InterviewSolution
Saved Bookmarks
| 1. |
A room has a window fixed with a pane of area `1.2m^(s)` The glass has thickness `2.2mm` If the temperature outside the room is `36^(@)C` and the temperature inside is `26^(@)C` (a) calculate the heat flowing into the room every hour (b) If the same single pane window is replaced by double paned window with an air gap of `0.50cm` between the two panes calculate the heat flowing into the room every hour `K_(g) =0.80 Wm^(-1)K^(-1),K_(air)=0.0234Wm^(-1)K^(-1)` . |
|
Answer» We assume the one side of the pane is at `36^(@)C` and the other side (inside the room) is at `26^(@)C` Given Thickness of the window pane `d = 2.2mm = 2.2 xx 10^(-3)m` Area of window pane ` A =1.2m^(2)` `(Q)/(t)=K(A(theta_(2)-theta_(1)))/(d)= (0.8xx1.2xx10)/(2.2xx10^(-3)) =4364J//s` Therefore heat flown into the room per hour is `Q = 4364 xx 3600 =1.57 xx 10^(2)J` (b) When single pane widow is replaced by a duble paned widow we have two layers of glss and one layer of air between them Thermal resistacne for glass `R_(a) =(d_(a))/(K_(a)A_(a)) = (0.5xx10^(-2))/(0.0234xx1.2) =178xx10^(3)K//W` Net thermal resistance `R_(T) =R_(g) +R_(a)+R_(g)` ` =(2.29 xx 10^(-3)+178 xx 10^(-3)+2.29 xx 10^(-3))` `=182.6 xx 10^(-3)K//W` `P=(Deltatheta)/(R_(T)) = (10)/(182.6xx10^(-3)) =55J//5` Therefore heat flown into the room per hour is `Q =P xx t =1.98 xx 10^(5) J` . |
|