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3 years ago

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Refrigerant-134a enters an adiabatic compressor at -30oC as a saturated vapor at a rate of 0.45 m3 /min and leaves at 900 kPa an

d 55oC. Determine (a) the power input to the compressor, (b) the isentropic efficiency of the compressor, and (c) the rate of exergy destruction and the second-law efficiency of the compressor. Take T0

1 answer:

saw5 [17]3 years ago

4 0

Answer:

a) 1.918 kw

b) 86.23%

c) 0.26 kw

Explanation:

Given data:

T1 = -30°C = 243 k , T0 = 27°C

using steam tables

h1 = 232.19 KJ/kg

s1 = 0.9559 Kj/Kgk

T2 = 55°C P2 = 900 kPa

Psat = 1492 kPa, h2 = 289.95 Kj/Kg, s2 = 0.9819 Kj/kgk , m = 0.0332 kg/s

<u>a) Determine the power input to the compressor</u>

power input = 1.918 kw

<u>b) Determine isentropic efficiency of compressor</u>

Isentropic efficiency = 86.23%

<u>c) Determine rate of exergy destruction</u>

rate = 0.26 kw

Attached below is the detailed solution of the given problems

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Liquid flows with a free surface around a bend. The liquid is inviscid and incompressible, and the flow is steady and irrotation

lions [1.4K]

Answer:

9 cm

Explanation:

The liquid on the bend will be affected by two accelerations: gravity and centripetal force.

Gravity will be of 9.81 m/s^2 pointing down at all points.

The centripetal acceleration will be of

ac = v^2/r

Pointing to the center of the bend (perpendicular to gravity).

The velocity will depend on the radius

v = (1 m^2/s) / r

Replacing:

ac = (1/r)^2 / r

ac = (1 m^4/s^2) / r^3

If we set up a cylindrical reference system with origin at the center of the bend, the total acceleration will be

a = (-1/r^3 * i - 9.81 * j)

The surface of the liquid will be an equipotential surface, this means all points on the surface have the same potential energy.

The potential energy of the gravity field is:

pg = g * h

The potential energy of the centripetal force is:

pc = ac * r

Then the potential field is:

p = -1/r^2 * - 9.81*h

Points on the surface at r = 1 m and r = 3 m have the same potential.

-1/1^2 * - 9.81*h1 = -1/3^2 * - 9.81*h2

-1 - 9.81*h1 = -1/9 - 9.81*h2

-1 + 1/9 = 9.81 * (h1 - h2)

h1 - h2 = (-8/9) / 9.81

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The outer part will be 9 cm higher than the inner part.

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3 years ago

Match each context to the type of the law that is most suitable for it.

Bas_tet [7]

Answer:

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Explanation:

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3 years ago

A 50 mm diameter shaft is subjected to a static axial load of 160 kN. If the yield stress of the material is 350 MPa, the ultima

zvonat [6]

In order to develop this problem it is necessary to take into account the concepts related to fatigue and compression effort and Goodman equation, i.e, an equation that can be used to quantify the interaction of mean and alternating stresses on the fatigue life of a materia.

With the given data we can proceed to calculate the compression stress:

$\sigma_c = \frac{P}{A}$

$\sigma_c = \frac{160*10^3}{\pi/4*0.05^2}$

$\sigma_c = 81.5MPa$

Through Goodman's equations the combined effort by fatigue and compression is expressed as:

$\frac{\sigma_a}{S_e}+\frac{\sigma_c}{\sigma_u}=\frac{1}{Fs}$

Where,

$\sigma_a=$ Fatigue limit for comined alternating and mean stress

$S_e =$ Fatigue Limit

$\sigma_c=$ Mean stress (due to static load)

$\sigma_u =$ Ultimate tensile stress

$Fs =$ Security Factor

We can replace the values and assume a security factor of 1, then

$\frac{\sigma_a}{320}+\frac{81.5}{400}=\frac{1}{1}$

Re-arrenge for $\sigma_a$

$\sigma_a = 254.8Mpa$

We know that the stress is representing as,

$\sigma_a = \frac{M_c}{I}$

Then,

Where $M_c$ =Max Moment

I= Intertia

The inertia for this object is

$I=\frac{\pi d^4}{64}$

Then replacing and re-arrenge for $M_c$

$M_c = \frac{\sigma_a*\pi*d^3}{32}$

$M_c = \frac{260.9*10^6*\pi*0.05^3}{32}$

$M_c = 3201.7N.m$

Thereforethe moment that can be applied to this shaft so that fatigue does not occur is 3.2kNm

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3 years ago

Which number is equivalent to the decimal number 149?

Sloan [31]

Answer:

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Explanation:

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3 years ago

Determine the critical load if the bottom is fixed and the top is pinned. ewew = 1. 6 ×(10)3ksi×(10)3ksi ,σyσy = 5 ksiksi

Katen [24]

<h3>What is a Critical Load?</h3>

Critical load Fcr or buckling load is the value of load that causes the phenomenon of change from stable to unstable equilibrium state.

With that beign said, first it is neessary to calculate the moment of inercia about the x-axis:

$Ix= \frac{db^3}{12}\\ Ix = \frac{2.(4)^3}{12} = 10.667in$

Then it is necessary to calculate the moment of inercia about the y-axis:

$Iy = \frac{db^3}{12}\\ Iy = \frac{4.(2)^3}{12} = 2.662in$

Comparing both moments of inercia it is possible to assume that the minimun moment of inercia is the y-axis, so the minimun moment of inercia is 2662in.

And so, it is possible to calculate the critical load:

$Pc\gamma = \frac{2046\pi ^2E.I}{L^2} \\Pc\gamma= \frac{2046.\pi ^2.(1,6.10^3.10^3).2662}{(10.12)^2} \\Pc\gamma= 5983,9db$

See more about critical load at: brainly.com/question/22020642

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1 year ago