Unit Four Quiz Solutions and Unit Four Quiz Solutions and Unit Five GoalsUnit Five Goals

Mechanical Engineering 370Thermodynamics

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

10 10 14 14 15 15 15 15 15 18 18

20 20 21 22 24 24 24 24 24 28

• 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

0

100

200

300

400

500

600

700

800

0.0 0.2 0.4 0.6 0.8 1.0

Volume (m3)

Pre

ssu

re (

kPa)

1

2

3

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

P-v Diagram

0

50

100

150

200

250

300

350

0 0.2 0.4 0.6 0.8 1 1.2

Volume (m3)

Pre

ssu

re (

kPa)

Point 1

Point 2

Point 4

Point 3

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 = U + 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)

• V < 0 means work will be negative

P

V

1

2

34

8

Finding the Answer

• Find P3-4 = P3 = P4 from state 3 defined by T3 = 400oC and v3 = v2 = V2/m

kg

Kkg

kJK

mkPa

kJmkPa

RT

VP

v

Vm 3781.0

4615.0)15.573(

1)1)(100(

33

1

11

1

1

• Use properties at the initial state to find the mass

kPa

mkPakJ

m

KKkgkJ

kg

V

mRT

V

mRTPP 8.146

1)8.0(

)15.673(4615.0

)3781.0(

332

3

3

343

9

Finding the Answer (cont’d)

• Now find heat and work

• Find T4 from m, P4 = P3 and V4 = 0.4 m3

K

KkgkJ

kg

mkPakJ

mkPa

mR

VPT 6.336

4615.0)3781.0(

1)4.0)(8.146( 3

3

444

)-()-(2

)-( 344-31221

14 VVPVVPP

TTmcWUQ v

10

Finding the Answer (concluded)

kJmmkPa

mmkPakPa

mkPa

kJ

KKKkg

kJkg

VVPVVPP

TTmc

WUQ

v

9.224.80-4.0)8.146(

1-8.02

3001001

15.573-6.3364108.1

)3781.0(

)-()-(2

)-(

33

333

344-31221

14

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

rate of energy change (ML2T-3)

uW the useful work rate or mechanical power (ML2T-3)

the mass flow rate (MT-1)m

gz the potential energy per unit mass (L2T-2)

the kinetic energy per unit mass (L2T-2)2

2V

systemsystem gzV

umE )2

(2

total energy (ML2T-2)

Q heat transfer rate (ML2T-3)

dt

dEsystem

14

Unit Five Goals Continued

– use the equation relating velocity, mass flow rate, flow area, A, and specific volume

– use the mass balance equation

outlet

iinlet

isystem mmdt

dm

v

AVm

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

0dt

dE

dt

dm systemsystem

inleti

iii

outleti

iiiu

system gzV

hmgzV

hmWQdt

dE

22

22

17

Steady-state equations

– Steady-state first law for open systems

– Steady-state mass balance for open systems

inleti

iii

outleti

iiiu gz

Vhmgz

VhmQW

22

22

outlet

iinlet

i mm

18

Unit Five Goals Continued

– recognize that kinetic and potential energies are usually negligible • A 1oC temperature change in air (ideal gas

with cp = 1.005 kJ/(kg∙K) has h = 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

– work with ratios q and w in simplest case

)hh(mQW inoutu

– handle simplest case: steady-state, one inlet, one outlet (one mass flow rate), negligible changes in kinetic and potential energies

m

Qq

m

Ww u

u

)( inoutu hhqw

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

)hh(mQW inoutu First law

21

Getting the Answer

• At Tin = 700oC and Pin = 10 MPa, hin = 3870.5 kJ/kg (p. 836)

• At Tout = 300oC and Pout = 500 kPa, hout = 3061.6 kJ/kg

• Heat loss is negative: Q = Qin - Qout

kWkJ

skW

kg

kJ

kg

kJ

s

kgkWWu 689,7

15.38706.3061

10)400(

22

Review Ideal Gases

• For ideal gases du = cvdT; dh = cpdT

• Ideal gas H = U + RT

• May have molar H 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 = 473.15 K and T2 = 673.15 K• Molar enthalpy change = 7229.3 kJ/kmol

TRM

h

M

TRh

M

uu

TTd

TTc

TTb

TTa

dTdTcTbTadTTchT

T

T

T

p

4

14

23

13

22

12

212

32

432

)(2

1

2

1

24

Getting u from molar h

• Use molar h just found, data on R and M, and T = 200oC = 200 K

TRM

hu

kg

kJK

Kkg

kJ

kmolkg

kmolkJ

u309

2004615.0

015.18

3.7229

25

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

• u = (17,490 kJ/kmol - 11,953 kJ/kmol) / (18.015 kg / kmol) = 307.4 kJ/kg