Refrigerators

Some preliminary work to start. Let's lay out all the information we know on a cartoon. As we learn more, we'll add onto it.

Convert the power consumption of the heat pump from watts (J/s) to megajoules per hour (MJ/h). Recall that the SI prefix mega means one million or 10⁶.

Add this information to our cartoon.

Although the diagram above is in megajoules per hour (a power unit) we will be doing our calculations in megajoules (an energy unit). The 1 hour time is just understood to always be there.

Begin calculating.

1. Heat pumps are basically refrigerators. They extract heat from somewhere cold (Q_C) and deposit it to somewhere hot (Q_H) by doing work (W) on a fluid called a refrigerant. From the law of conservation of energy we get…

   Q_H = Q_C + W

   This shows that the heat deposited on the hot side is greater than the heat removed from the cold side. Use this equation twice. In the winter, the hot side is indoors. In the summer, the hot side is outdoors.

   heating mode

   QH =	QC + W
   7.9 MJ =	QC + 2.67 MJ
   QC =	5.23 MJ

   cooling mode

   QH =	QC + W
   QH =	10.00 MJ + 2.92 MJ
   QH =	12.92 MJ

Let's update our cartoon.

2. The real coefficient of performance is defined by the equation…

   COPreal =	Q
   	W

   Where…

   Q =	the useful heat moved
   W =	the work done on the refrigerant

   In heating mode, the useful heat is the heat deposited into the home (Q_H). In cooling mode, it's the heat extracted from the home (Q_C). We'll assume that all of the energy from the electrical outlet went directly into work on the refrigerant. This isn't far from the truth since the compressor motor is what draws most of the current in a heat pump, air conditioner, or refrigerator.

   heating mode

   COPreal =	QH W
   COPreal =	7.90 MJ 2.67 MJ
   COPreal =	2.95

   cooling mode

   COPreal =	QC W
   COPreal =	10.00 MJ 2.92 MJ
   COPreal =	3.42

   These numbers make sense as they are both greater than 1. The COP for heating mode always has to be greater than 1 since the heat deposited on the hot side is always greater than the work done. The COP for cooling mode is almost always greater than 1 for devices you'd be willing to spend money on. (It's been like this since the 1950s, as far as I can tell.)

3. The ideal coefficient of performance is defined by the equation…

   COPideal =	T
   	TH − TC

   Where…

   T =	the temperature on the controlled side
   TH =	the temperature on the hotter side
   TC =	the temperature on the colder side

   Substitute carefully and compute. The temperatures need to be in kelvin for the math to work out right (12 °C = 290 K, 21 °C = 294 K, 27 °C = 300 K, 35 °C = 308 K).

   heating mode

   COPideal =	T TH − TC
   COPideal =	294 K 294 K − 283 K
   COPideal =	26.7

   cooling mode

   COPideal =	T TH − TC
   COPideal =	300 K 308 K − 300 K
   COPideal =	37.5

   I don't really know what to do with these numbers. They seem to imply that there's a lot more we could be doing to improve heat pumps.

4. The volume flow rate (q_V) of air passing through an opening is the area (A) of that opening multiplied by the speed of the air (v). Here's a little proof in case you forgot how it works.

   qV =	∆V	=	A∆s	= Av
   	∆t		∆t

   Solve for speed. Replace area with length times width.

   v =	qV	=	qV
   	A		ℓw

   Numbers in. (Watch the units. I recommend using just combinations of meters and seconds.)

   v =	8.1 m3/60 s
   	(0.66 m)(0.40 m)

   Answer out.

   v = 0.51 m/s

   Seems reasonable, like an unhurried walk.

5. To dehumidify is to remove at least some of the water vapor from the air contained in an interior space. A heat pump operating in cooling mode does this by condensation — changing the water in the air from a vapor to a liquid which then drains away into the environment. The heat associated with this phase change is the latent heat of vaporization (Q_v) It is distinct from the sensible heat needed to lower the temperature of the air (Q_C). There is no temperature change during a phase change so dehumidifying the air does not make it any cooler, only drier.

   1. Recall that…

      Q_v = mL_v

      Where…

      Qv =	the latent heat of vaporization (our goal)
      m =	the mass of water condensed each hour (0.9 kg, since one liter of water is one kilogram of mass by design)
      Lv =	the specific latent heat of vaporization of water (2.441 MJ/kg)

      Numbers in, answer out.

      Qv = mLv

      Qv = (0.9 kg)(2.441 MJ/kg)

      Qv = 2.197 MJ

      Although extracting this latent heat from the interior does not make any cooler inside, exhausting it to the environment probably will make it hotter outside. The heat that was latent on the inside will probably be sensible on the outside. Compute that new number next.

   2. Add the new latent heat to the old exhausted heat we calculated back in part a.ii.

      Q′H = QH + Qv

      Q′H = 12.92 MJ + 2.20 MJ

      Q′H = 15.12 MJ

      Or if you prefer to start from first principles, add it to the heat extracted from inside.

      Q′H = Q′C + W

      Q′H = (QC + Qv) + W

      Q′H = (10.00 MJ + 2.20 MJ) + 2.92 MJ

      Q′H = 15.12 MJ

      The second method isn't conceptually any better than the first, but it's what the diagram below shows.

Let's update our cartoon one last time.

3. Calculating the new real coefficient of performance is trivial. The only difference is that now the useful heat has two components — cooling (Q_C) and dehumidifying (Q_v).

   COPreal =	QC + Qv W
   COPreal =	10.00 MJ + 2.20 MJ 2.92 MJ
   COPreal =	4.17

   If you got 4.18 for this last answer, don't sweat it. Your work is probably still good. Although I reported all the energy values in this problem with two digits after the decimal point, I did all the calculations using un-rounded numbers (that is, using whatever precision my calculator uses).

   COPreal =	QC + Qv W
   COPreal =	10 MJ + 2.1969… MJ 2.9232… MJ
   COPreal =	4.172448…

   The important thing to note here is that the COP has improved (4.17 > 3.42). The heat pump does more useful work when it does two things (cooling and dehumidifying) instead of just one (only cooling).

Answer it.

Answer it.

Answer it.