Question

Refrigerant 22 undergoes a constant-pressure process within a piston–cylinder assembly from saturated vapor at 3.5 bar to a final temperature of 25°C. Kinetic and potential energy effects are negligible. Evaluate the work and the heat transfer, each in kJ per kg of refrigerant.

Answer 1

Answer:

The work for the process at constant pressure is 3.1 kj/kg and Heat transfer is 23.42 kj/kg.

Explanation:

We have a process with Refrigerant 22, which is a constant pressure P = 3.5 bar. We have to find the thermodynamic properties of Refrigerant 22 in Thermodynamic tables.  

According to the information presented in the problem, we can mention:  

1 - Refrigerant is a closed system.  

2- The process occurs at constant pressure

3- Work in this system will be a volume change mode.

4 - Variations for kinetic and potential energy are close to zero.  

To calculate the work in kJ/kg, we should study the constant pressure process. As we only will consider volume change mode work, it will be calculated as follows:

$W_1_2 =\int\limits^2_1 {P} \, dV = P(V2 - V1)$  

Values for V1 and V2 can be read from Thermodynamic tables for R-22.  

As the initial condition is P=3.5 bar saturated vapor. We should search the properties at P = 3.5 bar using saturated R-22 Table (See Table A-7 Attached).  

V1 is the specific volume for the initial condition, as it is satured vapor, V1 should be the specific volume for the gas phase in equilibrium.  

$V1 = Vg (P=3.5 bar) = 0.0652 \frac{m^3}{kg}$  

Final condition is the same pressure P=2.5 bar and T = 25°C. As Temperature is higher than Saturation Temperature at P = 2.5 bar. R-22 is superheated vapor. We can find the Volume (V2) in the table (See Table A-9 Attached).  

$V2 = V (P = 3.5 bar ; T = 25 C) = 0.0776 \frac{m^3}{kg}$.  

After that, we can replace the equation for the work and convert to adequate units:  

$W_1_2 = P (V2 - V1) = (2.5 bar) (0.0776 - 0.00652)\frac{m^3}{kg} \\W_1_2 = (2.5 bar) (0.0124)\frac{m^3}{kg} (\frac{1*10^5 N/m^2}{1 bar} ) (\frac{1 kJ}{1*10^3 Nm} ) = 3.1 \frac{kJ}{kg}$    

The work for the process at constante pressure is 3.1 kj/kg.  

The second part of the problem is to evaluate the heat transfer. We can use an energy balance, where Ek is kinetic energy, Ep is gravitational potential energy, U is internal energy, W is the work previously calculated and Q is the heat transfer.  

ΔEk + ΔEp +ΔU =Q - W

Kinetic and potential variations are negligible. So: ΔEk = ΔEp = 0.  

$U2 - U1 = Q - W_1_2$

Internal energies U2 and U1 are found in the table, as we did for the volume.  

U1 = Ug (P=3.bar) = 223.02 kj/kg.

U2 = U (P = 3.5 bar ; T = 25 C) = 243.34 kj/kg.

Replacing we can calculate the heat transfer Q.  

$Q = (U2 - U1) + W_1_2 = (243.34 - 223-02)\frac{kj}{kg} + 3.1 \frac{kj}{kg} \\Q= 23.42 \frac{kj}{kg}$

Heat transfer is 23.42 kj/kg.

Answer 2

The work done is "4.025 KJ/Kg" and heat transfer is "24.6 KJ/Kg".

At 3.5 bar,

• $T_{sat} = -10.38^{\circ} C$

Sat vapor,

• $h_1 =h_g = 401.0 \ KJ/Kg$

At, 25°C,

• $h_2 = 425.6 \ KJ/Kg$
• $v_1 = 0.0661 \ m^3/kg$
• $v_2 = 0.0776 \ m^3/Kg$

hence,

→ The heat transfer will be:

= $h_2 -h_1$

= $425.6-401$

= $24.6 \ KJ/Kg$

and,

→ The work done will be:

= $\int pdv$

= $p(v_2-v_1)$

= $3.5\times 100 (0.0776-0.0661)$

= $4.025 \ KJ/Kg$

Thus the above solution is appropriate.

Learn more:

https://brainly.com/question/23104192