1. Unit Four Quiz Solutions and Unit Five Goals
Mechanical Engineering 370
Thermodynamics
Larry Caretto
March 4, 2003

2. Outline Quiz three and four solutions are similar
Finding work (area under path) and u
Q = m (ufinal – uinitial) + W
Unit five – open systems
View first law as a rate equation
Have mass crossing system boundaries
Flows across boundaries have several energy forms including internal (u) , kinetic and potential energy plus flow work (Pv)

3. Quiz Three Results 21 students; max = 30; mean = 18.6
Q = m(ufinal – uinitial) + W
To compute u = h – Pv
Use specific volume in m3/kg
convert P to kPa for h and u in kJ/kg

4. Quiz Three Path Quiz three gave neon data near and in mixed region.
T1, P1, V1, P2, V2
T3, V4
The quiz three diagram is shown here.
This was not an ideal gas so we had to use property tables

5. Quiz Four Path
Quiz four has the same path with the same data items as quiz three.
T1, P1, V1, P2, V2
T3, V4
The quiz four path diagram is shown here.
This was an ideal gas so we use PV = mRT and du = cvdT

6. Quiz Four Solution Given: Water in three-step process
T1 = 300oC, V1 = 1 m3, P1 = 100 kPa
1-2 is a linear path to P2 = 300 kPa, V2 = 0.8 m3
2-3 is constant volume with T3 = 400oC
3-4 is constant pressure with V4 = 0.4 m3
Find the heat transfer, Q, using ideal gas
Find Q from first law (ideal gas with cv const):
Q = DU + W = m(u4 – u1) + W = mcv(T4 – T1) + W
Work is (directional) area under path

7. Path for This Process
Work = area under path = trapezoid area plus rectangle area
W = (P1 + P2)(V2 – V1)/2 + P3-4 (V4 – V3)
DV < 0 means work will be negative 2 P 3 4 1 V

8. Finding the Answer
Use properties at the initial state to find the mass
Find P3-4 = P3 = P4 from state 3 defined by T3 = 400oC and v3 = v2 = V2/m

9. Finding the Answer (cont’d)
Find T4 from m, P4 = P3 and V4 = 0.4 m3
Now find heat and work

10. Finding the Answer (concluded)

11. Future Quizzes Can use equation summary
Download from course web page (follow course notes link)
May have unannounced open book exams to allow use of tables
If you are late for a quiz you can
Come to class after quiz is over or
Start quiz and receive grade

12. Unit Five Goals
Topic is first law for open systems, i.e., systems in which mass flows across the boundary
Will look at general results and focus on steady-state systems.
As a result of studying this unit you should be able to
understand all the terms (and dimensions) in the first law for open systems:

13. Open System Concepts the useful work rate or mechanical power (ML2T-3)
the mass flow rate (MT-1)
the kinetic energy per unit mass (L2T-2)
gz the potential energy per unit mass (L2T-2)
total energy (ML2T-2)
heat transfer rate (ML2T-3)
rate of energy change (ML2T-3)

14. Unit Five Goals Continued
use the equation relating velocity, mass flow rate, flow area, A, and specific volume
use the mass balance equation

15. Flow Work
For open systems work is done on (or by) mass entering and leaving the system
Flow work is Pv times mass flow rate
Add this flow work to internal energy (times mass flow rate)
First law for mass flows has h = u + Pv (sum of internal energy plus flow work)

16. Unit Five Goals Continued
use the first law for open systems
use the steady-state assumptions and resulting equations

17. Steady-state equations
Steady-state first law for open systems
Steady-state mass balance for open systems

18. Unit Five Goals Continued
recognize that kinetic and potential energies are usually negligible
A 1oC temperature change in air (ideal gas with cp = kJ/(kg∙K) has Dh = 1005 J/kg
A similar kinetic energy change requires a velocity increase from zero to 45 m/s (~100 mph)
A similar potential energy change requires an elevation change of 102 m (336 ft)

19. Unit Five Goals Concluded
handle simplest case: steady-state, one inlet, one outlet (one mass flow rate), negligible changes in kinetic and potential energies
work with ratios q and w in simplest case

20. Example Calculation
Given: 10 kg/s of H2O at 10 MPa and 700oC renters a steam turbine; the outlet is at 500 kPa and 300oC. There is a heat loss of 400 kW.
Find: Useful work rate (power output)
Assumptions: Steady-state, negligible changes in kinetic and potential energies
Configuration: one inlet and one outlet
First law

21. Getting the Answer
At Tin = 700oC and Pin = 10 MPa, hin = kJ/kg (p. 836)
At Tout = 300oC and Pout = 500 kPa, hout = kJ/kg
Heat loss is negative: Q = Qin - Qout

22. Review Ideal Gases For ideal gases du = cvdT; dh = cpdT
Ideal gas DH = DU + RDT
May have molar DH data
Last week we looked at problem with H2O as an ideal gas, where T1 = 200oC and T2 = 400oC
How do we handle cv(T)?

23. Ideal Gas with cv(T) Find a, b, c and d from Table A-2(c), p 827
Use T1 = K and T2 = K
Molar enthalpy change = kJ/kmol

24. Getting Du from molar Dh
Use molar Dh just found, data on R and M, and DT = 200oC = 200 K

25. Ideal Gas Tables Find molar u(T) for H2O in Table A-23 on page 860
Ideal Gas Tables
Find molar u(T) for H2O in Table A-23 on page 860
Have to interpolate to find u1 = u(473.15 K) = 11,953 kJ/kmol and u2 = u( K) = 17,490 kJ/kmol
Du = (17,490 kJ/kmol - 11,953 kJ/kmol) / ( kg / kmol) = kJ/kg