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.

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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.

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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.

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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.

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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?

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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.