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ENGINEERING THERMODYNAMICS

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

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ENGINEERING THERMODYNAMICS

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  1. 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.

  4. 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.

  5. Applying I law of thermodynamics we get

  6. 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.

  7. Fig.1

  8. 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.

  9. 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

  10. 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

  11. From Steam Tables P1 = 2 MPa ,T1 = 400°C h1 = 3248.4 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 = [225.94 (0.9)(2372.3)] kJ/kg = 2361.01 kJ/kg

  12. ∆h = h2- h1 =2361.01 - 3248.42 kJ/kg =887.39 kJ/kg = 14.95 kJ/kg ∆p.e= g(z2-z1)/1000=9.81 (6-10)x10-3 = - 0.04 kJ/kg

  13. 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 = [887.39 14.95 0.04] kJ/kg = 872.48 kJ/kg

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

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