Chapter 6.11, Problem 118P

It is often stated that the refrigerator door should be opened as few times as possible for the shortest duration of time to save energy. Consider a household refrigerator whose interior volume is 0.9 m³ and average internal temperature is 4°C. At any given time, one-third of the refrigerated space is occupied by food items, and the remaining 0.6 m³ is filled with air. The average temperature and pressure in the kitchen are 20°C and 95 kPa, respectively. Also, the moisture contents of the air in the kitchen and the refrigerator are 0.010 and 0.004 kg per kg of air, respectively, and thus 0.006 kg of water vapor is condensed and removed for each kg of air that enters. The refrigerator door is opened an average of 20 times a day, and each time half of the air volume in the refrigerator is replaced by the warmer kitchen air. If the refrigerator has a coefficient of performance of 1.4 and the cost of electricity is $0.115/kWh, determine the cost of the energy wasted per year as a result of opening the refrigerator door. What would your answer be if the kitchen air were very dry and thus a negligible amount of water vapor condensed in the refrigerator?

To determine

The cost of the energy wasted per year.

The cost of the energy in the room of dry air.

Answer to Problem 118P

The cost of the energy wasted per year is $\underset{\_}{\text{\$1}\text{.85}/\text{year}}$.

The cost of the energy in the room of dry air is $\underset{\_}{\text{\$0}\text{.96}/\text{year}}$.

Explanation of Solution

Determine the total volume of refrigerated air replaced by room air per year.

${\nu }_{\text{air,replaced}}=\left(\text{total}\text{volume}\text{of}\text{refrigerated}\right)\times \left(\text{no}\text{.}\text{of}\text{day}\right)$ (I)

Determine the density of the air.

$\rho =\frac{P}{RT}$ (II)

Here, the air pressure is $P$, the universal gas constant of air is $R$, and the temperature of air is $T$.

Determine the mass of the air.

$m_{\text{air}}=\rho {\nu }_{\text{air}}$ (III)

Determine the amount of moisture condensed and removed by the refrigerator.

$m_{\text{moisture}}=m_{\text{air}}\times \left(\text{mositure}\text{removed}\text{per}\text{kg}\text{air}\right)$ (IV)

Determine the sensible heat gain of the refrigerated space.

$Q_{\text{gain,sensible}}=m_{\text{air}}{c}_p\left(T_{\text{room}}-T_{\text{refrig}}\right)$ (V)

Determine the latent heat gain of the refrigerated space.

$Q_{\text{gain,latent}}=m_{\text{moisture}}{h}_{fg}$ (VI)

Here, the heat of vaporization of water is $h_{fg}$

Determine the total heat gains of the refrigerated space.

$Q_{\text{gain,total}}=Q_{\text{gain,sensible}}+Q_{\text{gain,latent}}$ (VII)

Determine the amount of electrical energy the refrigerator will consume to remove this heat from the refrigerated space.

$\text{Electrical}\text{energy}\text{used}\left(\text{total}\right)=\frac{Q_{\text{gain,total}}}{\text{COP}}$ (VIII)

Determine the cost of energy wasted per year.

$\text{Cost}\text{of}\text{energy}\left(\text{total}\right)=\left[\begin{array}{l}\left(\text{Electrical energy}\text{used}\right)\\ \times \left(\text{unit cost}\text{of}\text{energy}\right)\end{array}\right]$ (IX)

Determine the amount of electrical energy the refrigerator will consume to remove this heat from the refrigerated space in dry air of the room.

$\text{Electrical}\text{energy}\text{used}\left(\text{sensible}\right)=\frac{Q_{\text{gain,sensible}}}{\text{COP}}$ (X)

Determine the cost of energy in the room of dry air.

$\text{Cost}\text{of}\text{energy}\left(\text{sensible}\right)=\left[\begin{array}{l}\left(\text{Electrical energy}\text{used}\right)\\ \times \left(\text{unit cost}\text{of}\text{energy}\right)\end{array}\right]$ (XI)

Conclusion:

From the Table A-1, “Ideal-gas specific heats of various common gases” to obtain the value of universal gas constant and specific heat of air at 300 K temperature as $0.2870\text{kJ}/\text{kg}\cdot \text{K}$ and $1.005\text{kJ}/\text{kg}\cdot °\text{C}$.

Refer to Table A-4, “Saturated water-Temperature”, to obtain the value heat of vaporization of water at $4°\text{C}$ of temperature using interpolation method of two variables.

Write the formula of interpolation method of two variables.

$y_2=\frac{\left(x_2-x_1\right)\left(y_3-y_1\right)}{\left(x_3-x_1\right)}+y_1$ (IV)

Here, the variables denote by x and y are temperature and enthalpy of vaporization.

Show the temperature at $0.01°\text{C}$ and $5°\text{C}$ as in Table (1).

S. No	Temperature, $°\text{C}$ $\left(x\right)$	enthalpy of vaporization  $\text{kJ}/\text{kg}$ $\left(y\right)$
1	$0.01°\text{C}$	$2500.9\text{kJ}/\text{kg}$
2	$4°\text{C}$	$y_2=?$
3	$5°\text{C}$	$2489.1\text{kJ}/\text{kg}$

Calculate heat of vaporization of water at $4°\text{C}$ of temperature for liquid phase using interpolation method.

Substitute $0.01°\text{C}$ for $x_1$, $4°\text{C}$ for $x_2$, $5°\text{C}$ for $x_3$, $2500.9\text{kJ}/\text{kg}$ for $y_1$, and $2489.1\text{kJ}/\text{kg}$ for $y_3$ in Equation (IV).

$\begin{array}{c}{y}_2=\frac{\left(4°\text{C}-0.01°\text{C}\right)\left(2500.9\text{kJ}/\text{kg}-2489.1\text{kJ}/\text{kg}\right)}{\left(5°\text{C}-0.01°\text{C}\right)}+2489.1\text{kJ}/\text{kg}\\ =2491.465\text{kJ}/\text{kg}\end{array}$

From above calculation the heat of vaporization of water at $4°\text{C}$ of temperature is $2491.465\text{kJ}/\text{kg}$.

Substitute $0.3{\text{m}}^{\text{3}}$ for total volume of refrigerated and $\text{20}/\text{day}$ for number of day in Equation (I).

$\begin{array}{c}{\nu }_{\text{air,replaced}}=\left(\text{0}\text{.3}{\text{m}}^3\right)\times \left(20/\text{day}\right)\\ =\left(\text{0}\text{.3}{\text{m}}^3\right)\times \left(20/\text{day}\right)\times \left(\text{356}\text{days}/\text{year}\right)\\ =2190{\text{m}}^{\text{3}}/\text{year}\end{array}$

Substitute 95 kPa for $P$, $0.287\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}$ for $R$, and $4°\text{C}$ for $T$ in Equation (I).

$\begin{array}{c}\rho =\frac{95\text{kPa}}{\left(0.287\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}\right)\left(4°\text{C}\right)}\\ =\frac{95\text{kPa}}{\left(0.287\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}\right)\left(4°\text{C+273}\right)}\\ =\frac{95\text{kPa}}{\left(0.287\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}\right)\left(\text{277}\text{K}\right)}\\ =1.1949\text{kg}/{\text{m}}^{\text{3}}\end{array}$

   $\cong 1.195\text{kg}/{\text{m}}^{\text{3}}$

Substitute $1.1949\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$ and $2190{\text{m}}^{\text{3}}/\text{year}$ for ${\nu }_{\text{air}}$ in Equation (II).

$\begin{array}{c}{m}_{\text{air}}=\left(1.1949\text{kg}/{\text{m}}^{\text{3}}\right)\times \left(2190{\text{m}}^{\text{3}}/\text{year}\right)\\ =2617\text{kg}/\text{year}\end{array}$

Substitute $2617\text{kg}/\text{year}$ for $m_{\text{air}}$ and $\text{0}\text{.006}\text{kg}/\text{kg}\text{air}$ for moisture removed per kg air in Equation (III).

$\begin{array}{c}{m}_{\text{moisture}}=\left(2617\text{kg}/\text{year}\right)\times \left(\text{0}\text{.006}\text{kg}/\text{kg}\text{air}\right)\\ =15.70\text{kg}/\text{year}\end{array}$

Substitute $2617\text{kg}/\text{year}$ for $m_{\text{air}}$, $1.005\text{kJ}/\text{kg}\cdot °\text{C}$ for $c_p$, $20°\text{C}$ for $T_{\text{room}}$, $4°\text{C}$ for $T_{\text{refrig}}$ in Equation (IV).

$\begin{array}{c}{Q}_{\text{gain,sensible}}=\left(2617\text{kg}/\text{year}\right)\left(1.005\text{kJ}/\text{kg}\cdot °\text{C}\right)\left(20-4\right)°\text{C}\\ =\left(2617\text{kg}/\text{year}\right)\left(1.005\text{kJ}/\text{kg}\cdot °\text{C}\right)\left(16°\text{C}\right)\\ =42081.36\text{kJ}/\text{year}\end{array}$

Substitute $15.70\text{kg}/\text{year}$ for $m_{\text{mositure}}$ and $2491.465\text{kJ}/\text{kg}$ for $h_{fg}$ in Equation (V).

$\begin{array}{c}{Q}_{\text{gain,latent}}=\left(15.70\text{kg}/\text{year}\right)\times \left(2491.465\text{kJ}/\text{kg}\right)\\ =\left(15.70\text{kg}/\text{year}\right)\times \left(2491.465\text{kJ}/\text{kg}\right)\\ =39116\text{kJ}/\text{year}\end{array}$

Substitute $42081.36\text{kJ}/\text{year}$ for $Q_{\text{gain,sensible}}$ and $39116\text{kJ}/\text{year}$ for $Q_{\text{gain,latent}}$ in Equation (VI).

$\begin{array}{c}{Q}_{\text{gain,total}}=\left(42081.36\text{kJ}/\text{year}\right)+\left(39116\text{kJ}/\text{year}\right)\\ =81197.36\text{kJ}/\text{year}\end{array}$

Substitute $81197.36\text{kJ}/\text{year}$ for $Q_{\text{gain,total}}$ and 1.4 for $\text{COP}$ in Equation (VII).

$\begin{array}{c}\text{Electrical}\text{energy}\text{used}\left(\text{total}\right)=\frac{\left(81197.36\text{kJ}/\text{year}\right)}{\text{1}\text{.4}}\\ =57998.11\text{kJ}/\text{year}\times \left(\frac{1\text{kWh}}{3600\text{kJ}}\right)\\ =16.11\text{kWh}/\text{year}\end{array}$

Substitute $16.11\text{kWh}/\text{year}$ for electrical energy used and $\text{\$0}\text{.115}/\text{kWh}$ for unit cost of energy in Equation (VIII).

$\begin{array}{c}\text{Cost}\text{of}\text{energy}\left(\text{total}\right)=\left[\begin{array}{l}\left(16.11\text{kWh}/\text{year}\right)\\ \times \left(\text{\$0}\text{.115}/\text{kWh}\right)\end{array}\right]\\ =\text{\$1}\text{.85}/\text{year}\end{array}$

Thus, the cost of the energy wasted per year is $\underset{\_}{\text{\$1}\text{.85}/\text{year}}$.

Substitute $42,081.36\text{kJ}/\text{year}$ for $Q_{\text{gain,total}}$ and 1.4 for $\text{COP}$ in Equation (VII).

$\begin{array}{c}\text{Electrical}\text{energy}\text{used}\left(\text{total}\right)=\frac{\left(42,081.36\text{kJ}/\text{year}\right)}{\text{1}\text{.4}}\\ =30058.11\text{kJ}/\text{year}\times \left(\frac{1\text{kWh}}{3600\text{kJ}}\right)\\ =8.349\text{kWh}/\text{year}\\ \cong 8.35\text{kWh}/\text{year}\end{array}$

Substitute $8.35\text{kWh}/\text{year}$ for electrical energy used and $\text{\$0}\text{.115}/\text{kWh}$ for unit cost of energy in Equation (VIII).

$\begin{array}{c}\text{Cost}\text{of}\text{energy}\left(\text{total}\right)=\left[\begin{array}{l}\left(8.35\text{kWh}/\text{year}\right)\\ \times \left(\text{\$0}\text{.115}/\text{kWh}\right)\end{array}\right]\\ =\text{\$0}\text{.96}/\text{year}\end{array}$

Thus, the cost of the energy in the room of dry air is $\underset{\_}{\text{\$0}\text{.96}/\text{year}}$.