Ask question

Ask question

All categories

3 years ago

10

A heat pump and a refrigerator are operating between the same two thermal reservoirs. Which one has a higher COP?

1 answer:

Mkey [24]3 years ago

4 0

Answer:

Heat pump

Explanation:

COP:

As we know that COP can be defines as the ratio of desired effect to the work in put.

The desired effect can be different for heat pump our desired effect to heat the space and for refrigeration our desired effect is cooled the space.

So

$COP_{pump}=\dfrac{T_H}{T_H-T_L}$

$COP_{refri}=\dfrac{T_L}{T_H-T_L}$

When heat pump and refrigerator are operating with same temperature limits then from expression we can say that

COP of heat pump = 1 + COP of refrigeration.

You might be interested in

25 points and brainliest is it A, B, C, D

Bogdan [553]

Answer:

SIR IT IS D HOPE THIS HELPS (☞ﾟヮﾟ)☞☜(ﾟヮﾟ☜)

Explanation:

4 0

3 years ago

Read 2 more answers

A saturated 1.5 ft3 clay sample has a natural water content of 25%, shrinkage limit (SL) of 12% and a specific gravity (GS) of 2

Svetllana [295]

$79 f t^{3}$ is the volume of the sample when the water content is 10%.

<u>Explanation:</u>

Given Data:

$V_{1}=100\ \mathrm{ft}^{3}$

First has a natural water content of 25% = $\frac{25}{100}$ = 0.25

Shrinkage limit, $w_{1}=12 \%=\frac{12}{100}=0.12$

$G_{s}=2.70$

We need to determine the volume of the sample when the water content is 10% (0.10). As we know,

$V \propto[1+e]$

$\frac{V_{2}}{V_{1}}=\frac{1+e_{2}}{1+e_{1}}$ ------> eq 1

$e_{1}=\frac{w_{1} \times G_{s}}{S_{r}}$

The above equation is at $S_{r}=1$ ,

$e_{1}=w_{1} \times G_{s}$

Applying the given values, we get

$e_{1}=0.25 \times 2.70=0.675$

Shrinkage limit is lowest water content

$e_{2}=w_{2} \times G_{s}$

Applying the given values, we get

$e_{2}=0.12 \times 2.70=0.324$

Applying the found values in eq 1, we get

$\frac{V_{2}}{100}=\frac{1+0.324}{1+0.675}=\frac{1.324}{1.675}=0.7904$

$V_{2}=0.7904 \times 100=79\ \mathrm{ft}^{3}$

7 0

3 years ago

what are the characteristics of an ideal fluid the general relation between shear stress and velocity gradient

Dafna11 [192]

Answer:

ideal fluid follow Newtonian law

that is, shear stress is directly proportional to rate change of shear strain.

watch handwritten explanation

6 0

3 years ago

Design circuits that demonstrate all of the principles listed below. Set up the circuits and take measurements to show that the

Nata [24]

<u>Explanation</u>:

For series

$\Delta V=V_{1}+V_{2}+\ldots+V_{n}=I R_{1}+I R_{2}+\ldots+I R_{n}(\text {voltages add to the batter } y)$

$\(I=I_{1}=I_{2}=I_{n}\) (current is the same)$

$V=I R(\text {voltage is directly proportional to } R)$

$R_{e q}=R_{1}+R_{2}+\ldots+R_{n} \quad \text { (resistance increase) }$

For parallel

$\Delta V=\Delta V_{1}=\Delta V_{2}=\Delta V_{n} \quad(\text { same voltage })$

$I=I_{1}+I_{2}+\ldots+I_{n}(\text {current adds})$

$I=\frac{\Delta V}{R_{e q}} \quad(R \text { inversal } y \text { proportional to } I)$

$\frac{1}{R_{e q}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\ldots+\frac{1}{R_{n}}$

3 0

3 years ago

A gearbox is needed to provide an exact 30:1 increase in speed, while minimizing the

alekssr [168]

Answer:

answer

Explanation:

4 0

2 years ago