Fall 1999, METBD330 - Thermodynamics
Procedure Sheet, Chapters 1-3

Prof. Dave Johnson, dhj1@psu.edu, Penn State - Erie, The Behrend College

Solving a thermodynamics problem:  Is the substance a liquid, a gas, or a mixture ?

If liquid: consider compressed liquid or saturated liquid.

Test the liquid using the saturation properties (T_SAT, P_SAT, etc.)

Common tests for compressed liquid:

• given P, is T < T_SAT ?
• given T, is P > P_SAT ?
• given T or P, is v < v_f, or is h < h_f, or is u < u_f ?

USE TABLES, remember compressed liquid properties can be approximated with saturated liquid properties (v_f, u_f, h_f) at the temperature of the liquid.

If gas: consider saturated vapor or superheated vapor - USE TABLES

Test the gas using the saturation properties (T_SAT, P_SAT, etc.)

Common tests for superheated vapor:

• given P, is T > T_SAT ?
• given T, is P < P_SAT ?
• given T or P, is v > v_g, or is h > h_g, or is u > u_g ?

OR, is it an ideal gas ? - USE Pv=RT or PV=mRT

Test the gas using the critical properties (T_CR, P_CR, etc.)

Should the Generalized Compressibility Chart be used ?

Only if a correction (Z) for non-ideal behavior is requested or is needed based on the reduced pressure and temperature, T_R, P_R.

If mixture: USE TABLES - you will need the quality, x, of the mixture.

v_avg = v_f + x(v_g - v_f)

u_avg = u_f + x(u_fg)

LOOK FOR A SPECIAL TYPE OF PROCESS:

• constant volume (isometric)
• constant pressure (isobaric)
• constant temperature (isothermal)

FIRST LAW OF THERMODYNAMICS for a closed system (mass = constant)

Q - W = DE

Q - W = DU + DKE + DPE

Q - W_boundary - W_other = DU + DKE + DPE

FIRST LAW OF THERMODYNAMICS for a stationary, closed system

(DKE = DPE = 0, mass = constant)

Q - W = DU

Q - W_boundary - W_other = DU

If the pressure is constant: Q - W_other  = DH

For a cycle: Q - W_boundary - W_other = 0

Properties of IDEAL GAS:

Du = Cv,avg (T2 - T1)
Dh = Cp,avg (T2 - T1)
Cp = Cv + R
Specific heat ratio: k = Cp/Cv (1.4 for air)

Properties of INCOMPRESSIBLE SUBSTANCES: (V = constant, v = constant)

Cp = Cv = C
Du = Cavg (T2 - T1)
Dh = Du + vDP

For IDEAL GAS problems using the FIRST LAW OF THERMODYNAMICS:

You need Du no boundary work occurs (or if the boundary work is not done at constant pressure).

You need Dh if the boundary work is done at constant pressure.

First, look up Cpo or Cvo at the initial and final state temperature in Table A-2(b). Calculate Cp,avg or Cv,avg and compute Dh or Du using the formulas above (based on T2 - T1).  You can also use the average temperature to find Cp,avg or Cv,avg.

Second, use Table A-2(a) to get either Cpo or Cvo if the initial and final state temperatures are very close to 300 K.

Last resort, use Table A-2(c) to evaluate Cpo at the initial and final state temperatures. Your may integrate over the temperature range to get Dh directly.   If you need Cvo, use Cp = Cv + R.