Internal energy vs. Enthalpy vs. Entropy

JSBeckton

Ok, I must admit that I am becoming a bit confused about these concepts. I understand that enthalpy is u + Pv, and entropy has something to do with molecular randomness. I was fine until we started to solve for heat transfer when dealing with entropy and now I am confused, sometimes we use:

Q=m(u2-u1) to describe heat transfer

and other times we use:

Q=m(h2-h1)

Can anyone please explain to me how to tell which should be used where? I know that its a very important concept that i must have missed.

Thanks

siddharth

Let's start with the first law. It says the change in the internal energy [itex]\Delta U[/itex] is

[tex] \Delta U = Q + W [/tex]

Similarly the change in enthalpy is

[tex] \Delta H = \Delta U + \Delta (PV) [/tex]

From this, you can calculate the heat exchange for a constant volume process and a constant pressure process.

So for a
1) Constant volume process
The first law reduces to
[tex] \Delta U = Q[/tex]
So, the heat transfer will be [itex] Q =m(u_2-u_1)[/itex]

2) Constant pressure process
In this case, the first law is
[tex] \Delta U + P\Delta V = Q [/tex]
But, from the definition of enthalpy, you also have
[tex]\Delta H = \Delta U + P\Delta V [/tex]

So, can you complete this and figure out how you calculate Q in each case?

If you need to calculate Q for a general process, try calculating the change in internal energy and the work, then use the first law.
Finally, in some cases you may be able to calculate the heat transferred if you know the change in entropy. (For example, a reversible isothermal process)

JSBeckton

Thanks a lot siddharth

[text]
\begin{array}{l}
Q = \Delta U + P\Delta V \\
{\rm where }\Delta H = U + P\Delta V \\
{\rm therefore }Q = \Delta H \\
\end{array}
[/text]

So, is it safe to say that I should use U for constant volume and H for constant pressure?

JSBeckton

Sorry for the double post, cant figure out this text editor

Thanks a lot siddharth

Q=delta H

So would it be safe to say that i should use U for constant volume and H for constant pressure?

siddharth

For an ideal gas, and ignoring changes in Kinetic energy and such, yes.

JSBeckton

When I have a steam turbine I use enthalpy not internal energy even though its not a constant pressure process. I know that steam is not an ideal gas but how do I choose which to use?

Is this true?

Q-W=U2-U1
W=(U1-U2)+Q
W=(H1-H2)

But if its adaibatic Q=0 so,
W=(U1-U2)

siddharth

Whoops, my error. You can use it for a non-ideal gas, but the system should be closed.

For a steam turbine, you have a flow process in an open system. In that case, you'll have to use the first law for open systems.

[tex] \frac{dE}{dt} = \dot{Q} - \dot{W} + \sum_i \dot{m_i} h_i - \sum_j \dot{m_j} h_j[/tex]

You'll find how you get this in any thermodynamics book. It's different from the first case, which is for closed systems.