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Nata [24]
4 years ago
7

In a simple ideal Rankine cycle, water is used as the working fluid. The cycle operates with pressures of 2000 psi in the boiler

and 4 psi in the condenser. What is the minimum temperature required at the turbine inlet, so that the quality of the steam at the turbine outlet is not less than 85%. What would be the thermal efficiency of the cycle?
Engineering
1 answer:
weqwewe [10]4 years ago
3 0

Answer:

Explanation:

The pressures given are relative

p1 = 2000 psi

P1 = 2014 psi = 13.9 MPa

p2 = 4 psi

P2 = 18.6 psi = 128 kPa

Values are taken from the steam pressure-enthalpy diagram

h2 = 2500 kJ/kg

If the output of the turbine has a quality of 85%:

t2 = 106 C

I consider the expansion in the turbine to adiabatic and reversible,  therefore, isentropic

s1 = s2 = 6.4 kJ/(kg K)

h1 = 3500 kJ/kg

t2 = 550 C

The work in the turbine is of

w = h1 - h2 = 3500 - 2500 = 1000 kJ/kg

The thermal efficiency of the cycle depends on the input heat.

η = w/q1

q1 is  not a given, so it cannot be calculated.

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Which traits are common in all four career pathways of the Information Technology field? Check all that apply.
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Answer:

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Explanation:

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2 years ago
. In one stroke of a reciprocating compressor, helium is isothermally and reversibly compressed in a piston + cylinder from 298
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Explanation:

Please see attachment for step by step guide

5 0
3 years ago
There are two identical oil tanks. The level of oil in Tank A is 12 ft and is drained at the rate of 0.5 ft/min. Tank B contains
Luba_88 [7]

Answer:

  16 minutes

Explanation:

This is an example of a class of problems in which two quantities start with different initial values and change at different rates. In such problems, the rates of change are generally ones that cause the values to converge.

The question usually asks when the values will be the same. The generic answer is, "when the difference in rates makes up the difference in initial values."

Here the tanks differ in initial fill height by 12 -8 = 4 ft. The rates of change differ by 0.5 -0.25 = 0.25 ft/min. The more filled tank is draining faster (important), so the fill heights will converge after ...

  (4 ft)/(0.25 ft/min) = 16 min

The level in the two tanks will be the same after 16 minutes.

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<em>Additional comment</em>

The oil levels at that time will be 4 ft.

You can write two equations for height:

  y = 12 -0.5x . . . . . . . height in feet after x minutes (tank A)

  y = 8 -0.25x . . . . . .  height in feet after x minutes (tank B)

These will be equal when ...

  y = y

  12 -0.5x = 8 -0.25x

  4 = 0.25x . . . . . . . . . . add 0.5x -8

  16 = x . . . . . . . . . . . . multiply by 4 . . . . time to equal height

The graph shows when the tanks will have equal heights and when they will be drained.

4 0
2 years ago
Why carbon is not used as a semiconductor material​
dmitriy555 [2]
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8 0
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