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
Answer:
The temperature drop is 61.1 °C
The final specific volume of the refrigerant is 1.236 m^3/kg
Explanation:
Initial pressure of refrigerant = 800 kPa = 800/100 = 8 bar
Final pressure of refrigerant = 140 kPa = 140/100 = 1.4 bar
From steam table
At 8 bar, initial saturated temperature is 170.4 °C
At 1.4 bar, final saturated temperature is 109.3 °C
Temperature drop = initial saturated temperature - final saturated temperature = 170.4 - 109.3 = 61.1 °C
Also, from steam table
At 1.4 bar, specific volume is 1.236 m^3/kg
Final specific volume of the refrigerant is 1.236 m^3/kg
Answer:
Explanation:
The complete detailed explanation which answer the question efficiently is shown in the attached files below.
I hope it helps a lot !
Answer:
Example for strengthening mechanism in single-phase material: Strain hardening- D.