diff options
Diffstat (limited to '2471/CH9')
| -rwxr-xr-x | 2471/CH9/EX9.1/Ex9_1.sce | 52 | ||||
| -rwxr-xr-x | 2471/CH9/EX9.2/Ex9_2.sce | 35 | ||||
| -rwxr-xr-x | 2471/CH9/EX9.3/Ex9_3.sce | 25 | ||||
| -rwxr-xr-x | 2471/CH9/EX9.4/Ex9_4.sce | 59 | ||||
| -rwxr-xr-x | 2471/CH9/EX9.5/Ex9_5.sce | 38 | ||||
| -rwxr-xr-x | 2471/CH9/EX9.6/Ex9_6.sce | 35 | ||||
| -rwxr-xr-x | 2471/CH9/EX9.7/Ex9_7.sce | 33 | ||||
| -rwxr-xr-x | 2471/CH9/EX9.8/Ex9_8.sce | 41 | ||||
| -rwxr-xr-x | 2471/CH9/EX9.9/Ex9_9.sce | 43 |
9 files changed, 361 insertions, 0 deletions
diff --git a/2471/CH9/EX9.1/Ex9_1.sce b/2471/CH9/EX9.1/Ex9_1.sce new file mode 100755 index 000000000..49f4ea6ab --- /dev/null +++ b/2471/CH9/EX9.1/Ex9_1.sce @@ -0,0 +1,52 @@ +clear;
+clc;
+// Example 9.1
+printf('Example 9.1\n\n');
+printf('Page No. 252\n\n');
+
+//given
+a = 40;// in m
+b = 25;// in m
+c = 20;// in m
+d = 10;// in m
+e = 5;// in m
+f = 2;// in m
+g = 3;// in m
+h = 6;// in m
+
+//(1) Production Area
+T1 = 21;// Temperature difference in degree celcius
+T2 = -3;// Temperature difference in degree celcius
+U1 = 1.2;// heat transfer coefficent in W/m-K
+U2 = 5.6;// heat transfer coefficent in W/m-K
+U3 = 2.0;// heat transfer coefficent in W/m-K
+U4 = 0.7;// heat transfer coefficent in W/m-K
+U5 = 0.9;// heat transfer coefficent in W/m-K
+// As Q = U*A*T
+Q1 = (b*h)*U1*T1;// Heat loss in W. wall in W
+Q2 = (((a-c)*h) + (d*h) + (d*f))*U1*T1;// Heat loss in N. wall in W
+Q3 = (c*f)*U2*T1;// Heat loss in N. window in W
+Q4 = (b*g)*U3*T2;// Heat loss in N. wall/internal in W
+Q5 = (b*g)*U1*T1;// Heat loss in E. wall/external in W
+Q6 = (((a-c)*h) + (d*h) + (d*f))*U1*T1;// Heat loss in S. wall in W
+Q7 = (c*f)*U2*T1;// Heat loss in S. window in W
+Q8 = (b*a)*U4*T1;// Heat loss in roof in W
+Q9 = (b*a)*U5*T1;// Heat loss in floor in W
+T_Q_P = Q1 + Q2 + Q3 + Q4 + Q5 + Q6 + Q7 + Q8 + Q9;// in W
+
+//For Office surface
+T3 = 24;// Temperature difference in degree celcius
+T4 = 3;// Temperature difference in degree celcius
+// As Q = U*A*T
+Q_1 = (b*g)*U3*T4;// Heat loss in W. wall in W
+Q_2 = (d*g)*U1*T3;// Heat loss in N. wall in W
+Q_3 = (((b-(2*e))*g) +(e*f))*U1*T3;// Heat loss in E. Wall in W
+Q_4 = (e*f)*U2*T3;// Heat loss in E. window in W
+Q_5 = (e*f)*U2*T3;// Heat loss in E. window in W
+Q_6 = (d*g)*U1*T3;// Heat loss in S. wall in W
+Q_7 = (b*d)*U4*T3;// Heat loss in S. roof in W
+Q_8 = (b*d)*U5*T3;// Heat loss in floor in W
+T_Q_O = Q_1 + Q_2 + Q_3 + Q_4 + Q_5 + Q_6 + Q_7 + Q_8; //in W
+
+T_Q = T_Q_P + T_Q_O;// in W
+printf('Total building fabric loss is %.0f W',T_Q)
diff --git a/2471/CH9/EX9.2/Ex9_2.sce b/2471/CH9/EX9.2/Ex9_2.sce new file mode 100755 index 000000000..4719ebd5d --- /dev/null +++ b/2471/CH9/EX9.2/Ex9_2.sce @@ -0,0 +1,35 @@ +clear ;
+clc;
+// Example 9.2
+printf('Example 9.2\n\n');
+printf('Page No. 255\n\n');
+
+//given
+L_Br = 0.105;// Length of brickwork in m
+L_Bl = 0.100;// Length of blockwork in m
+L_C = 0.05;// Length of cavity in m
+K_Br = 0.84;// Thermal conductivity of brickwork in W/m-K
+K_Bl = 0.22;// Thermal conductivity of blockwork in W/m-K
+K_C_in = 0.033;// Thermal conductivity of insulation in cavity in W/m-K
+R_Ex = 0.055;// Resistance of external surface in W/m^2-K
+
+//As R = L/K
+R_Br = (L_Br/K_Br);// Resistance of brickwork in W/m^2-K
+R_Bl = (L_Bl/K_Bl);// Resistance of blockwork in W/m^2-K
+R_C = 0.18;// Resistance of cavity in W/m^2-K
+
+//Without insulation of cavity
+R_T = 0.938;// Total Resistance in W/m^2-K
+// Thermal transmittance - U = (1/R_T)
+U = (1/R_T);// in W/m^2-K
+printf('The U-value of external wall is %.2f W/sq.m K \n',U)
+
+//With insulation of cavity
+//As R = L/K
+R_C_in = (L_C/K_C_in);// Resistance of insulation in cavity in W/m^2-K
+In = R_C_in - R_C;// Net increase in W/m^2-K
+R_T_New = R_T + In;// New total resistance in W/m^2-K
+// Thermal transmittance - U = (1/R_T)
+U_New = (1/R_T_New);// in W/m^2-K
+printf('The new U-value is with foamed insulation %.3f W/sq.m K',U_New)
+
diff --git a/2471/CH9/EX9.3/Ex9_3.sce b/2471/CH9/EX9.3/Ex9_3.sce new file mode 100755 index 000000000..88726abd9 --- /dev/null +++ b/2471/CH9/EX9.3/Ex9_3.sce @@ -0,0 +1,25 @@ +clear ;
+clc;
+// Example 9.3
+printf('Example 9.3\n\n');
+printf('Page No. 256\n\n');
+
+//given
+N_1 = 1.5;// Ventilation rate in the production area (air changes per hour)
+N_2 = 1.0;// Ventilation rate in the office suite (air changes per hour)
+
+//From example 9.1
+V_P = 6000;// Voulme of production area in m^3
+V_O = 750;// Voulme of office suite in m^3
+T1 = 21;// Temperature difference in degree celcius
+T2 = -3;// Temperature difference in degree celcius
+T_P = 18;// Temperature difference in degree celcius
+F_loss = 74.4*10^3;// Total fabric loss in W
+
+// As Q_vent = 0.33 * N * V * (T1 - T2)
+Q_vent_P = 0.33 * N_1 * V_P * (T_P - T2);// Ventilation loss in production area in W
+Q_vent_O = 0.33 * N_2 * V_O * (T1 - T2);// Ventilation loss in office suite in W
+V_loss = Q_vent_P + Q_vent_O;// Total ventilation loss in W
+T_loss = F_loss + V_loss;// Total heat loss in W
+p = (V_loss/T_loss)*100;
+printf('percentage of ventilation loss is %.0f percent',p)
diff --git a/2471/CH9/EX9.4/Ex9_4.sce b/2471/CH9/EX9.4/Ex9_4.sce new file mode 100755 index 000000000..642773d29 --- /dev/null +++ b/2471/CH9/EX9.4/Ex9_4.sce @@ -0,0 +1,59 @@ +clear ;
+clc;
+// Example 9.4
+printf('Example 9.4\n\n');
+printf('Page No. 260\n\n');
+
+//(a) Design loss
+T1 = 18;// Internal teemperature(specified as an Environmental temperature) in degree celcius
+//From example 9.1
+A = [150 200 40 75 75 200 40 1000 1000];// in m^2
+U = [1.2 1.2 5.6 2 1.2 1.2 5.6 0.7 0.9];// in W/m-K
+Qf = 58.3*10^3;// Fabric loss in production area in W
+T2 = -3;// in degree celcius
+s1 =0;
+s2 = 0;
+for i = [1:1:9]
+ s1 = s1+A(i);
+ s2 = s2+U(i)*A(i);
+end
+A_T = s1;// Total area in m^2
+UA_T = s2;// sum of U*A in W/m-K (answer wrongly calculated in the book)
+
+//From example 9.3
+N_1 = 1.5;// Ventilation rate in the production area (air changes per hour)
+V_P = 6000;// Voulme of production area in m^3
+
+//As Qvent = C * (T1 - T2) & C = 0.33*N*V*(1 + ((UA_T)/(4.8*A_T)))
+C = 0.33*N_1*V_P*(1 + ((UA_T)/(4.8*A_T)));
+Q_vent = C * (T1 - T2);// in W
+T_Q1 = Qf + Q_vent;// in W
+printf('The total design loss is %.0f W \n',T_Q1) // (deviation in answer is due to error in calculation in the book)
+
+//(b) Reduced heat loss
+// The heat transfer coeffieint in this problem has been changed as U1
+U1 = [0.44 0.44 2.8 2 0.44 0.44 2.8 0.44 0.9];//in W/m^2-K
+T = [21 21 21 -3 21 21 21 21 21];// Temperature difference in degree celcius
+s3 = 0;
+s4 = 0;
+for i = [1:1:9]
+ s3 = s3+U1(i)*A(i);
+ s4 = s4+U1(i)*A(i)*T(i);
+end
+U1A_T = s3;// in W/m-k (answer wrongly calculated in the book)
+Q_loss = s4// in W
+
+
+//As Qvent = C * (T1 - T2) & C = 0.33*N*V*(1 + ((UA_T)/(4.8*A_T)))
+C = 0.33*N_1*V_P*(1 + ((U1A_T)/(4.8*A_T)))
+Q_vent = C * (T1 - T2)// in W
+T_Q2 = Q_loss + Q_vent// in W
+
+Red = T_Q1 - T_Q2;// In W
+printf('The reduction in loss is %.0f W',Red) // (deviation in answer is due to error in calculation in the book)
+
+
+
+
+
+
diff --git a/2471/CH9/EX9.5/Ex9_5.sce b/2471/CH9/EX9.5/Ex9_5.sce new file mode 100755 index 000000000..e4078b062 --- /dev/null +++ b/2471/CH9/EX9.5/Ex9_5.sce @@ -0,0 +1,38 @@ +clear ;
+clc;
+// Example 9.5
+printf('Example 9.5\n\n');
+printf('Page No. 265\n\n');
+
+//given
+T = 21;// Temperature difference in degree celcius
+Deg_d = 2186;// Total degree-days base(15.5 deg C) September_April
+T_D = 18;// Design Temperature in degree celcius
+T_O = 4;// base offset temperature in degree celcius
+T_b = T_D - T_O;// Base temperature in degree celcius
+
+// From Table 9.11 Correction factor for base tempratures other than 15.5 deg C is obtained. So for 14 deg c its 0.82
+C = 0.82;// Correction factor
+Do = Deg_d * C// Corrected degree-days
+
+//(a) Original construction
+//from example 9.4
+Q_d_1 = 133.7*10^3;// Design heat loss in W
+
+H_1 = Q_d_1/T;
+//As E = 24 * H * Do - E = Energy consumption in (Wh)
+E1 = (24*H_1 *Do)/10^6;// in 10^6 Wh ( from this step 'Do' is mistakely taken as 1972 inplace of 1792 in the solution of the book, so there is deviation in answer)
+E_1 = (E1 * 3600)*10^6;// in J
+printf('The total energy consumption in original construction is %.0e J \n',E_1)// Deviation in the answer is due to some calculation error as mentioned above
+
+//(b) Improved insulation
+//from example 9.4
+Q_d_2 = 104.4*10^3;// Design heat loss in W
+
+H_2 = Q_d_2/T;
+//As E = 24 * H * Do - E = Energy consumption in (Wh)
+E2 = (24*H_2 *Do)/10^6;// in 10^6 Wh ( from this step 'Do' is mistakely taken as 2972 inplace of 2792 in the solution of the book, so there is deviation in answer)
+E_2 = (E2 * 3600)*10^6;// in J
+printf('The total energy consumption in improved insulation is %.4e J \n',E_2)// Deviation in the answer is due to some calculation error as mentioned above
+
+
diff --git a/2471/CH9/EX9.6/Ex9_6.sce b/2471/CH9/EX9.6/Ex9_6.sce new file mode 100755 index 000000000..2b3a80a06 --- /dev/null +++ b/2471/CH9/EX9.6/Ex9_6.sce @@ -0,0 +1,35 @@ +clear ;
+clc;
+// Example 9.6
+printf('Example 9.6\n\n');
+printf('Page No. 268\n\n');
+
+//given
+U1 = 5.6;// Single glazing heat transfer coefficient in W/m^2_K
+U2 = 2.8;// Double glazing heat transfer coefficient in W/m^2_K
+Ti = 21;// Internal Temperature in degree celcius
+To = -1;// External Temperature in degree celcius
+R_H = 0.5;// Relative humidity
+Rs_i = 0.123;// Surface resistance in (W/m^2-K)^-1
+
+// At 21 Degree celcius and R.H. = 0.5, the dew point is 10.5 degree celcius
+Dew_pt = 10.5;// Dew point in degree celcius
+//As Ts_i = Ti - (Rs_i * U *(Ti - To))
+
+//(a) Single Glazing
+Ts_i_S = Ti - (Rs_i * U1 *(Ti - To));// in degree celcius
+printf('The internal surface temperature for single glazing is %.1f deg C \n',Ts_i_S)
+if (Dew_pt > Ts_i_S) then
+ disp('Surface condensation will occur since it is less than 10.5 deg C.')
+else
+ disp('No surface condensation is expected as it is greater than 10.5 deg C.')
+end
+
+//(b) Double Glazing
+Ts_i_D = Ti - (Rs_i * U2 *(Ti - To));// in degree celcius
+printf('The internal surface temperature for single glazing is %.1f deg C \n',Ts_i_D)
+if (Dew_pt > Ts_i_D) then
+ disp('Surface condensation will occur since it is less than 10.5 deg C.')
+else
+ disp('No surface condensation is expected since it is greater than 10.5 deg C.')
+end
diff --git a/2471/CH9/EX9.7/Ex9_7.sce b/2471/CH9/EX9.7/Ex9_7.sce new file mode 100755 index 000000000..78cfb2b1e --- /dev/null +++ b/2471/CH9/EX9.7/Ex9_7.sce @@ -0,0 +1,33 @@ +clear ;
+clc;
+// Example 9.7
+printf('Example 9.7\n\n');
+printf('Page No. 269\n\n');
+
+//given
+l_1 = 240;// existing length of solid brick in mm
+l_u = 25;// upgraded internal lining in mm
+l_e = 9.5;// Expanded polystyrenne in mm
+T_i = 20;// Internal temperature in degre celcius
+R_H_i = 50;// Internal Relative humidity in percent
+T_e = 0;// External temperature in degre celcius
+R_H_e = 90;// External Relative humidity in percent
+
+K = [0.123 0.059 0.714 0.286 0.055];// Thermal resistance in W/m^2-K
+V_r = [0.0 0.475 3.57 9.60 0.0];// Vapour Resistance in 10^9 N-s/kg
+
+//Refer Figure 9.3
+//From Figure 9.3, the tempeature, dew point, vapour pressure for different interface are obtained
+T = [18.01 17.06 5.51 0.89];// Temperature in degree celcius
+V_p = [1170 1148 986 550];//Vapour pressure in N/m^2
+D_P = [9.5 9.2 7.1 -1.5];// Dew point in degree celcius
+
+h = (T_i - T_e)/sum(K);// in W/m^2
+printf('The heat flow is %.2f W/m^2 \n',h)
+V_p_i = V_p(1);// Internal vapour pressure in N/m^2
+V_p_e = V_p(4);// External vapour pressure in N/m^2
+m = ((V_p_i - V_p_e)/sum(V_r))*10^-9;// in kg/s
+printf('The vapour mass flow is %.1e kg/s',m)
+
+
+
diff --git a/2471/CH9/EX9.8/Ex9_8.sce b/2471/CH9/EX9.8/Ex9_8.sce new file mode 100755 index 000000000..f6a2111a2 --- /dev/null +++ b/2471/CH9/EX9.8/Ex9_8.sce @@ -0,0 +1,41 @@ +clear ;
+clc;
+// Example 9.8
+printf('Example 9.8\n\n');
+printf('Page No. 275\n\n');
+
+//given
+A = 10;// in m^2
+S = 0.77;
+Sa = 0.54;
+//for South
+printf('\t\t\t SOUTH \n')
+I1 = [200 185 165 155 165 185 200];// in W-m^2
+I2 = [500 455 405 385 405 455 500];// in W-m^2
+for i = [1:1:7]
+A_G_S (i) = (A*I1(i)*S) + (A*I2(i)*Sa)
+end
+
+printf('The monthly peak cooling loads for the month March is %.0f W \n',A_G_S(1))
+printf('The monthly peak cooling loads for the month April is %.0f W \n',A_G_S(2))
+printf('The monthly peak cooling loads for the month May is %.0f W \n',A_G_S(3))
+printf('The monthly peak cooling loads for the month June is %.0f W \n',A_G_S(4))
+printf('The monthly peak cooling loads for the month July is %.0f W \n',A_G_S(5))
+printf('The monthly peak cooling loads for the month Aug. is %.0f W \n',A_G_S(6))
+printf('The monthly peak cooling loads for the month Sept. is %.0f W \n\n',A_G_S(7))
+
+ //For east
+ printf('\t\t\t EAST \n')
+I3 = [110 150 180 190 180 150 110];// in W-m^2
+I4 = [435 510 515 505 515 510 435];// in W-m^2
+for j = [1:1:7]
+A_G_E(j) = (A*I3(j)*S) + (A*I4(j)*Sa);
+end
+printf('The monthly peak cooling loads for the month March is %.0f W \n',A_G_E(1))
+printf('The monthly peak cooling loads for the month April is %.0f W \n',A_G_E(2))
+printf('The monthly peak cooling loads for the month May is %.0f W \n',A_G_E(3))
+printf('The monthly peak cooling loads for the month June is %.0f W \n',A_G_E(4))
+printf('The monthly peak cooling loads for the month July is %.0f W \n',A_G_E(5))
+printf('The monthly peak cooling loads for the month Aug. is %.0f W \n',A_G_E(6))
+printf('The monthly peak cooling loads for the month Sept. is %.0f W \n\n',A_G_E(7))
+
diff --git a/2471/CH9/EX9.9/Ex9_9.sce b/2471/CH9/EX9.9/Ex9_9.sce new file mode 100755 index 000000000..893329a44 --- /dev/null +++ b/2471/CH9/EX9.9/Ex9_9.sce @@ -0,0 +1,43 @@ +clear ;
+clc;
+// Example 9.9
+printf('Example 9.9\n\n');
+printf('Page No. 277\n\n');
+
+//given
+A = 15;// glazing area in m^2
+l = 10;// Length of office in m
+h = 6;// height of office in m
+w = 3.5;// width of office in m
+Y_w = 4;// Admittance of wall in W/m^2-K
+Y_f = 3;// Admittance of floor in W/m^2-K
+Y_c = 3;// Admittance of ceiling in W/m^2-K
+N = 1.5;//Ventilation rate (air changes per hour)
+V = l*h*w;// Volume in m^3
+U_G = 5.6;// Transmittance in W/m^2-K
+
+//From table 9.18 and table 9.16
+To = 16.5;// External temperature of June in degree celcius
+T_O = 7.5;// Swing temperature in degre celcius
+I = 155; //Vertical S in W-m^2
+Is = 385;//Vertical S in W-m^2
+S = 0.77;// Solar gain factor
+Sa = 0.54;// Solar gain factor
+
+//As For the mean internal temperature -Ti = To + ((A*I*S)/((0.33*N*V) + (A*U_G)))
+Ti = To + ((A*I*S)/((0.33*N*V) + (A*U_G)));// in degree celcius
+printf('the mean internal temperature is %.1f deg C \n',Ti)
+
+A_G = (A*Is*Sa) + ((A*U_G) + (0.33*N*V))*T_O;// Swing in gain in W
+Net_A = 2*((w*h) + (l*w)) - A;// Net wall area in m^2
+A_f = l*h;// floor area in m^2
+A_c = l*h;//ceiling area in m^2
+A_Y_w = Net_A * Y_w;// Wall AY in W/K
+A_Y_f = A_f * Y_f;// Floor AY in W/K
+A_Y_c = A_c * Y_c;// ceiling AY in W/K
+A_Y_wi = 84;// Window AY in W/K
+Net_AY = A_Y_w + A_Y_f + A_Y_c + A_Y_wi// in W/K
+Ti_s = ((A_G)/((0.33*N*V) + (Net_AY)))// Internal Temperature swing in deg C
+T_p = Ti + Ti_s;// in deg C
+printf('Peak internal temperature is %.1f deg C',T_p) // Deviation in the answer is due to some calculation approximation in the book
+
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