Respuesta :
Answer:
a) v = 825 ft^3/min
b) T = 64.005 F
Explanation:
given data:
pressure ar airc conditioning system is 15 psia
volume flow rate is \dot V = 450 ft^3/min
duct diameter is D 10 inch
Q_{in} = 2 Btu /s
Inlet temp =50 F = 50 +460 = 510 R
From air table
gas constant of air R = 0.3704 psia ft^3/lbm R
specific heat constant [tex]c_p = 0.24 Btu/ lbm R[/tex]
velocity at inlet is [tex]= \frac{\dot V}{A} [/tex]
[tex]V_1 = \frac{\dot V}{\pi r^2}[/tex]
[tex]V_1 = \frac{450}{[\pi [\frac{5}{12}]^2} = 825 ft^3/min[/tex]
b) by ideal gas equation
[tex]PV =\dot RT[/tex]
[tex]15\times 450 = \dot m 0.3704 \times 510[/tex]
[tex]dot m = 35.73 lbm/min[/tex]
from energy equation we have
[tex]\dot m h_1 + Q =\dot m h_2[/tex]
[tex]h_2 = h_1 +\frac{Q}{\dot m}[/tex]
[tex]Cp T_2 = Cp T_1 +\frac{Q}{\dot m}[/tex]
[tex]T_2 =T_1 + \frac{Q}{\dot m Cp}[/tex]
[tex]= 510 + \frac{2}{0.595 \times 0.24}[/tex]
[tex]T_2 = 524.005 R = 64.005 F[/tex]
