Answer:
1. Mass flow rate = 0.621Kg/s
2. Velocity = 52.2m/s
3. Rate = +6.82kW
Explanation:
Given
Refrigerant = R134a
Diameter, d = 0.04m
T1 = 40°C
P1 = 300kPa
V1 = 40m/s
T2 = 50°C
P2 = 240kPa
a.
The specific volume of refrigerant R134a at T1 and P1 from Refrigerant table (A12) = v1 = 0.08089 m³/kg
The specific volume of refrigerant R134a at T2 and P2 from Refrigerant table (A12) = v2 = 0.10562 m³/kg.
Mass flow rate is calculated as follows...
= AV/v
Where A = Area =πd²/4
V = velocity = ,40
v = v1 = 0.08089 m³/kg.
So. Mass flow rate= (π * 0.04²/4 * 40 ) / 0.0809
Mass flow rate = 0.621Kg/s
b.
Calculating Velocity at the exit.
This is given as;
m * v2 / A
0.621 * 0.10562 / (π * 0.04²/4)
= 52.2m/s
c. Calculating the rate of heat transfer between the pipe and its surroundings
From energy equation, we have.
Q -> m * [(h2 - h1) + ((,V2² - V1²)/2)]
Where h2 is the specific enthalpy for refrigerant at T1 and P1
h2 = 234.05 Kj/Kg
And h1 is the specific enthalpy for refrigerant at T2 and P2
h2 = 244.47 Kj/Kg
Rate = the specific enthalpy for refrigerant at T1 and P1
h2 = 234.05 Kj/Kg
Rate = 0.621[(234.05 - 244.47) + (52.2² - 40²)/2]
Rate = 0.621[(234.05 - 244.47) +
562.42] --- Convert the kW
Rate = 0.621((244.47 - 234.05) +
562.42/1000)
Rate = +6.82kW
