A house heated by a heat pump requires QH= 25 kW of heat flow when acting between -5°C outside and 20 °C inside. A. Assuming the heat pump is ideal, how much work (in kW) is required to maintain this heat flow. B. Heat pumps are usually rated by their coefficient of performance 7= Q#/W. Find 7 for the ideal case. C. In practice 7× 4. For this 7, find the work actually required.
A house heated by a heat pump requires QH= 25 kW of heat flow when acting between -5°C outside and 20 °C inside. A. Assuming the heat pump is ideal, how much work (in kW) is required to maintain this heat flow. B. Heat pumps are usually rated by their coefficient of performance 7= Q#/W. Find 7 for the ideal case. C. In practice 7× 4. For this 7, find the work actually required.
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![### Problem 9.2
**A house heated by a heat pump requires \(Q_H = 25 \, \text{kW}\) of heat flow when acting between -5°C outside and 20°C inside.**
#### A. Assuming the heat pump is ideal, how much work (in kW) is required to maintain this heat flow?
#### B. Heat pumps are usually rated by their coefficient of performance \(\eta = \frac{Q_H}{W}\). Find \(\eta\) for the ideal case.
#### C. In practice \(\eta \approx 4\). For this \(\eta\), find the work actually required.
---
### Solution
1. **Ideal Work Calculation:**
The coefficient of performance (COP) for an ideal heat pump (Carnot heat pump) is given by:
\[
\eta_{\text{ideal}} = \frac{T_H}{T_H - T_C}
\]
Here, \(T_H\) and \(T_C\) must be in absolute temperatures (Kelvin).
\[
T_H = 20 + 273 = 293 \, \text{K}
\]
\[
T_C = -5 + 273 = 268 \, \text{K}
\]
Therefore,
\[
\eta_{\text{ideal}} = \frac{293}{293 - 268} = \frac{293}{25} \approx 11.72
\]
Using this COP, the work required \(W_{\text{ideal}}\) is calculated as:
\[
W_{\text{ideal}} = \frac{Q_H}{\eta_{\text{ideal}}} = \frac{25 \, \text{kW}}{11.72} \approx 2.13 \, \text{kW}
\]
2. **Actual Work Calculation:**
Given \(\eta_{\text{actual}} \approx 4\):
\[
W_{\text{actual}} = \frac{Q_H}{\eta_{\text{actual}}} = \frac{25 \, \text{kW}}{4} = 6.25 \, \text{kW}
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Faf192c4a-3623-40a7-8b1d-18d623c8a61a%2Fd93ae4ad-a402-4831-aaad-c939237c2aeb%2F0vbxpj5_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem 9.2
**A house heated by a heat pump requires \(Q_H = 25 \, \text{kW}\) of heat flow when acting between -5°C outside and 20°C inside.**
#### A. Assuming the heat pump is ideal, how much work (in kW) is required to maintain this heat flow?
#### B. Heat pumps are usually rated by their coefficient of performance \(\eta = \frac{Q_H}{W}\). Find \(\eta\) for the ideal case.
#### C. In practice \(\eta \approx 4\). For this \(\eta\), find the work actually required.
---
### Solution
1. **Ideal Work Calculation:**
The coefficient of performance (COP) for an ideal heat pump (Carnot heat pump) is given by:
\[
\eta_{\text{ideal}} = \frac{T_H}{T_H - T_C}
\]
Here, \(T_H\) and \(T_C\) must be in absolute temperatures (Kelvin).
\[
T_H = 20 + 273 = 293 \, \text{K}
\]
\[
T_C = -5 + 273 = 268 \, \text{K}
\]
Therefore,
\[
\eta_{\text{ideal}} = \frac{293}{293 - 268} = \frac{293}{25} \approx 11.72
\]
Using this COP, the work required \(W_{\text{ideal}}\) is calculated as:
\[
W_{\text{ideal}} = \frac{Q_H}{\eta_{\text{ideal}}} = \frac{25 \, \text{kW}}{11.72} \approx 2.13 \, \text{kW}
\]
2. **Actual Work Calculation:**
Given \(\eta_{\text{actual}} \approx 4\):
\[
W_{\text{actual}} = \frac{Q_H}{\eta_{\text{actual}}} = \frac{25 \, \text{kW}}{4} = 6.25 \, \text{kW}
\]
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