00:01
Hello students, in this question there is an office located at the top floor of an air conditioned building.
00:07
So this is the office location at the top floor.
00:12
And we are given the floor ceiling height is h is equal to 3 .8 meters.
00:20
Area of the floor is equal to 441 meter square.
00:25
Facade area is equal to 0 .21 times area.
00:31
Glazed, this is the facade area.
00:38
Now we can start by finding the sensible heat due to infiltration.
00:48
So we can first calculate the air density.
00:52
So air density, air density rho is equal to p atmospheric by specific gas constant c times t temperature of the outdoor.
01:08
Okay, so temperature of the outdoor is is we know the value.
01:17
So first p atmosphere is equal to 101 .325 kilo pascal divided by c, the specific heat.
01:28
So specific heat.
01:32
So that is equal to 101 .325 times 10 to the power 3 is there right kilo pascal divided by 287, which is the specific heat of air times temperature is 37 plus 273 .15 in kelvin.
01:51
So density will be equal to density will be equal to what density will be equal to this.
02:02
So delta right.
02:04
So delta delta t.
02:05
So delta t will be equal to delta t will be equal to 37 minus 24.
02:17
That is 13 kelvin.
02:19
So sensible infiltration sensible infiltration will be equal to 0 .75.
02:27
So this is approximately equal to 1 .13.
02:31
Okay, 0 .75 times 441 times 1 .13 times 1005 times 13.
02:43
So that is equal to 382, 417 .65 watts...