Problem 15
Question
In an absorption refrigerator, the energy driving the process is supplied not as work, but as heat from a gas flame. (Such refrigerators commonly use propane as fuel, and are used in locations where electricity is unavailable. \(^{*}\) ) Let us define the following symbols, all taken to be positive by definition: \(Q_{f}=\) heat input from flame \(Q_{c}=\) heat extracted from inside refrigerator \(Q_{r}=\) waste heat expelled to room \(T_{f}=\) temperature of flame \(T_{c}=\) temperature inside refrigerator \(T_{r}=\) room temperature (a) Explain why the "coefficient of performance" (COP) for an absorption refrigerator should be defined as \(Q_{c} / Q_{f}\) (b) What relation among \(Q_{f}, Q_{c},\) and \(Q_{r}\) is implied by energy conservation alone? Will energy conservation permit the COP to be greater than \(1 ?\) (c) Use the second law of thermodynamics to derive an upper limit on the \(\mathrm{COP},\) in terms of the temperatures \(T_{f}, T_{c},\) and \(T_{r}\) alone.
Step-by-Step Solution
VerifiedKey Concepts
Coefficient of Performance (COP)
The COP is defined as the ratio of the heat extracted from inside the refrigerator (\( Q_c \)) to the heat input from an external source, like a flame (\( Q_f \)). So, the formula for COP in this context is:
- \( \text{COP} = \frac{Q_c}{Q_f} \)
Understanding COP is like looking at the miles per gallon of a car; it helps compare different refrigerators' efficiency using the same input, encouraging smart energy use.
Energy Conservation
Here, the heat input from the flame (\( Q_f \)) is the total energy supply, and according to energy conservation, it splits into two parts:
- Heat extracted from inside the refrigerator (\( Q_c \))
- Waste heat expelled to the environment (\( Q_r \))
- \( Q_f = Q_c + Q_r \)
With this understanding, one might wonder if having the COP greater than 1 is possible. The equation alone suggests yes, because the heat extracted (\( Q_c \)) can indeed be higher than the heat supplied (\( Q_f \)), especially in systems that recycle energy efficiently.
Second Law of Thermodynamics
This law essentially places a limit on efficiency regarding energy transfers and heat flow among reservoirs at different temperatures—here, the flame (\( T_f \)), the refrigerator interior (\( T_c \)), and the environment (\( T_r \)). The law can be simplified into an inequality that governs the operation:
- \( \frac{Q_f}{T_f} - \frac{Q_c}{T_c} - \frac{Q_r}{T_r} \geq 0 \)
- \( \text{COP}_{max} = \frac{T_c}{T_f - T_r} \)