A heat pump needs to supply 20 kW of power to maintain the inside temperature of a house at 25 $\circ$ C while the outside temperature is 0 $\circ$ C.What is the savings in the electric bill per month by using the heat pump instead of direct electrical heating? Assume that the heat pump/direct heating is on for 4 hours per day and 1 kW.hr costs $0.10.

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Quizplus · Accepted Answer

To calculate the savings, we first need to determine the energy consumption and cost for both the heat pump and direct electrical heating.For direct electrical heating, the energy required is directly equal to the heat needed. So, for 4 hours per day:Energy required per day = 20 kW * 4 hours = 80 kWhEnergy required per month (assuming 30 days) = 80 kWh/day * 30 days = 2400 kWhCost for direct heating per month = 2400 kWh * $0.10/kWh = $240For the heat pump, we use the Coefficient of Performance (COP) to find its efficiency. The COP for a heat pump is given by:$$COP = \frac{T_{hot}}{T_{hot} - T_{cold}}$$where $T_{hot}$ and $T_{cold}$ are the temperatures inside and outside, respectively, in Kelvin.Converting the temperatures to Kelvin:$$T_{hot} = 25°C + 273.15 = 298.15 K$$$$T_{cold} = 0°C + 273.15 = 273.15 K$$$$COP = \frac{298.15}{298.15 - 273.15} = \frac{298.15}{25} = 11.926$$The effective energy consumption of the heat pump per day is:$$Energy = \frac{Energy required}{COP} = \frac{80 kWh}{11.926} \approx 6.71 kWh$$Energy required per month = 6.71 kWh/day * 30 days = 201.3 kWhCost for heat pump heating per month = 201.3 kWh * $0.10/kWh = $20.13Savings per month = Cost for direct heating - Cost for heat pump heatingSavings = $240 - $20.13 = $219.87Rounding to the nearest whole number, the savings are approximately $220 per month.

A Heat Pump Needs to Supply 20 KW of Power ∘\circ∘

A Heat Pump Needs to Supply 20 KW of Power $\circ$