What minimum work must be supplied to the heat pump per hour?

Tam · #1 23-07-2007, 08:41

A heat pump extracts 7.0 x 10^6 J of heat per hour from a well at 280 K and delivers its output heat into a house at 320 K. Assume if the heat pump uses an ideal Carnot cycle in its operation.

anilbakshi · #2 23-07-2007, 08:41

COP of pump = n = output power/input power =
Q1-dot/W-dot = T1/(T1-T2)

it can not extract heat per hour from (T2 - low temp) sink and deliver it to source (T1 high temp), so W work is being done on the pump per hour
Q2-dot + W-dot = Q1-dot >>> dot indicates time derivative

Q1-dot/W-dot = T1/(T1-T2)
(Q2-dot + W-dot )/W-dot = T1/(T1-T2)
Q2-dot/W-dot = T1/(T1-T2) - 1 = (T1-T1+T2)/(T1-T2)
Q2-dot/W-dot = (T2)/(T1-T2)
W-dot = Q2-dot (J/h) * (T1-T2)/T2 = 7.0 x 10^6 *[40/(280)]
W-dot = 10^6 Joule/hour

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