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Setler [38]
3 years ago
8

refrigerant 134a enters a compressor operating at steady state as saturated vapor at 0.12 MPa and exits at 1.2 MPa and 70 C at a

mass flow rate of 0.108kg/s. As the refrigerant passes through the compressor, heat transfer to the surroundings occurs at a rate of 0.32 kJ/s. Determine at steady state the power input to the compressor, in kW.
Physics
1 answer:
Afina-wow [57]3 years ago
7 0

Answer:

the power input to the compressor is 7.19Kw

Explanation:

Hello!

To solve this problem follow the steps below.

1. We will call 1 the refrigerant state at the compressor inlet and 2 at the outlet.

2. We use thermodynamic tables to determine enthalpies in states 1 and 2.

(note: Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)  

through prior knowledge of two other properties such as pressure and temperature.  )

h1[quality=1, P=0.12Mpa)=237KJ/Kg

h2(P=1.2Mpa, t=70C)=300.6KJ/kg

3. uses the first law of thermodynamics in the compressor that states that the energy that enters a system is the same that must come out

Q=heat=0.32kJ/s

W=power input to the compressor

m=mass flow=0.108kg/S

m(h1)+W=Q+m(h2)

solving for W

W=Q+m(h2-h1)

W=0.32+0.108(300.6-237)=7.19Kw

the power input to the compressor is 7.19Kw

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Answer:

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Explanation:

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  • The final momentum pf, can be written as follows:

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  • Since (2) and (3) are equal each other, we can solve for the only unknown that remains, v₂₀, as follows:

       v_{2o} = \frac{-6.0kg* 5m/s -8.0 kg*2.0m/s}{2.0kg}  = \frac{-46kg*m/s}{2.0kg} = -23.0 m/s  (4)

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6 0
4 years ago
A person pushes a block 4 m with a force of 25 N. How much work is being done?
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Answer:

<h2>100 J</h2>

Explanation:

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From the question we have

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Answer:

Explanation:

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