Refrigerant 134a enters the evaporator of a refrigeration system operating at st

Question

Refrigerant 134a enters the evaporator of a refrigeration system operating at steady state at -4 degree Celsius and a quality of 20% at a velocity of 7 m/s. At the exit, the refrigerant is saturated vapor at a temperature of -4 degree Celsius. The evaporator flow channel has constant diameter. If the mass flow rate of the entering refrigerant is 0.1 kg/s, determine:

(a) the diameter of the evaporator flow channel, in cm. (ANS: d = 1.732cm)

(b) the velocity at the exit, in m/s (ANS: v2 = 33.7 m/s)

*I have the answers, just not sure how to go about doing the problem. Hope you can help

Explanation / Answer

Given:

Inlet of Evaporator:

Tin = -4 0C

Quality , y1= 20 %

vf1= 0.0007644 m3/kg

vg1 = 0.0794 m3/kg

Velocity of flow, U = 7 m/s

Outlet of Evaporator:

Tout = -4 0C

Quality , y2= 100 %

vg2 = 0.0794 m3/kg

Mass flow rate , m= 0.1 kg/s

Solution:

At inlet

Volume of refrigerant, Vin = vf1 + 0.2(vg1 - vf1)

= 0.0007644 + 0.2( 0.0794 - 0.0007644)

  Vin = 0.0165 m3/kg

1 = 1/ Vin = 1/0.0165= 60.60 kg/m3  

At outlet

Volume of refrigerant, Vout = vg2 = 0.0794 m3/kg

Simillarly  2 = 1/Vout = 1/0.0794 = 12.59 kg/m3

We know by the law of conservation of mass that mass is conserved in any system.

1AU1 = 2AU2 = m = 0.1 kg/s    (area is constant at inlet and outlet)

Where A is the area of the pipe

A = 0.1/ (1xU1) = 0.1 / (60.60 x 7 ) = 2.36 x 10-4 m2

A = (/4)d2 = 2.36 x 10-4 m2

=> d= 17.33 mm

We know that

1U1 = 2U2 (Since area is constant at inlet and outlet)

U2 = 1U1/2 = 60.60x7/12.59 = 33.69 m/s

U2 = 33.69 m/s

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