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cestrela7 [59]
3 years ago
12

Air enters the compressor of a cold air-standard Brayton cycle with regeneration at 100 kPa, 300 K, with a mass flow rate of 6 k

g/s. The compressor pressure ratio is 10, and the turbine inlet temperature is 1400 K. The turbine and compressor each have isentropic efficiencies of 80% and the regenerator effectiveness is 80%. For k 5 1.4, calculate (a) the thermal efficiency of the cycle. (b) the back work ratio. (c) the net power developed, in kW. (d) the rate of entropy production in the regenerator, in kW/K
Engineering
1 answer:
SVETLANKA909090 [29]3 years ago
4 0
(a) The thermal efficiency of the cycle.

The thermal efficiency of the cycle is given by
NThermal= net work/ heat supplied
NThermal= WNet/QH
=2026.08/4547.82= 44.55%

(b) The backwork ratio:
compressor work/ turbine work
=1760.76/3786.84= 0.465

(c) The net power developed:
WNet= WT-WC
WNet= 3786.84- 1760.76= 2026.08 kW
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A steady, incompressible, two-dimensional velocity field is given by the following components in the x-y plane: u=1.85+2.05x+0.6
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1) a_x=4.287+2.772x\\a_y=-5.579+2.772y

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Explanation:

1)

The two components of the velocity field in x and y for the field in this problem are:

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v=0.754-2.18x-2.05y

The x-component and y-component of the acceleration field can be found using the following equations:

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The derivatives in this problem are:

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\frac{dv}{dx}=-2.18

\frac{dv}{dy}=-2.05

Substituting, we find:

a_x=0+(1.85+2.05x+0.656y)(2.05)+(0.754-2.18x-2.05y)(0.656)=\\a_x=4.287+2.772x

And

a_y=0+(1.85+2.05x+0.656y)(-2.18)+(0.754-2.18x-2.05y)(-2.05)=\\a_y=-5.579+2.772y

2)

In this part of the problem, we want to find the acceleration at the point

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So we have

x = -1

y = 5

First of all, we substitute these values of x and y into the expression for the components of the acceleration field:

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And so we find:

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