Chapter 6.11, Problem 150RP

To determine

The cost of energy “vented out” by the fans in 1 h.

Answer to Problem 150RP

The cost of energy “vented out” by the fans in 1 h is $\underset{\_}{\text{\$0}\text{.123}}$.

Explanation of Solution

Determine the density of air at the indoor conditions.

${\rho }_o=\frac{P_o}{R{T}_o}$ (I)

Here, the house maintain a pressure is $P_o$, the universal gas constant is $R$, and the house maintain a temperature is $T_0$.

Determine interior volume of the house per hour

${\dot{\nu }}_{\text{air}}=w\times l$ (II)

Here, the width of the house is $w$ and the length of the house is $l$.

Determine the mass flow rate of air vented out.

${\dot{m}}_{\text{air}}={\rho }_o{\dot{\nu }}_{\text{air}}$ (III)

Determine the rate of energy loss by the ventilating fans.

${\dot{Q}}_{\text{loss,fan}}={\dot{m}}_{\text{air}}{c}_p\left(T_{\text{indoor}}-T_{\text{outdoor}}\right)$ (IV)

Here, the specific heat of air at room temperature is $c_P$, the indoor air vented out at temperature is $T_{\text{indoor}}$, and the outdoor air temperature is $T_{\text{outdoor}}$.

Determine the amount of fuel energy loss by “vented out”.

$\text{Fuel energy loss}={\dot{Q}}_{\text{loss,fan}}\times \Delta t/{\eta }_{\text{furnace}}$ (V)

Here, the fan takes time is $\Delta t$ and the efficiency of the furnace is ${\eta }_{\text{furnace}}$.

Determine the amount of cost of the heat “vented out” per hour.

$\text{Money}\text{loss}=\left(\text{Fuel}\text{energy}\text{loss}\right)\times \left(\text{Unit}\text{cost}\text{of}\text{energy}\right)$ (VI)

Conclusion:

From the Table A-1, “Molar mass, gas constant, and critical-point properties” to obtain the value of gas constant of air as $0.287\text{kPa}\cdot {\text{m}}^3/\text{kg}\cdot \text{K}$.

From the Table A-2a, “Ideal-gas specific heats of various common gases” to obtain the value of specific heat of air at room temperature as $1.0\text{kJ}/\text{kg}\cdot °\text{C}$.

Substitute 92 kPa for $P_o$, $0.287\text{kPa}\cdot {\text{m}}^3/\text{kg}\cdot \text{K}$ for $R$, and $22°\text{C}$ for $T_o$ in Equation (I).

$\begin{array}{c}{\rho }_o=\frac{92\text{kPa}}{\left(0.287\text{kPa}\cdot {\text{m}}^3/\text{kg}\cdot \text{K}\right)\times \left(22°\text{C}\right)}\\ =\frac{92\text{kPa}}{\left(0.287\text{kPa}\cdot {\text{m}}^3/\text{kg}\cdot \text{K}\right)\times \left(22°\text{C+273}\right)}\\ =\frac{92\text{kPa}}{84.665\text{kPa}\cdot {\text{m}}^3/\text{kg}}\\ =1.0866\text{kg}/{\text{m}}^{\text{3}}\end{array}$

     $\cong 1.087\text{kg}/{\text{m}}^{\text{3}}$

Substitute $200{\text{m}}^{\text{2}}/\text{h}$ for $w$ and $2.8\text{m}/\text{h}$ for $l$ in Equation (II).

$\begin{array}{c}{\dot{\nu }}_{\text{air}}=\left(200{\text{m}}^{\text{2}}/\text{h}\right)\times \left(2.8\text{m}/\text{h}\right)\\ =\text{560}{\text{m}}^3/\text{h}\end{array}$

Substitute $1.087\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_o$ and $\text{560}{\text{m}}^3$ for ${\dot{\nu }}_{\text{air}}$ in Equation (III).

$\begin{array}{c}{\dot{m}}_{\text{air}}=\left(1.087\text{kg}/{\text{m}}^{\text{3}}\right)\times \left(\text{560}{\text{m}}^3/\text{h}\right)\\ =608.72\text{kg}/\text{h}\\ =608.72\text{kg}/\text{h}\times \left(\frac{1\text{kg}/\text{s}}{3600\text{kg}/\text{h}}\right)\\ =0.169089\text{kg}/\text{s}\end{array}$

Substitute $0.169089\text{kg}/\text{s}$ for ${\dot{m}}_{\text{air}}$, $1.0\text{kJ}/\text{kg}\cdot °\text{C}$ for $c_p$, $22°\text{C}$ for $T_{\text{indoor}}$ and $5°\text{C}$ for $T_{\text{outdoor}}$ in Equation (IV).

$\begin{array}{c}{\dot{Q}}_{\text{loss,fan}}=\left(0.169089\text{kg}/\text{s}\right)\times \left(1.0\text{kJ}/\text{kg}\cdot °\text{C}\right)\times \left(22°\text{C}-5°\text{C}\right)\\ =\left(0.169089\text{kg}/\text{s}\right)\times \left(1.0\text{kJ}/\text{kg}\cdot °\text{C}\right)\times \left(17°\text{C}\right)\\ =2.8745\text{kJ}/\text{s}\\ =2.8745\text{kJ}/\text{s}\times \left(\frac{1\text{kW}}{1\text{kJ}/\text{s}}\right)\end{array}$

          $=2.8745\text{kW}$

Substitute $2.8745\text{kW}$ for ${\dot{Q}}_{\text{loss,fan}}$, 1 h for $\Delta t$, and 0.96 for ${\eta }_{\text{furnace}}$ in Equation (V).

$\begin{array}{c}\text{Fuel energy loss}=\left(2.8745\text{kW}\right)\times \left(1\text{h}\right)/\left(0.96\right)\\ =\left(2.8745\text{kW}\right)\times \left(1.041667\text{h}\right)\\ =2.994\text{kWh}\end{array}$

Substitute $2.994\text{kWh}$ for fuel energy loss, $\text{\$1}\text{.20}/\text{therm}$ for unit cost of energy in Equation (VI).

$\begin{array}{c}\text{Money}\text{loss}=\left(2.994\text{kWh}\right)\times \left(\text{\$1}\text{.20}/\text{therm}\right)\\ =\left(2.994\text{kWh}\right)\times \left(\text{\$1}\text{.20}/\text{therm}\right)\times \left(\frac{1\text{therm}}{29.3\text{kWh}}\right)\\ =\$0.1226\\ \cong \$0.123\end{array}$

Thus, the cost of energy “vented out” by the fans in 1 h is $\underset{\_}{\text{\$0}\text{.123}}$.