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 --- 530/CH7/EX7.1/example_7_1.sce | 20 +++++++++ 530/CH7/EX7.2/example_7_2.sce | 48 +++++++++++++++++++++ 530/CH7/EX7.3/example_7_3.sce | 29 +++++++++++++ 530/CH7/EX7.4.a/example_7_4a.sce | 37 ++++++++++++++++ 530/CH7/EX7.4.b/example_7_4b.sce | 54 +++++++++++++++++++++++ 530/CH7/EX7.5/example_7_5.sce | 93 ++++++++++++++++++++++++++++++++++++++++ 6 files changed, 281 insertions(+) create mode 100755 530/CH7/EX7.1/example_7_1.sce create mode 100755 530/CH7/EX7.2/example_7_2.sce create mode 100755 530/CH7/EX7.3/example_7_3.sce create mode 100755 530/CH7/EX7.4.a/example_7_4a.sce create mode 100755 530/CH7/EX7.4.b/example_7_4b.sce create mode 100755 530/CH7/EX7.5/example_7_5.sce (limited to '530/CH7') diff --git a/530/CH7/EX7.1/example_7_1.sce b/530/CH7/EX7.1/example_7_1.sce new file mode 100755 index 000000000..9deb9d28d --- /dev/null +++ b/530/CH7/EX7.1/example_7_1.sce @@ -0,0 +1,20 @@ +clear; +clc; + +// A Textbook on HEAT TRANSFER by S P SUKHATME +// Chapter 7 +// Heat Exchangers + + +// Example 7.1 +// Page 285 +printf("Example 7.1, Page 285 \n \n"); + +h = 2000 ; // [W/m^2 K] +// From Table 7.1 +U_f = 0.0001 ; // fouling factor, m^2K/W +h_f = 1/[1/h+U_f]; +printf("Heat transfer coefficient including the effect of foulung = %f W/m^2 K \n",h_f); + +p = (h-h_f)/h*100; +printf("Percentage reduction = %f \n",p); \ No newline at end of file diff --git a/530/CH7/EX7.2/example_7_2.sce b/530/CH7/EX7.2/example_7_2.sce new file mode 100755 index 000000000..1bc9bc1b1 --- /dev/null +++ b/530/CH7/EX7.2/example_7_2.sce @@ -0,0 +1,48 @@ +clear; +clc; + +// A Textbook on HEAT TRANSFER by S P SUKHATME +// Chapter 7 +// Heat Exchangers + + +// Example 7.2 +// Page 294 +printf("Example 7.2, Page 294 \n \n"); + +m = 1000 ; // [kg/h] +Thi = 50 ; // [C] +The = 40 ; // [C] +Tci = 35 ; // [C] +Tce = 40 ; // [C] +U = 1000 ; // OHTC, W/m^2 K + +// Using Eqn 7.5.25 +q = m/3600*4174*(Thi-The) ; // [W] + +delta_T = ((Thi-Tce)-(The-Tci))/log((Thi-Tce)/(The-Tci)); // [C] +printf("delta T = %f \n\n",delta_T); + +// T1 = Th and T2 = Tc +R = (Thi-The)/(Tce-Tci) ; +S = (Tce-Tci)/(Thi-Tci) ; +// From fig 7.15, +F =0.91 ; + +printf("Taking T1 = Th and T2 = Tc \n") +printf("R = %f, S = %f \n",R,S); +printf("Hence, F = %f \n \n",F); + +// Alternatively, taking T1 = Tc and T2 = Th +R = (Tci-Tce)/(The-Thi); +S = (The-Thi)/(Tci-Thi); + +// Again from fig 7.15, +F =0.91 ; + +printf("Taking T1 = Tc and T2 = Th \n") +printf("R = %f, S = %f \n",R,S); +printf("Hence, F = %f \n",F); + +A = q/(U*F*delta_T); +printf("\nArea = %f m^2",A); diff --git a/530/CH7/EX7.3/example_7_3.sce b/530/CH7/EX7.3/example_7_3.sce new file mode 100755 index 000000000..617dd51de --- /dev/null +++ b/530/CH7/EX7.3/example_7_3.sce @@ -0,0 +1,29 @@ +clear; +clc; + +// A Textbook on HEAT TRANSFER by S P SUKHATME +// Chapter 7 +// Heat Exchangers + + +// Example 7.3 +// Page 295 +printf("Example 7.3, Page 295 \n \n"); + +// Because of change of phase , Thi = The +Thi = 100 ; // [C], Saturated steam +The = 100 ; // [C], Condensed steam +Tci = 30 ; // [C], Cooling water inlet +Tce = 70 ; // [C], cooling water outlet + +R = (Thi-The)/(Tce-Tci) ; +S = (Tce-Tci)/(Thi-Tci) ; + +// From fig 7.16 +F = 1; + +// For counter flow arrangement +Tm_counter = ((Thi-Tce)-(The-Tci))/log((Thi-Tce)/(The-Tci)); // [C] +// Therefore +Tm = F*Tm_counter ; +printf("Mean Temperaature Difference = %f C",Tm) diff --git a/530/CH7/EX7.4.a/example_7_4a.sce b/530/CH7/EX7.4.a/example_7_4a.sce new file mode 100755 index 000000000..c4880ad6c --- /dev/null +++ b/530/CH7/EX7.4.a/example_7_4a.sce @@ -0,0 +1,37 @@ +clear; +clc; + +// A Textbook on HEAT TRANSFER by S P SUKHATME +// Chapter 7 +// Heat Exchangers + + +// Example 7.4(a) +// Page 302 +printf("Example 7.4(a), Page 302 \n \n"); + +// (a) +printf("(a) \n"); +// Using Mean Temperature Difference approach +m_hot = 10 ; // [kg/min] +m_cold = 25 ; // [kg/min] +hh = 1600 ; // [W/m^2 K], Heat transfer coefficient on hot side +hc = 1600 ; // [W/m^2 K], Heat transfer coefficient on cold side + +Thi = 70 ; // [C] +Tci = 25 ; // [C] +The = 50 ; // [C] + +// Heat Transfer Rate, q +q = m_hot/60*4179*(Thi-The); // [W] + +// Heat gained by cold water = heat lost by the hot water +Tce = 25 + q*1/(m_cold/60*4174); // [C] + +// Using equation 7.5.13 +Tm = ((Thi-Tci)-(The-Tce))/log((Thi-Tci)/(The-Tce)); // [C] +printf("Mean Temperature Difference = %f C \n",Tm); + +U = 1/(1/hh + 1/hc); // [W/m^2 K] +A = q/(U*Tm); // Area, [m^2] +printf("Area of Heat Exchanger = %f m^2 \n",A); diff --git a/530/CH7/EX7.4.b/example_7_4b.sce b/530/CH7/EX7.4.b/example_7_4b.sce new file mode 100755 index 000000000..40873abec --- /dev/null +++ b/530/CH7/EX7.4.b/example_7_4b.sce @@ -0,0 +1,54 @@ +clear; +clc; + +// A Textbook on HEAT TRANSFER by S P SUKHATME +// Chapter 7 +// Heat Exchangers + + +// Example 7.4(b) +// Page 302 +printf("Example 7.4(b), Page 302 \n \n"); + +// Using Mean Temperature Difference approach +m_hot = 10 ; // [kg/min] +m_cold = 25 ; // [kg/min] +hh = 1600 ; // [W/m^2 K], Heat transfer coefficient on hot side +hc = 1600 ; // [W/m^2 K], Heat transfer coefficient on cold side + +Thi = 70 ; // [C] +Tci = 25 ; // [C] +The = 50 ; // [C] + +// Heat Transfer Rate, q +q = m_hot/60*4179*(Thi-The); // [W] + +// Heat gained by cold water = heat lost by the hot water +Tce = 25 + q*1/(m_cold/60*4174); // [C] + +// Using equation 7.5.13 +Tm = ((Thi-Tci)-(The-Tce))/log((Thi-Tci)/(The-Tce)); // [C] +U = 1/(1/hh + 1/hc); // [W/m^2 K] +A = q/(U*Tm); // Area, [m^2] + +m_hot = 20 ; // [kg/min] +// Flow rate on hot side i.e. 'hh' is doubled +hh = 1600*2^0.8 ; // [W/m^2 K] +U = 1/(1/hh + 1/hc); // [W/m^2 K] +m_hC_ph = m_hot/60*4179 ; // [W/K] +m_cC_pc = m_cold/60*4174 ; // [W/K] +// Therefore +C = m_hC_ph/m_cC_pc ; +NTU = U*A/m_hC_ph ; +printf("NTU = %f \n",NTU); + +// From equation 7.6.8 +e = [1 - exp(-(1+C)*NTU)]/(1+C) ; + +// Therefore (Thi - The)/(Thi - Tci) = e , we get +The = Thi - e*(Thi - Tci); // [C] + +// Equating the heat lost by water to heat gained by cold water , we get +Tce = Tci + [m_hC_ph*(Thi-The)]/m_cC_pc; +printf("Exit temperature of cold and hot stream are %f C and %f C respectively.",Tce,The); + diff --git a/530/CH7/EX7.5/example_7_5.sce b/530/CH7/EX7.5/example_7_5.sce new file mode 100755 index 000000000..a493c9186 --- /dev/null +++ b/530/CH7/EX7.5/example_7_5.sce @@ -0,0 +1,93 @@ +clear; +clc; + +// A Textbook on HEAT TRANSFER by S P SUKHATME +// Chapter 7 +// Heat Exchangers + + +// Example 7.5 +// Page 304 +printf("Example 7.5, Page 304 \n \n"); + +mc = 2000 ; // [kg/h] +Tce = 40 ; // [C] +Tci = 15 ; // [C] +Thi = 80 ; // [C] +U = 50 ; // OHTC, [W/m^2 K] +A = 10 ; // Area, [m^2] + +// Using effective NTU method +// Assuming m_c*C_pc = (m*C_p)s +NTU = U*A/(mc*1005/3600); +e = (Tce-Tci)/(Thi-Tci); +// From fig 7.23, no value of C is found corresponding to the above values, hence assumption was wrong. +// So, m_h*C_ph must be equal to (m*C_p)s, proceeding by trail and error method + + +printf("m_h(kg/h NTU C e T_he(C) T_he(C) (Heat Balance)"); + +mh = rand(1:5); +NTU = rand(1:5); +The = rand(1:5); +The2 = rand(1:5); + +mh(1) = 200 +NTU(1) = U*A/(mh(1)*1.161); +//Corresponding Values of C and e from fig 7.23 +C = .416; +e = .78; +//From Equation 7.6.2 Page 297 +The(1) = Thi - e*(Thi-Tci) +//From Heat Balance +The2(1) = Thi - mc*1005/3600*(Tce-Tci)/(mh(1)*1.161); +printf("\n\n %i %.3f %.3f %.3f %.2f %.2f",mh(1),NTU(1),C,e,The(1),The2(1)); + +mh(2) = 250 +NTU(2) = U*A/(mh(2)*1.161); +//Corresponding Values of C and e from fig 7.23 +C = .520; +e = .69; +//From Equation 7.6.2 Page 297 +The(2) = Thi - e*(Thi-Tci) +//From Heat Balance +The2(2) = Thi - mc*1005/3600*(Tce-Tci)/(mh(2)*1.161); +printf("\n\n %i %.3f %.3f %.3f %.2f %.2f",mh(2),NTU(2),C,e,The(2),The2(2)); + +mh(3) = 300 +NTU(3) = U*A/(mh(3)*1.161); +//Corresponding Values of C and e from fig 7.23 +C = .624; +e = .625; +//From Equation 7.6.2 Page 297 +The(3) = Thi - e*(Thi-Tci) +//From Heat Balance +The2(3) = Thi - mc*1005/3600*(Tce-Tci)/(mh(3)*1.161); +printf("\n\n %i %.3f %.3f %.3f %.2f %.2f",mh(3),NTU(3),C,e,The(3),The2(3)); + +mh(4) = 350 +NTU(4) = U*A/(mh(4)*1.161); +//Corresponding Values of C and e from fig 7.23 +C = .728; +e = .57; +//From Equation 7.6.2 Page 297 +The(4) = Thi - e*(Thi-Tci) +//From Heat Balance +The2(4) = Thi - mc*1005/3600*(Tce-Tci)/(mh(4)*1.161); +printf("\n\n %i %.3f %.3f %.3f %.2f %.2f",mh(4),NTU(4),C,e,The(4),The2(4)); + +mh(5) = 400 +NTU(5) = U*A/(mh(5)*1.161); +//Corresponding Values of C and e from fig 7.23 +C = .832; +e = .51; +//From Equation 7.6.2 Page 297 +The(5) = Thi - e*(Thi-Tci) +//From Heat Balance +The2(5) = Thi - mc*1005/3600*(Tce-Tci)/(mh(5)*1.161); +printf("\n\n %i %.3f %.3f %.3f %.2f %.2f",mh(5),NTU(5),C,e,The(5),The2(5)); + +clf(); +plot(mh,The,mh,The2,[295 295 200],[0 39.2 39.2]) +xtitle('The vs mh','mh (kg/hr)','The (C)'); +printf("\n\n From the plot, value of mh = 295 kg/hr and correspondingly The = 39.2 C") \ No newline at end of file -- cgit