A heat pump with a COP of 2.8 is used to heat an air-tight house. When running, the heat pump...

Question:

A heat pump with a COP of 2.8 is used to heat an air-tight house. When running, the heat pump consumes 5 kW of power. If the temperature in the house is 7{eq}^{\circ} {/eq}C when the heat pump is turned on, how long will it take for the heat pump to raise the temperature of the house to 22{eq}^{\circ} {/eq}C? Is this answer realistic or optimistic? Explain. Assume the entire mass within the house (air, furniture, etc.) is equivalent to 1500 kg of air

Heat Pump:

The heat pump is an electrical device whose performance is measured from the coefficient of performance. Heat pumps are more energy-efficient than other alternatives like air conditioners and furnace.

Answer and Explanation:


Given Data


  • COP of heat pump, {eq}CO{P_{HP}} = 2.8 {/eq}.
  • Power consumed, {eq}\dot W = 5\;{\rm{kW}} {/eq}.
  • Initial temperature, {eq}{T_i} = 7^\circ {\rm{C}} {/eq}.
  • Final temperature, {eq}{T_f} = 22^\circ {\rm{C}} {/eq}.
  • Entire mass in house, {eq}m = 1500\;{\rm{kg}} {/eq}.


The expression of COP of heat pump is,

{eq}\begin{align*} CO{P_{HP}} &= \dfrac{{{{\dot Q}_H}}}{{\dot W}}\\ {{\dot Q}_H} &= CO{P_{HP}} \times \dot W \end{align*} {/eq}


Here, {eq}{\dot Q_H} {/eq} is the amount of heat lost in the room.


Substituting the values in above expression.

{eq}\begin{align*} {{\dot Q}_H} &= 2.8 \times 5\\ {{\dot Q}_H} &= 14\;{\rm{kW}} \end{align*} {/eq}


The expression for total amount of heat to be supplied is,

{eq}Q = m{c_v}\Delta T {/eq}


Here, {eq}{c_v} {/eq} is the specific heat at constant volume for air whose value is {eq}0.718\;{\rm{kJ/kg}} \cdot ^\circ {\rm{C}} {/eq}.

{eq}\Delta T = \left( {{T_f} - {T_i}} \right) {/eq} is the change in temperature.


Substituting the values in above expression.

{eq}\begin{align*} Q &= 1500\left( {0.718} \right)\left( {22 - 7} \right)\\ Q &= 16,155\;{\rm{kJ}} \end{align*} {/eq}


The expression for time required to heat the pump is,

{eq}\Delta t = \dfrac{Q}{{{{\dot Q}_H}}} {/eq}


Substituting the values in above expression.

{eq}\begin{align*} \Delta t &= \dfrac{{16,155}}{{14}}\\ \Delta t &= 1153.928\;{\rm{s}}\\ \Delta t &= 19.23\;{\rm{min}} \end{align*} {/eq}


Thus, the time required for the heat pump to raise the temperature is {eq}19.23\;{\rm{min}} {/eq}.


The answer is optimistic because, some heat will be absorbed by the substance which is kept inside the room. So the pump will take more time to raise the temperature.


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