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. Outline. Quiz Two and Three Solutions Finding work as area under path Find internal energy then Q = m D u + W Unit four – first law for ideal gases

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### Unit Three Quiz Solutions and Unit Four Goals

Mechanical Engineering 370

Thermodynamics

Larry Caretto

February 25, 2003

Outline
• 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
• Since this is not an ideal gas we have to use property tables
• Requires trial and error solution
• Unit four considers ideal gas behavior
Quiz Three Solution
• 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
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

• 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
Unit Four Goals
• 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
Unit Four Goals Continued
• 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
Unit Three Goals Continued
• 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
Example Calculation
• 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
Using H2O Tables
• 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
Ideal Gas Calculations
• Q = DU + W = m(u2 – u1) + PdV
• Q = m(u2 – u1) + mPdv
• 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)
Work – Ideal Gas Assumption
• At T1 = 200oC and P1 = 100 kPa, v1 = RT1/P1 = (.4615 kJ/kgK)(473.15 K)/(100 kPa) = 2.1836 m3/kg
• At T2 = 400oC and P2 = P1 = 100 kPa, v2 = RT2/P2 = (.4615 kJ/kgK)(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
Ideal Gas Internal Energy
• 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
Constant cv Ideal Gas
• Get cv = 1.4108 kJ/kgK 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/kgK )(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
Ideal Gas with cv(T)
• Use kelvins for temperature
• Molar enthalpy change = 7229.3 kJ/kmol
• Q = mDu+ W = (10 kg)(309 kJ/kg) + 931 kJ
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(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
Assuming cv Constant
• 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