practice problem 1
- the heat…
- extracted from the environment per hour when the device is operated in heating mode
- exhausted to the environment per hour when the device is operated in cooling mode
- the real coefficient of performance…
- in heating mode
- in cooling mode
- the ideal coefficient of performance for a 22 °C inside temperature and…
- a −2 °C outside winter temperature
- a +34 °C outside summer temperature
- the velocity of the air blowing into the room when the device is operated in either mode
|heating mode||cooling mode||both modes|
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.
|743 J||1 MJ||3600 s||= 2.67 MJ/h|
|1 s||1,000,000 J||1 h|
|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.
We can now start computing answers.
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 cooling mode QH = QC + W QH = 10 MJ + 2.92 MJ
Let's update our cartoon with this new information.
The real coefficient of performance is defined by an equation…
COPreal = Q W
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.
The ideal coefficient of performance is defined by an equation…
COPideal = T TH − TC
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.
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)
v = 0.50 m/s
Seems reasonable, like an unhurried walk.
practice problem 2
practice problem 3
practice problem 4