A flow of steam enters a trubine at pressure P110Mpa and tem

A flow of steam enters a trubine at pressure P1=10Mpa and temperature T1=480 degrees celsius. The steam leaves the turbine at a pressure P2=100Kpa. The qualtiy of the exiting steam is X2=93%. The heat loss occuring during the process is Q12=850 Kj/s. Neglect the velocity effects of the flow at the inlet and at the exit. Also, both inlet and exit are located at the same height reltivety to ground. The mass flow of steam is 10Kg/s.

Determine:

a. turbine schematic with knowns and unknowns labeled

b. assumptions

c. full energy balance equation

d. enthalpy of inlet at h1

e. enthalpy of steam at exit h2

f. The power output in KJ then convert to MW

g. specific volume of steam(v1) at inlet

h. specific colume of steam(v2) at exit

i. tv diagram

Solution

The energy balance equation for level flow is simply change in enthalpy/kg = V*V/2g

d) From tables at 10MPA ( 100bar) enthalpy of steam is 2727 KJ/Kg

e) at 100 KPA ( .01 bar) enthalpy is 2675 KJ/Kg

the enthalpy used up is 52KJ/kg

From the energy equation the velocity equivalent for 93% steam would be

SQRT( 2* 52000*0.93) = 311 m/s

Power output = 52KJ/Kg/* 10Kg/s ( without heat losses)

= 520 K J/sec

g) sp vol from tables at inlet 0.018

h) from tables at outlet 1.69

Loss of 850 KJ/s is too high for this output.