The Physics
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Opus in profectus

Refrigerators

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Practice

practice problem 1

Determine the following quantities for a window mounted, residential heat pump described in the table below:
  1. the heat…
    1. extracted from the environment per hour when the device is operated in heating mode
    2. exhausted to the environment per hour when the device is operated in cooling mode
  2. the real coefficient of performance…
    1. in heating mode
    2. in cooling mode
  3. the ideal coefficient of performance for a 22 °C inside temperature and…
    1. a −2 °C outside winter temperature
    2. a +34 °C outside summer temperature
  4. the velocity of the air blowing into the room when the device is operated in either mode
Residential heat pump
heating mode cooling mode both modes
heating capacity
7.9 MJ/hour		 cooling capacity
10 MJ/hour		 volts rated
115 volts
heating amps
7.6 amps		 cooling amps
7.8 amps		 air circulation
8.1 m3/minute
heating watts
743 watts		 cooling watts
812 watts		 width
66 cm
moisture removal
0.90 liter/hour		 height
40 cm
depth
74 cm

solution

Start by converting 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.

heating mode
743 J   1 MJ   3600 s  = 2.67 MJ/h
1 s 1,000,000 J 1 h
cooling mode
812 J   1 MJ   3600 s  = 2.92 MJ/h
1 s 1,000,000 J 1 h

Let's lay out all the known information on a cartoon.

Cartoon house with heat going in and out

We can now start computing answers.

  1. Heat pumps are basically refrigerators. They extract heat from somewhere cold (QC) and deposit it to somewhere hot by doing work on a fluid called a refrigerant. The heat deposited to the hot side (QH) is greater than the heat removed from the cold side (QC) because of the conversion of mechanical work (W) to heat. From the law of conservation of energy we get…

    QH = QC + W

    Use this equation twice. In the winter, the hot side is indoors. In the summer, the hot side is outdoors. Since the time is one hour in each term, there's no reason to write that unit.

    heating mode
    QH =  QC + W
    7.9 MJ =  QC + 2.67 MJ
    QC =  5.23 MJ
    cooling mode
    QH =  QC + W
    QH =  10 MJ + 2.92 MJ
    QH =  12.92 MJ

Let's update our cartoon with this new information.

Cartoon house with heat going in and out

  1. The real coefficient of performance is defined by an 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 (QH). In cooling mode, it's the heat extracted from the home (QC).

    heating mode
    COPreal =  QH
    W
    COPreal =  7.9 MJ
    2.67 MJ
    COPreal = 2.95  
     
    cooling mode
    COPreal =  QC
    W
    COPreal =  10 MJ
    2.92 MJ
    COPreal = 3.42  
     

    These numbers make sense as they are both greater than one. The COP for heating mode always has to be greater than one since the heat deposited on the hot side is always greater than the work. The COP for cooling mode is almost always greater than one for devices you'd be willing to spend money on. It's been like this since the 1950s at least. That's just because of good engineering.

  2. The ideal coefficient of performance is defined by an 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 (22 °C = 295 K, −2 °C = 271 K, 34 °C = 307 K).

    heating mode
    COPideal =  T
    TH − TC
    COPideal =  295 K
    295 K − 271 K
    COPideal = 12.3  
     
    cooling mode
    COPideal =  T
    TH − TC
    COPideal =  295 K
    307 K − 295 K
    COPideal = 24.6  
     

    The ideal values are 4 and 7 times greater than the actual values. Heat pumps can still be improved and mechanical engineers are still needed.

  3. The volume flow rate (qV) of air passing through an opening is the area of that opening multiplied by the speed of the air (Av). 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 carefully. I think I would like the answer to be in meters per second. No fancy units for me, just good old bog standard SI.

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

    Answer out.

    v = 0.50 m/s

    Seems reasonable, like an unhurried walk.

practice problem 2

Write something else.

solution

Answer it.

practice problem 3

Write something different.

solution

Answer it.

practice problem 4

Write something completely different.

solution

Answer it.