1. Unit Eight Quiz Solutions and Unit Nine Goals Mechanical Engineering 370 Thermodynamics Larry Caretto April 1, 2003

2. 2 Outline Solution to quiz eight Revised schedule Review second law Goals for unit eight –Calculating entropy with ideal gases –Constant and variable heat capacity –Isentropic calculations

3. 3 Course Schedule Changes Midterm on April 24 (Review April 22) Move schedule for unit ten on April 22– 24 up to April 8–10 Unit ten quiz on April 22; no quiz on April 29 following midterm Writing assignment due April 11 Design project due May 16

4. 4 The Second Law There exists an extensive thermo- dynamic property called the entropy, S, defined as follows: dS = (dU + PdV)/T For any process dS ≥ dQ/T For an isolated system dS ≥ 0 T must be absolute temperature

5. 5 Entropy is a Property If we know the state of the system, we can find the entropy We can use the entropy as one of the properties to define the state Use the following if we are given a value of s and a value of T or P –if s compressed liquid –if s > s g (T or P) => gas (superheat) region –otherwise in mixed region

6. 6 Reversible Processes This is an idealization; we cannot do better than a reversible process Internal reversibility has dS = dQ/T Internal and external reversibility has dS isolated system = 0 It is possible to have dS = dQ/T for a system with dS isolated system > 0

7. 7 Maximum Work (Output) dS  dQ/T; if reversible dS = dQ/T Compare two processes with between same states (dU = dU rev ) dS = dQ rev /T = [dU rev + dW rev ]/T  dQ/T [dU rev + dW rev ]/T  [dU + dW]/T dW rev  dW Maximum work in a reversible process

8. 8 Maximum Adiabatic Work (  S = 0) From given inlet conditions, find the initial state properties including s initial The maximum work in an adiabatic process occurs when s final = s initial From s final = s initial and one other property of final state get all final state properties Find work from first law as done in previous quizzes and exercises

9. 9 First Unit Nine Goal As a result of studying this unit you should be able to compute entropy changes in ideal gases –using constant heat capacities –using equations that give heat capacities as a function of temperature –using ideal gas tables

10. 10 Second Unit Nine Goal As a result of studying this unit you should be able to compute the end states of isentropic processes in ideal gases –using constant heat capacities –using equations that give heat capacities as a function of temperature –using ideal gas tables

11. 11 Ideal Gas Entropy Entropy defined as ds = (du + Pdv)/T We can write Tds = du + Pdv Since du = d(h – Pv) = dh – Pdv – vdP We can also write Tds = dh - vdP Ideal gas: Pv = RT, du = c v dT, dh = c p dT For ideal gases ds = c v dT/T + Rdv/v For ideal gases ds = c p dT/T – RdP/P

12. 12 Ideal Gas Entropy Change Integrate ideal gas ds equations to get

13. 13 Ideal Gas Entropy Tables Define So that

14. 14 Example Air is heated from 300 K to 500 K at constant pressure. What is  s? From table A-17, page 849, s o (300 K) = 1.71865 kJ/kg ∙ K and s o (500 K) = 2.21952 kJ/kg ∙ K;  s = 0.50087 kJ/kg ∙ K Assuming constant c p = 1.005 kJ/kg ∙ K gives  s = c p ln(T 2 /T 1 ) = 1.005 ln(500/300) = 0.51338 kJ/kg ∙ K

15. 15 Ideal Gas Isentropic Processes Start with ds = c p dT/T – RdP/P For ds = 0, c p dT/T = RdP/P Integrate for ds = 0 and constant c p to get c p ln(T 2 /T 1 ) = R ln(P 2 /P 1 ) so that ln(T 2 /T 1 ) = R ln(P 2 /P 1 ) R/Cp or T 2 /T 1 = (P 2 /P 1 ) R/Cp R/c p = (c p – c v )/c p = (c p /c v – 1)/ (c p /c v ) R/c p = (k – 1)/k, where k = c p /c v

16. 16 Ideal Gas Isentropic Processes Final result: T 2 /T 1 = (P 2 /P 1 ) (k-1)/k Can derive similar equations for other variables –T 2 /T 1 = (v 2 /v 1 ) (k-1) –P 2 /P 1 = (v 2 /v 1 ) k Apply only to ideal gases with constant heat capacities in isentropic processes

17. 17 Variable Heat Capacity Set s 2 – s 1 = 0 in equations below for ideal gas isentropic process

18. 18 Variable Heat Capacity Can solve for volume or pressure ratios if T 1 and T 2 are given A trial-and-error solution is required if T 1 or T 2 are unknown

19. 19 Variable Heat Capacity Tables For ideal gas, isentropic processes, with variable heat capacities we can define P r (T), and v r (T) such that –v 2 /v 1 = v r (T 2 )/ v r (T 1 ) –P 2 /P 1 = P r (T 2 )/ P r (T 1 ) Values of P r (T), and v r (T) given in ideal gas tables We still use Pv = RT at points

20. 20 Example Problem Adiabatic, steady-flow air compressor used to compress 10 kg/s of air from 300 K, 100 kPa to 1 MPa. What is the minimum work? Minimum work in adiabatic process is when process is isentropic First law:

21. 21 Example Continued What is outlet state for maximum work? Use ideal gas tables for air on page 849 –P r (300 K) = 1.3860 –P 2 /P 1 = P r (T 2 )/ P r (T 1 ) so that – P r (T 2 ) = P r (T 1 ) P 2 /P 1 = 1.3860(1000/100) –What is T with P r = 13.860 –P r (570 K) = 13.50; P r (580 K) = 14.38 –Interpolate to get P r = 13.86 at T = 574.1 K

22. 22 Example Concluded Use enthalpy data from ideal gas tables –h in = h(300 K) = 300.19 kJ/kg –h out = h(574.1 K) = 579.86 kJ/kg Negative value shows work input Minimum input in absolute value

23. 23 Repeat Example with Constant c p Here we use data for air that k = 1.4 and c p = 1.005 kJ/kg ∙K T 2 = T 1 (P 2 /P 1 ) (k – 1)/k = (300 K)[(1000 kPa)/(100 kPa)] (1.4 – 1)/1.4 = 579.2 K