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 m 3 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 m 3 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?

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Answer 1

Textbook Question

Chapter 6.11, Problem 124P

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?

Expert Solution & Answer

To determine

The cost of the energy wasted per year.

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

Answer to Problem 124P

The cost of the energy wasted per year is $1.85/year_.

The cost of the energy in the room of dry air is $0.96/year_.

Explanation of Solution

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

νair,replaced=(total volume of refrigerated)×(no. of day) (I)

Determine the density of the air.

ρ=PRT (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.

mair=ρνair (III)

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

mmoisture=mair×(mositure removed per kg air) (IV)

Determine the sensible heat gain of the refrigerated space.

Qgain,sensible=maircp(Troom−Trefrig) (V)

Determine the latent heat gain of the refrigerated space.

Qgain,latent=mmoisturehfg (VI)

Here, the heat of vaporization of water is hfg

Determine the total heat gains of the refrigerated space.

Qgain,total=Qgain,sensible+Qgain,latent (VII)

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

Electrical energy used (total)=Qgain,totalCOP (VIII)

Determine the cost of energy wasted per year.

Cost of energy (total)=[(Electrical energy used)×(unit cost of energy)] (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.

Electrical energy used(sensible)=Qgain,sensibleCOP (X)

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

Cost of energy(sensible)=[(Electrical energy used)×(unit cost of energy)] (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 kJ/kg⋅K and 1.005 kJ/kg⋅°C.

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

Write the formula of interpolation method of two variables.

y2=(x2−x1)(y3−y1)(x3−x1)+y1 (IV)

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

Show the temperature at 0.01°C and 5°C as in Table (1).

S. No	Temperature, °C (x)	enthalpy of vaporization kJ/kg (y)
1	0.01°C	2500.9 kJ/kg
2	4°C	y2=?
3	5°C	2489.1 kJ/kg

Calculate heat of vaporization of water at 4°C of temperature for liquid phase using interpolation method.

Substitute 0.01°C for x1, 4°C for x2, 5°C for x3, 2500.9 kJ/kg for y1, and 2489.1 kJ/kg for y3 in Equation (IV).

y2=(4°C−0.01°C)(2500.9 kJ/kg−2489.1 kJ/kg)(5°C−0.01°C)+2489.1 kJ/kg=2491.465 kJ/kg

From above calculation the heat of vaporization of water at 4°C of temperature is 2491.465 kJ/kg.

Substitute 0.3 m3 for total volume of refrigerated and 20/day for number of day in Equation (I).

νair,replaced=(0.3 m3)×(20/day)=(0.3 m3)×(20/day)×(356 days/year)=2190 m3/year

Substitute 95 kPa for P, 0.287 kPa⋅m3/kg⋅K for R, and 4°C for T in Equation (I).

ρ=95 kPa(0.287 kPa⋅m3/kg⋅K)(4°C)=95 kPa(0.287 kPa⋅m3/kg⋅K)(4°C+273)=95 kPa(0.287 kPa⋅m3/kg⋅K)(277 K)=1.1949 kg/m3

≅1.195 kg/m3

Substitute 1.1949 kg/m3 for ρ and 2190 m3/year for νair in Equation (II).

mair=(1.1949 kg/m3)×(2190 m3/year)=2617 kg/year

Substitute 2617 kg/year for mair and 0.006 kg/kg air for moisture removed per kg air in Equation (III).

mmoisture=(2617 kg/year)×(0.006 kg/kg air)=15.70 kg/year

Substitute 2617 kg/year for mair, 1.005 kJ/kg⋅°C for cp, 20°C for Troom, 4°C for Trefrig in Equation (IV).

Qgain,sensible=(2617 kg/year)(1.005 kJ/kg⋅°C)(20−4)°C=(2617 kg/year)(1.005 kJ/kg⋅°C)(16°C)=42081.36 kJ/year

Substitute 15.70 kg/year for mmositure and 2491.465 kJ/kg for hfg in Equation (V).

Qgain,latent=(15.70 kg/year)×(2491.465 kJ/kg)=(15.70 kg/year)×(2491.465 kJ/kg)=39116 kJ/year

Substitute 42081.36 kJ/year for Qgain,sensible and 39116 kJ/year for Qgain,latent in Equation (VI).

Qgain,total=(42081.36 kJ/year)+(39116 kJ/year)=81197.36 kJ/year

Substitute 81197.36 kJ/year for Qgain,total and 1.4 for COP in Equation (VII).

Electrical energy used (total)=(81197.36 kJ/year)1.4=57998.11 kJ/year×(1 kWh3600 kJ)=16.11 kWh/year

Substitute 16.11 kWh/year for electrical energy used and $0.115/kWh for unit cost of energy in Equation (VIII).

Cost of energy (total)=[(16.11 kWh/year)×($0.115/kWh)]=$1.85/year

Thus, the cost of the energy wasted per year is $1.85/year_.

Substitute 42,081.36 kJ/year for Qgain,total and 1.4 for COP in Equation (VII).

Electrical energy used (total)=(42,081.36 kJ/year)1.4=30058.11 kJ/year×(1 kWh3600 kJ)=8.349 kWh/year≅8.35 kWh/year

Substitute 8.35 kWh/year for electrical energy used and $0.115/kWh for unit cost of energy in Equation (VIII).

Cost of energy (total)=[(8.35 kWh/year)×($0.115/kWh)]=$0.96/year

Thus, the cost of the energy in the room of dry air is $0.96/year_.

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Textbook Question

Chapter 6.11, Problem 124P

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?

Expert Solution & Answer

To determine

The cost of the energy wasted per year.

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

Answer to Problem 124P

The cost of the energy wasted per year is $1.85/year_.

The cost of the energy in the room of dry air is $0.96/year_.

Explanation of Solution

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

νair,replaced=(total volume of refrigerated)×(no. of day) (I)

Determine the density of the air.

ρ=PRT (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.

mair=ρνair (III)

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

mmoisture=mair×(mositure removed per kg air) (IV)

Determine the sensible heat gain of the refrigerated space.

Qgain,sensible=maircp(Troom−Trefrig) (V)

Determine the latent heat gain of the refrigerated space.

Qgain,latent=mmoisturehfg (VI)

Here, the heat of vaporization of water is hfg

Determine the total heat gains of the refrigerated space.

Qgain,total=Qgain,sensible+Qgain,latent (VII)

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

Electrical energy used (total)=Qgain,totalCOP (VIII)

Determine the cost of energy wasted per year.

Cost of energy (total)=[(Electrical energy used)×(unit cost of energy)] (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.

Electrical energy used(sensible)=Qgain,sensibleCOP (X)

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

Cost of energy(sensible)=[(Electrical energy used)×(unit cost of energy)] (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 kJ/kg⋅K and 1.005 kJ/kg⋅°C.

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

Write the formula of interpolation method of two variables.

y2=(x2−x1)(y3−y1)(x3−x1)+y1 (IV)

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

Show the temperature at 0.01°C and 5°C as in Table (1).

S. No	Temperature, °C (x)	enthalpy of vaporization kJ/kg (y)
1	0.01°C	2500.9 kJ/kg
2	4°C	y2=?
3	5°C	2489.1 kJ/kg