Question

dark stormy, cold Mississippi winter night, Fido is watching television in front of a hre place that is adding 10,000 W of heat to the room. The house losses heat to the winter nignt a Problem-nFinLLaw and Carno.cycles4S0% On a dark, stormy, cold Mississippi winter night, Fido is watching rate of Qlo..(k//sec)-0.5(Thou"-Toet-ide); while the television utilizes 500 watts or electricity supplied from a power line entering the house. Ignore the heat added by Fido's y the Carnot Cyele maintains the house at a constant temperature. Qloso(k.J/sec) 0.5(Thouse Toutside Assumptions: Thouseconstant Toutside 5 QB Thouse 25 Work Carnot Cycle House Energy Calculations a) Assuming a constant house temperature of 25°C, perform an energy balance for the House (excluding the Carnot Cycle) and Calculate the value of Qa in kJ/sec b) Using your result from part (a), state the direction of Qa transfer relative to the house (into or out of the house). c) Calculate the Carnot COP (Coefficient of Performance), B', assuming the Carnot Cycle is a Heat Pump, with Qa being Qu in the Cycle (DO NOT calculate the refrigeration COP) d) The manufacture of the heat pump claims it has a Coefficient of Performance 25. Does this manufacturer's value of B' violate the 2nd Law of Thermodynamics, based on your answer in part (c)? Irrelevant of your answers in parts (a) through (d), use the following to calculate the required work for the heat pump: QA QH-+10 kW (in to the house) and P-5.

Answer 1

Inside temperature of home = 25⁰C

Outside temperature of house = 5⁰C

Heat loss from the house, Q_loss = 0.5 (T_house - T_outside)

= 0.5 (25 - 5)

= 10 kW

Heat coming into the room in the form of of electricity = 500 watts

Heat generated inside the room due to television = 10000 watts

By doing heat balance across the room:

Heat coming into the room + Heat generated inside the room - Heat loss from the house= Net heat inside the room

500 + 10000 - 10000 = Net heat inside the room

500 watts = Net heat inside the room

Here, net heat going into the room (in the form of electricity)

Q_A = 500 watts

b)

As it is given that Carnot cycle can be considered as heat pump, therefore:

The direction of Q_A should be into the house.

c)

For a heat pump:

COP = QH

For carnot heat pump, we can write:

キCOP = TH

Here:

TL = 278 ( K )

Ту-298 (K)

$\Rightarrow COP=\frac{1}{1-\frac{278}{298}}$

COP-14.9

d)

The manufacture of heat pump claims that it has COP = 25. But as per calculations done in section 'c', none of the process can have more COP than carnot cycle. Hence manufacturer claim is wrong as per second law of thermodynamics.

e)

Here:

10000 (W)

COP 5

1-0 10000

2000

$\Rightarrow 4=\frac{Q_{L}}{2000}$

QL-8000 ( W )

By doing heat balance across the carnot cycle,we can write:

$Q_{L}+W=Q_{H}$

$\Rightarrow Q_{H}-Q_{L}=W$

$\Rightarrow 10000-8000=W$

, = 2000 ( Watts)

Therefore, required work for heat pump = 2000 watts