To solve this problem, we will use the T-s (Temperature-entropy) and P-h (Pressure-enthalpy) diagrams for the vapor-compression refrigeration cycle.

First, let's label the states of the cycle:

State 1: Saturated vapor entering the compressor at 3 bar
State 2: Compressed vapor exiting the compressor
State 3: Saturated liquid leaving the condenser at 8 bar
State 4: Expanded vapor returning to the evaporator

Given information:
Pressure at State 1 (P1) = 3 bar
Pressure at State 3 (P3) = 8 bar
Mass flow rate (m) = 0.1 kg/s
Compressor efficiency (η) = 80%

a. To find the compressor power (Pc), we need to find the enthalpy changes in the cycle. We can use the enthalpy at State 1 (h1) and the isentropic efficiency (η) of the compressor.

First, we need to determine the properties at State 1. We can use a refrigerant table for Refrigerant 134a to find the enthalpy (h1) and entropy (s1) at State 1 at 3 bar (assuming it's saturated vapor).

From the table, we find:
h1 = 285.1 kJ/kg (enthalpy at State 1)
s1 = 1.017 kJ/kg·K (entropy at State 1)

Next, we need to determine the entropy at State 2 (s2). Since the compression process is isentropic, the entropy remains constant. Therefore:

s2 = s1 = 1.017 kJ/kg·K

Using the compressor efficiency (η = 80%), we can calculate the actual enthalpy change during compression (h2_actual) as:

h2_actual = h1 + (h2_actual(isentropic) - h1) / η

We can find the isentropic enthalpy change during compression (h2_isentropic) by using the formula:

h2_isentropic = h1 + (s2 - s1) * (T2 - T1)

Since the process is reversible and adiabatic, the entropy change (Δs) and temperature change (ΔT) are related by:

Δs = c_p * ln(T2/T1) - R * ln(P2/P1)

Using the approximate values of cp = 1.32 kJ/kg·K and R = 0.246 kJ/kg·K for Refrigerant 134a, we can solve for ΔT:

1.017 = 1.32 * ln(T2/338.15) - 0.246 * ln(P2/300)

Rearranging the equation, we have:

ln(T2/338.15) = (1.017 + 0.246 * ln(P2/300)) / 1.32

Solving for T2, we find:
T2 = 373.6 K

Now we can calculate h2_isentropic:

h2_isentropic = h1 + (s2 - s1) * (T2 - T1)
= 285.1 + (1.017 - 1.017) * (373.6 - 338.15)
= 285.1 kJ/kg

And finally, we can calculate h2_actual:

h2_actual = h1 + (h2_isentropic - h1) / η
= 285.1 + (285.1 - 285.1) / 0.8
= 285.1 kJ/kg

Now we can calculate the compressor power (Pc):

Pc = m * (h2_actual - h1)
= 0.1 * (285.1 - 285.1)
= 0 kJ/s (or 0 kW)

Therefore, the compressor power is 0 kW.

b. The refrigerating capacity (Q) is given by:

Q = m * (h1 - h3)

To find the enthalpy at State 3 (h3), we can use the refrigerant table for Refrigerant 134a at 8 bar (assuming saturated liquid). From the table, we find:

h3 = 82.73 kJ/kg

Now we can calculate the refrigerating capacity:

Q = 0.1 * (285.1 - 82.73)
= 20.678 W

Converting to tons:

1 ton of refrigeration = 211 kJ/min = 3.517 W

Refrigerating capacity = 20.678 / 3.517
= 5.87 tons

Therefore, the refrigerating capacity is approximately 5.87 tons.

c. The coefficient of performance (COP) is given by:

COP = Q / Pc

Substituting the values we calculated in parts a and b:

COP = (20.678 W) / (0 W)
= undefined

Because the compressor power is 0 W, the COP is undefined.

d. To create the T-s and P-h plots, we need to determine the state properties for each state (1, 2, 3, and 4) using the refrigerant table for Refrigerant 134a:

State 1: P1 = 3 bar
h1 = 285.1 kJ/kg
s1 = 1.017 kJ/kg·K
T1 = (from the table)

State 2: P2 = (from the problem statement)
h2 = (from the calculations in part a)
s2 = (from the calculations in part a)
T2 = (from the table)

State 3: P3 = 8 bar
h3 = 82.73 kJ/kg
s3 = (from the table)
T3 = (from the table)

State 4: P4 = P1 = 3 bar
h4 = (from the table)
s4 = (from the table)
T4 = (from the table)

Using the calculated state properties, we can plot the points on the T-s and P-h diagrams, and connect State 4 and State 1 to close the cycle.

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