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A heat pump receives heat from a lake that has an average winter time temperature of 6o C and supplies heat into a house having an average temperature of 25o C. (a) If the house loses heat to the atmosphere at the rate of 60,000 kJ/h, determine the minimum power supplied to the heat pump (in kW) that can maintain the interior temperature of the house at 25o C. (b) Suppose the heat pump absorbs energy from the ground (10o C) instead of the lake, Calculate the minimum power (in kW) required for the heat pump.

Respuesta :

Answer:

Explanation:

60000 kJ / h = 60000 x 1000 / (60 x 60 )

= 16667 J /s

a)

[tex]\frac{Q_H}{W} =\frac{T_H}{T_H-T_C}[/tex]

where [tex]Q_H[/tex] and [tex]W[/tex] are heat supplied and heat given from  outside source .

Here [tex]Q_H[/tex] = 16667 J/s

[tex]\frac{T_H}{T_H-T_C}=\frac{298}{298-279}[/tex]

= 15.684

[tex]Q_H = 16667 J/s[/tex]

16667 / W = 15.684

W= minimum power supplied  = 1062.7 W.

b )

If  [tex]T_C=283K[/tex]

[tex]\frac{T_H}{T_H-T_C}=\frac{298}{298-283}[/tex]

=19.87

[tex]\frac{Q_H}{W} = 19.87[/tex]

[tex]W=\frac{Q_H}{19.87}[/tex]

[tex]W=\frac{16667}{19.87}[/tex]

= 838.8 W .

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