1. Unit One Quiz Solutions and Unit Two Goals Mechanical Engineering 370 Thermodynamics Larry Caretto February 11, 2003

2. 2 Outline Solution to quiz –Look at individual problems –Solutions available on line Unit two – work and paths –Unit goals available on line –Group exercise on Thursday –Quiz on Tuesday, February 18

3. 3 Problem One Solution Given: P = 2 MPa, T = 200 K, V = 2 m 3 Find the mass, m, using tables First, find the specific volume from superheat tables v(2 MPa,200 K) = 0.04164 m 3 /kg m = 48.03 kg

4. 4 Searching the Table T, K P,MPa 200 v 2 h s 0.0416 269.92 4.2172

5. 5 Problem Two Solution Given: P = 2 MPa, T = 200 K, V = 2 m 3 Find the mass, m, using ideal gas equation Still have m = V/v with v = RT/P m = 48.55 kg, a 1.3% error

6. 6 Problem Three Solution Given: Neon at 200 K and 2 MPa cooled at constant volume to 30 K Find: Final state For constant volume process initial specific volume = final specific volume = 0. 04164 m 3 /kg from problem one Final state is v = 0. 04164 m 3 /kg and T = 30 K

7. 7 Problem Three Continued Check v f and v g at T = 30 K Find v f = 0.000869 m 3 /kg < v = 0.4164 m 3 /kg < v g = 0.05016 m 3 /kg Mixed, so P = P sat (30 K) = 0.2238 MPa

8. 8 Unit Two Goals As a result of studying this unit you should be able to –find properties more easily than you were able to do after completing unit one –describe the path for a process –find the work as the integral of PdV –find the work as the area under the path on a P-V diagram

9. 9 Unit Two Goals Continued –understand the difference between the applied force (or pressure) and the system force (or pressure). –recognize that work is always the integral of the applied force (or pressure) over distance (or volume) –recognize when the applied pressure and the system pressure are the same

10. 10 More Unit Two Goals –use the description of the path as an equation –recognize the difference between the path equation and the equation of state –use the path equation and the equation of state in a trial-and-error procedure to find the final state –use the path equation and the equation of state in a trial-and-error procedure to find the final state when the "equation of state" is a set of tables

11. 11 Analysis of Work Start with dW = Fdx dW = (PA)d(V/A) = PdV Note that dimensions are (F/L 2 ) times L 3 E. g., pascal-m 3 gives N-m or joules psia-ft 3 times 144 in 2 /ft 2 gives ft-lb f W =  PdV over path Work depends on path

12. 12 Simple Path Here we have an initial point (1), a final point (2) Work = area under path = trapezoid area = (V 2 – V 1 ) (P 1 + P 2 )/2 Work is positive P V 1 2

13. 13 Integrating a Line Work = area under path = trapezoid area = (V 2 – V 1 ) (P 1 + P 2 )/2 Work is positive because if V 2 > V 1 (and pressure is always positive) Work is positive when system expands (system does work on surroundings) 1

14. 14 Reverse Path Here the initial point (1) and final point (2) are reversed Work = area under path = trapezoid area = (V 2 – V 1 ) (P 1 + P 2 )/2 Work is negative P V 1 2

15. 15 Integrating a Reverse Line In the previous chart the initial point (1) and final point (2) are reversed Work = area under path = trapezoid area = (V 2 – V 1 ) (P 1 + P 2 )/2 Work is negative because V 2 < V 1 Work is negative when system is compressed (work done on system) 1

16. 16 More Complex Path Here we have an initial point (1), a final point (4) and two intermediate points (2 and 3) Work = area under path = trapezoid area plus rectangle area P V 1 23 4

17. 17 How Do We Specify State To integrate PdV, we need path equation, P(V) Path equation is integrated along path in terms of P and V Can specify states in terms of T Need to use equation-of-state or tables to get v from (P,T) and V = m v

18. 18 Example Calculation Given: 10 kg of H 2 O at 200 o C and 50% quality is expanded to 400 o C at constant pressure Find: Work Equation: W =  PdV = P(V 2 – V 1 ) for constant pressure How do we find P, V 1 and V 2 from data given?

19. 19 Example Continued Use tables for H 2 O to find P and v Get total volume as V = m v Initial state is in mixed region so the constant pressure, P = P sat (200 o C) = 1.5538 MPa and v = v f + x (v g – v f ) At 200 o C, v f and v g, respectively, = 0.001157 and 0.12736 m 3 /kg For x 1 = 0.5, v 1 = 0.06426 m 3 /kg

20. 20 Example Concluded V 1 = m v 1 = 0.6426 m 3 for m = 10 kg Point 2 has T = 400 o C and P = initial P = 1.5538 MPa From superheat table interpolation this final state has v = 0.19646 m 3 /kg Since m = 10 kg, V 2 = 1.9646 m 3 W = (1.5538 MPa)(1.9646 – 0.6426) m 3

21. 21 Units for Work Basic idea is that Pam 3 gives J, kPam 3 gives kJ, MPam 3 gives MJ, For other pressure units need to convert into kJ; here W = 2,054 kJ Without unit conversion we could get the work as 2.054 MJ