1. Internal Combustion Engineering Division
ENGINEERING THERMODYNAMICS
M.R.SWAMINATHAN
Lecturer
Internal Combustion Engineering Division
Department of Mechanical Engineering
ANNA UNIVERSITY CHENNAI
CHENNAI-25.

2. ENGINEERING THERMODYNAMICS
Dr. M.R.SWAMINATHAN
Assistant Professor
Internal Combustion Engineering Division
Department of Mechanical Engineering
ANNA UNIVERSITY
CHENNAI-25.

3. STEADY FLOW ENERGY EQN. Under steady-flow conditions, the
mass and energy contents of a control volume remain constant.

6. The fluid entering the system will have its own internal, kinetic and potential energies.
Let u1be the specific internal energy of the fluid entering
C1be the velocity of the fluid while entering
Z1be the potential energy of the fluid while entering
Similarly let u2 ,C2 and Z2 be respective entities while leaving.

7. Applying I law of thermodynamics we get

9. PROBLEM 1
The power output of an adiabatic steam turbine is 5 MW, and the inlet and the exit conditions of the steam are as indicated in Fig.1
(a) Compare the magnitudes of ∆h, ∆ke & ∆ pe.
(b) Determine the work done per unit mass of the steam flowing through the turbine.
(c) Calculate the mass flow rate of the steam.

10. Fig.1

11. Solution
The inlet and exit conditions of a steam turbine and its power output are given.
The changes in kinetic energy, potential energy,
enthalpy of steam,
the work done per unit mass and the mass flow rate of steam are to be determined.

12. Assumptions
This is a steady-flow process since there is no change with time at any point and thus mCV = 0 and ECV = 0.
The system is adiabatic and thus there is no heat transfer

13. Analysis
We take the turbine as the system. This is a control volume since mass crosses the system boundary during the process. There is only one inlet and one exit and thus m1=m2=m
Also, work is done by the system. The inlet and exit velocities and elevations are given
Thus the kinetic and potential energies are to be considered

14. From Steam Tables
P1 = 2 MPa ,T1 = 400°C h1 = kJ/kg
At the turbine exit, we have a saturated liquid–vapor mixture at 15-kPa pressure.
The enthalpy at this state is
h2 = hf +x2hfg
= [ (0.9)(2372.3)] kJ/kg
= kJ/kg

15. ∆h = h2- h1
= kJ/kg
= kJ/kg
= kJ/kg
∆p.e= g(z2-z1)/1000=9.81 (6-10)x10-3
= kJ/kg

16. Making an energy balance for the steady flow system Ein= E out
Dividing by the mass flow rate, the work done by the unit mass of the system is
= [ ] kJ/kg
= kJ/kg

17. For a 5 MW power output we have
Mass flow rate =
Mass flow rate = 5.73 kg/s