Question

Question 5 (10 marks) A perfect gas, initially at high pressure, flows steadily into an adiabatic, reversible turbine at a mass flow rate of 1.0 kg s!. It emerges at low pressure. The velocities at inlet and outlet are 50 m stand 75 m s', respectively, and the outlet pipe is 2.5 m above the inlet pipe. The power output of the turbine is 966 kW. Calculate: (a) the change in kinetic energy across the turbine per kg of gas; (2 marks) (b) the change in potential energy across the turbine per kg of gas; (2 marks) (c) the work done per kg of gas; (1 mark) (d) the enthalpy difference across the turbine per kg of gas; (3 marks) (e) the temperature difference across the turbine, if the gas has a constant specific heat capacity of 5.2 kJ kg K-1 (2 marks)

Answer 1

(a) change of kinetic energy per unit mass across the turbine is given by

AK.E = 0.5(2-2)

= 0.5+ (752-502)

  
= 1562.5

(b) Change in potential energy per unit mass

AP.E= g(Zoulet – Zinlet)

=9.81 +2.5

= 24.525J/kg

(c) Work done per unit mass(J/kg) = Power(J/s) / mass flow rate(kg/s)

  

966 * 1000

  
= 966000
  

(d)  change in enthalpy is given by :

general equation for our flow device

+ minlet(hin +0.5 * un + gzinlet) – moutlet (hout +0.5 * =Q - W dt uut + gzoutlet)

where

E_{CV} - energy content of the control volume enclosing the turbine

Q dot is rate of heat transfer

W dot is power

h being enthalpy

u in flow velocity

z is the elevation

Now

steady flow device LHS term zero

adiabatic assumpton gives  Q dot zero

W dot is given which is 966 kW .

Also mass flow rate is same (inlet = outlet) as steady flow

So we have

Houtlet – hinlet = - +0.5 * (uin - uput) +g* (zinlet – Zoutlet)

Using above derived results in {a}, {b} , {c}

= -966000 - 1562.5 - 24.525

= -967587.025 J/kg

(e) Now

Ah= CAT

So

AT =

= -186 deg kelvin (for difference unit doesn't matter kelvin or celsius)

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