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3 years ago

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A heat pump with refrigerant-134a as the working uid is used to keep a space at 25C by absorbing heat from geothermal water that

enters the evaporator at 60C at a rate of 0.065 kg/s and leaves at 40C. Refrigerant enters the evaporator at 12C with a quality of 15 percent and leaves at the same pressure as saturated vapor. If the compressor consumes 1.6 kW of power, determine (a) the mass ow rate of the refrigerant, (b) the rate of heat supply, (c) the COP, and (d) the minimum power input to the compressor for the same rate of heat supply.

1 answer:

Snezhnost [94]3 years ago

8 0

Answer:

A) 0.03382 kg/s

B) 7.0372 Kw

C) 4.3982

D) 0.7396 kw

Explanation:

Given data:

Evaporator at 60 C

Space temperature = 25 C

power consumed by compressor = 1.6 kw

T1( evaporator temperature ) = 12°C

attached below is the detailed solution

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Answer:

47.91 sec

Explanation:

it is given that $\alpha =\frac{1}{4v^{2}}$

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we have to find time at which velocity will be 3.3 $\frac{m}{sec^{2}}$

we know that $\alpha =\frac{dv}{dt}=\frac{1}{4v^{2}}$

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$\frac{4v^{3}}{3}=t+c$ ---------------eqn 1

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4 years ago

Disconnecting a circuit while in operation can create a voltage blank

zlopas [31]

Answer:

what is the question

Explanation:

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3 years ago

What is the maximal coefficient of performance of a refrigerator which cools down 10 kg of water (and then ice) to -6C. Upper he

inysia [295]

Given:

Temperature of water, $T_{1}$ = $-6^{\circ}C$ =273 +(-6) =267 K

Temperature surrounding refrigerator, $T_{2}$ = $21^{\circ}C$ =273 + 21 =294 K

Specific heat given for water, $C_{w}$ = 4.19 KJ/kg/K

Specific heat given for ice, $C_{ice}$ = 2.1 KJ/kg/K

Latent heat of fusion, $L_{fusion}$ = 335KJ/kg

Solution:

Coefficient of Performance (COP) for refrigerator is given by:

Max $COP_{refrigerator}$ = $\frac{T_{2}}{T_{2} - T_{1}}$

= $\frac{267}{294 - 267}$ = 9.89

Coefficient of Performance (COP) for heat pump is given by:

Max $COP_{heat pump}$ = $\frac{T_{1}}{T_{2} - T_{1}}$ $\frac{294}{294 - 267}$ = 10.89

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4 years ago

Determine the dimensions for W if W = P L^3 / (M V^2) where P is a pressure, L is a length, M is a mass, and V is a velocity.

Eva8 [605]

Correct answer is option E. No dimensions

As we know formula Pressure (P) is $\frac{F}{A}$

also,

Dimensional formula of <em>Pressure is </em> $M^{1}L^{-1}T^{-2}$
Dimensional formula of <em>length is L </em>
Dimensional formula of <em>mass is M</em>
Dimensional formula of <em>velocity is </em> $L^{1} T^{-1}$

So, as given W= $\frac{P*L^{3} }{M*V^{2} }$

Dimensional formula of W = $\frac{M^{1}L^{-1}T^{-2} L^{3} }{M^{1} L^{2}T^{-2} }$

since all terms get cancelled

Work is dimensionless i.e no dimensions

Learn more about dimensions here brainly.com/question/20351712

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2 years ago

In a home, air infiltrates from the outside through cracks around doors and windows. Consider a residence where the total length

masya89 [10]

Answer:

Time period = 41654.08 s

Explanation:

Given data:

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Rate of air infiltration $9.4 \times 10^{-5} kg/s$

length of cracks 62 m

air density = 1.186 kg/m^3

Total rate of air infiltration $= 9.4\times 10^{-5} \times 62 = 582.8\times 10{-5} kg/s$

total volume of air infiltration $= \frac{582.8\times 10{-5}}{1.156} = 5.04\times 10^{-3} m^3/s$

Time period $= \frac{210}{5.04\times 10^{-3}} = 41654.08 s$

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4 years ago