Air conditioning systems in large building are required to bring in 20 cubic fee

Question

Air conditioning systems in large building are required to bring in 20 cubic feet per minute (cfm) of outdoor air per person, according to ASHRAE Standard 62. In a particular case, the heating, ventilating and air conditioning (HVAC) system for a building is being designed for a capacity of N_p = 200 persons. Outdoor air at T_1 = 92 degree F and atmospheric pressure is brought into the building through a D_1 = 2.5 ft diameter circular duct and mixed with chilled air at T_2 = 55 degree F and atmospheric pressure that is flowing through a D_2 = 4 ft diameter duct at velocity v_2= 600 ft/min. The mixing box is shown in Figure 1. The mixture is discharged at atmospheric pressure and velocity v_out = 330 ft/min through 4 identical exit ducts (this configuration reduces duct noise). Assume that the mixing process to be adiabatic and steady-state and neglect the humidity in the air. Do not neglect the kinetic energy of the streams. What is the temperature of the discharge, mixed air in Fahrenheit? Determine the necessary diameter of the four discharge ducts in inches.

Explanation / Answer

solution:

1)here are two air stream one is relatively hot at temperature of T1=92 F or 33.333 C and other is cold at temperature of T2=55 F or 12.7778 C and it will logically gives stream of air at temperature in between this two temperature limit based on enthalphy and kinetic energy content of two streams.

2)here first we solved for case b

first stream flow rate=Q1=20*200=4000 ft3/min or 1.858 m3/s

second air stream=Q2=A2*V2=(pi/4)D2^2*V2=(pi/4)*4^2*600=7539.8223 ft3/min or 3.558 m3/s

so resultant flow rate by conservation of mass we can write as for flow rate as

Qr=Q1+Q2

where

Qr=4000+7539.8223=11539.8223 ft3/min

where there are four identicle exits so we have to devide it in four stream of equal flow rate

Qr'=Qr/4=2884.95 ft3/min

Qr'=(pi/4)*d^2*Vout

2884.95=(pi/4)*d^2*330

d=3.3363 ft or 40.0356 ''

hence diameter of exit openings are d=40.0356 ''

3)for case a now

here for staedy flow process equation is

Q'-W'=m(dH+dKE+dPE)

as here no heat,work is supplied so we removed those terms as well PE change is negligible so we can write equation as

m(dh+dKE)=0

here are two air stream which forms single output so we can write as

m1*h1+m1*(v1^2/2)+m2*h2+m2*(v2^2/2)=mr[hr+4*(Vout^2/2)]

as there are four air stream at exit

h=cp*T

here density for stream 1 and 2 are given by density=pressure/(287*temperature)

density1=1.1524 kg/m3

density2=1.235 kg/m3

where velocity of first stream as

Q1=(pi/4)*D1^2*v1

4000=(pi/4)*2.5^2*v1

v1=814.87 ft/min or 4.14 m/s

for second air stream as

v2=600 ft/min or 3.05 m/s

vout=330 ft/min or 1.68 m/s

Cp=1.005

T1=273+33.333

T2=273+12.7778

we get m1=density1*Q1=2.1413 kg/s

m2=density2*Q2=4.3941 kg/s

mr=m1+m2=6.5354 kg/s

m1*h1+m1*(v1^2/2)+m2*h2+m2*(v2^2/2)=mr[hr+4*(Vout^2/2)]

m1*Cp*T1+m1*(v1^2/2)+m2*Cp*T2+m2*(v2^2/2)=mr[Cp*Tout+4*(Vout^2/2)]

2.1413*1.005*(273+33.33)+2.1413*(4.14^2/2)+4.3941*1.005*(273+12.7778)+4.3941*(3.05^2/2)=6.5354*(1.005*Tout+4*(1.68^2/2))

Tout=292.80 K or 67.64 F

hence temperature of output stream is

Tout=67.64 F

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