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author | priyanka | 2015-06-24 15:03:17 +0530 |
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committer | priyanka | 2015-06-24 15:03:17 +0530 |
commit | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (patch) | |
tree | ab291cffc65280e58ac82470ba63fbcca7805165 /2507/CH6 | |
download | Scilab-TBC-Uploads-b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b.tar.gz Scilab-TBC-Uploads-b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b.tar.bz2 Scilab-TBC-Uploads-b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b.zip |
initial commit / add all books
Diffstat (limited to '2507/CH6')
-rwxr-xr-x | 2507/CH6/EX6.1/Ex6_1.sce | 15 | ||||
-rwxr-xr-x | 2507/CH6/EX6.2/Ex6_2.sce | 22 | ||||
-rwxr-xr-x | 2507/CH6/EX6.3/Ex6_3.sce | 23 |
3 files changed, 60 insertions, 0 deletions
diff --git a/2507/CH6/EX6.1/Ex6_1.sce b/2507/CH6/EX6.1/Ex6_1.sce new file mode 100755 index 000000000..6854abbe9 --- /dev/null +++ b/2507/CH6/EX6.1/Ex6_1.sce @@ -0,0 +1,15 @@ +clc
+clear
+printf("Example 6.1 | Page number 148 \n\n");
+//Find the heat rejected at the condensor and thermal efficiency of the plant
+
+//Given Data
+Q1 = 300 //kJ //heat supplied at the boiler
+Wt = 100 //kJ //work output of turbine
+Wp = 0.5 //kJ //work input to pump
+
+//Solution
+Q2 = Q1 - (Wt - Wp) //kJ //heat rejected at the condensor
+printf("Heat rejected at the condensor = %.1f kJ\n",Q2);
+efficiency = 1 - (Q2/Q1)
+printf("The thermal efficiency of plant = %.2f ",efficiency)
diff --git a/2507/CH6/EX6.2/Ex6_2.sce b/2507/CH6/EX6.2/Ex6_2.sce new file mode 100755 index 000000000..8df32ca89 --- /dev/null +++ b/2507/CH6/EX6.2/Ex6_2.sce @@ -0,0 +1,22 @@ +clc
+clear
+printf("Example 6.2 | Page number 153 \n\n");
+//Part(a)Find the rate of heat transfer in the evaporator and the power input to the compressor.
+//Part(b)Calculate COP if refrigerator were to operate as heat pump
+
+//Given Data
+COP_ref = 4 //COP of refrigerator
+Q1 = 0.5 //kJ/s //rate of heat transfer at the condensor
+
+//Solution
+//Part(a)
+printf("Part(a)\n");
+Wc = Q1/(COP_ref+1) //kJ/s //Power input to compressor
+Q2 = COP_ref*Wc //kJ/s //Rate of heat transfer in evaporator
+printf("Rate of heat transfer in evaporator = %.1f kJ/s\n",Q2)
+printf("Power input to compressor = %.1f kJ/s\n\n",Wc)
+
+//Part(b)
+printf("Part(b)\n");
+COP_hp = 1 + COP_ref //COP of heat pump
+printf("COP of heat pump = %.1f",COP_hp)
diff --git a/2507/CH6/EX6.3/Ex6_3.sce b/2507/CH6/EX6.3/Ex6_3.sce new file mode 100755 index 000000000..611da8e93 --- /dev/null +++ b/2507/CH6/EX6.3/Ex6_3.sce @@ -0,0 +1,23 @@ +clc
+clear
+printf("Example 6.3 | Page number 154 \n\n");
+//Part(a)Find the additional monthly electricity bill
+//Part(b)What would be the saving in monthly electricity bill
+
+//Part(a)
+printf("Part(a)\n");
+I = 4.5 //Amp //Current drawn
+V = 220 //V
+Electricity_consumption = I*V //Watts
+ElectricityUnitPerDay = Electricity_consumption/1000*8 //kWh
+MonthlyBill_part_a = ElectricityUnitPerDay * 5 * 30
+printf("The additional monthly electricity bill = Rs. %.2f\n\n",MonthlyBill_part_a);
+
+//Part(b)
+printf("Part(b)\n");
+Q1 = Electricity_consumption*.001 //kW //Rate of heat transfer from heat pump
+COP_hp = 4 //COP of heat pump
+W = Q1/COP_hp //kW //rate at which energy is consumed
+ElectricityUnitPerDay = W*8
+MonthlyBill_part_b = ElectricityUnitPerDay * 5 * 30
+printf("Saving in monthly electricity bill = Rs. %.2f",MonthlyBill_part_a - MonthlyBill_part_b)
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