Transcript Thermal Cycles - Rankine Cycle with Reheat - plaza

Fundamentals Exam
UNIVERSITY OF
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Fundamentals Exam
Thermodynamics Review
Fundamentals Exam
I assume you have applied?!?
Morning
Session:
UNIVERSITY OF
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Afternoon Session:
General or Discipline Specific- 60
questions, 1 pt each
Fundamentals Exam
What to do in the afternoon?
Your
call...pass/fail
is
about
the
same.
UNIVERSITY OF
FLORIDA Preparation easier for general.
Fundamentals Exam
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Morning Session:
11 out of 120 thermo questions
Afternoon General Session:
6 out of 60 thermo questions
Fundamentals Exam
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NCEES Reference Handbook
Have you got it?
Why not?
How do you get it?
How do you use it?
www.ncees.org
Fundamentals Exam
Morning session
Generally
unrelated.
About
2
minutes
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per question.
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Fast recall essential.
Use a marking system to keep track
of your progress.
Afternoon: 4 minutes per question.
Fundamentals Exam
Process of Elimination
Cross
out
wrong
answers
first.
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Wrong answers are sometimes easier
to find than right ones! Units on
answer is sometimes a clue.
Answers are seldom given with more
than 3 sig figs, your choice should be
the closest to your solution.
Fundamentals Exam
Guessing
No
penalty
for
guessing.
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Leave 10 minutes for each session for
“educated” guessing.
Fundamentals Exam
Hint:
Write correct answer in the margin of
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your test booklet beside the question
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and wait to you get to the end of the
page before transferring to the answer
key.
Lookout for those long, drawn out
questions… questions with four
paragraphs for answers!!
Fundamentals Exam
Try working the following problem (you have two minutes
for each type problem like this):
If a sample experiencing a change of temperature from 23
deg C to 46 deg C also experiences a change in specific
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OF
of 120 kJ/kg, of what material is the sample most
likely to be composed? You can use the data in the NCEES
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Supplied Reference Handbook or the tables in the back of
your book. Better to practice these problems with your
NCEES handbook…..remember to be at one with this
book!!
Did you get Helium?
There are sample tests on the NCEES website. Try these out.
There are also sample tests in books like Barron’s How to
Prepare for the Fundamentals of Engineering FE/EIT Exam. The
bookstores have books like this to help you review for the exam.
Fundamentals Exam
Select the best response for an isolated
system.
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a. The entropy of the system remains
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constant
b. The heat transfer equals the work done
c. The heat transfer equals the internal
energy change
d. The heat transfer is zero.
Fundamentals Exam
Select the best response for an isolated system.
a. The entropy of the system remains constant
b. The heat transfer equals the work done
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c. The heat transfer equals the internal energy
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d. The heat transfer is zero.
For a closed thermodynamic system:
Q – w = DU + DKE + DPE, isolated implies
Q = W = 0, (d) is the answer, (b) is close but not
complete
Fundamentals Exam
Two kilograms of air are contained in a
cylinder. If 80 kJ of heat are added to
the air, estimate the temperature rise if
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FLORIDA the pressure is held constant. Cp = 1.0
kJ/kgK, Cv = 0.716kJ/kgK and k = 1.4.
a. 56 deg C
b. 40 deg C
c. 33 deg C
d. 28 deg C
Fundamentals Exam
Two kilograms of air are contained in a cylinder.
If 80 kJ of heat are added to the air, estimate
the temperature rise if the pressure is held
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constant. Cp = 1.0 kJ/kgK, Cv =
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a. 56 deg C
b. 40 deg C
c. 33 deg C
d. 28 deg C
Answer is (b) Q = mDh = m CpDT, 80 =
2*1.0*DT, DT= 40 deg C
Fundamentals Exam
Steam at high temperature and pressure
passes through a half open globe valve.
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Select the property that remains
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a. enthalpy
b. temperature
c. pressure
d. entropy
Fundamentals Exam
Steam at high temperature and pressure passes
through a half open globe valve. Select the
property that remains constant through the
valve.
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a. enthalpy
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b. temperature
c. pressure
d. entropy
Answer is (a), energy equation q-ws = Dh + Dpe +
Dke, q = 0, ws = 0, Dpe = 0, Dke=0 therefore
Dh = 0, an isenthalpic process, enthalpy is
constant
Fundamentals Exam
For an isentropic process of an ideal gas (k= 1.4),
with an initial pressure of 50 pounds per
square inch absolute, an initial specific
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volume of 8.2 cubic feet per pound mass, and
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value of the specific volume?
a. 8.2 cubic feet/lbm
b. 3.42 cubic feet/lbm
c. 19.7 cubic feet/lbm
d. 4.39 cubic feet/lbm
Fundamentals Exam
For an isentropic process of an ideal gas (k= 1.4), with an
initial pressure of 50 pounds per square inch absolute,
an initial specific volume of 8.2 cubic feet per pound
mass, and a final pressure of 120 psia, what is the final
value of the specific volume?
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a. 8.2 cubic feet/lbm
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b. 3.42 cubic feet/lbm
c. 19.7 cubic feet/lbm
d. 4.39 cubic feet/lbm
Answer is (d). P1v1k = P2v2k, v2 = v1(P1/P2)1/k
= (8.2)*(50/120)1/1.4 = 4.39 ft3/lbm
Fundamentals Exam
How much energy must be transferred
through heat interaction to raise the
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temperature of a 4 kilogram sample of
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deg C to 35 deg C?
a. 34.8 kJ
b. 45 kJ
c. 139 kJ
d. 180 kJ
Fundamentals Exam
How much energy must be transferred through
heat interaction to raise the temperature of a 4
kilogram sample of methane in a closed
system from 15 deg C to 35 deg C?
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a. 34.8 kJ
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b. 45 kJ
c. 139 kJ
d. 180 kJ
Answer is ( c). Q- w = DU + DKE + DPE Closed
system. Q = DU = mDu = mCvDT =
(4kg)*(1.74kJ/kg K)*(35-15 deg C)=
139.2 kJ
Fundamentals Exam
A tank contains 0.02m3 of liquid and 1.98 m3 of
vapor. If the density of the liquid is 960
3 and that of the vapor is 0.5kg/m3, what
kg/m
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FLORIDA is the quality of the mixture?
a.
b.
c.
d.
5.2%
4.9%
2.04%
1.01%
Fundamentals Exam
A tank contains 0.02m3 of liquid and 1.98 m3 of
vapor. If the density of the liquid is 960
kg/m3 and that of the vapor is 0.5kg/m3, what
is the quality of the mixture?
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a. 5.2%
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b. 4.9%
c. 2.04%
d. 1.01%
Answer is (b). x = mg/(mg + mf) =
(1.98)*(.5)/((1.98*0.5)+(0.02*960)) = 0.049
or 4.9%
Fundamentals Exam
Which of the following is an intensive
property?
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OF Pressure
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b. Entropy
c. Internal Energy
d. Enthalpy
Answer is (a) Pressure does not depend on
mass , extensive properties are proportional to
mass
Fundamentals Exam
Which of the following devices is possible?
a. A cyclic machine that will experience no other interaction than to produce energy
through a work interaction, while transferring energy from a high-temperature
reservoir to a low-temperature reservoir through heat interactions.
b. A cyclic machine that will experience no other interaction than to transfer to a
thermal reservoir an amount of energy equal to the amount of energy it
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receives from a work interaction.
c. A device that will change the thermodynamic state of a material from on
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equilibrium state to another without experiencing a change in the amount of
energy contained in the material, in the amount of material, or in the external
forces placed on the material.
d. A cyclic machine that will experience no other interaction than to accept from a
heat interaction with a high-temperature reservoir an amount of energy equal
to the amount of energy it receives from a work interaction.
Answer is (a). Note that this problem takes about a minute to read!! You better
understand this one as you read it or you won’t do this in 2 minutes!! (b) is
not right because entropy decreases continuously, (c) is not right because you
can’t be in two equilibrium states, (d) is not correct because energy is
increasing continuously
Fundamentals Exam
Energy is added in the amount of 50 kJ in a heat interaction
to a closed system while 30 kJ of work is done by the
system. The change of the internal energy of the
system is:
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OF 80 kJ
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b. 20 kJ
c. –20kJ
d. –80kJ
Answer is (b). DE = Q-W = +50kJ-(+30kJ) = 20 kJ
This is an example of how you need to know the first law and
know the correct sign conventions for work and energy.
Q into a system is +
W done by a system is + (remember it is a positive thing to
get work out of a student!!)
Fundamentals Exam
Devices:
Turbines
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Compressors
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Diffusers
Nozzles
Throttling Devices
Heat Exchangers
Fundamentals Exam
Devices:
Turbines
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Compressors
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mi’(hi +Vi2/2 + gzi) –me’ (he +Ve2/2 +
gze)+Q’ –W’ = 0
Assume well insulated, assume Vi=Ve
and steady flow mi’ = me’,and zi = ze
Then: m’(hi – he) = W’
Compressors: W’ is –
Turbines: W’ is +
Fundamentals Exam
Devices:
Nozzles and Diffusers
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2/2 + gz ) –m ’ (h +V 2/2 +
m
’(h
+V
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i
i
i
e
e
e
gze)+Q’ –W’ = 0
Assume well insulated, no shaft
work, and steady flow mi’ =
me’,and zi = ze
Then: (hi – he +Vi2/2 -Ve2/2 ) = 0
Ve > Vi for nozzles and Ve<Vi for
diffusers
Fundamentals Exam
Devices:
Throttling Device
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2/2 + gz ) –m ’ (h +V 2/2 +
m
’(h
+V
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i
i
i
e
e
e
gze)+Q’ –W’ = 0
Assume well insulated, assume Vi=Ve
and steady flow mi’ = me’, zi = ze and
no shaft work
Then: (hi – he) = 0 isenthalpic
Fundamentals Exam
Devices:
Heat Exchangers
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2/2 + gz ) –m ’ (h +V 2/2 +
m
’(h
+V
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i
i
i
e
e
e
gze)+Q’ –W’ = 0
Assume well insulated, assume Vi=Ve
and steady flow mi’ = me’, no shaft
work and zi = ze
Then: m’(hi – he) = Q’
Fundamentals Exam
Cycles:
Carnot
Reversed Carnot
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Otto
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Rankine
Refrigeration
Afternoon: if mechanical: two stage refrig
cycle, air refrigeration, psychrometric
cycles, Brayton cycle, Brayton cycle with
regeneration, etc.
Fundamentals Exam
Review state functions, specifically the
concept of quality, x
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u = xug + (1-x)uf
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h = xhg + (1-x)hf
s = xsg + (1-x)sf
v = xvg +(1-x)vf
Look at Rankine cycle, steam quality out of
a turbine
Thermodynamics Review
A Carnot engine operates between 300°C and 40 °C.
What is the efficiency of the engine?
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(A)
(B)
(C)
(D)
87%
65%
45%
30%
Thermodynamics Review
Solution: (C)
See Page 49 in Reference Handbook
Carnot Cycle efficiency:
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c = (TH  TL ) / TH = 1  TL / TH
( 40  273) K
c = 1 
= .45 = 45%
(300  273) K
Not (A):
If you got (A) you didn’t use the absolute temperature
units.
40o C
c = 1 
= .8666 = 86.7%  45%
o
300 C
“When in doubt use absolute.”
Thermodynamics Review
Refrigerant 134a is isentropically compressed in a
compressor from a saturated vapor state at 0.4 MPa
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OF
pressure
to 2 MPa pressure. The work required to run
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the compressor is:
(A) 130 kJ/kg
(B) -100 kJ/kg
(C) 100kJ/kg
(D) -60kJ/kg
Thermodynamics Review
Solution: (B)
See Pages 48 and 55 in Reference Handbook for First Law
(energy OF
balance) and p-h table for Refrigerant 134a
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Use the Chart:
The question states that the refrigerant is in a saturated vapor
state, therefore the enthalpy can be obtained by finding the
intersection line for the given pressures and the right side of the
“dome” respectively.
 hi  330(kJ / kg ) and he  430(kJ / kg)
The exit enthalpy has the same s as the inlet but at a higher P.
Energy balance:
W = hi  he = 330 (kJ / kg )  430( kJ / kg ) = 100 (kJ / kg )
The answer is negative because work is put into the system.
Thermodynamics
The Air Standard Assumptions
• for gas powered cycles
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• for Otto cycle, diesel cycle and Brayton cycle
• the working fluid is air which continuously circulates
through the system, acts as an ideal gas
• all processes are internally reversible
• combustion process is approximated by a heat
addition process from external source
• exhaust process is approximated by a heat rejection
process
Thermodynamics
The Cold Air Standard Assumptions
• all of above assumptions
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• assumes specific heats are constant and are
evaluated at 25 deg C or 77 deg F
• all of the above assumptions are made to
simplify a very complex cycle
Thermodynamics
The Otto Cycle, aka SIE, Spark Ignition Engine
• This cycle applies to two stroke and four stoke
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engines. It is an ideal representation of the
process.
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1-2: isentropic compression (compression
stroke)
2-3: constant volume heat addition (power
stroke)
3-4: isentropic expansion (exhaust stroke)
4-1: constant volume heat rejection (intake
stroke)
Thermodynamics
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Thermodynamics
Model this as a closed system. No major
changes in kinetic or potential energy.
Energy equation becomes:
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(q -q )+(wOF-w
in
out
in
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out)
= Du
But more importantly, let’s look at the
processes from 2 to 3 and 4 to 1:
qin = u3-u2 = Cv(T3-T2)
qout = u4 – u1 = Cv(T4 – T1)
(no work done during these processes)
th,Otto = wnet/qin
Looking at the first law, in = out
wnet = qin – qout
Therefore,
th,Otto = wnet/qin = (qin-qout)/qout
Thermodynamics
th,Otto = wnet/qin = (qin-qout)/qout
Substituting for qin and qout
th,Otto = 1 – ((T4-T1)/(T3-T2)
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Since 1 to2 and 3 to 4 are isentropic
processes, we can substitute our isentropic
relationships for T and v.
Therefore, (see derivation on page 360)
thOtto = 1 
1
r
k 1
Where the compression ratio r = V1/V2 = v1/v2
And k = specific heat ratio = Cp/Cv
Thermodynamics
Let’s look at an example:
Assume a compression ratio of 8.
Assume
airOF
at the beginning of the process
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is
at 17 deg C and 100KPa.
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Assume 800 kJ/kg of heat is added in the
heat addition process (this would be
equivalent to the fuel added to the cycle).
Assume constant specific heats.
The efficiency of this cycle would equal:
thOtto = 1 
1
r
k 1
= 1
1
8
(1.4 1)
= 0.565or 56.5%
Thermodynamics
Let’s find the temperature at 3. This would be
the maximum temperature our equipment would
have to tolerate.
qin= 800kJ/kg = Cv(T3-T2)=0.718*(T3 – T2)
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We
need T2 to solve for T3.
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From the isentropic relationships,
T1/T2 = (V2/v1)k-1
Note that r = v1/v2,
Therefore,
T1/T2 = (1/r)k-1
And T2 = 290K/(1/8) (1.4-1)
T2 = 666 K
So T3 = (800/0.718)+666 = 1780 K or 2745deg F
Can also do this assuming variable specific
heats.
Thermodynamics
Let’s find the pressure at 3. This would be the
maximum pressure our equipment would have
to tolerate.
P1v1/T1 = P2v2/T2
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Therefore
P2 = P1v1T2/v2 T1
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P2 = 100kPa(666K/290K)(8) = 1837 kPa
And
P2v2/T2 = P3v3/T3
P3 = P2T3v2/T2v3
P3 = (1837kPa)(1780K/666K)(1)
Note v2 = v3
P3 = 4910 kPa or 712 psia!!
Thermodynamics
The Brayton Cycle
http://travel.howstuffworks.com/turbine3.htm
• Used to analyze gas turbines (where compression and
expansion occur using rotating machinery)
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• Gas
Turbines usually operate as an open cycle
• Used in aircraft propulsion and electric power generation
Exhaust propels craft or used to
generate steam.
Thermodynamics
Simple Ideal Brayton Cycle
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Modeled as a closed cycle. Air
standard assumptions are
applied. Air is the working fluid.
1-2: isentropic compression
2-3: constant pressure heat addition
3-4: isentropic expansion
4-1: constant pressure heat rejection
Thermodynamics
Model this as a closed system. No major changes
in kinetic or potential energy.
qin = h3 – h2 = Cp (T3 – T2)
-q
out = h1 – h
4 = Cp(T1 – T4)
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OF
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Or
qout = Cp(T4 – T1)
th,Brayton = wnet/qin
Looking at the first law, in = out
wnet = qin – qout
Therefore,
th,Brayton = wnet/qin = (qin-qout)/qin
th,Brayton = 1 – T1(T4/T1-1)/T2(T3/T2-1)
Substituting the isentropic relationships for T as a
function of P and realizing that P2=P3 and P1 = P4,
th,Brayton = 1 – 1/rp (k-1)/k
where rp = pressure ratio = P2/P1
Thermodynamics
• For gas turbine engines, the rp ranges from 5 t0 20.
• Since some of the turbine work goes to run the compressor, there is another term
used to describe this cycle:
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the back work ratio = Compressor work/turbine work.
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• Usually more than half the turbine work goes to run the compressor.
• The back work ratio for steam power plants is very low in comparison.
• Gas turbines used in power plants can be brought on line very quickly whereas the
Rankine cycle steam cycles take a lot of time to bring up to speed. This is
why gas turbine engines are used as “peaking” units.
• With improvements in firing temperature, turbomachinery efficiency and heat
recovery, the gas turbine power generating systems are now comparable to
steam plants in performance, especially when the waste heat is combined
with a Rankine cycle plant (bottoming cycle).
Thermodynamics
Let’s look at an example:
Assume a pressure ratio of 8.
Assume
theOFair at the compressor inlet (pt
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1)is
at 300K (room temperature).
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Assume the air at the turbine inlet(pt 3) is
at 1300 K. (1880 deg F)
Find the gas temperatures at the exits of
the turbine and compressor.
Find the back work ratio.
Find the thermal efficiency.
Thermodynamics
Using the cold air standard assumptions
and assuming negligible changes in
kinetic and potential energy:
R air = 0.3704 psia ft3/lbm R
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CFLORIDA
p = 0.24 Btu/lbm R
K = 1.4
th,Brayton = 1 – 1/r (k-1)/k
th,Brayton = 1 – 1/(8) (1.4-1)/1.4
th,Brayton = 0.448 or 44.8%
T2/T1 = (8) (1.4-1)/1.4= 1.811
T2 = (300)( 1.811 ) = 543.4K
And
T4/T3 = (1/8) (1.4-1)/1.4 = .552
T4 = (1300)(0.522) = 717.6K
Thermodynamics
qin = h3 – h2 = Cp (T3 – T2)
qin = ( 1.005 kJ/kg K)(1300-543.4)
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q = C (T
out
p
4
– T1) = (1.005 kJ/kg K)(717.6-300)
th,Brayton = (qin-qout)/qin =
0.448 or 44.8%
Back work ratio = Cp(T2-T1)/Cp(T3-T4) =
(543.5-300)/(1300-717.6) =
0.42
Thermodynamics
Using air standard assumptions (see text):
Back work ratio = 0.402
Thermal
efficiency
= 42.6%
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OF
TFLORIDA
2 = 540K
T4 = 770 K
Our cold air standard assumptions worked
well.
Thermodynamics
The Rankine or Vapor Power Cycle
• Used to steam power plant operations
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Thermodynamics
Simple Ideal Rankine Cycle
1-2 isentropic compression with a pump
3-4 constant pressure heat addition in a boiler
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5-6FLORIDA
isentropic expansion in a turbine
6-1 constant pressure heat rejection in
condenser
Other points can be used to
describe pipe
Losses (thermal and pressure)
Thermodynamics
We will consider water(steam) as our motive fluid.
The steam leaving the boiler (pts. 4 and 5) is
usually superheated steam.
The
steam leaving
the turbine (pt 6) is
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OF
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usually
high quality steam. Remember
that s5 = s6
The water leaving the condenser is either
saturated liquid or subcooled liquid water.
We usually assume that h2 = h1.
Or we can estimate h2 = h1 + v(P2-P1).
Let’s work an example problem to
illustrate how this works.
Thermodynamics
Consider a steam power plant and assume an
ideal Rankine cycle to model the system.
P
The steam enters the turbine at 3MPa and 350 C
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Condenser
pressure is at 75 kPa.
4
1
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What
is the thermal efficiency for this cycle?
th,Rankine = wnet/qin= (qin – qout)/qin
qin = h1 – h4
3
2
qout = h3-h2
h1 = h of superheated steam = 3115.3 kJ/kg
s1 = s2 = 6.7428 kJ/kg K
s2 = sf75kPa + x sfg75kPa = 6.7428 kJ/kg K
x = (6.7428-1.213)/6.2434 = 0.88
Using x and hf75kPa and hfg75kPa, we can calc.
h2 = 384.39+(0.88)(2278.6)= 2402.6 kJ/kg
h
Thermodynamics
h3 = hf75kPa = 384.39 kJ/kg
h4 = h3 approx.
qin = h1 – h4= 3115.3 – 384.39 kJ/kg = 2730.9 kJ/kg
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P
4
1
qout = h3-h2 = 384.39 - 2402.6 = -2018.2 kJ/kg
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th,Rankine = (2730.9-2018.2)/2730.9 =
26%
Note that I calculated the same efficiency
neglecting any change in h across the pump. I
assumed the pump was isenthalpic instead of
isentropic.
If we do consider the pump Dh = 0, then the pump
work is approx. zero and the back work ratio = 0.
If the work of the pump is calculated we find that
the back work ratio is 0.004 or 0.4%. Compare to
the 0.42 for the Brayton cycle.
3
2
h