# When are work and heat calculated by differences in enthalpy or internal energy

1. Oct 6, 2009

### Yellow Fang

Hi,

This is not exactly homework, but I'm not sure it belongs in any other section either. I've been reading through Applied Thermodynamics for Engineering Technologists by Eastop & McConkey. In general it's pretty good I think. However, in the first few chapters, it explained that Q+W = U2-U1, i.e. Heat plus work equals the difference in internal energy (in steam). Then I got to the chapter on steam cycles where it started calculating heat and work in differences in enthalpy. I haven't really got to the bottom of why this is. I know that enthalpy is internal energy plus the product of pressure and volume (h = u + pv). I looked up one book in the library which refered to 'shaft work' when work is done with no change in pressure, but I can't see how this could apply with a turbine because there in change in pressure during the compression and expansion processes.

It seems that enthalpies are used for calculations where mass flow is involved, and that the equation in question is

m(u1 + v12/2 + gh1 + p1v1) + Q + W = m(u2 + v22/2 + gh2 + p2v2)

Am I along the right lines? Would Q+W=U2-U1 still apply for a reciprocating steam engine?

2. Oct 7, 2009

### defunc

Q+W=U2-U1 is applicable when you have a system (fixed mass). For a control volume (or open system) the correct energy balance is Q+W=H2-H1 .