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BabaBlast [244]
3 years ago
9

An inventor claims to have developed a power cycle operating between hot and cold reservoirs at 1175 K and 295 K, respectively,

that provides a steady-state power output of (a) 28 kW, (b) 31.2 kW, while receiving energy by heat transfer from the hot reservoir at the rate 150,000 kJ/h. Evaluate each claim
Engineering
1 answer:
Alinara [238K]3 years ago
4 0

Answer:

<em>a) efficiency is equal to 67.2%, and this is lesser than the maximum obtainable efficiency, so this power output is possible.</em>

<em>b) efficiency is 75%, this is approximately equal to the maximum obtainable efficiency, but not more that it. This power output is also possible.</em>

<em></em>

Explanation:

Cold reservoir temperature Tc = 295 K

Hot reservoir temperature Th = 1175 K

Energy input Q = 150000 K/h

Converting to kJ/s, Q  = 150000/3600 = 41.66 kJ/s

Maximum efficiency that can be obtained from this cycle = 1 - \frac{Tc}{Th}

==> 1 - \frac{295}{1175} = 0.748 ≅ 75%

also recall that actual cycle efficiency = \frac{W}{Q}

Where W is the energy output or work

a) for work of 28 kW,

eff =  \frac{W}{Q} =  \frac{28}{41.66} = 0.672 ≅ 67.2%

this is lesser than the maximum obtainable efficiency, so this power output is possible.

b) for work of 31.2 kW

eff =  \frac{W}{Q} =  \frac{31.2}{41.66} = 0.748 ≅ 75%

this is approximately equal to the maximum obtainable efficiency, but not more that it. This power output is possible.

NB: kW is also equal to kJ/S

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Answer:

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What are the optical properties of steel
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Answer:

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Explanation:

3 0
3 years ago
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
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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

3 0
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False

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In a Rankine cycle, superheated steam that enters the turbine at 1273.15 K and 1.8 MPa is then expanded to a vapor at 0.1 MPa. W
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Answer:

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Given:

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From the table superheated,

h_{i} = 4635 \frac{K J}{Kg} and  h_{f} = 2706.54 \frac{K J}{Kg}

Work done by shaft is,

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Therefore, the shaft work generated per kilogram is -1.3 \frac{MJ}{kg}

6 0
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