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Chapter 6.16

Chapter 6.16#

Question #16: Heat Pump Savings#

Consider a house with an average heating load of \(5\:kW\) over the three months on winter in Vancouver.

a) calculate the heating cost over the winter assuming an electric heater is used (12.21 cents per \(kWh\) for electricity in Vancouver).

Now consider using a heat pump with a \(COP=2.5\) to supply heat.

b) calculate the average electrivity consumption over the winter

c) calculate the heating cost over the winter assuming a similar heat pump is used

d) assuming a unit price of \(250\:CAD\) for a \(5\:kW\) electric heater and a unit price of \(1300\:CAD\) for a heat pump with the same heating load, how many years will take to justify the application of a heat pump over an electric heater?

Solution Approach for a)#

assuming a 90-day peried for the three months of winter and 24-hours per day

#define variables
P = 12.21   #electricity price in cents per kWh
Q_H_avg = 5E+3   #average heating load in W

Q_H = Q_H_avg * 90 * 24 / 1000  #total heating load over the winter in kWh
P_Q_H = Q_H * P /100   #total heating price in winter in CAD

print('The total heating price using an electric heater is:', f"{P_Q_H:.2f}", 'CAD')

The total heating price using an electric heater is: 1318.68 CAD

Solution Approach for b)#

for a heat pump

\(COP=Q_H/W\)

therefore

\(W=Q_H/COP\)

#define variables
COP = 2.5   #heat pump COP

W_avg = Q_H_avg / COP   #average heating electrivity usage over the winter in W

print('The average heating electrivity consumption over the winter using a heat pump is:', f"{W_avg/1000:.2f}", 'kW')

The average heating electrivity consumption over the winter using a heat pump is: 2.00 kW

Solution Approach for c)#

W = W_avg * 90 * 24 / 1000  #total electricity consumption over the winter in kWh
P_W = W * P /100   #total heating electrivity price in winter in CAD

print('The total heating electrivity price using a heat pump is:', f"{P_W:.2f}", 'CAD')

The total heating electrivity price using a heat pump is: 527.47 CAD

Solution Approach for d)#

#define prices
U_E = 250   #unit price for a electric heater in CAD
U_H = 1300   #unit price for a heat pump in CAD

U_D = U_H - U_E   #unit price difference in CAD
U_P = P_Q_H - P_W   #savings per winter using a heat pump in CAD

Y = U_D / U_P   #number of years to make up the price difference

print('It will take', f"{Y:.1f}", 'years for the heat pump to save the unit price difference in electrivity consumption')

It will take 1.3 years for the heat pump to save the unit price difference in electrivity consumption
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