Unit One Quiz Solutions and Unit One Quiz Solutions and Unit Two GoalsUnit Two Goals

Mechanical Engineering 370Thermodynamics

Larry Caretto

February 11, 2003

2

Outline

• Solution to quiz – Look at individual problems– Solutions available on line

• Unit two – work and paths– Unit goals available on line– Group exercise on Thursday– Quiz on Tuesday, February 18

3

Problem One Solution

• Given: P = 2 MPa, T = 200 K, V = 2 m3

• Find the mass, m, using tables• First, find the specific volume from superheat tables v(2 MPa,200 K)

= 0.04164 m3/kg

kg

m

m

v

Vm

3

3

04164.0

2

• m = 48.03 kg

4

Searching the Table

  T, K

P,MPa       200

         

         

v2 h s

   

   

   

0.0416269.924.2172

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Problem Two Solution

• Given: P = 2 MPa, T = 200 K, V = 2 m3

• Find the mass, m, using ideal gas equation• Still have m = V/v with v = RT/P

• m = 48.55 kg, a 1.3% error

)200(4119344.0

1000)2)(2(

33

KKkg

kJmMPa

kJmMPa

RT

PVm

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Problem Three Solution

• Given: Neon at 200 K and 2 MPa cooled at constant volume to 30 K

• Find: Final state• For constant volume process initial

specific volume = final specific volume = 0. 04164 m3/kg from problem one

• Final state is v = 0. 04164 m3/kg and T = 30 K

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Problem Three Continued

• Check vf and vg at T = 30 K

• Find vf = 0.000869 m3/kg < v = 0.4164

m3/kg < vg = 0.05016 m3/kg

• Mixed, so P = Psat(30 K) = 0.2238 MPa

7946.0

kgm00869.0kg

m05016.0

kgm00869.0kg

m04164.0

vv

vvx

33

33

fg

f

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Unit Two Goals

• As a result of studying this unit you should be able to– find properties more easily than you were

able to do after completing unit one – describe the path for a process – find the work as the integral of PdV– find the work as the area under the path on

a P-V diagram

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Unit Two Goals Continued

– understand the difference between the applied force (or pressure) and the system force (or pressure).

– recognize that work is always the integral of the applied force (or pressure) over distance (or volume)

– recognize when the applied pressure and the system pressure are the same

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More Unit Two Goals

– use the description of the path as an equation – recognize the difference between the path

equation and the equation of state – use the path equation and the equation of

state in a trial-and-error procedure to find the final state

– use the path equation and the equation of state in a trial-and-error procedure to find the final state when the "equation of state" is a set of tables

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Analysis of Work

• Start with dW = Fdx

• dW = (PA)d(V/A) = PdV

• Note that dimensions are (F/L2) times L3

• E. g., pascal-m3 gives N-m or joules

• psia-ft3 times 144 in2/ft2 gives ft-lbf

• W = PdV over path

• Work depends on path

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Simple Path

• Here we have an initial point (1), a final point (2)

• Work = area under path = trapezoid area = (V2 – V1) (P1 + P2)/2

• Work is positive

P

V

1

2

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Integrating a Line

• Work = area under path = trapezoid area = (V2 – V1) (P1 + P2)/2

• Work is positive because if V2 > V1 (and pressure is always positive)

• Work is positive when system expands (system does work on surroundings)

1

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Reverse Path

• Here the initial point (1) and final point (2) are reversed

• Work = area under path = trapezoid area = (V2 – V1) (P1 + P2)/2

• Work is negative

P

V

1

2

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Integrating a Reverse Line

• In the previous chart the initial point (1) and final point (2) are reversed

• Work = area under path = trapezoid area = (V2 – V1) (P1 + P2)/2

• Work is negative because V2 < V1

• Work is negative when system is compressed (work done on system)

1

16

More Complex Path

• Here we have an initial point (1), a final point (4) and two intermediate points (2 and 3)

• Work = area under path = trapezoid area plus rectangle area

P

V

1

2 3

4

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How Do We Specify State

• To integrate PdV, we need path equation, P(V)

• Path equation is integrated along path in terms of P and V

• Can specify states in terms of T

• Need to use equation-of-state or tables to get v from (P,T) and V = m v

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Example Calculation

• Given: 10 kg of H2O at 200oC and 50% quality is expanded to 400oC at constant pressure

• Find: Work• Equation: W = PdV = P(V2 – V1) for

constant pressure• How do we find P, V1 and V2 from data

given?

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Example Continued

• Use tables for H2O to find P and v• Get total volume as V = m v• Initial state is in mixed region so the

constant pressure, P = Psat(200oC) = 1.5538 MPa and v = vf + x (vg – vf)

• At 200oC, vf and vg, respectively, = 0.001157 and 0.12736 m3/kg

• For x1 = 0.5, v1 = 0.06426 m3/kg

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Example Concluded

• V1 = m v1 = 0.6426 m3 for m = 10 kg

• Point 2 has T = 400oC and P = initial P = 1.5538 MPa

• From superheat table interpolation this final state has v = 0.19646 m3/kg

• Since m = 10 kg, V2 = 1.9646 m3

• W = (1.5538 MPa)(1.9646 – 0.6426) m3

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Units for Work

33 1000

)6426.09646.1)(5538.1(mMPa

kJmMPaW

• Basic idea is that Pa•m3 gives J, kPa•m3 gives kJ, MPa•m3 gives MJ,

• For other pressure units need to convert into kJ; here W = 2,054 kJ

• Without unit conversion we could get the work as 2.054 MJ