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A heat pump is used to heat a house in the winter and to cool it in the summer. During the winter, the outside air serves as a low-temperature heat source; during the summer, it acts as a high-temperature heat sink. The heat-transfer rate through the walls and roof of the house is 0.75 kJ s—l for each C of temperature difference between the inside and outside of the house, summer and winter. The heat-pump motor is rated at 1.5 kW. Determine the minimum outside temperature for which the house can be maintained at 200C during the winter and the maximum outside temperature for which the house can be maintained at 250C during the summer.

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aaregbe

Answer:

[tex]T_{C}[/tex] = -4.2°C

[tex]T_{H}[/tex] = 49.4°C

Explanation:

A Carnot cycle is known as an ideal cycle in thermodynamic. Therefore, in theory, we have:

|[tex]\frac{Q_{C} }{Q_{H} }[/tex]| = [tex]\frac{T_{C} }{T_{H} }[/tex]

Similarly,

|[tex]Q_{H}[/tex]| = |[tex]Q_{C}[/tex]| + |[tex]W_{S}[/tex]|

During winter, the value of |[tex]T_{H}[/tex]| = 20°C = 273.15 + 20 = 293.15 K and |[tex]W_{S}[/tex]| = 1.5 kW. Therefore,

|[tex]Q_{H}[/tex]| = 0.75([tex]T_{H}[/tex] -  [tex]T_{C}[/tex])

Similarly,

|[tex]\frac{W_{S} }{Q_{H} }[/tex]| = 1 - [tex]\frac{T_{C} }{T_{H} }[/tex]

1.5/0.75*(293.15-[tex]T_{C}[/tex]) = 1 - ([tex]T_{C}[/tex]/293.15

Further simplification,

[tex]T_{C}[/tex] = -4.2°C

During summer, [tex]T_{C}[/tex] = 25°C = 273.15+25 = 298.15 K, and |[tex]W_{S}[/tex]| = 1.5 kW. Therefore,

|[tex]Q_{C}[/tex]| = 0.75([tex]T_{H}[/tex] -  [tex]T_{C}[/tex])

Similarly,

|[tex]\frac{W_{S} }{Q_{C} }[/tex]| = [tex]\frac{T_{H} }{T_{C} }[/tex] - 1

1.5/0.75*([tex]T_{H}[/tex] - 298.15) = ([tex]T_{H}[/tex]/298.15

Further simplification,

[tex]T_{H}[/tex] = 49.4°C

A) The minimum outside temperature for which the house can be maintained at 20°C is; T_L = -4.21 °C

B) The maximum outside temperature for which the house can be maintained at 20°C is; T_H = 37.21 °C

A) Heat transfer rate; Q'_H = 0.75 kJ/s/°C

Rating of heat pump motor; W'_in = 1.5 kW = 1.5 kJ/s

Temperature inside house; T_H = 20 °C = 293.15 K

Formula to find the minimum outside temperature is;

[tex]\frac{Q_{H}}{W'_{in}} = \frac{T_{H}}{T_{H} - T_{L}}[/tex]

Where [tex]T_{L}[/tex] is the minimum outside temperature.

[tex]{Q_{H}} = {Q'_{H}}({T_{H} - T_{L}})[/tex]

Thus, plugging in the relevant values gives;

(0.75/1.5)(293.15 - [tex]T_{L}[/tex]) = 293.15/(293.15 - [tex]T_{L}[/tex])

0.5(293.15 - [tex]T_{L}[/tex]) = 293.15/(293.15 - [tex]T_{L}[/tex])

(293.15 - [tex]T_{L}[/tex])² = 293.15/0.5

(293.15 - [tex]T_{L}[/tex]) = √586.3

[tex]T_{L}[/tex] = 293.15 - 24.21

[tex]T_{L}[/tex] = 268.94 K

Converting to °C gives

[tex]T_{L}[/tex] = -4.21°C

B) Now, we want to find the maximum outside temperature when T_L = 25°C = 298.15 K

Thus, we will use the formula;

[tex]\frac{Q_{H}}{W'_{in}} = \frac{T_{L}}{T_{H} - T_{L}}[/tex]

(0.75/1.5) * ([tex]T_{H} - T_{L}[/tex]) = [tex]\frac{T_{L}}{T_{H} - T_{L}}[/tex]

0.5 * ([tex]T_{H} - 298.15[/tex]) = 298.15/([tex]T_{H} - 298.15[/tex])

149.075 = ([tex]T_{H} - 298.15[/tex])²

√149.075 = ([tex]T_{H} - 298.15[/tex])

12.21 + 298.15 = [tex]T_{H}[/tex]

T_H = 310.36 K

Converting to °C gives;

T_H = 37.21 °C

Read more about carnot engine at; https://brainly.com/question/14983940

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