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Naddika [18.5K]
4 years ago
11

Refrigerant 134a enters a horizontal pipe operating at steady state at 40°C, 300 kPa, and a velocity of 25 m/s. At the exit, the

temperature is 90°C and the pressure is 240 kPa. The pipe diameter is 0.1 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.
Engineering
1 answer:
salantis [7]4 years ago
3 0

Answer:

a) 2.42 kg/s

b) 37.20 m/s

c) 120.56 kW

Explanation:

Given that:

The fluid in the Refrigerant = R-134a

Diameter (d) = 0.1 m

In the Inlet:

Temperature T_1 = 40^0C

Pressure P_1= 300kPa

Velocity V_1 = 25 m/s

At the exit:

Temperature T_2 = 90^0C

Pressure P_2 = 240 kPa

From the  Table A-12 for Refrigerant R-134a at T_1 = 40^0C and P_1= 300kPa

Specific Volume v_1 = 0.0809 m^3/kg

From the  Table A-12 for Refrigerant R-134a at T_2 = 90^0C and P_2 = 240 kPa

Specific Volume v_2 = 0.12038 kJ/kg

Their corresponding Enthalpy h_1 and h_2 are as follows:

Enthalpy h_1  =284.05 kJ/kg

Enthalpy h_2 = 333 kJ/kg

a) The mass flow rate of the refrigerant can be calculated as :

m_1 = \frac{AV_1}{v_1}

m_1 = \frac{\frac{\pi (0.1)^2}{4}*25}{0.08089}

m_1 = 2.42 kg/s

b) The velocity at the exit point:

we knew that:

m=m_1 =m_2

∴

\frac{AV_1}{v_1} =\frac{AV_2}{v_2}

V_2 = \frac{v_2}{v_1} V_1

V_2 = \frac{0.12038}{0.08089} *25

V_2 = 37.20 m/s

c) Expression for calculating heat transfer (as long as there is no work that is said to be done and the pipe is horizontal) can be represented as:

Q_{cv}= m[(h_2-h_1)+\frac{1}{2}(V_2^2-V_1^2)]

Q_{cv}= 2.42*[(333.49-284.05)+\frac{1}{2}(37.20^2-25^2)]

Q_{cv}= 2.42*[49.44+379.42]

Q_{cv}= 119.6448kW+918.19W(\frac{1kW}{1000W} )

Q_{cv}= 119.6448kW+0.92 kW

Q_{cv} = 120.56 kW

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