Dear Dr. Flynn:
I am having trouble convincing myself that the first law of thermodynamics can reduce itself to dH = dQ - dW for steady flow systems. I start with the basic equations of dU = dQ - dW and dH = dU+PdV+VdP. dH must be equal to dU which means PdV must = -VdP. This can only be true for an isothermal system with an ideal gas since PV = nRT and d(PV) = PdV + VdP = 0. Are these the correct assumptions for the derivation of dH = dQ -dW? I would appreciate seeing the derivation.
You will note I may be using different sign conventions than what some of you are used to, but I am sure we can all get over that.
The equation dH = dQ - dW holds true for a steady flow system.
The general equation is
dH = dQ - dW + pdV + Vdp
Now, we need to define what “steady flow” means. It means you have a pipe where everything that goes in one end comes out the other.
First, since we are talking about a pipe, you know that volume isn’t going to change. dV = 0.
So, what about Vdp? Well, if you increase the pressure at one end of the pipe, water shoots out the other end faster. It has to. If the pressure against the walls of the pipe didn’t remain the same, the system wouldn’t remain STEADY. More water would be going in one end than coming out the other.
Therefore, since you can’t just go around increasing or decreasing the pressure inside the pipe, dp = 0.
Therefore dH = dQ - DW.
As a consequence of all this, you know that, for ideal gasses, the system is isothermal because pV = nRT and pdV + Vdp = RTdn + nRdT + nTdR = 0. Since dR = 0 and dn = 0, dT must = 0.
Call me this cynic, but I am quite the realist, too: I believe in practical application - not theory. Your first thermo class will get so bogged down with all these dthis and dthat - infintesimal pieces of internal energy, enthalpy, etc. It’s ridiculous!
Things boil down to simple equations - especially when analyzing a steady flow system. Unless you’re really out to reinvent the wheel, the thermo tables for numerous working fluids are out there. And, it’s all the same set of equations, over and over again, as you analyze the cycle you’re applying.
I think they teach it this way as a weeder course. It’s so unnecessary! You don’t have to start from scratch and re-derive the thermo basics every time you just want to analyze a cycle!
Why make a mountain out of a mole hill?
Jinx, Victim of the rigorous approach to every damn thring college tried to teach… get these profs iout of their ivory towers!
Jinx, eventually the time will come when you will need to go back to basics. For instance, it might come when you first look at a new system of a different type. Sure, you know thermo of ideal gasses. What if the gasses aren’t ideal? What if you’re not dealing with a gas at all, but domains in a magnetic material? These things can also be studied thermodynamically, but they might not use the same simplifications that work for ideal gasses. The only things that do work for all systems are the basic, ivory tower, “dthis and dthat”.
I originally addressed this question to a Dr. Flynn, but never received a reply. His name appeared on this site by error.
Pencil Pusher makes a good case and confirms that the equation is true for an isothermal system. That being the case, then dQ = 0 except for phase change situations. Since dW = VdP, and since PP established dP = 0, that means dH = 0 except for phase changes. Nope, I’m not convinced. Something is missing here.
I agree with Chronos: you’ve got to go back to basics to understand the foundations. How do you explain enthalpy? From dH = dU + PdV + VdP, that’s how. Now put it into words: it is an energy change associated with a change in the internal energy of a system plus the change in it’s PV energy.