How can enthalpy values be calculated for an isothermal process involving steam?

[Jameseyboy]

What equations are relevant to finding the initial and final enthalpy values of an isothermal process?

There is an initial dryness fraction and a heat value is also added.

I know how to use steam tables to find the initial value of H but not the final. Does H even change?

Thanks

 

[sanka]

You need to provide a bit more detail about the specific problem. Suppose you are referring to condensation which is essentially isothermal heat rejection. So if you know the inital value of h (from an assumption of isentropic expansion through the turbine, for example), the final value of h will depend on the condition of the working fluid at the end. If you are told saturated water is present at the exit, that just means you can find the value of hf at the same temperature as that at the inlet (isothermal, T=constant).

 

[CFXMSC]

I think this equation might solve your problem.

\frac{dE}{dt}=\dot{Q}-\dot{W}+\overbrace{\sum_{i}\dot{m_i}\left(h_i+\frac{v_{i}^{2}}{2}+z_i \right)}^{Inlet flow}-\overbrace{\sum_{j}\dot{m_j}\left(h_j+\frac{v_{j}^{2}}{2}+z_j \right)}^{Outlet flow}

 

[Ritz_physics]

I assume you know how to find the enthalpy of a saturated liquid- vapour mixture. If heat is transferred at constant pressure, ie. at saturation pressure, just add the amount of heat to the initial enthalpy value, to get to the final enthalpy value.
If heat addition is not isobaric, but isothermal, the enthalpy should increase till the mixture gets converted into saturated vapour. Beyond that, it would remain constant. You will see a pressure drop and increase in volume.