Question

There are 300 occupants dancing in a large ballroom where a condition of 24 °C db and 60% RH is being maintained. Other heat gains are estimated as 371 kW and are all sensible. Conditioned air is supplied to the room through several registers at 14.5 °C db. The (CLF)oc.= 1.0. Calculate the followings:

(a) Total cooling load.

(b) Mass of moisture that must be removed from the ballroom per hour

Answer 1

The problem is solved below. Two assumptions are there that the air which is supplied is having 100% RH and mass flow rate of dry air is taken as 1 kg/s.

First sensible heat is evaluated, then latent heat is evaluated. Total heat is the sum of sensible heat, latent heat and other sensible gains. Mass of water vapour accumulated is calculated for the flow rate of 1 kg/s of dry air.

No. of occupants - 300 Room Condition = 24 c & 600 TH. (0 - 371 kw (etter Sectible Heat Garu) other Total Heating Supply Air = 14.50 (CEF) = 1.0 a) Assuming Supply Air is two%. RH & mass flow Rate of air is 1kg16 Asip - MaQa (AT) (Sensible Heat Den Person) fp = 1*1005x(24-14.5) Sie = 9.5475 KW Q = (No. of People) x (Sendible Heat Pen Person) x (CLP) Q = 300 x 9.547561 Q = 2864.25 KW Qs = Qs + Comer ta Q. - ma -w,) h oad (hatent Heating Load) For &=6or. or ¢=0.6 Q = 1x (0.0115 -0.01011 ) x 2501 = 4 (W) 0.621 627) P = 3.476 xw (1) P= I bar an W = (0.68 0.0191 hy = 2500 KJ 1 Kg (Latent Heat of R = 0.0298 bar at 24°C. vaporizatien fer W2 = 0.0115 kg kg dia (W) = 0.0191 Kgling da. water) -= 2864.25 + 371 + 3.476 For & = 100% d=1 6 = 3238.72 KW. (Total Heating (W) = 0.622 () Load) W, =(,) = 0.01011 Puc = 0.016 bar at 14.50 mass of water vapour (ww.) = 0.0104 kg / kg. da m m (W-w) ma = 1 kgls ( Arsening mass flow rate of dry air to be ligla) m = 1% (0.0015 0-01011) - m. = (-39x10*° Kyls. m = 1:39x10** * 3600 Kylie m = 5 kglm TP-Tus Pelbar bir