TITLE.PM5

128 ENGINEERING THERMODYNAMICS

dharm

M-therm/th4-2.pm5

The completed table is given below :

Process Q(kJ/min) W(kJ/min) ∆ E(kJ/min)

1—2 0 4340 – 4340

2—3 42000 0 42000

3—4 – 4200 69000 – 73200

4—1 – 105800 – 141340 35540

Since cycleΣΣQW=cycle

Rate of work output = – 68000 kJ/min = –^6800060 kJ/s or kW

= 1133.33 kW. (Ans.)

Example 4.13. The power developed by a turbine in a certain steam plant is 1200 kW. The

heat supplied to the steam in the boiler is 3360 kJ/kg, the heat rejected by the system to cooling

water in the condenser is 2520 kJ/kg and the feed pump work required to pump the condensate

back into the boiler is 6 kW.

Calculate the steam flow round the cycle in kg/s.

Solution. The power developed by the turbine = 1200 kW

The heat supplied to the steam in the boiler = 3360 kJ/kg

The heat rejected by the system to cooling water = 2520 kJ/kg

Feed pump work = 6 kW

Wout

Qout

Condenser

Boundary

Feed pump

Win

Qin Boiler

Turbine

Fig. 4.19
Fig. 4.19 shows the cycle. A boundary is shown which encompasses the entire plant. Strictly,
this boundary should be thought of as encompassing the working fluid only.

zdQ = 3360 – 2520 = 840 kJ/kg