Question

Refrigerant 134a enters the evaporator of a refrigeration system operating at steady state at -8oC and a quality of 20% at a velocity of 5 m/s. At the exit, the refrigerant is a saturated vapor at -8oC. The evaporator flow channel has constant diameter of 1.7 cm. Determine the mass flow rate of the refrigerant, in kg/s, and the velocity at the exit, in m/s.

Answer 1

Refrigerant 134a enters the evaporator of a refrigeration system operating at a steady-state at -8oC and a quality of 20% at a velocity of 5 m/s. The mass flow rate of the refrigerant is 0.0594 kg/s  and the velocity at the exit is 24.191  m/s.

From the information given, we can use the properties table of refrigerant-134a to determine the values of the specific volume of saturated liquid and vapor at -8° C.

• $\mathbf{v_f= 0.0007570 \ m^3/kg}$
• $\mathbf{v_g= 0.092438 \ m^3/kg}$

Now the specific volume of the refrigerant at the inlet of the evaporator can be computed by using the formula;

$\mathbf{v_1=v_f +x(v_g-v_f)}$

where;

• x (quality of refrigerant) = 20% = 0.20

$\mathbf{v_1=0.0007570 +0.2(0.092438-0.0007570)}$

$\mathbf{v_1=0.0007570 +0.2(0.091681)}$

$\mathbf{v_1=0.0007570 +0.0183362}$

v₁ = 0.0191 m³/kg

The density of the refrigerant at the inlet of the evaporator is:

$\mathbf{\rho_1 = \dfrac{1}{v_1}}$

$\mathbf{\rho_1 = \dfrac{1}{0.0191} }$

$\mathbf{\rho_1={52.356 \ kg/m^3}}$

However, the density of the  refrigerant at the outlet of the evaporator is:

$\mathbf{\rho_g = \dfrac{1}{v_g}}$

$\mathbf{\rho_g= \dfrac{1}{0.0092438} }$

$\mathbf{\rho_g={10.818 \ kg/m^3}}$

Recall that:

• the diameter of the evaporation flow channel = 1.7 cm = 0.017 m
• velocity of refrigerant at the entrance of evaporation $v_1$ = 5 m/s

Now, the mass flow rate of the refrigerant can be computed by using the formula:

$\mathbf{m = \rho_1 \Big[ \dfrac{\pi}{4} \times d^2 \Big] v_1 }$

$\mathbf{m = 52.356 \Big[ \dfrac{\pi}{4} \times (0.017)^2 \Big] 5 }$

mass flow rate (m) = 0.0594 kg/s

Also, the velocity of the refrigerant at the exit of the evaporator is determined by using the formula:

$\mathbf{m=\rho_2 \Big[ \dfrac{\pi}{4} \times d^2 \Big] v_2}$

$\mathbf{ 0.0594 kg/s = 10.818 \Big[ \dfrac{\pi}{4} \times (0.017)^2 \Big] v_2 }$

$\mathbf{v_2 = 24.191 \ m/s}$

Therefore, we can conclude that the mass flow rate of the refrigerant is 0.0594 kg/s  and the velocity at the exit is 24.191  m/s.

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