Question

A closed, rigid tank contains Refrigerant 134a initially at 100°C. The refrigerant is cooled until it becomes saturated vapor at 20°C. Kinetic and potential energy effects can be ignored. For the refrigerant:

a) Show the process on a T-v diagram

b) Determine the initial and final pressures, each in bar

c) Determine the heat transfer, in kJ/kg.

Answer 1

The process of a closed, rigid tank that contains Refrigerant 134a initially at 100°C is shown in the T-v diagram attached below.

Given that:

• initial temperature T₁ = 100° C

• final temperature T₂ = 20° C

At 20° C from the property tables of R-134a

Final pressure p₂ =  5.176 bar

u₂ = 237.91 kJ/kg

T₁ is present in its vapor state since it is 100⁰ C

As such, by iteration from superheated property tables

$\mathbf{p_1 = 7+(8-7) (\dfrac{0.0358-0.04064}{0.03519 -0.04064})}$

Initial pressure $\mathbf{p_1 = 7.89 \ bar}$

Similarly;

$\mathbf{u_1 = 309.74 +(308.93-309.74) (\dfrac{7.89-7}{8-7})}$

$\mathbf{u_1 = 309.74 +(-0.81) (\dfrac{0.89}{1})}$

$\mathbf{u_1 = 309.019 \ kJ/kg }$

(c) According to the first law of thermodynamics, heat energy is the change in the internal energy of an object.

Q - W = Δu      

Q = u₂ - u₁             (since K.E and P.E are ignored, Volume = constant)

Q = 309.019 kJ/kg - 237.91 kJ/kg

Q = -71.109 kJ/kg

Thus, heat is lost in the process.

Therefore, we can conclude that the initial and final pressures are 7.89 bar and 5.176 bar respectively.

Learn more about Heat transfer here:

https://brainly.com/question/18337974?referrer=searchResults