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OleMash [197]
3 years ago
15

Consider a simple ideal Rankine cycle and an ideal regenerative Rankine cycle with one open feedwater heater. The two cycles are

very much alike, except the feedwater in the regenerative cycle is heated by extracting some steam just before it enters the turbine. Hence, the simple ideal Rankine cycle is more efficient than the ideal regenerative Rankine cycle. How would you compare the efficiencies of of these two cycles?
Engineering
1 answer:
borishaifa [10]3 years ago
4 0

Answer:

They both have the same efficiency.

Explanation:

The simple ideal Rankine cycle and an ideal regenerative Rankine cycle with one open feedwater heater would both have the same efficiency because the extraction steam would just create a mini cycle that recirculates. The energy given to the feedwater heater is proportional to the added heat in the boiler to the feedwater in the simple cycle to raise its temperature to the same boiler inlet condition.

Therefore in comparison, the efficiency is the same for both.

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At what depth in water is the increased pressure five times greater than atmospheric pressure (101 kPa)?​
umka21 [38]

Explanation:

40.4m

Explanation:

Pressure at depth is given as

P = P, + pgh

Final pressure at depth h= 5 Po

5Po= Po + pgh

pgh = 4Po = 4 x 1.01 x 10^5

h = (4.04×10^5)/ (1000x10)

h=40.4m

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As a human resources professional for a large company, Tommy is involved with the ___________ process. His goal is to find the r
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Read 2 more answers
A thick steel slab (rho= 7800 kg/m3 , cp= 480 J/kg K, k= 50 W/m K) is initially at 300 °C and is cooled by water jets impinging
dimaraw [331]

Answer:

t = 2244.3 sec

Explanation:

calculate the thermal diffusivity

\alpha = \frac{k}{\rho c}

           = \frac{50}{7800\times 480} = 1.34 \times 10^{-5} m^2/s

                   

Temperature at 28 mm distance after t time  = =  50 degree C

we know that

\frac[ T_{28} - T_s}{T_i -T_s} = erf(\frac{x}{2\sqrt{at}})

\frac{ 50 -25}{300-25} = erf [\frac{28\times 10^{-3}}{2\sqrt{1.34\times 10^{-5}\times t}}]

0.909 = erf{\frac{3.8245}{\sqrt{t}}}

from gaussian error function table , similarity variable w calculated as

erf w = 0.909

it is lie between erf w = 0.9008  and erf w = 0.11246 so by interpolation we have

w = 0.08073

erf 0.08073 = erf[\frac{3.8245}{\sqrt{t}}]

0.08073 = \frac{3.8245}{\sqrt{t}}

solving fot t we get

t = 2244.3 sec

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3 years ago
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