Unit Four Quiz Solutions and Unit Five Goals
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Transcript Unit Four Quiz Solutions and Unit Five Goals
Quiz Twelve Solutions and
Review for Final Examination
Mechanical Engineering 370
Thermodynamics
Larry Caretto
May 13, 2003
Outline
• Quiz twelve – simple refrigeration cycle
• Review for final
– Property tables and ideal gases
– First law for closed and open (steady and
unsteady) systems
– Entropy and maximum work calculations
– Isentropic efficiencies
– Cycle calculations (Rankine, refrigeration, air
standard) including mass flow rate ratios
2
Quiz Solution
• Pevap = 100 kPa; Pcond 800 kPa;
Refrigerant load = 200 kJ/min = 3.333 kW
• Find: Power Input
h1 = hg(100 kPa) = 231.35 kJ/kg 3
Condenser
s1 = sg(100 kPa) = 0.9395 kJ/kgK
h2 = h(P2 = Pcond,s2 = s1)
Throttling
Valv e
= 274.33 kJ/kg
h3 = hf(800 kPa) = 93.42 kJ/kg
Ev aporator
4
h4 = h3= 93.42 kJ/kg
2
Compressor
1
3
Quiz Solution Continued
• cop = |qevap|/|wcomp| = |h1 – h4| /|h1 – h2| =
|231.35 kJ/kg – 93.42 kJ/kg| / |
231.35 kJ/kg – 274.33 kJ/kg| = 3.209
Wcomp
Q evap
cop
3.333 kW
3.209
1.039 kW
• For hs 1, (cop)hs = hs(cop)hs = 1
• For hs 80%, cop = 2.052(80%) = 2.567
Wcomp
Q evap
cop
3.333 kW
2.567
1.298 kW
4
Quiz Solution – Third Part
• For isentropic expansion instead of throttling
valve h4 = h(Pcond, s4 = s3 = sf(Pcond) )
• Find x4 = 31.90% and h4 = 84.90 kJ/kg
• |wxpndr| = h3 – h4 = 8.52 kJ/kg
• |wcomp| = 42.98 kJ/kg from first part
• |wnet| = 34.46 kJ/kg
• |qevap| = h1 – h4 = 231.35 - 84.90 = 146.45 kJ/kg
• cop = |qevap| / |wnet| = 146.45 / 34.46 = 4.250
Wnet
Q evap
cop
3
.
333
kW
4.250
0.784 kW
5
Review for Final
• Properties from tables and ideal gases
• First law for closed and open systems
– steady flow and unsteady systems
• Second law and entropy calculations
– Basic cycle quantities: h and cop
– Entropy as a property, tables and ideal
gases, maximum work calculations, hs
• Rankine, refrigeration and air standard
cycles
6
But, first a work about units
• Units and dimensions
• SI units and engineering units
• Extensive, intensive and specific
– E is extensive, e.g., V, U, H, S, Q, W
– T and P are intensive
– e = E/m is specific (e.g. kJ/kg, Btu/lbm)
• Unit conversions (kPam3 = kJ) (m2/s2 =
J/kg) (lbf & lbm) (psia ft3,Btu, lbmft2 /s2)
7
Property Data and Relations I
• Find properties from tables
– Given T and P, T < Tsat(P) or P > Psat(T) is
liquid; T > Tsat(P) or P < Psat(T) is gas
– Liquid at P, T approximately saturated liquid
at given T
– When given P or T and e where e may be v,
u, h, s, compare e to saturation properties
• e < ef(P or T) is liquid; e > eg (P or T) is gas
• otherwise compute x = ( e – ef ) / (eg – ef)
8
Property Data and Relations II
• Ideal gas equations and properties
– Pv = RT, du = cvdT, dh = cpdT, ds = cvdT/T + Rdv/v
= cpdT/T – RdP/P, cp = cv + R, h = u + RT
– u, h, cv and cp = f(T) only (k = cp/cv)
– Pick constant heat capacity at average T
– Handle variable heat capacities by equations or
use ideal gas tables for u(T), h(T) and so(T)
– Isentropic relations for constant and variable heat
capacities, e.g P1v1k = P2v2k, P2/P1 = Pr(T2)/Pr/(T1)
9
Basic First Law Terms
• Energy terms include internal energy, u,
kinetic energy, V2/2 and potential energy, gz
• Heat, Q, is energy in transit due only to a
temperature difference
• Work, W, is action of force over displacement
• Heat added to a system is positive, heat
removed from a system is negative
• Work done by a system is positive, work done
on a system is negative
10
Energy Balances
• System energy change = Heat added to
system – work done by system +
Energy from inflows – Energy outflows
• Usually in kJ (or Btu), but open systems
can use power (kW or Btu/hr)
• Can use q = Q/m and w = W/m or
/m
equivalent rates: q = Q / m ; w = W
• Flowing stream terms include flow work
to give h = u + Pv
11
Closed Systems
• Q = DU + W = m(ufinal – uinitial) + PdV
• Integral is area under path
– Path equation gives P(V) for process
– Integrate equation or find area
– Watch sign
• Internal energy depends on state
– Tables, may have to use u = h – Pv
– Ideal gases: du = cvdT or u(T) in tables
12
Work as Area Under Path
P-v Diagram
250
Point 2
200
Pressure (kPa)
• This works if the path
has a simple shape
• Here we have a path
with three components
• W = W1-2 + W2-3 + W3-4
• W = (P1 + P2)(V2- V1)/2 +
0 + P3-4(V4 – V3)
• W is zero if V is constant
and is negative when
volume decreases
Point
3
150
Point 4
Point 1
100
50
0
0
0.2
0.4
0.6
0.8
1
1.2
3
Volume (m )
13
Formal Integration of Path
• Analytical path equation examples
– Isothermal ideal gas: P = RT/v
– Polytropic process: Pvn = const (n k)
– Arbitrary: P = P1 + a(V – V1)2 + ...
• Evaluate PdV from V1 to V2
• Use P(V)dV for work in kJ (or Btu) or
use P(v)dv for kJ/kg (or Btu/lbm)
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Open Systems/Assumptions
• General energy and mass balances
dE system
dt
dmsystem
2
2
Vi
Vi
Q Wu m i hi
gzi m i hi
gzi
2
2
outlet
inlet
• Steady flow:
dt
m i
m
inlet
outlet
i
dm system
dt
=
dE system
dt
=0
• One inlet and one outlet
• Negligible kinetic and potential energies
15
Steady-Flow Systems
2
2
Vi
Vi
∑m i hi
gzi ∑m i hi
gzi
2
2
outlet
inlet
Q Wu
Mass balance
∑m ∑m
i
inlet
i
outlet
First law for DKE = DPE = 0
Q Wu
∑m i hi - ∑m i hi
outlet
inlet
For DKE = DPE = 0, one inlet and one outlet
Q Wu m hout - hin
q wu hout - hin
16
Unsteady Flow Equations
2
V
2
V
gz m1 u
gz
Q Wu
m2 u
2
2
2
1 system
2
2
Vi
Vi
mi hi
gzi mi hi
gzi
2
2
outlet
inlet
m2 m1 system mi mi
inlet
outlet
17
The Second Law
• There exists an extensive thermodynamic property called the entropy, S,
defined as follows:
dS = (dU + PdV)/T
• For any process dS ≥ dQ/T
• For an isolated system dS ≥ 0
• T must be absolute temperature
18
Entropy as a Property
• Dimensions of entropy are energy
divided by temperature (kJ/K or Btu/R
for S, kJ/kgK or Btu/lbm R for s = S/m)
• If we know the state we can find the
entropy (tables or ideal gas relations)
• If we know the entropy, we can use it to
find the state (tables or ideal gases)
• Use in tables similar to specific volume
19
Cycles with |QH| = |QL| + |W|
High Temperature
Heat Sink
Temperature
High Temperature
Heat
Source Temp
Temperature
sourcesourc
erature
e
e
Temperature
• Engine cycle
converts heat
|QH |
to work
• Refrigeration
cycle
transfers
|QL |
heat from low
to high
Low Temperature
Sink
TTemperature
atur
temperature Heat
Sink
e
Engine Cycle
Schematic
Source
|QH |
|W|
|W|
|QL |
Low Temperature
Heat
Source
Temperature
Sink
Refrigeration Cycle
Schematic
20
Cycle Parameters
• Engine cycle efficiency
h=
W
QH
QL
• Refrigeration cycle COP
=
(coefficient of performance)
W
• General definitions, valid for any cycle
• Engine efficiency always less than one
• COP can be greater than one
21
Reversible Process
• Idealization (the = of the sign), cannot
do better than a reversible process
• Internal reversibility dS = dQ/T
• External reversibility dSisolated system = 0
• Maximum work in a reversible process
– Minimum work input for work input device
– For adiabatic process Ds = 0 for maximum
22
Isentropic Efficiencies
|Ws|
P
2
h
2s
|w|
P
|W|
|Ws|
2
Low P
Hi
gh
h
gh
Hi 1
1
2s
w
Lo
s
Output hs = |w|/|ws|
P
s
Input hs = |ws|/|w|
23
Isentropic Efficiency Problems
• Find ideal work from given inlet state
and one outlet state property: Ds = 0
• e.g., w = hin – hout,s
• Actual work = hs |w| for work output or
|w|/ hs for work input
• Actual outlet state: hout = hin - w
• Note that hout is different from hout,s
24
Cycle Idealizations
• Use these idealizations in lieu of data
– No line losses (output state of one
device is input to the next device)
– Work devices are isentropic
– Heat transfer has no work and DP = 0
– Exit from two-phase device is saturated
– Air standard cycles assume air as
working fluid with heat transfer into fluid
25
Rankine Cycle
Compute Rankine
cycle efficiency given
only T3, P3 and Pcond
h
h4 h1 h2 h1
h3 h2
h1 = hf(Pcond)
h2 = h1 + v1(P3 – P1)
h3 = h(T3,P3); s3 = s(T3,P3)
h4 = h(Pcond, s4 = s3)
3
Turbine
4
Steam
Generator
Condenser
2
1
Pump
26
Modified Rankine Cycle
• Different mass flows
rates in different parts
• Results depend on
ratio of mass flows
• Get mass flow rate
ratios from analysis of
devices where all h
values are known
4
High Pressure
Turbine (T1)
Low Pressure
Turbine (T2)
5
6
7
Steam
Generator
8
Feedwater
Heater
3
2
Condenser
1
Pump
(P1)
Pump
(P2)
27
Refrigeration Cycles
• Pevaporator = P1 = P4 = Psat(T4 = T1)
• Pcondenser = P2 = P3 = Psat(T3 < T2)
State 1: h1 = hg(P1)
State 2: h2 = h(P2, s2 =
s1= sg(P1)
State 3: h3 = hf(P3)
State 4: h4 = h1
cop = (h1 – h4) / (h2 – h1)
3
2
Condenser
Throttling
Valv e
Compressor
Ev aporator
4
1
28
Air-Standard Cycle Analysis
• Use air properties as ideal gas with
variable or constant heat capacity
• Model chemical energy release as heat
addition (~1,200 Btu/lbm or 2,800 kJ/kg)
• Heat addition at constant pressure,
volume or temperature
• Isentropic work
• Closed system except Brayton Cycle
29
Air-Standard Cycle Example
1,200
1,000
800
Presure (kPa)
• Brayton Cycle
• Given: PR, P1, T1,
qH, Find: h
• P2 = P1/PR
• Isentropic compression to P2
• T2 = T1(PR)(k - 1) / k
• T3 = T2 + qH / cp
600
400
200
0
0
0.5
1
1.5
2
2.5
3
3.5
Specific volume (m3/kg)
Compression
Heat addition
Expansion
Heat rejection
30
Brayton Cycle Example
2,500
2,000
Temperature (K)
• Isentropic expansion from P3 = P2
to P4 = P1
• T2 = T1/(PR)(k - 1) / k
• |qL| = cp|T1 – T4|
• h = 1 - |qL| / |qH|
• Can show that h =
1 – 1 / (PR)(k - 1) / k
for constant cp
1,500
1,000
500
0
0
0.5
1
1.5
2
Specific entropy(kJ/kg-K)
Compression
Expansion
Heat addition
Heat rejection
31
And, in conclusion
• Need to know property relations (tables
and ideal gases) to work problems
• First law energy balances in a variety of
systems (closed, steady and unsteady)
• Main application of second law is
isentropic work and efficiencies
• Cycle analysis looks at groups of
devices to get overall efficiency or cop
32
What’s on the Final?
•
•
•
•
Similar to the midterm and quizzes
Open book and notes
Group work on sample final next time
Put more time on explaining how you
will solve the problem than on details of
solution
• Any questions?
33