Unit Four Quiz Solutions and Unit Five Goals Mechanical Engineering 370 Thermodynamics Larry Caretto...
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Transcript of 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
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