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xxTIMURxx [149]
3 years ago
13

Consider a power plant with a heat rate of 10,800 kJ/kWh burning bituminous coal with 75 percent carbon and a heating value (ene

rgy released when it is burned) of 27,300 kJ/kg. About 15% of thermal losses are up the stack, and the remaining 85% are taken away by cooling water.a. Find the efficiency of the plant.b. Find the mass of coal that must be provided per kWh delivered. c. Find the rate of carbon and CO2 emissions from the plant in kg/kWh.d. Find the min flow of cooling water per kWh if its temperature is only allowed to increase by 10◦C.
Engineering
1 answer:
grandymaker [24]3 years ago
8 0

Answer:

1. \eta_{p} = 33.34%

2. m_{coal} = 0.396 kg/kWh

3. carbon emission = 0.297 kg/kWh[/tex]

4. carbon-dioxide emission = 1.089 kg/kWh

5. Min flow of cooling water =  = 1457.14 kg/kWh

Given:

Heat Rate = 10,800 kJ/kWh

Heating value = 27,300 kJ/kg

Solution:

In order to calculate the plant efficiency, [rex]\eta_{p}[/tex]

Heat rate in coal fired steam power plant = \frac{3600}{\eta_{p}}

Therefore,

\eta_{p} = \frac{3600}{Heat rate}

\eta_{p} = \frac{3600}{10800}\times 100 = 33.34%

Now,

To calculate mass of coal per kWh, m_{coal}:

m_{coal} = \frac{Required heat rate}{heating value per kg}

m_{coal} = \frac{10800}{27300} = 0.396 kg/kWh

Now,

Rate of emission carbon and carbon-dioxide from the plant:

In accordance to the question, 75% of the bituminous coal being burned is carbon, thus carbon emission is given by:

carbon emission = 0.75\times m_{coal}

carbon emission = 0.75\times 0.396 = 0.297 kg/kWh = 29.7%

Now, for carbon-dioxide emission:

Since, molecular weight of carbon-dioxide = 44 kg

Thus carbon-dioxide emission:

0.297\times \frac{44}{12} = 1.089 kg/kWh

The minimum flow of cooling water per kWh if the allowance in temperature is 10^{\circ}:

\frac{0.85\times \frac{2}{3}\times 10800}{4.2} = 1457.14 kg/kWh

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fredd [130]

Answer:

T_o = 141.81 ^0C

Explanation:

Given that;

Thermal diffusivity \alpha = 5 \times 10 ^{-6} m^2/s

Thermal conductivity k = 20 \ W/m.K

Heat transfer coefficient h = ( we are to assume the imposed surface temperature ) = 20 W/m².K

Initial temperature = 150 ° C = (150+273) K = 423 K

Then coolant temperature with which the casting is exposed to = 20° C = (20+273)K = 293 K

Time = 40 seconds

Length = 20mm = 0.02 m

The objective is to determine the  temperature at the surface  at a depth of 20 mm after 40 seconds.

Bi = \dfrac{hL}{k}

Bi = \dfrac{20*0.02}{20}

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\tau = \dfrac{\alpha t}{L^2}

\tau=  \dfrac{5*10^{-6 }* 40}{0.020^2}

\tau = 0.5

For a wall at 0.2 Bi

A_1 = 1.0311

\lambda _1 = 0.4328

Therefore;

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\dfrac{T_o - 293 }{423 - 293}= 1.0311 \times e ^{-( 0.438^2 \times 0.5 )

\dfrac{T_o - 293 }{423 - 293}= 1.0311 \times e ^{-( 0.0959 )

\dfrac{T_o - 293 }{130}= 1.0311 \times 0.9085

\dfrac{T_o - 293 }{130}= 0.937

T_o - 293= 0.937 \times 130

T_o - 293= 121.81

T_o = 121.81+ 293

T_o = 414.81 \ K

T_o = (414.81 - 273)^0C

T_o = 141.81 ^0C

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Related Questions (More Answers Below)
7 0
3 years ago
What is the thermal efficiency of this reheat cycle in terms of enthalpies?
schepotkina [342]

Answer:

   \eta =\dfrac{\left (h_3-h_4\right )+(h_5-h_6)-(h_2-h_1)}{(h_3-h_2)+(h_5-h_4)}

Explanation:

For close gas turbine:

       Gas turbine works on Brayton cycle.Gas turbine have lots of applications like ,it is use in aircraft,in land applications etc.

Reheating is the method to improve the efficiency of the gas turbine.In reheating gas is expanding in two turbine instead of one turbine alone.Two turbine like high pressure turbine and low pressure turbine are used for expansion.

In the above diagram 1-2 is a compressor,2-3 heat addition,3-4 high pressure turbine,4-5 reheating of cycle 5-6 low pressure turbine,6-1 heat rejection,

We know that    \eta =\frac{W_{net}}{Q_{s}}

Now take h_{1},h_{2},,h_{3},h_{4},h_{5},h_{6} represent the enthalpy of point 1,2,3,4,5,6 in the cycle respectively.

So total heat supplied Q_S=

\left (h_3-h_2\right )+\left (h_5-h_4\right )

Net work out put

W_{net}=\left (h_5-h_6\right )-\left (h_2-h_1\right )

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      \eta =\dfrac{\left (h_3-h_4\right )+(h_5-h_6)-(h_2-h_1)}{(h_3-h_2)+(h_5-h_4)}

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