Dear Madhav,

Kindly calculate the Mach No. for your flow. If is less then 0.3 (hopefully as mentioned by you that your flow is laminar), then whatever energy is given to fluid can only do two things, (1) increase/decrease the internal energy of the and fluid, (2) work done by the fluid/work done on the fluid. Simple thermodynamic laws with equation of state can be used to derive the relationship.

However, no clear cut information is provided by you on the type of process happening, if we assume a steady state and apply the conservation of mass by in the defined control volume needed to be studied we can get change in velocity in terms of change in density.

Simply apply density*flow velocity*cross-sectional area = constant (C) .......{1}

This constant (C) can be obtained from the boundary condition. By assuming constant area duct and differentiating the equation you will get change in velocity in terms of density,change in density and velocity as du = -ud(rho)/rho

Now, K.E is given as = 1/2*rho*u*u.............(2)

Similarly, differentiate equation (2) to obtain change in K.E.

Hopefully things are clear now.

The key of the link between velocity and heat is the dissipation term. It is mu*S:S where S is the symmetric part of the velocity gradient. This term appears both in the kinetic energy equation and in the internal energy equation with opposite sign. The relevance of the dissipation is intrinsically due to the viscosity oif the fluid and the magnitude of the gradient and, consequently on the lenght scale. For example in turbulence the dissipation acts on the highest wavenumber components of the flow and convert kinetic energy into internal energy. Similar is the energy transfer in presence of shocks.

static the kinetic energy depends on the variation of heat rate in other words on density or total pressure

Thank you Syed Mughees Ali Got the clear idea.

I am happy I could help. Best of luck for your research