Question

Bananas are to be cooled from 28C28^{\circ} \mathrm{C} to 12C12^{\circ} \mathrm{C}at a rate of 1140 kg/h by a refrigerator that operates on a vapor-compression refrigeration cycle. The power input to the refrigerator is 8.6 kW. Determine (a) the rate of heat absorbed from the bananas, in kJ/h, and the COP, (b) the minimum power input to the refrigerator, and (c) the second-law efficiency and the exergy destruction for the cycle. The specific heat of bananas above freezing is 3.35kJ/kgC3.35 \mathrm{kJ} / \mathrm{kg} \cdot^{\circ} \mathrm{C}.

Solution

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Part A\textbf{\large Part A}

The rate of heat taken away from the bananas is determined from an energy balance on them and their change of temperature:

Q˙L=m˙cpΔT=11403.3516kJh=61100kJh\begin{align*} \dot Q_{L}&=\dot mc_{p}\Delta T\\ &=1140\cdot3.35\cdot16\:\dfrac{\text{kJ}}{\text{h}}\\ &=\boxed{61100\:\dfrac{\text{kJ}}{\text{h}}} \end{align*}

The COP is then easily determined with the given power input to the refrigerator:

COP=Q˙LW˙=6110060608.6=1.97\begin{align*} \text{COP}&=\dfrac{\dot Q_{L}}{\dot W}\\ &=\dfrac{61100}{60\cdot60\cdot8.6}\\ &=\boxed{1.97} \end{align*}

Part B\textbf{\large Part B}

The minimum power input is determined from the rate of exergy transferred from the bananas where the source temperature is taken to be the initial temperature of the bananas and the surroundings temperature is the average of the initial and final temperature of the bananas:

W˙min=Q˙L(1T0TL)=611006060(1293301)kW=0.45kW\begin{align*} \dot W_{\text{min}}&=\dot Q_{L}\bigg(1-\dfrac{T_{0}}{T_{L}}\bigg)\\ &=\dfrac{61100}{60\cdot60}\bigg(1-\dfrac{293}{301}\bigg)\:\text{kW}\\ &=\boxed{0.45\:\text{kW}} \end{align*}

Part C\textbf{\large Part C}

The second-law efficiency is determined from the minimum and actual power input:

ηII=W˙minW˙=0.458.6=0.052\begin{align*} \eta_{\text{II}}&=\dfrac{\dot W_{\text{min}}}{\dot W}\\ &=\dfrac{0.45}{8.6}\\ &=\boxed{0.052} \end{align*}

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