Friday, April 14, 2006

HW #6 - P #8 - Cost of Opening the Refrigerator Door - 8 pts

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 4oC. 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 20oC and 95 kPa, respectively. Also, the absolute humidity of the air in the kitchen and the air inthe refrigerator are 0.010 and 0.004 kg water / kg dry air (BDA), respectively. Thus, 0.006 kg of water vapor is condensed and removed from the air in the refrigerator for each kg of BDA that enters the refrigerator.

The refrigerator door is opened an average of 8 times a day. Each time the door is opened, half of the air volume in the refrigerator is replaced by the warmer, more humid kitchen air. If the refrigerator has a COP of 1.4 and the cost of electricity is $0.075 / kW-h, determine :
a.) the cost of the energy wasted per year as a result of opening the refrigerator door.
b.) What would your answer be if the kitchen air were very dry and thus a negligible amount of water condensed in the refrigerator ?

6 comments:

Anonymous said...

Dr.B

We are somewhat confused about the the process you described in the hint. First of all, in case to condense the water vapor in the air, it takes Q to do it. Second of all, it takes Q to cool the humid air down. Am I right about this?

First, with IG law, we found molar V of air in the kitchen. From that, we solved mass of air and mass of water in the humid air that's coming into the fridge. Mass of air we got is .3389 Kg, and .00203 kg of water. Then we did: Qh= dt*Cp*mass of water. From the Qh and COP, we solved for work. The work we got was 0.0136 KJ. Then we did: W*8/24hr to get KJ/hr. Where did we go wrong?

Dr. B said...

Kaifu 3:12 PM
You are correct. The total heat load that the refrigeration cycle must remove from the air is the Sum of the heat required to cool the air AND the latent heat that must beremoved to condense the water vapor in the air.

Did you get 0.3389 kg of humid air EACH time the fridge is opened ?

The Q you are trying to calculate is Qc (it is removed from the cold reservoir). This is NOT QH. QH is rejected to the kitchen. This Qc, as you pointed out is more than just the sensible heat of cooling down the humid air. You musst also consider the energy that must be removed from the humid air to condense the water out of it. Look up the heat of vaporization of water.

Dr. B said...

I received some very general questions about this problem by email. Here are some hints.

When you open the fridge door, 0.3 m^3 of warm humid air comes inside. Use the IG EOS to determine the mass of humid air that enters the fridge. Because the mass fraction of water in the air is small, you can just use the MW of air here. This air must be cooled down to 4 degC. This is a sensible heat. Use the Cp of humid air given in the problem statement. In addition, water must be condensed out of the air to reduce the absolute humidity. This is a latent heat. You can use the latent heat of vaporization of water at 5 degC or you caninterpolate to determine the value at 4 degC if you want to be more accurate.

Add the sensible and latent heats to determine the total heat duty each time the fridge door is open. Use the COP to determine the required W input. Figure out how many times the fridge is opened each year. Keep track of your units and you should get this one, no problem. Keep in mind that 1 kWh = 3600 kJ.

Anonymous said...

what numbers do we use to show we can use the IG EOS?

Anonymous said...

Anonymous 10:31 AM

Use the worst case T (the one that yields the smallest molar volume. That would be 4 degC. The pressure is 95 kPA everywhere in this problem. Don't worry about the small amount of water in the air when determining whether the ideal gas EOS is valid.

Anonymous said...

HW 7 is sooo hard.. post a blog Dr B!!!