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12345 [234]
2 years ago
5

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.
Engineering
1 answer:
Snezhnost [94]2 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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The stagnation chamber of a wind tunnel is connected to a high-pressure airbottle farm which is outside the laboratory building.
Natasha2012 [34]

This question is not complete, the complete question is;

The stagnation chamber of a wind tunnel is connected to a high-pressure air bottle farm which is outside the laboratory building. The two are connected by a long pipe of 4-in inside diameter. If the static pressure ratio between the bottle farm and the stagnation chamber is 10, and the bottle-farm static pressure is 100 atm, how long can the pipe be without choking? Assume adiabatic, subsonic, one-dimensional flow with a friction coefficient of 0.005

Answer:

the length of the pipe is 11583 in or 965.25 ft

Explanation:

Given the data in the question;

Static pressure ratio; p1/p2 = 10

friction coefficient f = 0.005

diameter of pipe, D =4 inch

first we obtain the value from FANN0 FLOW TABLE for pressure ratio of ( p1/p2 = 10 )so

4fL_{max} / D = 57.915

we substitute

(4×0.005×L_{max}) / 4  = 57.915

0.005L_{max} = 57.915

L_{max} = 57.915 / 0.005

L_{max}  = 11583 in

Therefore, the length of the pipe is 11583 in or 965.25 ft

6 0
3 years ago
In normal operation, a paper mill generates excess steam at 20 bar and 400◦C. It is planned to use this steam as the feed to a t
Keith_Richards [23]

Answer:

The maximum power that can be generated is 127.788 kW

Explanation:

Using the steam table

Enthalpy at 20 bar = 2799 kJ/kg

Enthalpy at 2 bar = 2707 kJ/kg

Change in enthalpy = 2799 - 2707 = 92 kJ/kg

Mass flow rate of steam = 5000 kg/hr = 5000 kJ/hr × 1 hr/3600 s = 1.389 kg/s

Maximum power generated = change in enthalpy × mass flow rate = 92 kJ/kg × 1.389 kg/s = 127.788 kJ/s = 127.788 kW

6 0
3 years ago
Physical properties of minerals
lord [1]

Most minerals can be characterized and classified by their unique physical properties: hardness, luster, color, streak, specific gravity, cleavage, fracture, and tenacity.

<em>-</em><em> </em><em>BRAINLIEST</em><em> answerer</em><em> ❤️</em><em>✌</em>

6 0
2 years ago
Read 2 more answers
Water flows at low speed through a circular tube with inside diameter of 2 in. A smoothly contoured body of 1.5 in. diameter is
Art [367]

Answer:

Pressure = 11.38 psi

Force = 13.981 Ibf

Explanation:

Step by step solution is in the attached document.

5 0
3 years ago
What makes building an airplane while flying so difficult?
Naddik [55]

Answer: Because if something goes wrong while you are flying it it will crash

Explanation:

7 0
3 years ago
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