### Heat Pump Cycle Analysis#### Description:The heat pump cycle depicted in the figure operates at a steady state and provides energy by heat transfer at a rate of 15 kW to maintain a dwelling at 22°C when the outside temperature is 5°C. The manufacturer claims that the power input required for this operating condition is 4 kW.#### Diagram Explanation:- **House Interior Temperature (\( T_{\text{inside}} \)):** 22°C- **Outside Temperature (\( T_{\text{outside}} \)):** 5°C- **Heat Transfer Rate to Maintain Temperature (\( \dot{Q}_H \)):** 15 kW- **Power Input for Heat Pump Cycle (\( \dot{W}_{\text{cycle}} \)):** 4 kWIn the diagram:- The inside of the house is maintained at 22°C.- The outside temperature is 5°C.- The heat pump cycle receives power (\( \dot{W}_{\text{cycle}} \)) and provides heat to the inside (\( \dot{Q}_H = 15 \, \text{kW} \)).- The heat pump extracts energy from the outside (\( \dot{Q}_C \)).#### Objective:Applying energy and entropy rate balances, determine the rate of entropy production in kW/K.#### Calculation:Entropy Production Rate (σ):\[ \sigma = \, \boxed{kW/K} \]#### Question:Is this cycle possible?- **Options:** Yes / NoPlease input your answer in the space provided and select Yes or No to determine if the cycle is theoretically feasible based on the given parameters and balances.

Answered: Image /qna-images/answer/82855f58-9055-4b84-bebd-3b0235da41bf.jpg

The heat pump cycle shown in the figure below operates at steady state and provides energy by heat transfer at a rate of 15 kW to maintain a dwelling at 22°C when the outside temperature is Toutside = 5°C. The manufacturer claims that the power input required for this operating condition is W cycle = 4 kW. QH=15 kW 0 = i Is this cycle possible? Tinside = 22°C kW/K Heat pump cycle Toutside W Applying energy and entropy rate balances determine the rate of entropy production, in kW/K. cycle

The heat pump cycle shown in the figure below operates at steady state and provides energy by heat transfer at a rate of 15 kW to maintain a dwelling at 22°C when the outside temperature is Toutside = 5°C. The manufacturer claims that the power input required for this operating condition is W cycle = 4 kW. QH=15 kW 0 = i Is this cycle possible? Tinside = 22°C kW/K Heat pump cycle Toutside W Applying energy and entropy rate balances determine the rate of entropy production, in kW/K. cycle

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Question

### Heat Pump Cycle Analysis

#### Description:
The heat pump cycle depicted in the figure operates at a steady state and provides energy by heat transfer at a rate of 15 kW to maintain a dwelling at 22°C when the outside temperature is 5°C. The manufacturer claims that the power input required for this operating condition is 4 kW.

#### Diagram Explanation:
- **House Interior Temperature (\( T_{\text{inside}} \)):** 22°C
- **Outside Temperature (\( T_{\text{outside}} \)):** 5°C
- **Heat Transfer Rate to Maintain Temperature (\( \dot{Q}_H \)):** 15 kW
- **Power Input for Heat Pump Cycle (\( \dot{W}_{\text{cycle}} \)):** 4 kW

In the diagram:
- The inside of the house is maintained at 22°C.
- The outside temperature is 5°C.
- The heat pump cycle receives power (\( \dot{W}_{\text{cycle}} \)) and provides heat to the inside (\( \dot{Q}_H = 15 \, \text{kW} \)).

- The heat pump extracts energy from the outside (\( \dot{Q}_C \)).

#### Objective:
Applying energy and entropy rate balances, determine the rate of entropy production in kW/K.

#### Calculation:
Entropy Production Rate (σ):
\[ \sigma = \, \boxed{kW/K} \]

#### Question:
Is this cycle possible?
- **Options:** Yes / No

Please input your answer in the space provided and select Yes or No to determine if the cycle is theoretically feasible based on the given parameters and balances.

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