10.40 Thermodynamics: Problem Set 5, #1 (Solutions)

Problem Set 5, #1 10.40 Thermodynamics Problem Set 5 Fall 2003 Problem #1 Let’s compare the pressure-temperature behavior of two important expansion processes that are used for refrigeration and cooling applications: (i) expansion across a Joule-Thompson valve (ii) expansion in an adiabatic turbine To carryout this comparison it is helpful to construct an equation to express the ratio of how temperature varies with pressure for each process. Defining this ratio as follows: α≡ (∂T / ∂P) process (i ) (∂T / ∂P) process (ii ) We will also need to specify the type of fluid to be expanded. To keep things simple, let’s limit our analysis to pure CO2 which can be represented by four constitutive property models whose accuracy will depend on the PVT region where the expansion is carried out. The four models are: (1) ideal gas (2) The Law of Corresponding States (Hint: Express α in terms of Z and its derivatives) (3) PR EOS (4) Thermodynamic charts given in Figures 8.12 a,b (a) What is the initial value of α for each model above for the expansion of pure CO2 from an initial state of 100 bar and 37oC? (b) Which process would you expect to liquefy a greater fraction of the entering CO2 gas if the final pressure state for the expansion is 10 bar? (c) What technological and economic issues would be important in selecting a process for practical use? Physical property data for carbon dioxide are given at the end of the problem set and in Appendix G of the text. 10.40 Fall 2003 Problem Set 5 Solutions Page 1 of 6 Problem Set 5, #1 Solution: (a) For case (i), expansion across a Joule-Thompson valve is isenthalpic, so that: ⎛ ∂T ⎞ ⎛ ∂T ⎞ =⎜ ⎜ ⎟ ⎟ = αH ⎝ ∂P ⎠ process ( i ) ⎝ ∂P ⎠H (1) For case (ii), we must assume that expansion across an adiabatic turbine is reversible in order to proceed. Then, the expansion becomes reversible, so we can say that it is an isentropic expansion: ⎛ ∂T ⎞ ⎛ ∂T ⎞ (2) =⎜ ⎜ ⎟ ⎟ = αS ⎝ ∂P ⎠ process ( ii ) ⎝ ∂P ⎠S From Example 5.2 or Example 8.1 in the book, we get: T ∂V −V ∂T P αH = Cp ( αS = ( ) T ∂V ∂T ) (3) P Cp Therefore, taking the ratio of the expressions in equation (3): T ∂V −V V ⎛ ∂T ⎞ ∂T P = 1− ⎜ α = ⎟ T ⎝ ∂V ⎠P T ∂V ∂T P ( ( ) (4) ) We can now determine α for each of the given models: (i) Ideal Gas Model PV=RT For an ideal gas: R ⎛ ∂V ⎞ ⎜ ⎟ = ⎝ ∂T ⎠P P RT − V V −V P = =0 α = RT V P (5) (6) Therefore, α = 0 always for an ideal gas. This should be obvious since for an ideal gas, H = H(T) only, so for an isenthalpic process where H is constant, T must also be constant and αΗ = 0. (ii) Law of Corresponding States Z=PV/RT We can re-write Z in terms of V: RTZ V = P 10.40 Fall 2003 Problem Set 5 Solutions (7) Page 2 of 6 Problem Set 5, #1 Plugging this into the derivative in our expression for α: ⎡ ⎤ R ⎛ ∂ (TZ ) ⎞ R ⎢ ⎛ ∂Z ⎞ ⎛ ∂V ⎞ ⎛ ∂T ⎞ ⎥ ⎟ = ⎢T ⎜ ⎜ ⎟ = ⎜ ⎟ +Z⎜ ⎟ ⎥ ∂T ⎠P ⎥ ⎝ ∂T ⎠P P ⎝ ∂T ⎠P P ⎢ ⎝ ∂T ⎠P ⎝1 424 3 =1 ⎣ ⎦ ⎤ R ⎡ ⎛ ∂Z ⎞ ⎛ ∂V ⎞ ⎜ ⎟ = ⎢T ⎜ ⎟ + Z⎥ ⎝ ∂T ⎠P P ⎣ ⎝ ∂T ⎠P ⎦ (8) (9) Substituting equation (9) into α and multiplying the numerator and denominator by P/RT gives: RT ⎡ ∂Z + Z ⎥⎤ − V T ∂Z T +Z −Z ⎢ ∂ T ∂T P P ⎣ ⎦ = α = P (10) RT ⎡ ∂Z ⎤ +Z T ∂Z + T Z ∂T P ∂T P ⎦⎥ P ⎣⎢ T ∂Z ∂T P α = (11) +Z T ∂Z ∂T P ( ( ( ) ( ) ) ( ) ( ) ) To calculate α, we need values of Z and its derivative with respect to T at our conditions of interest, namely P = 100 bar = 10 MPa and T = 37oC = 310 K. Using the handout included in the problem statement, we can use the PVT data given to calculate Z and then estimate its derivative using a finite differences argument. Data is given in the table below. P (MPa) 10 10 10 T (K) 305 310 315 ρ (kg/m3) 751.67 685.77 586.02 Z 0.2200 0.2412 0.2822 Thus: ∆Z 0.282 − 0.220 ⎛ ∂Z ⎞ = = 0.00622 K -1 ⎜ ⎟ ≈ 10 K ⎝ ∂T ⎠P ∆T (12) ( 310 K ) ( 0.00622 K -1 ) α= = 0.889 0.2412 + ( 310 K ) ( 0.00622 K -1 ) The isentropic process cools down the CO2 more during expansion initially. (iii) Peng-Robinson EOS RT a P= − V − b V (V + b ) + b (V − b ) where: ⎛ ⎛ ⎞⎞ a = ac ⎜ 1 + κ ⎜ 1 − T ⎟ ⎟ T C ⎠⎠ ⎝ ⎝ 10.40 Fall 2003 Problem Set 5 Solutions (13) 2 (14) Page 3 of 6 Problem Set 5, #1 κ = 0.37464 + 1.54226ω − 0.26992ω 2 (15) and b and ac are defined in terms of the critical constants (see Equation (8-47)-(8-51) in the text). Since we are working with a pressure-explicit equation of state, we can not evaluate (∂V/∂T)P directly. However, making use of the triple-product rule: ⎛ ∂V ⎞ ⎛ ∂T ⎞ ⎛ ∂P ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ = −1 ⎝ ∂T ⎠P ⎝ ∂P ⎠V ⎝ ∂V ⎠T (16) − ∂P ⎛ ∂V ⎞ ∂T V ⎜ ⎟ = ∂P ⎝ ∂T ⎠P ∂V T ( ( ) ) Now we have two derivatives of the pressure explicit EOS. Analytically evaluating the numerator and denominator of equation (16): ∂a R ⎛ ∂P ⎞ ∂T V − ⎜ ⎟ = ⎝ ∂T ⎠V V − b V (V + b ) + b (V − b ) where: −aκ ⎛ ∂a ⎞ ⎜ ⎟ = 12 ⎝ ∂T ⎠V (αTTC ) ( ) and 2a (V + b ) −RT ⎛ ∂P ⎞ + ⎜ ⎟ = 2 ⎝ ∂V ⎠T (V − b ) V 2 + 2bV − b 2 ( ) (17) (18) (19) 2 We can now proceed with determining values for the derivatives in equations (17-19) at the temperature and pressure of interest. Values for the critical constants, accentric factor, and intermediate calculation values are given in the table below. The volume at this condition is found using equation (13) by finding the roots of the cubic equation. Since at 100 bar and 37oC we are above the critical point of CO2, there should be only one real root. Conditions T = 37oC = 310 K P = 100 bar = 10 MPa V = 7.099 x 10-5 m3/mol Constants TC = 304.2 K PC = 73.8 bar = 73.8 x 105 Pa ac = 0.3963 Pa(m3/mol)2 ω = 0.225 b = 2.666 x 10-5 m3/mol Intermediate Calcs κ = 0.707984 a = 0.39086 Pa(m3/mol)2 (∂a/∂T)V = -9.07 x 10-4 (Pa/K)(m3/mol)2 (∂P/∂T)V = 2.993 x 105 Pa/K (∂P/∂V)T = -1.528 x 1011 Pa-mol/m3 Finally, after plugging in all our calculations, we get α = 0.883 Once again, we find that the isentropic process cools more rapidly initially. 10.40 Fall 2003 Problem Set 5 Solutions Page 4 of 6 Problem Set 5, #1 (iv) Thermodynamic Charts in Fig. 8.12 a and b We can calculate (∂T/∂V)P by following an isobar, or a line of constant P, and reading values for the temperature and volume just before and after the initial starting point. Then we estimate the derivative using the same method employed for case (ii). Notice that we can not graphically find the derivative at the initial point because the figure axes are not in our units of interest. We must use Fig. 8.12b since Fig. 8.12a does not have isochors that extend over to the initial point. The values read from the figure are given in the table below. P (MPa) 10 10 10 T (K) 301 310 315 ρ (kg/m3) 600 675 800 Z 0.25 ∆T 315 K − 301 K K ⋅ kg ⎛ ∂T ⎞ = = 33600 ⎜ ⎟ ≈ 3 m3 ⎝ ∂V ⎠P ∆V ⎛ 1 − 1 ⎞ m ⎜ ⎟ ⎝ 600 800 ⎠ kg (20) 1 m ( 675 ) ( kg ) K ⋅ kg = 0.839 = α α = 1− 33600 3 310 K (21) m3 This is consistent with the calculations from the previous EOS. The reason for the discrepancy is most probably due to difficulties in reading the figure. (b) We can use Figures 18.12a and b to answer this question. For the isenthalpic process, we use Fig. 18.12a and follow a line of constant enthalpy vertically down from the initial point at 1450 psia and about 100oF to the final pressure of 14.5 psia. We end up in the vapor-liquid region, and from the chart we can see that the mixture is 42% liquid (58% vapor). For the isentropic process, we use Fig. 18.12b and follow a line of constant entropy vertically down from the initial point at 10 MPa and 310 K down to 1 MPa. Once again, we are in the vapor-liquid region, and from the chart see that mixture is about 53% liquid (47% vapor). We conclude that the isentropic process produces more liquid CO2. This conclusion is supported by our models and makes sense practically. A turbine would extract work form the gas as it expands, whereas the valve does not, thus lowering its total internal energy more. (c) Economically, a simple valve is less expensive than a turbine and all its moving parts (although you could generate electricity from the turbine, which may defray some of its costs). Technically, the turbine will not be reversible as we have assumed, so we will not get as great a (∂T/∂V)P as predicted. Plus, if the CO2 were to liquefy in the turbine, it would cause cavitation and corrosion, thus quickly destroying the turbine as well as greatly reducing its efficiency. 10.40 Fall 2003 Problem Set 5 Solutions Page 5 of 6 Problem Set 5, #1 Image removed due to copyright considerations. Please see “Span, and Wagner. J. Phys. Chem. Ref. Data 25, no. 6 (1996).” 10.40 Fall 2003 Problem Set 5 Solutions Page 6 of 6