Unit Eight Quiz Solutions and Unit Nine Goals

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

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. 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. 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. 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 < sf(T or P) => compressed liquid • if s > sg(T or P) => gas (superheat) region • otherwise in mixed region

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 dSisolated system = 0 • It is possible to have dS =  dQ/T for a system with dSisolated system > 0

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

8. Maximum Adiabatic Work (DS = 0) • From given inlet conditions, find the initial state properties including sinitial • The maximum work in an adiabatic process occurs when sfinal = sinitial • From sfinal = sinitial 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. 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. 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. 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 = cvdT, dh = cpdT • For ideal gases ds = cvdT/T + Rdv/v • For ideal gases ds = cpdT/T – RdP/P

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

13. Ideal Gas Entropy Tables • Define • So that

14. Example • Air is heated from 300 K to 500 K at constant pressure. What is Ds? • From table A-17, page 849, so(300 K) = 1.71865 kJ/kg∙K andso(500 K) = 2.21952 kJ/kg∙K; Ds = 0.50087 kJ/kg∙K • Assuming constant cp = 1.005 kJ/kg∙K gives Ds = cpln(T2/T1) = 1.005 ln(500/300) = 0.51338 kJ/kg∙K

15. Ideal Gas Isentropic Processes • Start with ds = cpdT/T – RdP/P • For ds = 0, cpdT/T = RdP/P • Integrate for ds = 0 and constant cp to get cpln(T2/T1) = R ln(P2/P1) so that ln(T2/T1) = R ln(P2/P1)R/Cp or T2/T1 = (P2/P1)R/Cp • R/cp = (cp – cv)/cp = (cp/cv – 1)/ (cp/cv) • R/cp = (k – 1)/k, where k = cp/cv

16. Ideal Gas Isentropic Processes • Final result: T2/T1 = (P2/P1)(k-1)/k • Can derive similar equations for other variables • T2/T1 = (v2/v1)(k-1) • P2/P1 = (v2/v1)k • Apply only to ideal gases with constant heat capacities in isentropic processes

17. Variable Heat Capacity • Set s2 – s1 = 0 in equations below for ideal gas isentropic process

18. Variable Heat Capacity • Can solve for volume or pressure ratios if T1 and T2 are given • A trial-and-error solution is required if T1 or T2 are unknown

19. Variable Heat Capacity Tables • For ideal gas, isentropic processes, with variable heat capacities we can define Pr(T), and vr(T) such that • v2/v1 = vr(T2)/ vr(T1) • P2/P1 = Pr(T2)/ Pr(T1) • Values of Pr(T), and vr(T) given in ideal gas tables • We still use Pv = RT at points

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. Example Continued • What is outlet state for maximum work? Use ideal gas tables for air on page 849 • Pr(300 K) = 1.3860 • P2/P1 = Pr(T2)/ Pr(T1) so that • Pr(T2) = Pr(T1) P2/P1 = 1.3860(1000/100) • What is T with Pr = 13.860 • Pr(570 K) = 13.50; Pr(580 K) = 14.38 • Interpolate to get Pr = 13.86 at T = 574.1 K

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

23. Repeat Example with Constant cp • Here we use data for air that k = 1.4 and cp = 1.005 kJ/kg∙K • T2 = T1(P2/P1)(k – 1)/k = (300 K)[(1000 kPa)/(100 kPa)](1.4 – 1)/1.4 = 579.2 K