# 9.5 Applications of thermodynamics: heat pumps and refrigerators  (Page 3/7)

 Page 3 / 7

## The best COPhp Of a heat pump for home use

A heat pump used to warm a home must employ a cycle that produces a working fluid at temperatures greater than typical indoor temperature so that heat transfer to the inside can take place. Similarly, it must produce a working fluid at temperatures that are colder than the outdoor temperature so that heat transfer occurs from outside. Its hot and cold reservoir temperatures therefore cannot be too close, placing a limit on its ${\text{COP}}_{\text{hp}}$ . (See [link] .) What is the best coefficient of performance possible for such a heat pump, if it has a hot reservoir temperature of $\text{45}\text{.}0\text{º}\text{C}$ and a cold reservoir temperature of $-\text{15}\text{.}0\text{º}\text{C}$ ?

Strategy

A Carnot engine reversed will give the best possible performance as a heat pump. As noted above, ${\text{COP}}_{\text{hp}}=1/\text{Eff}$ , so that we need to first calculate the Carnot efficiency to solve this problem.

Solution

Carnot efficiency in terms of absolute temperature is given by :

${\text{Eff}}_{\text{C}}=1-\frac{{T}_{\text{c}}}{{T}_{\text{h}}}\text{.}$

The temperatures in kelvins are ${T}_{\text{h}}=\text{318 K}$ and ${T}_{\text{c}}=\text{258 K}$ , so that

${\text{Eff}}_{\text{C}}=1-\frac{\text{258 K}}{\text{318 K}}=0\text{.}\text{1887}\text{.}$

Thus, from the discussion above,

${\text{COP}}_{\text{hp}}=\frac{1}{\text{Eff}}=\frac{1}{0\text{.}\text{1887}}=5\text{.}\text{30},$

or

${\text{COP}}_{\text{hp}}=\frac{{Q}_{\text{h}}}{W}=5\text{.}\text{30}\text{,}$

so that

${Q}_{\text{h}}=5.30 W\text{.}$

Discussion

This result means that the heat transfer by the heat pump is 5.30 times as much as the work put into it. It would cost 5.30 times as much for the same heat transfer by an electric room heater as it does for that produced by this heat pump. This is not a violation of conservation of energy. Cold ambient air provides 4.3 J per 1 J of work from the electrical outlet.

Real heat pumps do not perform quite as well as the ideal one in the previous example; their values of ${\text{COP}}_{\text{hp}}$ range from about 2 to 4. This range means that the heat transfer ${Q}_{\text{h}}$ from the heat pumps is 2 to 4 times as great as the work $W$ put into them. Their economical feasibility is still limited, however, since $W$ is usually supplied by electrical energy that costs more per joule than heat transfer by burning fuels like natural gas. Furthermore, the initial cost of a heat pump is greater than that of many furnaces, so that a heat pump must last longer for its cost to be recovered. Heat pumps are most likely to be economically superior where winter temperatures are mild, electricity is relatively cheap, and other fuels are relatively expensive. Also, since they can cool as well as heat a space, they have advantages where cooling in summer months is also desired. Thus some of the best locations for heat pumps are in warm summer climates with cool winters. [link] shows a heat pump, called a “ reverse cycle” or “ split-system cooler” in some countries.

a perfect square v²+2v+_
kkk nice
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
or infinite solutions?
Kim
y=10×
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
rolling four fair dice and getting an even number an all four dice
Kristine 2*2*2=8
Differences Between Laspeyres and Paasche Indices
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
is it 3×y ?
J, combine like terms 7x-4y
im not good at math so would this help me
how did I we'll learn this
f(x)= 2|x+5| find f(-6)
f(n)= 2n + 1
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
preparation of nanomaterial
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
can nanotechnology change the direction of the face of the world
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
how did you get the value of 2000N.What calculations are needed to arrive at it
Got questions? Join the online conversation and get instant answers!