What Is the Maximum Work Done and Thermal Efficiency of a Steam Plant?

[steamer]

A steam plant burns 6000 kg/hr coal, heat value 34 MJ/kg
18kg of air at 15 C is used/kg coal burned. Flue gas tem is 250 C.
46000 L/min cooling water used for condenser - inlet at 8 C and outlet 20 C

Find max work done kW/h and thermal efficiency.

I tried the following formula : Thermal Eff = heat energy converted to work / heat energy supplied or Themal Eff = heat energy equivalent (MJ/Kwh) / spec fuel cons. x calorific value.

but I must be missing something here ??

Any direction would be appreciated.
Steamer

 

[Andrew Mason]

A steam plant burns 6000 kg/hr coal, heat value 34 MJ/kg
18kg of air at 15 C is used/kg coal burned. Flue gas tem is 250 C.
46000 L/min cooling water used for condenser - inlet at 8 C and outlet 20 C

Find max work done kW/h and thermal efficiency.

I tried the following formula : Thermal Eff = heat energy converted to work / heat energy supplied or Themal Eff = heat energy equivalent (MJ/Kwh) / spec fuel cons. x calorific value.

but I must be missing something here ??

Any direction would be appreciated.
Steamer

Your approach is correct. The work done per unit of heat supplied is not given. Instead they give you information from which you can determine the heat supplied and the heat not converted to work (ie Q_C - the heat delivered to the cold reservoir or surroundings):

$$\eta = \frac{W}{Q_H} = \frac{Q_H - Q_C}{Q_H} = 1-\frac{Q_C}{Q_H}$$

AM

 

[thepatientmental]

To calculate the maximum work done and thermal efficiency of the steam plant, we need to use the following formula:

Thermal Efficiency = (Work Done / Heat Energy Supplied) x 100%

First, we need to calculate the heat energy supplied by the coal. The heat value of the coal is 34 MJ/kg, and 6000 kg of coal is burned per hour. So, the total heat energy supplied by the coal is 34 MJ/kg x 6000 kg/hr = 204000 MJ/hr.

Next, we need to calculate the heat energy converted to work. This can be done by using the formula: Work Done = Mass Flow Rate x Specific Heat x Temperature Difference. In this case, we have two sources of heat energy converted to work - the flue gas and the cooling water.

For the flue gas, we have a mass flow rate of 18 kg air/kg coal burned, and a temperature difference of (250-15) = 235 C. The specific heat of air is 1.005 kJ/kgK. So, the heat energy converted to work by the flue gas is 18 kg x 1.005 kJ/kgK x 235 C = 4232.1 kJ/hr.

For the cooling water, we have a mass flow rate of 46000 L/min, which is equivalent to 46000 kg/min. The temperature difference is (20-8) = 12 C. The specific heat of water is 4.186 kJ/kgK. So, the heat energy converted to work by the cooling water is 46000 kg/min x 4.186 kJ/kgK x 12 C = 2192608 kJ/hr.

Therefore, the total heat energy converted to work is 4232.1 kJ/hr + 2192608 kJ/hr = 2196839.1 kJ/hr.

Now, we can plug these values into the formula for thermal efficiency:

Thermal Efficiency = (2196839.1 kJ/hr / 204000 MJ/hr) x 100% = 1.077%

To find the maximum work done, we need to convert the heat energy converted to work from kJ/hr to kW/hr, which is done by dividing by 3600 (since 1 kW = 3600 kJ). So, the maximum work done is 2196839.1 kJ/hr /