Question

Refrigerant 134a enters a horizontal pipe operating at steady state at 40oC, 300 kPaand a velocity of 40 m/s. At the exit, the temperature is 50oC and the pressure is 240 kPa. The pipe diameter is 0.04 m. Determine (a) the mass flow rate of the refrigerant, in kg/s, (b) the velocity at the exit, in m/s, and (c) the rate of heat transfer between the pipe and its surroundings, in kW

Answer 1

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