1 a Define specific enthalpy in terms of specific internal e

1. a). Define specific enthalpy in terms of specific internal energy, pressure and density.

b). write down conservation of energy equation for steady-state steady flow process, for a single input and output, including thermal, kenetic, and potential energy, define all symbols and subscripts

c). write down conservation of mass equation for a stedy, one dimensional, constant-density flow in a duct whose cross sectional area varies with location [A= A(x)]

d). how can (P/) + (V²/2) + gz, represent the combined specific enthalpy, kenetic and potential energy for fluid flowing through a pipe or duct?

e). How do you obtain \"head\" from (P/) + (V²/2) + gz, specifying units in either meters or feet?

Solution

1(a)     specific enthalpy is the total energy per unit mass of the fluid when it is flowing. It is the sum of specific internal energy of the fluid and the specific flow work done by the fluid while it is flowing. It can be represented by the expression of

                                               h = u + pv

h = u + p/

1(b)    conservation of energy equation for steady state steady flow process can be expressed as

h₁ + v₁²/2 + gz₁ + q^. = h₂ + v₂²/2 + gz₂  + w

where

h= specific enthalpy

v²/2 = specific kinetic energy

gz = specific potential energy

q and w are the heat transfere and the work involve in the process

1(c) consevation of mass equation can be expressed as ....with varying the area

  ₁A₁V1 = ₂A₂V₂

  where = density of the fgluid

   A_{1 and} A₂ are the varying area at the inlet and the outlet

   v is the velocity

1(d)     the equation ) + (V²/2) + gz   this equation represents the whole specific enthalpy, kinetic and potential energy. Because the first term represents the specific enthalpy of the fluid and the second term represents the kinetic and the third term represents the potential form of energy of the fluid. All in the per unit mass i.e. specific energy term.

1(e)    the given term is a specific energy equation. And the specific energy can be converted to mt5he head term by just dividing the term by “g” (acceleration due to gravity)