Steam flow calculation 3

[flinana]

I have a calculation based on enthalpy so as to attemperate steam to a turbine. The calculation has the following formula which I can´t understand.

Flow = m x(Enthalpy (HPT)- Enthalpy (CRH) /Enthalpy (CRH) - 764.8) kg/s

Enthalpy (HPT) = High pressure turbine. Enthalpy calculated from presssure and temperature before HP turbine

Enthalpy (CRH) = Enthalpy calculated from pressure and temperature before reheater, to where the HP steam is bypassed.

m = mass flow according to Steam valve opening, using pressure and temperature before valve.

Find logic attached.

The part I dont get is (Enthalpy (HPT)- Enthalpy (CRH) /Enthalpy (CRH) - 764.8).

Can someone explain.

thanks

 

[ione]

Finana,

You have stream IN with enthalpy equal to enthalpy (HPT).
You have stream OUT with enthalpy equal to enthalpy (CRH)

The value 764.8, I presume expressed in kJ/kg, should be the specific water enthalpy at saturation pressure (9.08 barg) or saturation temperature (180.344 °C) before reheater. What works here is the enthalpy of evaporation, which is the difference between the specific enthalpy of steam and the specific enthalpy of water.

 

[flinana]

thanks, but i dont get is why the enthalpy out does not include the enthalpy from the ST outlet. Enthalpy in = enthalpy out, i.e.Enthalpy of steam = enthalpy of steam after water has been sprayed. But the pressure reading is taken after the connection from the ST hP exhaust!! see my dwg attached.

 

[hacksaw]

sounds like they've taken the worst case (no work by the turbine)

 

[flinana]

Good point hacksaw. But why attemperate in that case?

 

[davefitz]

Just rearrange the terms and end up with the first law balance across the spray station:

(Whpt+Ws)*Hcrh = Whpt*Hhpt+Ws*Hs

where Hs= spray water enthalpy