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lara31 [8.8K]
3 years ago
8

Problem 4.041 SI Refrigerant 134a enters an insulated compressor operating at steady state as saturated vapor at -26oC with a vo

lumetric flow rate of 0.18 m3/s. Refrigerant exits at 8 bar, 70oC. Changes in kinetic and potential energy from inlet to exit can be ignored. Determine the volumetric flow rate at the exit, in m3/s, and the compressor power, in kW
Engineering
1 answer:
Rom4ik [11]3 years ago
4 0

Answer:

0.0297M^3/s

W=68.48kW

Explanation:

Hello! To solve this problem, we must first find all the thermodynamic properties at the input (state 1) and the compressor output (state 2), using the thermodynamic tables

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.  

state 1

X=quality=1

T=-26C

density 1=α1=5.27kg/m^3  

entalpy1=h1=234.7KJ/kg

state 2

T2=70

P2=8bar=800kPa

density 2=α2=31.91kg/m^3  

entalpy2=h2=306.9KJ/kg

Now to find the flow at the outlet of the compressor, we remember the continuity equation that states that the mass flow is equal to the input and output.

m1=m2

(Q1)(α1)=(Q2)(α2)

\frac{(Q1)(\alpha 1) }{\alpha 2} =Q2\\Q2=\frac{(0.18)(5.27) }{31.91} =0.0297M^3/s

the volumetric flow rate at the exit is 0.0297M^3/s

To find the power of the compressor we use the first law of thermodynamics that says that the energy that enters must be equal to the energy that comes out, in this order of ideas we have the following equation

W=m(h2-h1)

m=Qα

W=(0.18)(5.27)(306.9-234.7)

W=68.48kW

the compressor power is 68.48kW

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A 1020 CD steel shaft is to transmit 15 kW while rotating at 1750 rpm. Determine the minimum diameter for the shaft to provide a
vladimir2022 [97]

Answer:

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Explanation:

given data

power = 15 kW

rotation N = 1750 rpm

factor of safety = 3

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minimum diameter

solution

we will apply here power formula to find T that is

power = 2π×N×T / 60    .................1

put here value

15 ×10^{3} = 2π×1750×T / 60

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here Z is section modulus i.e = πd³/ 16

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416.75 / d³ = 530 × 10^{6} / 3

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3 years ago
Four kilograms of carbon monoxide (CO) is contained in a rigid tank with a volume of 1 m3. The tank is fitted with a paddle whee
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Answer:

a) 1 m^3/Kg  

b) 504 kJ

c) 514 kJ

Explanation:

<u>Given  </u>

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<u>Required  </u>

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Assumption  

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(a) The volume and the mass doesn't change then, the specific volume is constant.

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The heat have (+) sign the n it is added to the system.

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