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Kinetic Energy and Bertz Limit

See the attached file. In the file, you will see a diagram that shows a steam-tube in which the rate of flow of mass is dm/dt. Write down an expression for dm/dt at the turbine in terms of the density of air p.

The power P generated is equal to the rate of change of kinetic energy
P = 1/2 dm/dt (u0^2-u2^2) (i)

Show that an alternative expression is given by
P = dm/dt (u0-u2)u1 (ii)

Hence show that: u2 = 2u1-u0

Substituting in (ii) and writing u1 = u0 (1-a) show that
P = 1/2pu0^3A1Cp where the power coefficient Cp = 4a(1-a)^2


Solution Preview

Please refer to the attachment for the complete solution with diagrams.

In case of an incompressible liquid flowing in the tube (such as water), the density ρ can be assumed to be constant at all sections of the tube. However, the gases being conpressible, the density will vary at sections of different cross sectional areas in case of a gas flowing in the tube. In either case (liquid or gas), the rate of mass inflow of fluid must equal the rate of mass outflow if we assume no accumulation of the fluid within the tube. In fact, rate of flow of mass (dm/dt) any where within the tube must be same (hence dm/dt = constant).
Let us consider a cylinderical element of fluid of cross sectional area A0 and an infinitessimally small width dx at the entrance of the tube.
Volume of the fluid element dV = A0dx
Mass of the fluid element dm = A0 dx ρ0 (where ρ0 = fluid density at the entrance)
Rate of mass inflow = dm/dt = A0 (dx/dt) ρ0 = A0 u0 ρ0 (where u0 = velocity of ...

Solution Summary

The following posting helps with problems involving kinetic energy and bertz limit.