unit four quiz solutions and unit five goals mechanical engineering 370 thermodynamics larry caretto...

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