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

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

answer: copyright is extremely important to authors because it constitutes the ownership of one's original work. having ownership means you can protect your work as intellectual property and control who makes money from it—which is you and those to whom you assign rights.

answer:

my guess it 8 i'm probably wrong