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