Question

In an evaporator refrigerant R-22 enters at steady state at -4 °C and quality=0.2 with a velocity 7 m/s and mass flow rate 0.1 kg/s. The exit condition of R-22 is saturated vapor at -4°C. The diameter of the evaporator flow channel is uniform. Find the diameter of the evaporator flow channel and the exit velocity.

Answer 1

To find the diameter of the evaporator flow channel and the exit velocity of the refrigerant R-22, use the principle of continuity and the equation Po1A1v1 = Po2A2v2. The diameter of the evaporator flow channel can be found using the equation d = √((4(mdot/Po1)) / (pi * v1)), and the exit velocity can be calculated using the equation v2 = (mdot / (Po2 * A2)).

Step-by-step explanation:

To find the diameter of the evaporator flow channel and the exit velocity of the refrigerant R-22, we can apply the principle of continuity, which states that the mass flow rate of a fluid is constant through a pipe. We can use the equation:

Po1A1v1 = Po2A2v2

where Po is the density, A is the cross-sectional area, and v is the velocity. We also need to use the fact that the refrigerant R-22 is incompressible and at steady state.

Solving the equation for the diameter of the evaporator flow channel, we get:

d = √((4(mdot/Po1)) / (pi * v1))

where mdot is the mass flow rate and v1 is the velocity at the inlet of the evaporator flow channel. Plugging in the values and solving, we find the diameter of the evaporator flow channel.

For the exit velocity, we can calculate it using the equation:

v2 = (mdot / (Po2 * A2))

where A2 is the cross-sectional area at the exit of the evaporator flow channel. Plugging in the values and solving, we find the exit velocity of the refrigerant R-22.