Answer:
Z= 0.868
Explanation:
Given that
P= 3 MPa
T = 160 K
We know that
P v= Z R T
P= Pressure
v = specific volume
R= gas constant
T = Absolute temperature
Z= Compressibility factor
Here specific volume of gas is not given so we assume that specific volume gas
We know that for oxygen gas constant
R = 0.259 KJ/kg.K
Now by putting the values
P v = Z R T
3000 x 0.012 = Z x 0.259 x 160
Z= 0.868
So Compressibility factor is 0.868.
Answer:
A) EXIT TEMPERATURE = 14⁰C
b) rate of heat transfer of air = - 13475.78 = - 13.5 kw
Explanation:
Given data :
diameter of duct = 20-cm = 0.2 m
length of duct = 12-m
temperature of air at inlet= 50⁰c
pressure = 1 atm
mean velocity = 7 m/s
average heat transfer coefficient = 85 w/m^2⁰c
water temperature = 5⁰c
surface temperature ( Ts) = 5⁰c
properties of air at 50⁰c and at 1 atm
= 1.092 kg/m^3
Cp = 1007 j/kg⁰c
k = 0.02735 W/m⁰c
Pr = 0.7228
v = 1.798 * 10^-5 m^2/s
determine the exit temperature of air and the rate of heat transfer
attached below is the detailed solution
Calculate the mass flow rate
= p*Ac*Vmean
= 1.092 * 0.0314 * 7 = 0.24 kg/s
