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Unit Three Quiz Solutions and Unit Four GoalsPowerPoint Presentation

Unit Three Quiz Solutions and Unit Four Goals

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Unit Three Quiz Solutions and Unit Four Goals

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Unit Three Quiz Solutions and Unit Four Goals

Mechanical Engineering 370

Thermodynamics

Larry Caretto

February 25, 2003

- Quiz Two and Three Solutions
- Finding work as area under path
- Find internal energy then Q = m Du + W

- Unit four – first law for ideal gases
- Heat capacities, cv and cp are properties
- For ideal gases du = cvdT and dh = cpdT, regardless of path
- For ideal gases u = u(T) only and h = u + Pv = u + RT = h(T) only

- In quiz two the final state lies along path and along 320 K iso-therm.
- Since this is not an ideal gas we have to use property tables
- Requires trial and error solution
- Unit four considers ideal gas behavior

- Given: Neon in three-step process
- T1 = 280 K, V1 = 1 m3, P1 = 200 kPa
- 1-2 is a linear path to P2 = 700 kPa, V2 = 0.08 m3
- 2-3 is constant volume with T3 = 30 K
- 3-4 is constant pressure with V4 = 0.04 m3

- Find the heat transfer, Q, using tables
- Find Q from first law:
- Q = DU + W = m(u4 – u1) + W

- Work is (directional) area under path

- 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

- Properties at the initial state
- From T1 = 280 K and P1 = 200 kPa, find v1 and m = V1/v1 = 1.732 kg; also u1 = h1 – P1v1 = 237 kJ/kg

- State 3 defined by T3 = 80 K, v3 = v2 = V2/m
- Find P3 = Psat(30 K) = 223.8 kPa

- State 4 defined by P4 = P3, v4 = V4/m
- This is in mixed region with u4 = 42.7 kJ/kg

- Q = (1.732 kg)(237 – 42.7) kJ/kg – 414 kJ = –751 kJ

- As a result of studying this unit you should be able to
- describe the path for a process and determine the work with greater confidence than you had after completing unit 3
- understand the heat capacities Cx (e.g. Cp and Cv) as dQ = Cx dT in a “constant-x” process
- use the property relations for ideal gases du = cv dT and dh = cp dT for any process

- find changes in internal energy and enthalpy for an ideal gas where the heat capacity is constant or a function of temperature .
- use ideal gas tables to find changes in internal energy and enthalpy where the heat capacities are functions of temperature
- find internal energy changes for ideal gases as Dh = Du - R DT
- convert results from a per-unit-mole basis to a per-unit-mass basis and vice versa

- be able to find other properties about a state when you know (or able to calculate) the internal energy or enthalpy
- be able to work problems using the first law, PV = RT, du = cv dT, and a path equation (may be iterative)
- use the equation cp - cv = R to find cp from cv (and vice versa), which also applies to equations; if cp = a + bT + cT2, then cv = (a-R) + bT + cT2

- Given: 10 kg of H2O at 100k Pa and 200oC is expanded to 400oC at constant pressure
- Find: Heat Transfer
- using H2O tables
- using ideal gas with constant heat capacity
- using ideal gas with variable heat capacity

- First Law: Q = DU + W = m(u2 – u1) + W
- Path: W = PdV = P1-2 (V2 – V1) for constant pressure, P1-2 = P1 = P2
- u2 – u1 = cvdT for ideal gas

- At T1 = 200oC and P1 = 100 kPa, v1 = 2.172 m3/kg and u1 = 2658.1 kJ/kg
- At T2 = 400oC and P2 = P1 = 100 kPa, v2 = 3.103 m3/kg and u2 = 2967.9 kJ/kg
- W = P1-2 (V2 – V1) = P1-2 m(v2 – v1) = (10 kg)(100 kPa)(3.103 - 2.172) m3/kg =
- Q = m(u2 - u1) + W = (10 kg)(2967.9 - 2658.1) kJ/kg + 931 kJ = 4,029 kJ

- Q = DU + W = m(u2 – u1) + PdV
- Q = m(u2 – u1) + mPdv
- PV = mRT Pv = RT
- We use PV = mRT to determine mass and specific volume from P and T
- The work calculation does not depend on assumptions about cv (or cp = cv + R)

- At T1 = 200oC and P1 = 100 kPa, v1 = RT1/P1 = (.4615 kJ/kgK)(473.15 K)/(100 kPa) = 2.1836 m3/kg
- At T2 = 400oC and P2 = P1 = 100 kPa, v2 = RT2/P2 = (.4615 kJ/kgK)(673.15 K)/(100 kPa) = 3.1066 m3/kg
- W = P1-2 (V2 – V1) = P1-2 m(v2 – v1) = (10 kg)(100 kPa)(3.106 - 2.184) m3/kg = 923 kJ

- u2 – u1 = cv(T)dT = cp(T)dT - RDT
- Possible calculations for cv (or cp)
- Assume constant (easiest) Du = cDT
- Integrate equation giving cv or cp as a function of temperature (Table A-2, p 827)
- Use ideal gas tables giving u(T) and h(T) (Tables A-17 to A-26, pp 849-863)
- Last two give molar properties

- Get cv = 1.4108 kJ/kgK for water from Table A-2, p828
- DU = mDu = m cv(T)dT = cv(T2 – T1) = mcvDT, if cv is constant
- Here, DU = mcv(T2 - T1) = (10 kg) (1.4108 kJ/kgK )(673.15 K - 473.15 K) = 2,822 kJ
- Q = DU + W = 2,822 kJ + 923 kJ = 3,745 kJ, a 7% error compared to actual properties

- Use kelvins for temperature
- Molar enthalpy change = 7229.3 kJ/kmol
- Q = mDu+ W = (10 kg)(309 kJ/kg) + 931 kJ

- 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(673.15 K) = 17,490 kJ/kmol
- DU = (10 kg)(17,490 kJ/kmol - 11,953 kJ/kmol) / (18.015 kg / kmol) = 3,074 kJ
- Q = DU+ W = 3,074 kJ + 923 kJ

- Assumption of constant heat capacity introduces about a 7% error
- Accounting for temperature variation of heat capacity reduces error to <= 0.8%
- Constant heat capacity assumption is best for noble gases (e.g., argon, neon) and reasonable for diatomic molecules at ambient temperatures
- Assumption worsens as the temper-ature range increases