO+N2*mN2+O2*mO2;// total mass of gas/kg fuel, [kg] +x1 = CO2*mCO2/mgas;// composition of CO2 by mass +x2 = CO*mCO/mgas;// composition of CO by mass +x3 = O2*mO2/mgas;// composition of O2 by mass +x4 = N2*mN2/mgas;// composition of N2 by mass + +m1 = x1*mC/mCO2+x2*mC/mCO;// mass of C/kg of dry flue gas, [kg] +m2 = C;// mass of C/kg fuel, [kg] +mf = m2/m1;// mass of dry flue gas/kg fuel, [kg] +mo2 = mf*x3;// mass of excess O2/kg fuel, [kg] +mair = mo2/aO2;// mass of excess air/kg fuel, [kg] +m = ma+mair;// mass of excess air supplied/kg fuel, [kg] + +mprintf('\n The mass of air supplied per/kg of fuel burnt is = %f kg\n',m); + +// End diff --git a/2705/CH9/EX9.1/Ex9_1.sce b/2705/CH9/EX9.1/Ex9_1.sce new file mode 100755 index 000000000..4a04d7b87 --- /dev/null +++ b/2705/CH9/EX9.1/Ex9_1.sce @@ -0,0 +1,32 @@ +clear; +clc; +disp('Example 9.1'); + +// aim : To determine +// the heat loss per hour through the wall and interface temperature + +// Given values +x1 = .25;// thickness of brick,[m] +x2 = .05;// thickness of concrete,[m] +t1 = 30;// brick face temperature,[C] +t3 = 5;// concrete face temperature,[C] +l = 10;// length of the wall, [m] +h = 5;// height of the wall, [m] +k1 = .69;// thermal conductivity of brick,[W/m/K] +k2 = .93;// thermal conductivity of concrete,[W/m/K] + +// solution +A = l*h;// area of heat transfer,[m^2] +Q_dot = A*(t1-t3)/(x1/k1+x2/k2);// heat transferred, [J/s] + +// so heat loss per hour is +Q = Q_dot*3600*10^-3;// [kJ] +mprintf('\n The heat lost per hour is = %f kJ\n',Q); + +// interface temperature calculation +// for the brick wall, Q_dot=k1*A*(t1-t2)/x1; +// hence +t2 = t1-Q_dot*x1/k1/A;// [C] +mprintf('\n The interface temperature is = %f C\n',t2); + +// End diff --git a/2705/CH9/EX9.2/Ex9_2.sce b/2705/CH9/EX9.2/Ex9_2.sce new file mode 100755 index 000000000..1a60b1763 --- /dev/null +++ b/2705/CH9/EX9.2/Ex9_2.sce @@ -0,0 +1,35 @@ +clear; +clc; +disp('Example 9.2'); + +// aim : To determine +// the minimum +// thickness of the lagging required + +// Given values +r1 = 75/2;// external radious of the pipe,[mm] +L = 80;// length of the pipe,[m] +m_dot = 1000;// flow of steam, [kg/h] +P = 2;// pressure, [MN/m^2] +x1 = .98;// inlet dryness fraction +x2 = .96;// outlet dryness fraction +k = .08;// thermal conductivity of of pipe, [W/m/K] +t2 = 27;// outside temperature,[C] + +// solution +// using steam table at 2 MN/m^2 the enthalpy of evaporation of steam is, +hfg = 1888.6;// [kJ/kg] +// so heat loss through the pipe is +Q_dot = m_dot*(x1-x2)*hfg/3600;// [kJ] + +// also from steam table saturation temperature of steam at 2 MN/m^2 is, +t1 = 212.4;// [C] +// and for thick pipe, Q_dot=k*2*%pi*L*(t1-t2)/log(r2/r1) +// hence +r2 = r1*exp(k*2*%pi*L*(t1-t2)*10^-3/Q_dot);// [mm] + +t = r2-r1;// thickness, [mm] + +mprintf('\n The minimum thickness of the lagging required is = %f mm\n',t); + +// End diff --git a/2705/CH9/EX9.3/Ex9_3.sce b/2705/CH9/EX9.3/Ex9_3.sce new file mode 100755 index 000000000..0901fd861 --- /dev/null +++ b/2705/CH9/EX9.3/Ex9_3.sce @@ -0,0 +1,41 @@ +clear; +clc; +disp('Example 9.3'); + +// aim : To determine the +// (a) heat loss per hour +// (b) interface temperature og lagging + +// Given values +r1 = 50; // radious of steam main,[mm] +r2 = 90;// radious with first lagging,[mm] +r3 = 115;// outside radious os steam main with lagging,[mm] +k1 = .07;// thermal conductivity of 1st lagging,[W/m/K] +k2 = .1;// thermal conductivity of 2nd lagging, [W/m/K] +P = 1.7;// steam pressure,[MN/m^2] +t_superheat = 30;// superheat of steam, [K] +t3 = 24;// outside temperature of the lagging,[C] +L = 20;// length of the steam main,[m] + +// solution +// (a) +// using steam table saturation temperature of steam at 1.7 MN/m^2 is +t_sat = 204.3;// [C] +// hence +t1 = t_sat+t_superheat;// temperature of steam,[C] + +Q_dot = 2*%pi*L*(t1-t3)/(log(r2/r1)/k1+log(r3/r2)/k2);// heat loss,[W] +// heat loss in hour is +Q = Q_dot*3600*10^-3;// [kJ] + +mprintf('\n (a) The heat lost per hour is = %f kJ\n',Q); + +// (b) +// using Q_dot=2*%pi*k1*(t1-t1)/log(r2/r1) +t2 = t1-Q_dot*log(r2/r1)/(2*%pi*k1*L);// interface temperature of lagging,[C] + +mprintf('\n (b) The interface temperature of the lagging is = %f C\n',t2); + +// There is some calculation mistake in the book so answer is not matching + +// End diff --git a/2705/CH9/EX9.4/Ex9_4.sce b/2705/CH9/EX9.4/Ex9_4.sce new file mode 100755 index 000000000..b6f5f4106 --- /dev/null +++ b/2705/CH9/EX9.4/Ex9_4.sce @@ -0,0 +1,22 @@ +clear; +clc; +disp('Example 9.4'); + +// aim : To determine +// the energy emetted from the surface + +// Given values +h = 3;// height of surface, [m] +b = 4;// width of surface, [m] +epsilon_s = .9;// emissivity of the surface +T = 273+600;// surface temperature ,[K] +sigma = 5.67*10^-8;// [W/m^2/K^4] + +// solution +As = h*b;// area of the surface, [m^2] + +Q_dot = epsilon_s*sigma*As*T^4*10^-3;// energy emitted, [kW] + +mprintf('\n The energy emitted from the surface is = %f kW\n',Q_dot); + +// End diff --git a/2705/CH9/EX9.5/Ex9_5.sce b/2705/CH9/EX9.5/Ex9_5.sce new file mode 100755 index 000000000..e916c7e40 --- /dev/null +++ b/2705/CH9/EX9.5/Ex9_5.sce @@ -0,0 +1,34 @@ +clear; +clc; +disp('Example 9.5'); + +// aim : To determine +// the rate of energy transfer between furnace and the sphere and its direction + +// Given values +l = 1.25;// internal side of cubical furnace, [m] +ti = 800+273;// internal surface temperature of the furnace,[K] +r = .2;// sphere radious, [m] +epsilon = .6;// emissivity of sphere +ts = 300+273;// surface temperature of sphere, [K] +sigma = 5.67*10^-8;// [W/m^2/K^4] + +// Solution +Af = 6*l^2;// internal surface area of furnace, [m^2] +As =4 *%pi*r^2;// surface area of sphere, [m^2] + +// considering internal furnace to be black +Qf = sigma*Af*ti^4*10^-3;// [kW] + +// radiation emitted by sphere is +Qs = epsilon*sigma*As*ts^4*10^-3; // [kW] + +// Hence transfer of energy is +Q = Qf-Qs;// [kW] + +mprintf('\n The transfer of energy will be from furnace to sphere and transfer rate is = %f kW\n',Q); + +// There is some calculation mistake in the book so answer is not matching + +// End + diff --git a/2705/CH9/EX9.6/Ex9_6.sce b/2705/CH9/EX9.6/Ex9_6.sce new file mode 100755 index 000000000..07386e7e2 --- /dev/null +++ b/2705/CH9/EX9.6/Ex9_6.sce @@ -0,0 +1,34 @@ +clear; +clc; +disp('Example 9.6'); + +// aim : To determine +// the overall transfer coefficient and the heat loss per hour + +// Given values +x1 = 25*10^-3;// Thickness of insulating board, [m] +x2 = 75*10^-3;// Thickness of fibreglass, [m] +x3 = 110*10^-3;// Thickness of brickwork, [m] +k1 = .06;// Thermal conductivity of insulating board, [W/m K] +k2 = .04;// Thermal conductivity of fibreglass, [W/m K] +k3 = .6;// Thermal conductivity of brickwork, [W/m K] +Us1 = 2.5;// surface heat transfer coefficient of the inside wall,[W/m^2 K] +Us2 = 3.1;// surface heat transfer coefficient of the outside wall,[W/m^2 K] +ta1 = 27;// internal ambient temperature, [C] +ta2 = 10;// external ambient temperature, [C] +h = 6;// height of the wall, [m] +l = 10;// length of the wall, [m] + +// solution +U = 1/(1/Us1+x1/k1+x2/k2+x3/k3+1/Us2);// overall heta transfer coefficient,[W/m^2 K] + +A = l*h;// area ,[m^2] + +Q_dot = U*A*(ta1-ta2);// heat loss [W] + +// so heat loss per hour is +Q = Q_dot*3600*10^-3;// [kJ] +mprintf('\n The overall heat transfer coefficient for the wall is = %f W/m^2 K\n',U); +mprintf('\n The heat loss per hour through the wall is = %f kJ\n',Q); + +// End diff --git a/2705/CH9/EX9.7/Ex9_7.sce b/2705/CH9/EX9.7/Ex9_7.sce new file mode 100755 index 000000000..4af2d8450 --- /dev/null +++ b/2705/CH9/EX9.7/Ex9_7.sce @@ -0,0 +1,44 @@ +clear; +clc; +disp('Example 9.7'); + +// aim : To determine +// the heat loss per hour and the surface temperature of the lagging + +// Given values +r1 = 75*10^-3;// External radiou of the pipe, [m] +t_l1 = 40*10^-3;// Thickness of lagging1, [m] +t_l2 = t_l1; +k1 = .07;// thermal conductivity of lagging1, [W/m K] +k2 = .1;// thermal conductivity of lagging2, [W/m K] +Us = 7;// surface transfer coefficient for outer surface, [W/m^2 K] +L = 50;// length of the pipe, [m] +ta = 27;// ambient temperature, [C] +P = 3.6;// wet steam pressure, [MN/m^2] + +// solution +// from steam table saturation temperature of the steam at given pressure is, +t1 = 244.2;// [C] +r2 = r1+t_l1;// radious of pipe with lagging1,[m] +r3 = r2+t_l2;// radious of pipe with both the lagging, [m] + +R1 = log(r2/r1)/(2*%pi*L*k1);// resistance due to lagging1,[C/W] +R2 = log(r3/r2)/(2*%pi*L*k2);// resistance due to lagging2,[C/W] +R3 = 1/(Us*2*%pi*r3*L);// ambient resistance, [C/W] + +// hence overall resistance is, +Req = R1+R2+R3;// [C/W] +tdf = t1-ta;// temperature driving force, [C] +Q_dot = tdf/Req;// rate of heat loss, [W] +// so heat loss per hour is, +Q = Q_dot*3600*10^-3;// heat loss per hour, [kJ] + +// using eqn [3] +t3 = ta+Q_dot*R3;// surface temperature of the lagging, [C] + +mprintf('\n The heat loss per hour is = %f kJ\n',Q); +mprintf('\n The surface temperature of the lagging is = %f C\n',t3); + +// there is minor variation in the answer + +// End -- cgit