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Enthalpy Property and the First Law: "When Do I Use Enthalpy and Why?"

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Enthalpy Property and the First Law: Enthalpy Property and the First Law: "When do I use enthalpy Professor Amanda D. Smith and why?" 1 What is this enthalpy thing? Start with First Law Closed, general form: Enthalpy is a just collection of other properties: dE = dU + dKE + dP E = δQ − δW H = U + PV (or, per unit mass: h = u + P v) 1. No significant kinetic or potential energy changes (although the boundary moves, the system as a whole stays in place) Collecting internal energy, pressure and volume together in this way is convenient because these properties often wind up together dU = δQ − δW when we simplify the First Law. Let’s heat a few things up to illus- trate. 2. Integrate over the duration of the process 1 to 2 2 First Law for a Closed System, Constant U2 − U1 = Q12 − W12 Mass and Volume 3. Note that the only work interaction is boundary work (the fluid Ex. A brick sitting on a hot plate is heated up expands and exerts work on its environment) Z 2 U2 − U1 = Q12 − P dV 1 Start with First Law Closed, general form: 4. If the piston is exposed to atmospheric pressure, and the pro- dE = dU + dKE + dP E = δQ − δW cess from 1 to 2 takes place slowly enough that we can con- 1. No significant kinetic or potential energy changes (brick just sider it to be a series of quasi-equilibrium states (i.e. the sits there) piston does not shoot up when the gas heats up), we can dU = δQ − δW pull pressure out of the integral. 2. No work done on or by the brick U2 − U1 = Q12 − P (V2 − V1) dU = δQ 3. Integrate over the duration of the process 1 to 2 5. Now let’s gather the terms for state 1 and the terms for state U2 − U1 = Q12 2, and put Q by itself. 4. Per unit mass (U2 + PV2) − (U1 + PV1) = Q12 u2 − u1 = q12 Note that here the system is at rest with a consistent boundary, Recognize anyone here? and the change in internal energy of the system gives us the heat transferred into the system. We don’t run into enthalpy. H2 − H1 = Q12 3 First Law for a Closed System, Constant 6. Per unit mass Mass, Changing Volume h2 − h1 = q12 Ex. Gas in a piston-cylinder is heated Now the system has a moving boundary, and we find that the change in enthalpy gives us the heat transferred into the system. Note that ∆h does not represent the ∆e for the system; rather it is a collection of internal energy changes and PdV work. This works for any simple compressible system with boundary work occurring at constant pressure. © 2016 A.D. Smith 4 Ex. First Law for an Open System, Steady Proceed through steps 1 through 6, as we did on the previous State page . h2 − h1 = q12 Ex. Fluid flowing through a pipe is heated at a steady rate This means that the change in enthalpy during this process will tell us how much heat we have to add to boil the water (going from x=0 to x=1). So, if we wanted to take water from its saturated liquid state to its saturated vapor state by heating it up, we would need to add energy in the amount of hg −hf . This quantity, hfg, is called the en- Start with First Law Open, rate form: thalpy of vaporization (also known as latent heat of vaporization), dE X 1 2 X 1 2 = Q_ − W_ + m_ (h + V + gz) − m_ (h + V + gz) dt in 2 out 2 telling us the amount of energy it takes to vaporize the water. Note that the enthalpy of vaporization can only have meaning if 1. Steady state (no change in energy within control volume if we specify its corresponding pressure (or temperature—these are fluid enters at a constant temperature) dependent properties at saturation). X 1 2 X 1 2 0 = Q_ −W_ + m_ (h+ V +gz)− m_ (h+ V +gz) in 2 out 2 Think about this process on our T-V diagram with saturation dome. 2. Steady mass flow (one inlet, one exit, no accumulation) Say that the atmospheric pressure is P1 here: 1 1 0 = Q_ − W_ +m _ (h + V 2 + gz ) − m_ (h + V 2 + gz ) i 2 i i e 2 e e 3. No significant kinetic energy changes (mass flow rate is the same at inlet and exit; if the entrance and exit are the same size, fluid velocity should be the same) 0 = Q_ − W_ +m _ (hi + gzi) − m_ (he + gze) 4. No potential energy changes (horizontal pipe) 0 = Q_ − W_ +m _ (hi − he) 5. Per unit mass (divide through by m_ ) 0 = q − w + hi − he 6. No work done on or by the system 1 he − hi = q The enthalpy of vaporization is the energy it takes to get from So here, ∆h represents the heat transferred to the fluid again. the left side of the dome to the right side of the dome across this Note that enthalpy change is exit – inlet, which makes sense be- straight horizontal line (that is, at constant pressure). cause we expect the fluid to have a higher specific enthalpy after being heated. Note that there is another isobar (line of constant pressure) shown above ours, P2, and the distance from left to right is shorter at this 5 Enthalpy and Phase Changes higher pressure. • What do you think this indicates about the energy it takes to Let’s look at the example from section 3 again, but start the pro- boil water at different atmospheric pressures? cess with liquid water as the working fluid. • Can you see why cooking directions might be different for Ex. Saturated liquid is heated in a piston-cylinder at constant high altitudes? pressure until it becomes a saturated vapor Start with First Law Closed, general form: dE = dU + dKE + dP E = δQ − δW 1Remember that flow work (pushing fluid through the control volume) is accounted for in the enthalpy term and not pulled out separately. © 2016 A.D. Smith.
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