11.6. Exercise 6 Solutions#
#1
# A window-mounted air conditioner removes 3.5 kJ from the inside of a home using 1.75 kJ work input. How much energy is released outside and what is its coefficient of performance?
Ql = 3.5 #kJ
Wnet = 1.75 #kJ
# air conditioner (refr) W = Qh - Ql
# W = deltaU - Qout
deltaU = Wnet + Ql
# coeff = Ql/W
coeff = Ql/Wnet
print(f"The energy released outside is {deltaU} kJ and the coefficient of performance is {coeff}")
The energy released outside is 5.25 kJ and the coefficient of performance is 2.0
#2
# A room is heated with a 1500-W electric heater. How much power can be saved if a heat pump with a COP of 2.5 is used instead?
Qdoth = 1500 # = Wdot
COP = 2.5 # = Qh/(Qh - Ql) = Qh/Wnet
Wdotnet = Qdoth/COP
Wdot_save = Qdoth - Wdotnet
print(f"{Wdot_save} W of power can be saved")
900.0 W of power can be saved
#3
# A large coal fired power plant has an efficiency of 45% and produces net 1,500 MW of electricity. Coal releases 25 000 kJ/kg as it burns so how much coal is used per hour and what is the heat rejection?
eta = 0.45
Wnet = 1500e3 # J/s (W)
heat_comb_CO2 = 25e3 # J/kg
# how much coal is used per hour
mass = 3600*Wnet/heat_comb_CO2/eta
print(f"{mass} kg/s of coal is used per hour")
Qdoth = Wnet/eta
Qdotl = Qdoth - Wnet
print(f"The heat rejection is {round(Qdotl/1000,1)} MW")
480000.0 kg/s of coal is used per hour
The heat rejection is 1833.3 MW
#4
# Find the power output and the low T heat rejection rate for a Carnot cycle heat engine that receives 7 kW at 250°C and rejects heat at 30°C.
Kc = 273.15 #correction from C to K
QdotH = 7 # kW
TL = 30 + Kc
TH = 250 + Kc
eta = 1 - TL/TH
Wdotnet = eta*QdotH
QdotL = QdotH - Wdotnet
print(f"The power output is {round(Wdotnet,1)} kW and the low T heat rejection rate is {round(QdotL,1)} kW")
The power output is 2.9 kW and the low T heat rejection rate is 4.1 kW
#5
# A heat pump is used to heat a house during the winter. The house is to be maintained at 20°C at all times. When the ambient temperature outside drops to −10°C, the rate at which heat is lost from the house is estimated to be 25 kW. What is the minimum electrical power required to drive the heat pump?
QdotH = 25 # kW
TH = 20 + Kc
TL = -10 + Kc
COP = TH/(TH - TL)
Wdot = QdotH/COP
print(f"The minimum electrical power required to drive the heat pump is {round(Wdot,1)} kW")
The minimum electrical power required to drive the heat pump is 2.6 kW
#6
# An inventor claims to have developed a heat engine that receives 700 kJ of heat from a source at 500 K and produces 300 kJ of net work while rejecting the waste heat to a sink at 290 K. Is this a reasonable claim? Why?
TH = 500 # K
TL = 290 # K
Wnet = 300 # kJ
QdotH = 700 # kJ
etamax = 1 - TL/TH
eta_actual = Wnet/QdotH
print(f"The actual thermal efficiency, {round(eta_actual,4)}, is higher than the max theoretical thermal efficiency, {round(etamax,4)}, so this is not possible")
The actual thermal efficiency, 0.4286, is higher than the max theoretical thermal efficiency, 0.42, so this is not possible
#7
# A Carnot heat pump is to be used to heat a house and maintain it at 25 ºC in the winter. On a day when the average outdoor temperature is about 2 ºC, the house is estimated to lose heat at a rate of 55,000 kJ/hr. If the heat pump consumes 4.8 kW of power while operating, determine (a) how long the heat pump ran for that day, (b) the total heating costs, assuming average price of $0.11/kWh for electricity, and (c) the heating cost for the same day if resistance heating is used instead of a heat pump.
TH = 25 + Kc
TL = 2 + Kc
QdotH = 55e3 # kJ/hr
Wdotnet = 4.8 # kJ/s
COP = TH/(TH - TL)
QH = QdotH*24
Wnet = QH/COP
time = Wnet/Wdotnet/3600
print(f"a) The heat pump was on for {round(time,2)} hrs")
price_electricity = 0.11 #dollars/kWh
cost = Wdotnet*time*price_electricity
print(f"b) The cost of heating is {round(cost,2)} dollars")
new_cost = QH/3600*price_electricity
print(f"c) The cost of heating using a resistance heater would be {round(new_cost,2)} dollars")
a) The heat pump was on for 5.89 hrs
b) The cost of heating is 3.11 dollars
c) The cost of heating using a resistance heater would be 40.33 dollars
#8
# A heat pump with R-134a as the working fluid is used to keep a space at 25 ºC by absorbing heat from geothermal water that enters the evaporator at 60 ºC at a rate of 0.065 kg/s and leaves at 40 ºC. Refrigerant enters the evaporator at 12 ºC with a quality of 15 percent and leaves at the same pressure as saturated vapor. If the compressor consumes 1.6 kW of power, determine
#(a) the mass flow rate of the refrigerant
#(b) the rate of heat supply
#(c) the COP
#(d) the minimum power input to the compressor for the same rate of heat supply
import cantera as ct
T1 = 60 + Kc
T2 = 40 + Kc
x1 = 0
x2 = 0
Win = 1.6 #kW
mdot_w = 0.065 # kg/s
state1 = ct.Water()
state1.TQ = T1, x1
state2 = ct.Water()
state2.TQ = T2, x2
hw1 = state1.HP[0]/1000
hw2 = state2.HP[0]/1000
Qf = mdot_w*(hw1 - hw2)
# water
T1 = 12 + Kc
x1 = 0.15
state1 = ct.Hfc134a()
state1.TQ = T1, x1
P1 = state1.PQ[0]
P2 = P1
x2 = 1
state2 = ct.Hfc134a()
state2.PQ = P2, x2
h1 = state1.HP[0]/1000
h2 = state2.HP[0]/1000
mdot_r = Qf/(h2 - h1)
print(f"The mass flow rate of the refrigerant is {round(mdot_r,3)} kg/s")
QdotH = Qf + Win
print(f"The heat load is {round(QdotH,2)} kW")
COP = QdotH/Win
print(f"The COP is {round(COP,2)}")
TH = 60 + Kc
TL = 25 + Kc
COPmax = TH/(TH-TL)
Wmin = Qf/COPmax
print(f"The minimum power input to the compressor for the same rate of heat supply is {round(Wmin,2)} kW")
The mass flow rate of the refrigerant is 0.034 kg/s
The heat load is 7.03 kW
The COP is 4.39
The minimum power input to the compressor for the same rate of heat supply is 0.57 kW
#9
# Cold water at 10 ºC enters a water heater at the rate of 0.02 m3/min and leaves the water heater at 50 ºC. The water heater receives heat from a heat pump that receives heat from a heat source at 0 ºC
# a) Assuming water to be incompressible liquid that does not change phase during heat addition, determine the rate of heat supplied to the water in kJ/s.
# b) Assuming the water heater acts as a heat sink having an average temperature of 30 ºC, determine the minimum power supplied to the heat pump in kW
Vdot = 0.02 #m3/min
Ti = 10 + Kc
Te = 50 + Kc
TL = 0 + Kc
T_amb = 300 #K
P_amb = 100e3
water = ct.Water()
water.TP = T_amb, P_amb
cp = water.cp/1000
v = water.TV[1]
mdot = Vdot/60/v
QdotH = mdot*cp*(Te - Ti)
print(f"The rate of heat supplied to the water is {round(QdotH,2)} kW")
TH = 30 + Kc
COPmax = TH/(TH -TL)
Wnet = QdotH/COPmax
print(f"The minimum power supplied to the heat pump is {round(Wnet,2)} kW")
The rate of heat supplied to the water is 55.56 kW
The minimum power supplied to the heat pump is 5.5 kW