Window Design 5.2 Answers

CALCULATION 5.1 The data for EAHE system are taken as below for selected space which has to cooled:-
Parameter Symbol Value Unit
Length L 2.555 m
Height H 2.403 m
Width W 2.428 m
Thickness of Wall 1 t1 0.312 m
Thickness of Wall 2 t2 0.25 m
Thickness of Wall 3 t3 0.25 m
Thickness of Wall 4 t4 0.25 m
Thickness of plaster tp 0.0125 m
Thickness of brick (For Outside wall) tb 0.1 m
Thickness of concrete (For Outside wall) tco 0.2 m
Thickness of concrete (For inside wall) tci 0.14 m
Thickness of thermocol sheet tth 0.01 m
Area of wall 1 A1 3.413 m2
Area of wall 2 A2 6.203 m2
Area of wall 3 A3 5.194 m2
Area of wall 4 A4 3.677 m2
Area of ceiling A5 6.201 m2
Area of Door D 2.0047 m2
Area of Window 1 W1 2.726 m2
Area of Window 2 W2 0.944 m2
Ambient
below ground level T2 20-25 oC
Thermal conductivity of glass Kg 0.78 W/moC
Thermal conductivity of concrete Kc 1.73 W/moC
Thermal conductivity of brick Kb 1.32 W/moC
Thermal conductivity of plaster Kp 8.65 W/moC
Thermal conductivity of thermocol Kth 0.02 W/m2oC
Thermal conductivity of aluminum Kalu 250 W/m2oC
Heat Transfer Co-efficient for Glass Ug 4.59 W/m2oC
Convective factor Fc 1.42 -
Radiative factor Fr 1.58 -
Film co-efficient of outer side Fo 23 W/m2oC
Film co-efficient of inner side Fi 7 W/m2oC
No. of occupants N 2 -
Sensible heat factor Rs 53
For Light Colour :- ∆te = ∆tes + 0.78(Rs/Rm)(∆tem - ∆tes) = (32) + 0.78(1.2)(0) = 32 ℃

5.3.3 Heat Transfer through Partition wall(Wall 3) (Qp3):- Qp3 = UpA3[(TETD)P (Fc) + (TETD)A(FR)] = (2.923)(5.194)[(32)(1.42) + (32)(1.58)] = 1.457 kW 5.3.4 Heat Transfer through Partition wall(Wall 4) (Qp4):- Qp4 = UpA4[(TETD)P (Fc) + (TETD)A(FR)] = (2.923)(3.677)[(32)(1.42) + (32)(1.58)] = 1.031 kW 5.3.5 Heat Transfer through Thermocol Sheets (Qth):- Qth = UthA5∆t = (1.459)(9.152)(2) = 0.026 kW 5.3.6 Heat Transfer through Glass (Qg):- Qg = UgW1∆t = (4.59)(2.726)(5) = 0.062 kW

5.3.7 Heat Transfer through Internal People (Qpeople):- 5.3.7.1 Sensible Heat Transfer (QS) QS = (No. of occupants)*(Sensible heat gain