Engr Help

Finite Control Volume – Conservation of Energy (continued)

In a Nut Shell: There are differing versions of the integral form for conservation of energy

depending on the conditions of the flow and of the fluid.

If the flow is steady (no change with time), then ∂/∂t ∫ e ρ dV is zero.

Steady flow exists when the flow is established and doesn’t change with time.

For steady flow the integral form for conservation of energy becomes:

∫ e ρ V . n dS = dQ/dt _{net in} + dW/dt_shaft _{net
in}

∫ [ u + p/ρ + ½ V² + gz ] ρ V . n dS = dQ/dt _{net in} + dW/dt_shaft _{net
in}

Steady, Incompressible, Uniform Flow (uniform velocity profile across cs)

Σ [ u + p/ρ + ½ V² + gz ] dm/dt _out - Σ [ u + p/ρ + ½ V² + gz ] dm/dt _in =

dQ/dt _{net
in} + dW/dt_shaft _{net in}

If in addition there is only one stream entering and one stream leaving the cv

{ [ u + p/ρ + ½ V² + gz ]_out- [ u + p/ρ + ½ V² + gz ]_in } dm/dt = dQ/dt _{net in} + dW/dt_shaft _{net in}

Note: h = u + p/ρ = enthalpy per unit mass Note: u, p/ρ, V², gz are in ft lb/slug, m N / kg

dm/dt is the mass flow rate in slugs/sec, kg/sec Note: Each term is in ft lb / sec, N m / sec

The One-Dimensional Energy Equation (in terms of head)

[ p/ρg + V² /2g + z ]_in + h_s = [ p/ρg + V² /2g + z ]_out + h_L

p/ρg – pressure head, V²/ 2g – velocity head, z – elevation head (units are ft or meters)

h_s - shaft work head (as in a pump); turbine would be negative work head (units are ft or meters)

h_L – head loss due to friction (head loss) (units are ft or meters)