Question

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

Answer 1

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 $10^{\circ}$:

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