Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 15 - The Laws of Thermodynamics - General Problems - Page 442: 77

Answer

a) $-4^\circ \;\rm C$ b) $29\%$

Work Step by Step

a) We need to maintain the heat temperature inside the house constant, so the heat added by the heat pump must be equal to the heat loss by conduction. We are given the rate of heat loss by conduction and we know that the rate of heat added by the pump is $Q_h/t$. $$\dfrac{Q_H}{t}=650(T_{in}-T_{out})$$ Recall that $Q_H=W+Q_L$; $$\dfrac{W+Q_L}{t}=650(T_{in}-T_{out})\tag 1$$ The author mentioned that the pump is ideal, so its efficiency is given by $$e=1-\dfrac{T_L}{T_H}=\dfrac{W}{Q_H}=\dfrac{W}{Q_L+W}$$ noting that $T_H$ is the inside temperature since we are heating the house and $T_L$ is the outside temperature. So that $$ 1-\dfrac{T_{out}}{T_{in}} =\dfrac{W}{Q_L+W}$$ Thus, $$Q_L+W=\dfrac{W}{\left[1-\dfrac{T_{out}}{T_{in}} \right]}=W\left[\dfrac{T_{in}}{T_{in}-T_{out}}\right]$$ Plugging into (1); $$\dfrac{W}{t}\left[\dfrac{T_{in}}{T_{in}-T_{out}}\right]=650(T_{in}-T_{out})$$ $$\left(\dfrac{W}{t}\right)\;T_{in}=650(T_{in}-T_{out})^2$$ Solving for $T_{out}$; $$T_{in}-T_{out} =\sqrt{\dfrac{\left(\dfrac{W}{t}\right)\;T_{in}}{650}}$$ $$T_{out}=T_{in}- \sqrt{\dfrac{\left(\dfrac{W}{t}\right)\;T_{in}}{650}}$$ Plugging the known; $$T_{out}=22+273- \sqrt{\dfrac{1500\cdot (22 +273) }{650}}=269\;\rm K$$ $$\boxed{T_{out}=\color{red}{\bf-4}^\circ \;\rm C}$$ ---- b) Since we know the outside temperature of $8^\circ$ C, so we can find the heat loss by conduction from the given formula of $$\dfrac{Q_H}{t}=650\left(T_{in}-T_{out}\right)=650(22-8)=\bf 9100\;\rm W$$ This amount of heat loss is the same amount provided by the heat pump to the house since we need to maintain a constant temperature. Noting that the total power provided by the heat pump which is given by $$P=\dfrac{Q_L+W}{t}=9100 $$ Plugging from (2); $$\dfrac{W\left[\dfrac{T_{in}}{T_{in}-T_{out}}\right]}{t}=9100 $$ Thus, $$\dfrac{W}{t}=9100\cdot \left[\dfrac{T_{in}-T_{out}}{T_{in}}\right] $$ Plugging the known; $$\dfrac{W}{t}=9100\cdot \left[\dfrac{22+272-(8+273)}{22+273}\right] =\bf432\;\rm W$$ So, the power needed is 432 W while the maximum power the plant can apply is 1500 W. So, the time percentage to work at this low power is $$\dfrac{t}{t_{max}}=\dfrac{P}{P_{max}}\times100\%=\dfrac{432}{1500}\times100\%$$ $$t\approx \color{red}{\bf29\%}\;t_{max} $$ It is about 30$\%$ of the time needed to work on full power.
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