Power Plant Engineering

FUNDAMENTAL OF POWER PLANT 31

On solving, we get m = 0.1304 kg

Total work done = 1 × (H 1 – H 2 ) + (1 – m)(H 2 – H 3 )

= (3314 – 2460) + (1 – 0.1304)(2560 – 2100)

= 1154 kJ/kg

(1) Specific steam consumption =

3600

1154

= 3.12Kg/kW-hr. Ans.

(2) Thermal efficiency =

1154

(3314 439.43)−

= 40.15%. Ans.

(3) With out regeneration feed water heating the work done will be

H 1 – H 2 = 3314 – 2100 = 1214 kJ/kg

Steam consumption =

3600

1214

= 2.94 kg/kW-hr

Without regeneration heating, the thermal efficiency

η =^13

13

(H H )

(H H )

−

− w

Now from the steam tables

Hw 3 = 121.42 kJ/kg

η =

(3314 2100)

(3314 121.42)

−

−

= 0.38

Increase in thermal efficiency due to regeneration feed water heating is

=

(0.4015 0.38)

0.4015

−

= 5.5 % Ans.

THEORETICAL QUESTIONS

1.What is the concept of Power plant?

2.Define the types of energy.

3.What are the resources for power development in India?

4.What is the present position of power in India?

5.What is the future planning for power generation?

6.Define different types of power cycle.

7.Write short notes on fuel and combustion.

8.Write short notes on steam generators.

9.Write short notes on steam condenser.

10.Briefly describe water turbine.

11.Differentiate between impulse and reaction turbine.