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A power company desires to use groundwater from a hot spring to power a heat engine. If the groundwater is at 95 deg C and the atmosphere is at 20 deg C, what is the maximum power output for a mass flow of 0.2 kg/s? Assume the water is cooled to atmospheric temperature.

Answer :

Answer:

W  = 12.8 KW

Explanation:

given data:

mass flow rate = 0.2 kg/s

Engine recieve heat from ground water at 95 degree ( 368 K)  and reject that heat to atmosphere  at 20 degree (293K)

we know that maximum possible efficiency is given as

[tex]\eta = 1- \frac{T_L}{T_H}[/tex]

[tex]\eta = 1 - \frac{ 293}{368}[/tex]

[tex]\eta = 0.2038[/tex]

rate of heat transfer is given as

[tex]Q_H = \dot m C_p \Delta T[/tex]

[tex]Q_H = 0.2 * 4.18 8(95 - 20)[/tex]

[tex]Q_H = 62.7 kW[/tex]

Maximuim power is given as

[tex]W = \eta Q_H[/tex]

W = 0.2038 * 62.7

W  = 12.8 KW

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