A refrigeration cycle is used to keep the chilled food section of a supermarket at 4oC (see figure).
Heat in the cold food is given up to refrigerant fluid in the evaporator from state 4 to state 1. A
compressor adds work to the refrigerant from state 1 to state 2, and a condenser enables heat to be
released to the atmosphere from state 2 to state 3. The high pressure condensed refrigerant is then
passed through an expansion valve to reduce the pressure while also keeping the enthalpy the same
(i.e. h 3 = h 4 ).
Paper Code: ENGEN180-22B
8 END
a. Given the Coefficient of Performance (COP) of the refrigeration cycle is 2.70, the
compressor work (W in) is 19 kW, determine the rate of heat removal from the
refrigerated space?
(1 mark)
b. If the mass flow of the refrigerant fluid is 0.4 kg/s, determine the enthalpy (kJ/kg) of
state 4.
(1 mark)
c. Calculate the rate of heat rejected to atmosphere (QH)?
(1 mark)
d. How much milk in kg/s can be chilled to T f = 4 oC by the refrigeration system cooling
load of QL if the initial temperature of the milk is T i = 14°C? The heat capacity of milk is
3.89 kJ/kg.°C

Let's tackle each part of the problem step by step, using thermodynamic principles for refrigeration cycles.

### a) Rate of heat removal from the refrigerated space

The **Coefficient of Performance (COP)** of a refrigeration cycle is defined as:

\[
COP = \frac{Q_L}{W_{\text{in}}}
\]

Where:
- \( COP = 2.70 \)
- \( W_{\text{in}} = 19 \text{ kW} \) (compressor work)
- \( Q_L \) is the rate of heat removal from the refrigerated space (what we're solving for).

Rearranging the equation:

\[
Q_L = COP \times W_{\text{in}} = 2.70 \times 19 \text{ kW}
\]

\[
Q_L = 51.3 \text{ kW}
\]

So, the rate of heat removal from the refrigerated space is **51.3 kW**.

---

### b) Enthalpy at state 4

To determine the enthalpy at state 4, we can use the first law of thermodynamics for the evaporator. The rate of heat removal from the refrigerated space is:

\[
Q_L = \dot{m} \times (h_1 - h_4)
\]

Where:
- \( \dot{m} = 0.4 \text{ kg/s} \) (mass flow rate)
- \( h_1 \) and \( h_4 \) are the enthalpies at states 1 and 4, respectively.
- \( Q_L = 51.3 \text{ kW} \) or \( 51.3 \text{ kJ/s} \) (from part a)

Rearranging the equation:

\[
h_4 = h_1 - \frac{Q_L}{\dot{m}} = h_1 - \frac{51.3}{0.4} = h_1 - 128.25 \text{ kJ/kg}
\]

So, the enthalpy at state 4 is **128.25 kJ/kg** lower than the enthalpy at state 1.

Without knowing the value of \( h_1 \), we leave this expression in terms of \( h_1 \). If additional data for \( h_1 \) were provided (from tables, charts, or other known values for the refrigerant), we could compute \( h_4 \) directly.

---

### c) Rate of heat rejected to the atmosphere

The rate of heat rejected to the atmosphere is given by:

\[
Q_H = Q_L + W_{\text{in}}
\]

Where:
- \( Q_L = 51.3 \text{ kW} \)
- \( W_{\text{in}} = 19 \text{ kW} \)

\[
Q_H = 51.3 \text{ kW} + 19 \text{ kW} = 70.3 \text{ kW}
\]

So, the rate of heat rejected to the atmosphere is **70.3 kW**.

---

### d) Mass flow rate of milk being chilled

To find the mass flow rate of milk, we use the equation for heat transfer:

\[
Q_L = \dot{m}_{\text{milk}} \times C_p \times \Delta T
\]

Where:
- \( Q_L = 51.3 \text{ kW} = 51.3 \text{ kJ/s} \)
- \( C_p = 3.89 \text{ kJ/kg}^\circ C \) (specific heat capacity of milk)
- \( \Delta T = T_i - T_f = 14^\circ C - 4^\circ C = 10^\circ C \)
- \( \dot{m}_{\text{milk}} \) is the mass flow rate of milk (what we're solving for).

Rearranging the equation:

\[
\dot{m}_{\text{milk}} = \frac{Q_L}{C_p \times \Delta T} = \frac{51.3}{3.89 \times 10} = \frac{51.3}{38.9} \approx 1.318 \text{ kg/s}
\]

So, the refrigeration system can chill approximately **1.32 kg/s** of milk.

---

### Summary:
- a) The rate of heat removal is **51.3 kW**.
- b) The enthalpy at state 4 is **\( h_1 - 128.25 \text{ kJ/kg} \)**.
- c) The rate of heat rejected to the atmosphere is **70.3 kW**.
- d) The mass flow rate of milk is **1.32 kg/s**.

Would you like further details on any of these steps?

---

Here are 5 questions related to this topic:
1. How does increasing the COP of a refrigeration cycle affect the system's efficiency?
2. What factors influence the enthalpy at different states in a refrigeration cycle?
3. Why is the expansion valve used to reduce pressure while keeping the enthalpy constant?
4. How does the choice of refrigerant affect the performance of a refrigeration cycle?
5. What are the environmental implications of heat rejection in refrigeration systems?

**Tip**: In refrigeration cycles, reducing the compressor work \(W_{\text{in}}\) can significantly improve system efficiency without compromising cooling capacity.

A refrigeration cycle is used to keep the chilled food section of a supermarket at 4°C. Heat in the cold food is given up to refrigerant fluid in the evaporator from state 4 to state 1. A compressor adds work to the refrigerant from state 1 to state 2, and a condenser enables heat to be released to the atmosphere from state 2 to state 3. The high pressure condensed refrigerant is then passed through an expansion valve to reduce the pressure while also keeping the enthalpy the same (i.e. h3 = h4). Paper Code: ENGEN180-22B. a. Given the Coefficient of Performance (COP) of the refrigeration cycle is 2.70, the compressor work (Win) is 19 kW, determine the rate of heat removal from the refrigerated space? b. If the mass flow of the refrigerant fluid is 0.4 kg/s, determine the enthalpy (kJ/kg) of state 4. c. Calculate the rate of heat rejected to atmosphere (QH)? d. How much milk in kg/s can be chilled to Tf = 4°C by the refrigeration system cooling load of QL if the initial temperature of the milk is Ti = 14°C? The heat capacity of milk is 3.89 kJ/kg°C.

Explore how to calculate the rate of heat removal, enthalpy, and heat rejection in a refrigeration cycle with a COP of 2.70 and compressor work of 19 kW. This problem also covers the mass flow rate of milk being chilled. Suitable for undergraduate engineering students.

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