500 kghr of steam drives turbine the steam enters the turbin

500 kg/hr of steam drives turbine. the steam enters the turbine at 44 atm and 450 C at a velocity of 60 m/sec and leaves at point 5 m below the turbine inlet at atmospheric pressure and velocity of 360 m/sec. the turbine drives shaft work at a rate of 70 kW, and heat loss from the turbine is estimated to be 11.6 kW. Calculate the specific enthalpy change associated with the process ?

Solution

The energy eqn is ( H + V^2/2 + gZ)*m\' = Q\' -W\'

where H = enthalpy = U+PV

V = velocity

Z equal reference height above datum

Q\' = rate of heat transfer

W\' = rate of work

evaluating m = 500 kg/hr = .139 kg/s

Contribution of energy of steam velocity (m\'/2)( V2^2 -V1^2) = .139 *(360^2-60^2) = 8.76 KW

Contribution of potential energy change m\'gDZ = .139 *g* (-5) = -6.82 *10^-3 KW

Heat input - work output = -11.6 -70 = -81.6 KW

Summing all the contributions, get -90.35 KW

Enthalpy change per unit mass = specific enthalpy = -90.35/.139 = -650 KJ/kg