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BASIC EQUATIONS IN INTEGRAL FORM FOR A CONTROL VOLUME – CH 4

BASIC EQUATIONS IN INTEGRAL FORM FOR A CONTROL VOLUME – CH 4. CONTROL VOLUME APPROACH. t. t + t. X t = total amount of mass, momentum or energy of fluid particles in control volume (V 1 and V 2 ) at time t. At t + t same fluid particles now in V 2 and V 3. Divide by t.

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BASIC EQUATIONS IN INTEGRAL FORM FOR A CONTROL VOLUME – CH 4

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  1. BASIC EQUATIONS IN INTEGRAL FORM FOR A CONTROL VOLUME – CH 4

  2. CONTROL VOLUME APPROACH t t + t

  3. Xt = total amount of mass, momentum or energy of fluid particles in control volume (V1 and V2) at time t. At t + t same fluid particles now in V2 and V3. Divide by t

  4. t +t XV1(t) XV3(t+t) V2 t In the limit as t goes to zero V2 approaches that of the control volume. Rate of change of X within control volume Difference between rate at which X leaves control volume to that at which it enters.

  5. In the limit as t goes to zero, V2 approaches that of the control volume. t +t V2 t Rate of change of X within control volume Difference between rate at which X leaves control volume to that at which it enters. Rate of mass flow through surface x = X per unit mass

  6. 0 t V(x,y,z) Rate of mass flow through surface x = X per unit mass

  7. x = X per unit mass • X is any property of the fluid • (mass, momentum, energy) • x is the amount of X per unit mass • (in any small portion of the fluid) • The total amount of X in the control volume • =    CV xdV • dV = differential mass of fluid, dm, so x = dX/dm

  8. x = dX/dm Relates properties of a fixed mass system of fluid particles to the properties of the fluid inside of and crossing through the boundaries of a control volume.

  9. x = X per unit mass • The total amount of X in the control volume • =    CV xVol • dVol = differential mass of fluid, dm, so x = dX/dm • Msystem =    CV dm =    CV dVol, so x = 1 • Psystem =    CV Vdm =    CV VdVol,so x = V • Esystem =    CV edm =    CV edVol, so x = e

  10. Conservation of Mass X = total mass M of system of fluid particles dM/dt = 0 since mass can neither be created or destroyed x = 1 steady

  11. VA cos  VA if 1-D + –

  12. Example 100 m3 300oK Air dp/dt = ?

  13. Example 100 m3 300oK Air dp/dt = ? = ? = ?

  14. Example 100 m3 300oK Air dp/dt = ? 100 /t (kg/s)– 8 (kg/s) = 0 p = RT p/t = RT/t = 0.287 (kJ/kg-K)300(K) (8 (kg/s)/100 (m3)) = 6.89 kJ/(s-m3) = 6.89 kPa/s

  15. Conservation of Momentum X = total lin. mom. p of system of fluid particles dp/dt is the rate of change of lin. mom. which equals the sum of the forces x = V

  16. Rate of increase of linear momentum within control volume Forces may be pressure, viscous, gravity, magnetic, electric, surface tension, …. Note – control volume can not be accelerating Net rate of efflux of linear momentum through the control volume

  17. steady steady

  18. What is force on plate to keep in place? Air stream, 2 cm in diameter and 100 m/s and density of 1.2 kg/m3

  19. Air stream, 2 cm in diameter and 100 m/s and density of 1.2 kg/m3 What is force on plate ? Fplate= (100 m/s) [(1.2 kg/m3)(100 m/s)(/4)(0.022m2)] Fplate = 3.77 kg-m/s2 = 3.77 N

  20. Conservation of Energy Q - W’ = dE  indicates interactions across system boundary d indicates change of property within system Q = heat W’ = all work including mechanical, electric and magnetic E = energy such as internal, U, kinetic energy, KE, and various forms of potential energy, PE.

  21. Conservation of Energy E = total energy of fluid particles X = e e refers to the total energy E of the fixed mass system per unit mass

  22. If E can be assumed to be U + KE + mgh then e = u + V2/2 + gz

  23. W’ = flow work (pressure) + W (viscous, shaft, electric, magnetic) Work that acts against the external pressure at boundaries ~ if volume of mass is V then work required is p V or since the density  = m / V, (so  m =  V) the flow work (p V) per unit mass (m) = p/ + flow work - flow work

  24. It is convenient to combine flow work per unit mass, p/, with the internal energy per unit mass, u, into the thermodynamic property enthalpy, h: Energy equation Energy equation

  25. Determine Power Output of Steam Turbine = ? h = 3kJ/kg h = 2.6kJ/kg 100 m/s 30 m/s 0.1 kg/s = 0.6 kJ/s

  26. 0 {(2600 – 3000)kJ/kg + (1002- 302)/2m2/s2} 0.1 kg/s = -0.6kJ/s – dW/dt

  27. 0 {(2600 – 3000)kJ/kg + (1002- 302)/2 m2/s2} 0.1 kg/s = -0.6kJ/s – dW/dt 4550m2/s2 = 4550kgm2/[kgs2] = 4550Nm/kg = 4.550kJ/kg -395.45 kJ/kg x 0.1 kg/s = -39.545 kJ/s -39.545 kJ/s = -0.6 kJ/s – dW/dt dW/dt = 38.945 kJ/s = 38.945 Watts

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