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3 years ago

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

Engineering

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.

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

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

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.

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An Ideal gas is being heated in a circular duct as while flowing over an electric heater of 130 kW. The diameter of duct is 500

Rashid [163]

Answer:

Exit temperature = 32 °C

Explanation:

We are given;

Initial Pressure;P1 = 100 KPa

Cp =1000 J/kg.K = 1 KJ/kg.k

R = 500 J/kg.K = 0.5 Kj/Kg.k

Initial temperature;T1 = 27°C = 273 + 27K = 300 K

volume flow rate;V' = 15 m³/s

W = 130 Kw

Q = 80 Kw

Using ideal gas equation,

PV' = m'RT

Where m' is mass flow rate.

Thus;making m' the subject, we have;

m' = PV'/RT

So at inlet,

m' = P1•V1'/(R•T1)

m' = (100 × 15)/(0.5 × 300)

m' = 10 kg/s

From steady flow energy equation, we know that;

m'•h1 + Q = m'h2 + W

Dividing through by m', we have;

h1 + Q/m' = h2 + W/m'

h = Cp•T

Thus,

Cp•T1 + Q/m' = Cp•T2 + W/m'

Plugging in the relevant values, we have;

(1*300) - (80/10) = (1*T2) - (130/10)

Q and M negative because heat is being lost.

300 - 8 + 13 = T2

T2 = 305 K = 305 - 273 °C = 32 °C

13000 + 300 - 8000 = T2

6 0

3 years ago

For some metal alloy, the following engineering stresses produce the corresponding engineering plastic strains prior to necking.

kirza4 [7]

Answer:

203.0160

Explanation:

Because you add then subtract then multiply buy 7 the subtract then divide then you add that to the other numbers you got than boom

7 0

2 years ago

Design for human-fit strategies include:

andreev551 [17]

Answer:

B- extreme fit, close fit, adjustable fit

Explanation:

A human-fit design typically involves the process of manufacturing or producing products (tools) that are easy to use by the end users. Therefore, human-fit designs mainly deals with creating ideas that makes the use of a particular product comfortable and convenient for the end users.

The design for human-fit strategies include; extreme fit, close fit and adjustable fit.

Hence, when the aforementioned strategies are properly integrated into a design process, it helps to ensure the ease of use of products and guarantees comfort for the end users.

5 0

3 years ago

A distillation column is initially designed to separate a mixture of toluene and xylene at around ambient temperature (say, 100°

weeeeeb [17]

Answer:

Purchase cost= 87056

Bar module cost= 292725

Explanation:

solution is attached below

6 0

3 years ago

At an axial load of 22 kN, a 15-mm-thick × 40-mm-wide polyimide polymer bar elongates 4.1 mm while the bar width contracts 0.15

Alenkasestr [34]

Answer:

The Poisson's Ratio of the bar is 0.247

Explanation:

The Poisson's ratio is got by using the formula

Lateral strain / longitudinal strain

Lateral strain = elongation / original width (since we are given the change in width as a result of compession)

Lateral strain = 0.15mm / 40 mm =0.00375

Please note that strain is a dimensionless quantity, hence it has no unit.

The Longitudinal strain is the ratio of the elongation to the original length in the longitudinal direction.

Longitudinal strain = 4.1 mm / 270 mm = 0.015185

Hence, the Poisson's ratio of the bar is 0.00375/0.015185 = 0.247

The Poisson's Ratio of the bar is 0.247

Please note also that this quantity also does not have a dimension

3 0

3 years ago