{ "metadata": { "name": "", "signature": "sha256:ecf6c25d050cb67c2d6108384bdb4d395e1d33e1a117952976bf67a230d01af6" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter9-Building construction\n" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex1-pg252" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "## Example 9.1\n", "print('Example 9.1\\n\\n');\n", "print('Page No. 252\\n\\n');\n", "\n", "##given\n", "a = 40.;## in m\n", "b = 25.;## in m\n", "c = 20.;## in m\n", "d = 10.;## in m\n", "e = 5.;## in m\n", "f = 2.;## in m\n", "g = 3.;## in m\n", "h = 6.;## in m\n", "\n", "##(1) Production Area\n", "T1 = 21.;## Temperature difference in degree celcius\n", "T2 = -3.;## Temperature difference in degree celcius\n", "U1 = 1.2;## heat transfer coefficent in W/m-K\n", "U2 = 5.6;## heat transfer coefficent in W/m-K\n", "U3 = 2.0;## heat transfer coefficent in W/m-K\n", "U4 = 0.7;## heat transfer coefficent in W/m-K\n", "U5 = 0.9;## heat transfer coefficent in W/m-K\n", "## As Q = U*A*T\n", "Q1 = (b*h)*U1*T1;## Heat loss in W. wall in W\n", "Q2 = (((a-c)*h) + (d*h) + (d*f))*U1*T1;## Heat loss in N. wall in W\n", "Q3 = (c*f)*U2*T1;## Heat loss in N. window in W\n", "Q4 = (b*g)*U3*T2;## Heat loss in N. wall/internal in W\n", "Q5 = (b*g)*U1*T1;## Heat loss in E. wall/external in W\n", "Q6 = (((a-c)*h) + (d*h) + (d*f))*U1*T1;## Heat loss in S. wall in W\n", "Q7 = (c*f)*U2*T1;## Heat loss in S. window in W\n", "Q8 = (b*a)*U4*T1;## Heat loss in roof in W\n", "Q9 = (b*a)*U5*T1;## Heat loss in floor in W\n", "T_Q_P = Q1 + Q2 + Q3 + Q4 + Q5 + Q6 + Q7 + Q8 + Q9;## in W\n", "\n", "##For Office surface\n", "T3 = 24;## Temperature difference in degree celcius\n", "T4 = 3;## Temperature difference in degree celcius\n", "## As Q = U*A*T\n", "Q_1 = (b*g)*U3*T4;## Heat loss in W. wall in W\n", "Q_2 = (d*g)*U1*T3;## Heat loss in N. wall in W\n", "Q_3 = (((b-(2*e))*g) +(e*f))*U1*T3;## Heat loss in E. Wall in W\n", "Q_4 = (e*f)*U2*T3;## Heat loss in E. window in W\n", "Q_5 = (e*f)*U2*T3;## Heat loss in E. window in W\n", "Q_6 = (d*g)*U1*T3;## Heat loss in S. wall in W\n", "Q_7 = (b*d)*U4*T3;## Heat loss in S. roof in W\n", "Q_8 = (b*d)*U5*T3;## Heat loss in floor in W\n", "T_Q_O = Q_1 + Q_2 + Q_3 + Q_4 + Q_5 + Q_6 + Q_7 + Q_8; ##in W\n", "\n", "T_Q = T_Q_P + T_Q_O;## in W\n", "print'%s %.2f %s'%('Total building fabric loss is ',T_Q,' W')\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Example 9.1\n", "\n", "\n", "Page No. 252\n", "\n", "\n", "Total building fabric loss is 74358.00 W\n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex2-pg255" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "## Example 9.2\n", "print('Example 9.2\\n\\n');\n", "print('Page No. 255\\n\\n');\n", "\n", "##given\n", "L_Br = 0.105;## Length of brickwork in m\n", "L_Bl = 0.100;## Length of blockwork in m\n", "L_C = 0.05;## Length of cavity in m\n", "K_Br = 0.84;## Thermal conductivity of brickwork in W/m-K\n", "K_Bl = 0.22;## Thermal conductivity of blockwork in W/m-K\n", "K_C_in = 0.033;## Thermal conductivity of insulation in cavity in W/m-K\n", "R_Ex = 0.055;## Resistance of external surface in W/m^2-K\n", "\n", "##As R = L/K\n", "R_Br = (L_Br/K_Br);## Resistance of brickwork in W/m^2-K\n", "R_Bl = (L_Bl/K_Bl);## Resistance of blockwork in W/m^2-K\n", "R_C = 0.18;## Resistance of cavity in W/m^2-K\n", "\n", "##Without insulation of cavity\n", "R_T = 0.938;## Total Resistance in W/m^2-K\n", "## Thermal transmittance - U = (1/R_T)\n", "U = (1./R_T);## in W/m^2-K\n", "print'%s %.2f %s'%('The U-value of external wall is ',U,' W/sq.m K \\n')\n", "\n", "##With insulation of cavity\n", "##As R = L/K\n", "R_C_in = (L_C/K_C_in);## Resistance of insulation in cavity in W/m^2-K\n", "In = R_C_in - R_C;## Net increase in W/m^2-K\n", "R_T_New = R_T + In;## New total resistance in W/m^2-K\n", "## Thermal transmittance - U = (1/R_T)\n", "U_New = (1./R_T_New);## in W/m^2-K\n", "print'%s %.2f %s'%('The new U-value is with foamed insulation ',U_New,' W/sq.m K')\n", " \n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Example 9.2\n", "\n", "\n", "Page No. 255\n", "\n", "\n", "The U-value of external wall is 1.07 W/sq.m K \n", "\n", "The new U-value is with foamed insulation 0.44 W/sq.m K\n" ] } ], "prompt_number": 2 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex3-pg256" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "## Example 9.3\n", "print('Example 9.3\\n\\n');\n", "print('Page No. 256\\n\\n');\n", "\n", "##given\n", "N_1 = 1.5;## Ventilation rate in the production area (air changes per hour)\n", "N_2 = 1.0;## Ventilation rate in the office suite (air changes per hour)\n", "\n", "##From example 9.1\n", "V_P = 6000.;## Voulme of production area in m^3\n", "V_O = 750.;## Voulme of office suite in m^3\n", "T1 = 21.;## Temperature difference in degree celcius\n", "T2 = -3.;## Temperature difference in degree celcius\n", "T_P = 18.;## Temperature difference in degree celcius\n", "F_loss = 74.4*10**3;## Total fabric loss in W\n", "\n", "## As Q_vent = 0.33 * N * V * (T1 - T2)\n", "Q_vent_P = 0.33 * N_1 * V_P * (T_P - T2);## Ventilation loss in production area in W\n", "Q_vent_O = 0.33 * N_2 * V_O * (T1 - T2);## Ventilation loss in office suite in W\n", "V_loss = Q_vent_P + Q_vent_O;## Total ventilation loss in W\n", "T_loss = F_loss + V_loss;## Total heat loss in W\n", "p = (V_loss/T_loss)*100.;\n", "print'%s %.2f %s'%('percentage of ventilation loss is ',p,' percent',)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Example 9.3\n", "\n", "\n", "Page No. 256\n", "\n", "\n", "percentage of ventilation loss is 47.87 percent\n" ] } ], "prompt_number": 3 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex4-pg260" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "import numpy\n", "## Example 9.4\n", "print('Example 9.4\\n\\n');\n", "print('Page No. 260\\n\\n');\n", "\n", "##(a) Design loss\n", "T1 = 18.;## Internal teemperature(specified as an Environmental temperature) in degree celcius\n", "##From example 9.1\n", "A = numpy.array([150, 200 ,40 ,75 ,75 ,200 ,40 ,1000, 1000]);## in m^2\n", "U = numpy.array([1.2, 1.2 ,5.6, 2, 1.2 ,1.2 ,5.6 ,0.7 ,0.9]);## in W/m-K\n", "Qf = 58.3*10**3;## Fabric loss in production area in W\n", "T2 = -3.;## in degree celcius\n", "s1 =0.;\n", "s2 = 0.;\n", "for i in range (0,9):\n", " s1 = s1+A[i];\n", " s2 = s2+U[i]*A[i];\n", "\n", "A_T = s1;## Total area in m^2\n", "UA_T = s2;## sum of U*A in W/m-K (answer wrongly calculated in the book)\n", "\n", "##From example 9.3\n", "N_1 = 1.5;## Ventilation rate in the production area (air changes per hour)\n", "V_P = 6000;## Voulme of production area in m^3\n", "\n", "##As Qvent = C * (T1 - T2) & C = 0.33*N*V*(1 + ((UA_T)/(4.8*A_T)))\n", "C = 0.33*N_1*V_P*(1. + ((UA_T)/(4.8*A_T)));\n", "Q_vent = C * (T1 - T2);## in W\n", "T_Q1 = Qf + Q_vent;## in W\n", "print'%s %.2f %s'%('The total design loss is ',T_Q1,' W \\n') ## (deviation in answer is due to error in calculation in the book)\n", "\n", "##(b) Reduced heat loss\n", "## The heat transfer coeffieint in this problem has been changed as U1\n", "U1 = [0.44, 0.44, 2.8, 2, 0.44,0.44, 2.8, 0.44, 0.9];##in W/m^2-K\n", "T = [21 ,21 ,21 ,-3 ,21 ,21 ,21 ,21 ,21];## Temperature difference in degree celcius\n", "s3 = 0;\n", "s4 = 0;\n", "for i in range (0,9):\n", "\ts3 = s3+U1[i]*A[i];\n", "\ts4 = s4+U1[i]*A[i]*T[i];\n", "U1A_T = s3;## in W/m-k (answer wrongly calculated in the book)\n", "Q_loss = s4## in W\n", "\n", "\n", "##As Qvent = C * (T1 - T2) & C = 0.33*N*V*(1 + ((UA_T)/(4.8*A_T)))\n", "C = 0.33*N_1*V_P*(1 + ((U1A_T)/(4.8*A_T)))\n", "Q_vent = C * (T1 - T2)## in W\n", "T_Q2 = Q_loss + Q_vent## in W\n", "\n", "Red = T_Q1 - T_Q2;## In W\n", "print'%s %.2f %s'%('The reduction in loss is ',Red,' W') ## (deviation in answer is due to error in calculation in the book)\n", "\n", "\n", "\n", "\n", "\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Example 9.4\n", "\n", "\n", "Page No. 260\n", "\n", "\n", "The total design loss is 134448.98 W \n", "\n", "The reduction in loss is 24613.38 W\n" ] } ], "prompt_number": 4 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex5-pg265" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "## Example 9.5\n", "print('Example 9.5\\n\\n');\n", "print('Page No. 265\\n\\n');\n", "\n", "##given\n", "T = 21.;## Temperature difference in degree celcius\n", "Deg_d = 2186.;## Total degree-days base(15.5 deg C) September_April\n", "T_D = 18.;## Design Temperature in degree celcius\n", "T_O = 4.;## base offset temperature in degree celcius\n", "T_b = T_D - T_O;## Base temperature in degree celcius\n", "\n", "## 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\n", "C = 0.82;## Correction factor\n", "Do = Deg_d * C## Corrected degree-days \n", "\n", "##(a) Original construction\n", "##from example 9.4\n", "Q_d_1 = 133.7*10**3;## Design heat loss in W\n", "\n", "H_1 = Q_d_1/T;\n", "##As E = 24 * H * Do - E = Energy consumption in (Wh)\n", "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)\n", "E_1 = (E1 * 3600.)*10**6;## in J\n", "print'%s %.2e %s'%('The total energy consumption in original construction is ',E_1,' J \\n')## Deviation in the answer is due to some calculation error as mentioned above\n", "\n", "##(b) Improved insulation\n", "##from example 9.4\n", "Q_d_2 = 104.4*10**3;## Design heat loss in W\n", "\n", "H_2 = Q_d_2/T;\n", "##As E = 24 * H * Do - E = Energy consumption in (Wh)\n", "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)\n", "E_2 = (E2 * 3600.)*10**6;## in J\n", "print'%s %.2e%s'%('The total energy consumption in improved insulation is ',E_2,' J \\n')## Deviation in the answer is due to some calculation error as mentioned above\n", "\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Example 9.5\n", "\n", "\n", "Page No. 265\n", "\n", "\n", "The total energy consumption in original construction is 9.86e+11 J \n", "\n", "The total energy consumption in improved insulation is 7.70e+11 J \n", "\n" ] } ], "prompt_number": 5 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex6-pg268" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "## Example 9.6\n", "print('Example 9.6\\n\\n');\n", "print('Page No. 268\\n\\n');\n", "\n", "##given\n", "U1 = 5.6;## Single glazing heat transfer coefficient in W/m^2_K\n", "U2 = 2.8;## Double glazing heat transfer coefficient in W/m^2_K\n", "Ti = 21.;## Internal Temperature in degree celcius\n", "To = -1.;## External Temperature in degree celcius\n", "R_H = 0.5;## Relative humidity\n", "Rs_i = 0.123;## Surface resistance in (W/m^2-K)^-1\n", "\n", "## At 21 Degree celcius and R.H. = 0.5, the dew point is 10.5 degree celcius\n", "Dew_pt = 10.5;## Dew point in degree celcius\n", "##As Ts_i = Ti - (Rs_i * U *(Ti - To))\n", "\n", "##(a) Single Glazing\n", "Ts_i_S = Ti - (Rs_i * U1 *(Ti - To));## in degree celcius\n", "print'%s %.2f %s'%('The internal surface temperature for single glazing is ',Ts_i_S,' deg C \\n')\n", "if (Dew_pt > Ts_i_S):\n", "\tprint('Surface condensation will occur since it is less than 10.5 deg C.')\n", "else:\n", " print('No surface condensation is expected as it is greater than 10.5 deg C.')\n", "\n", "\n", "##(b) Double Glazing\n", "Ts_i_D = Ti - (Rs_i * U2 *(Ti - To));## in degree celcius\n", "print'%s %.2f %s'%('The internal surface temperature for single glazing is ',Ts_i_D,' deg C \\n')\n", "if (Dew_pt > Ts_i_D):\n", "\tprint('Surface condensation will occur since it is less than 10.5 deg C.')\n", "else:\n", "\tprint('No surface condensation is expected since it is greater than 10.5 deg C.')\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Example 9.6\n", "\n", "\n", "Page No. 268\n", "\n", "\n", "The internal surface temperature for single glazing is 5.85 deg C \n", "\n", "Surface condensation will occur since it is less than 10.5 deg C.\n", "The internal surface temperature for single glazing is 13.42 deg C \n", "\n", "No surface condensation is expected since it is greater than 10.5 deg C.\n" ] } ], "prompt_number": 6 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex7-pg269" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "## Example 9.7\n", "print('Example 9.7\\n\\n');\n", "print('Page No. 269\\n\\n');\n", "\n", "##given\n", "l_1 = 240.;## existing length of solid brick in mm\n", "l_u = 25.;## upgraded internal lining in mm\n", "l_e = 9.5;## Expanded polystyrenne in mm\n", "T_i = 20.;## Internal temperature in degre celcius\n", "R_H_i = 50.;## Internal Relative humidity in percent\n", "T_e = 0.;## External temperature in degre celcius\n", "R_H_e = 90.;## External Relative humidity in percent\n", "\n", "K = ([0.123, 0.059 ,0.714, 0.286, 0.055]);## Thermal resistance in W/m^2-K\n", "V_r = ([0.0, 0.475 ,3.57 ,9.60 ,0.0]);## Vapour Resistance in 10^9 N-s/kg\n", "\n", "##Refer Figure 9.3\n", "##From Figure 9.3, the tempeature, dew point, vapour pressure for different interface are obtained\n", "T = ([18.01, 17.06, 5.51 ,0.89]);## Temperature in degree celcius\n", "V_p = ([1170, 1148 ,986 ,550]);##Vapour pressure in N/m^2\n", "D_P = ([9.5, 9.2, 7.1, -1.5]);## Dew point in degree celcius\n", "\n", "h = (T_i - T_e)/sum (K);## in W/m^2\n", "print'%s %.2f %s'%('The heat flow is ',h,' W/m^2 \\n')\n", "V_p_i = V_p[0];## Internal vapour pressure in N/m^2\n", "V_p_e = V_p[3];## External vapour pressure in N/m^2\n", "m = ((V_p_i - V_p_e)/sum(V_r))*10**-9;## in kg/s\n", "print'%s %.2e %s'%('The vapour mass flow is ',m,' kg/s')\n", "\n", "\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Example 9.7\n", "\n", "\n", "Page No. 269\n", "\n", "\n", "The heat flow is 16.17 W/m^2 \n", "\n", "The vapour mass flow is 4.54e-08 kg/s\n" ] } ], "prompt_number": 7 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex8-pg275" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "import numpy\n", "## Example 9.8\n", "print('Example 9.8\\n\\n');\n", "print('Page No. 275\\n\\n');\n", "\n", "A_G_S=numpy.zeros\n", "##given\n", "A = 10.;## in m^2\n", "S = 0.77;\n", "Sa = 0.54;\n", "##for South\n", "print('\\t\\t\\t SOUTH \\n')\n", "I1 = ([200 ,185 ,165 ,155 ,165 ,185 ,200]);## in W-m^2\n", "I2 = ([500, 455 ,405 ,385, 405 ,455, 500]);## in W-m^2\n", "A_G_S0 = (A*200*S) + (A*500*Sa)\n", "A_G_S1 = (A*185*S) + (A*455*Sa)\n", "A_G_S2 = (A*165*S) + (A*405*Sa)\n", "A_G_S3 = (A*155*S) + (A*385*Sa)\n", "A_G_S4 = (A*165*S) + (A*405*Sa)\n", "A_G_S5 = (A*185*S) + (A*455*Sa)\n", "A_G_S6 = (A*200*S) + (A*500*Sa)\n", "\n", "\n", "print'%s %.2f %s'%('The monthly peak cooling loads for the month April is ',A_G_S0,' W \\n')\n", "print'%s %.2f %s'%('The monthly peak cooling loads for the month May is ',A_G_S1,' W \\n')\n", "print'%s %.2f %s'%('The monthly peak cooling loads for the month June is ',A_G_S2,' W \\n')\n", "print'%s %.2f %s'%('The monthly peak cooling loads for the month July is ',A_G_S3,' W \\n')\n", "print'%s %.2f %s'%('The monthly peak cooling loads for the month Aug. is ',A_G_S4,' W \\n')\n", "print'%s %.2f %s'%('The monthly peak cooling loads for the month Sept. is ',A_G_S5,' W \\n\\n')\n", "\n", " ##For east\n", "print('\\t\\t\\t EAST \\n')\n", "I3 = ([110, 150, 180 ,190 ,180 ,150 ,110]);## in W-m^2\n", "I4 = ([435 ,510, 515, 505, 515 ,510 ,435]);## in W-m^2\n", "A_G_E0 = (A*110*S) + (A*435*Sa)\n", "A_G_E1 = (A*150*S) + (A*510*Sa)\n", "A_G_E2 = (A*180*S) + (A*515*Sa)\n", "A_G_E3 = (A*190*S) + (A*505*Sa)\n", "A_G_E4 = (A*180*S) + (A*515*Sa)\n", "A_G_E5 = (A*150*S) + (A*510*Sa)\n", "A_G_E6 = (A*110*S) + (A*435*Sa)\n", "\n", "\n", "\n", "print'%s %.2f %s'%('The monthly peak cooling loads for the month March is ',A_G_E0,' W \\n')\n", "print'%s %.2f %s'%('The monthly peak cooling loads for the month April is ',A_G_E1,' W \\n')\n", "print'%s %.2f %s'%('The monthly peak cooling loads for the month May is ',A_G_E2,' W \\n')\n", "print'%s %.2f %s'%('The monthly peak cooling loads for the month June is ',A_G_E3,' W \\n')\n", "print'%s %.2f %s'%('The monthly peak cooling loads for the month July is ',A_G_E4,' W \\n')\n", "print'%s %.2f %s'%('The monthly peak cooling loads for the month Aug. is ',A_G_E5,' W \\n')\n", "print'%s %.2f %s'%('The monthly peak cooling loads for the month Sept. is ',A_G_E6,' W \\n\\n')\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Example 9.8\n", "\n", "\n", "Page No. 275\n", "\n", "\n", "\t\t\t SOUTH \n", "\n", "The monthly peak cooling loads for the month April is 4240.00 W \n", "\n", "The monthly peak cooling loads for the month May is 3881.50 W \n", "\n", "The monthly peak cooling loads for the month June is 3457.50 W \n", "\n", "The monthly peak cooling loads for the month July is 3272.50 W \n", "\n", "The monthly peak cooling loads for the month Aug. is 3457.50 W \n", "\n", "The monthly peak cooling loads for the month Sept. is 3881.50 W \n", "\n", "\n", "\t\t\t EAST \n", "\n", "The monthly peak cooling loads for the month March is 3196.00 W \n", "\n", "The monthly peak cooling loads for the month April is 3909.00 W \n", "\n", "The monthly peak cooling loads for the month May is 4167.00 W \n", "\n", "The monthly peak cooling loads for the month June is 4190.00 W \n", "\n", "The monthly peak cooling loads for the month July is 4167.00 W \n", "\n", "The monthly peak cooling loads for the month Aug. is 3909.00 W \n", "\n", "The monthly peak cooling loads for the month Sept. is 3196.00 W \n", "\n", "\n" ] } ], "prompt_number": 7 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex9-pg277" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "## Example 9.9\n", "print('Example 9.9\\n\\n');\n", "print('Page No. 277\\n\\n');\n", "\n", "##given\n", "A = 15.;## glazing area in m^2\n", "l = 10.;## Length of office in m\n", "h = 6.;## height of office in m\n", "w = 3.5;## width of office in m\n", "Y_w = 4.;## Admittance of wall in W/m^2-K\n", "Y_f = 3.;## Admittance of floor in W/m^2-K\n", "Y_c = 3.;## Admittance of ceiling in W/m^2-K\n", "N = 1.5;##Ventilation rate (air changes per hour)\n", "V = l*h*w;## Volume in m^3\n", "U_G = 5.6;## Transmittance in W/m^2-K\n", "\n", "##From table 9.18 and table 9.16\n", "To = 16.5;## External temperature of June in degree celcius\n", "T_O = 7.5;## Swing temperature in degre celcius\n", "I = 155.; ##Vertical S in W-m^2\n", "Is = 385.;##Vertical S in W-m^2\n", "S = 0.77;## Solar gain factor\n", "Sa = 0.54;## Solar gain factor\n", "\n", "##As For the mean internal temperature -Ti = To + ((A*I*S)/((0.33*N*V) + (A*U_G))) \n", "Ti = To + ((A*I*S)/((0.33*N*V) + (A*U_G)));## in degree celcius\n", "print'%s %.2f %s'%('the mean internal temperature is ',Ti,' deg C \\n')\n", "\n", "A_G = (A*Is*Sa) + ((A*U_G) + (0.33*N*V))*T_O;## Swing in gain in W\n", "Net_A = 2*((w*h) + (l*w)) - A;## Net wall area in m^2\n", "A_f = l*h;## floor area in m^2\n", "A_c = l*h;##ceiling area in m^2\n", "A_Y_w = Net_A * Y_w;## Wall AY in W/K\n", "A_Y_f = A_f * Y_f;## Floor AY in W/K \n", "A_Y_c = A_c * Y_c;## ceiling AY in W/K \n", "A_Y_wi = 84.;## Window AY in W/K\n", "Net_AY = A_Y_w + A_Y_f + A_Y_c + A_Y_wi## in W/K\n", "Ti_s = ((A_G)/((0.33*N*V) + (Net_AY)))## Internal Temperature swing in deg C\n", "T_p = Ti + Ti_s;## in deg C\n", "print'%s %.2f %s'%('Peak internal temperature is ',T_p,' deg C') ## Deviation in the answer is due to some calculation approximation in the book\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Example 9.9\n", "\n", "\n", "Page No. 277\n", "\n", "\n", "the mean internal temperature is 26.03 deg C \n", "\n", "Peak internal temperature is 30.86 deg C\n" ] } ], "prompt_number": 11 } ], "metadata": {} } ] }