Fundamentals Exam

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Thermodynamics Review

Fundamentals Exam

Morning Session:

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11 out of 120 thermo questionsAfternoon General Session:6 out of 60 thermo questions

Fundamentals Exam

Process of Elimination

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Cross out wrong answers first.Wrong answers are sometimes easier g

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 bethan 3 sig figs, your choice should be

the closest to your solution.

Fundamentals ExamSelect the best response for an isolated system.a. The entropy of the system remains constant

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py yb. The heat transfer equals the work donec. The heat transfer equals the internal energy

changechanged. The heat transfer is zero.For a closed thermodynamic system(NCEES p 74)(NCEES p. 74)Q – w = ∆U + ∆KE + ∆PE, isolated impliesQ = W = 0, (d) is the answer, (b) is close but not

l tcomplete

Ideal Gas Law

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Ideal Gas Law

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Ideal Gas Law NCEES p. 73

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Constant EntropyNCEES p. 73

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Fundamentals ExamTwo 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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, pthe pressure is held constant. Cp = 1.0 kJ/kgK, Cv = 0.716kJ/kgK and k = 1.4.

a 56 deg Ca. 56 deg Cb. 40 deg Cc. 33 deg Cd. 28 deg CAnswer is (b) Q = m∆h = m Cp∆T, 80 = 2*1.0*∆T, ∆T= 40 deg C∆T= 40 deg C

Fundamentals ExamSteam at high temperature and pressure passes

through a half open globe valve. Select the h i h h h

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property that remains constant through the valve.

a. enthalpyb. temperaturec. pressured. entropypyAnswer is (a), energy equation q-ws = ∆h + ∆pe +

∆ke, q = 0, ws = 0, ∆pe = 0, ∆κε=0 therefore ∆h = 0, an isenthalpic process, enthalpy is , p p , pyconstant

Fundamentals ExamFor 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

d fi l f 120 i h t i th fi lUNIVERSITY OF

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mass, and a final pressure of 120 psia, what is the final value of the specific volume?

a. 8.2 cubic feet/lbmb. 3.42 cubic feet/lbmb. 3.42 cubic feet/lbmc. 19.7 cubic feet/lbmd. 4.39 cubic feet/lbmAnswer is (d). P1v1

k = P2v2k, v2 = v1(P1/P2)1/k ( ) 1 1 2 2 2 1( 1 2)

= (8.2)*(50/120)1/1.4 = 4.39 ft3/lbm

Fundamentals ExamHow much energy must be transferred through

heat interaction to raise the temperature of a 4 kil l f h i l d

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kilogram sample of methane in a closed system from 15 deg C to 35 deg C?

a. 34.8 kJb. 45 kJc. 139 kJd. 180 kJAnswer is ( c). Q- w = ∆U + ∆KE + ∆PE Closed

system. Q = ∆U = m∆u = mCv∆T = (4kg)*(1.74kJ/kg K)*(35-15 deg C)= ( g) ( g ) ( g )139.2 kJ

Fundamentals ExamA tank contains 0.02m3 of liquid and 1.98 m3 of

vapor. If the density of the liquid is 960 k / 3 d h f h i 0 5k / 3 h

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kg/m3 and that of the vapor is 0.5kg/m3, what is the quality of the mixture?

a. 5.2%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%

Quality x (applies to liquid-vapor systems at saturation) is defined as the mass fraction of the

h NCEES 73vapor phase NCEES p. 73

Fundamentals ExamWhich of the following is an intensive

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property?a. Pressureb. Entropyb. Entropyc. Internal Energyd. Enthalpypy

Answer is (a) Pressure does not depend on ( ) pmass , extensive properties are proportional to mass

Fundamentals ExamWhich 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-

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gy g , g gy gtemperature 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 itthermal reservoir an amount of energy equal to the amount of energy it receives from a work interaction.

c. A device that will change the thermodynamic state of a material from on equilibrium state to another without experiencing a change in the amount of energy contained in the material in the amount of material or in theenergy 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

l t th t f it i f k i t tiequal 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 g py y ( ) gyou can’t be in two equilibrium states, (d) is not correct because energy is increasing continuously

Fundamentals ExamEnergy 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

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system. The change of the internal energy of the system is:

a. 80 kJb. 20 kJc. –20kJd. –80kJ

Answer is (b). ∆E = Q-W = +50kJ-(+30kJ) = 20 kJThis 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!!)1st Law of Thermodynamics NCEES p 741st Law of Thermodynamics NCEES p. 74

Thermodynamics Review

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Thermodynamics Review

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Thermodynamics Review

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Power Cycles Review

A Carnot engine operates between 300°C and 40 °C.Wh i h ffi i f h i ?

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What is the efficiency of the engine?

(A) 87%( )(B) 65%(C) 45%(D) 30%(D) 30%

Power Cycles ReviewSolution: (C)See Page 75 in Reference Handbook

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See Page 75 in Reference HandbookCarnot Cycle efficiency:

/1/)( TTTTT

%4545.)273300()27340(1

/1/)(

==+

−=∴

−=−=

KK

TTTTT

c

HLHLHc

η

η

)273300( + Kcη

Not (A): If t (A) did ’t th b l t t tIf you got (A) you didn’t use the absolute temperature units.

%45%7868666401 ≠Co

“When in doubt use absolute.”

%45%7.868666.300

1 ≠==−=Cocη

Power Cycles Review

Refrigerant 134a is isentropically compressed in a f d 0 4 MP

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compressor from a saturated vapor state at 0.4 MPa pressure to 2 MPa pressure. The work required to run the compressor is:p

(A)130 kJ/kg

(B) -100 kJ/kg(B) -100 kJ/kg

(C) 100kJ/kg

(D) 60kJ/k(D)-60kJ/kg

Power Cycles ReviewSolution: (B)See Pages 75 and 81 in Reference Handbook for First Law

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See Pages 75 and 81 in Reference Handbook for First Law (energy balance) and p-h table for Refrigerant 134aUse 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 theintersection line for the given pressures and the right side of the “dome” respectively.

and )/(330 kgkJhi ≈∴ )/(430 kgkJhe ≈The exit enthalpy has the same s as the inlet but at a higher P.Energy balance:

i )( ge

The answer is negative because work is put into the system.)/(100)/(430)/(330 kgkJkgkJkgkJhhW ei −=−=−=

Power Cycles Review

The Otto Cycle, aka SIE, Spark Ignition Engine

UNIVERSITY OF

FLORIDA• This cycle applies to two stroke and four stoke engines. It is an ideal representation of the g pprocess.

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 (intake4-1: constant volume heat rejection (intake stroke)

Power Cycles Review

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Power Cycles ReviewModel this as a closed system. No major changes in kinetic or potential energy.

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Energy equation becomes:

(qin-qout)+(win-wout) = ∆u

But more importantly let’s look at theBut 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)

ηth Otto = wnet/qinηth,Otto wnet/qin

Looking at the first law, in = out

wnet = qin – qout

Therefore,

ηth,Otto = wnet/qin = (qin-qout)/qout

Power Cycles Reviewηth,Otto = wnet/qin = (qin-qout)/qout

Substituting for qin and qout

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ηth,Otto = 1 – ((T4-T1)/(T3-T2)

Since 1 to2 and 3 to 4 are isentropic processes, we can substitute our isentropic relationships for T and v.

Therefore,

1

11 −−= kthOtto rη

rWhere the compression ratio r = V1/V2 = v1/v2

A d k ifi h t ti C /CAnd k = specific heat ratio = Cp/Cv

Power Cycles ReviewLet’s look at an example:

Assume a compression ratio of 8

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Assume a compression ratio of 8.

Assume air at the beginning of the process is at 17 deg C and 100KPa.

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:

%5.56565.08

1111 )14.1(1 orr kthOtto =−=−= −−η

Power Cycles ReviewLet’s find the temperature at 3. This would be the maximum temperature our equipment would have to tolerate.

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have to tolerate.

qin= 800kJ/kg = Cv(T3-T2)=0.718*(T3 – T2)

We need T2 to solve for T3.

From the isentropic relationships,

T1/T2 = (V2/v1)k-1

Note that r = v /vNote that r = v1/v2,

Therefore,

T1/T2 = (1/r)k-11 2

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.

Power Cycles ReviewLet’s find the pressure at 3. This would be the maximum pressure our equipment would have to tolerate.

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to tolerate.

P1v1/T1 = P2v2/T2

Therefore P2 = P1v1T2/v2 T1

P2 = 100kPa(666K/290K)(8) = 1837 kPa

And

P v /T = P v /TP2v2/T2 = P3v3/T3

P3 = P2T3v2/T2v3

P3 = (1837kPa)(1780K/666K)(1)3

Note v2 = v3

P3 = 4910 kPa or 712 psia!!

Power Cycles ReviewThe Brayton Cycle• Used to analyze gas turbines (where compression and

http://travel.howstuffworks.com/turbine3.htm

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Used to analyze gas turbines (where compression and expansion occur using rotating machinery)

• Gas Turbines usually operate as an open cycley p p y

• Used in aircraft propulsion and electric power generation

Exhaust propels craft or used to generate steam.

Power Cycles ReviewSimple 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

applied. Air is the working fluid.

3-4: isentropic expansion

4-1: constant pressure heat rejection

Power Cycles ReviewModel this as a closed system. No major changes in kinetic or potential energy.

q = h – h = C (T – T )

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qin = h3 – h2 = Cp (T3 – T2)

-qout = h1 – h4 = Cp(T1 – T4)

Or qout = Cp(T4 – T1)

ηth,Brayton = wnet/qin

Looking at the first law, in = out

wnet = qin – qout

Therefore,

ηth Brayton = wnet/qin = (qin-qout)/qinηth,Brayton wnet/qin (qin qout)/qin

ηth,Brayton = 1 – T1(T4/T1-1)/T2(T3/T2-1)

Substituting the isentropic relationships for T as a f ti f P d li i th t P P d P Pfunction of P and realizing that P2=P3 and P1 = P4, ηth,Brayton = 1 – 1/rp

(k-1)/k

where rp = pressure ratio = P2/P1

Power Cycles Review• 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

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• Since some of the turbine work goes to run the compressor, there is another term used to describe this cycle:

the back work ratio = Compressor work/turbine work.

• 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• 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 heatWith 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).

Power Cycles ReviewLet’s look at an example:

Assume a pressure ratio of 8

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Assume a pressure ratio of 8.

Assume the air at the compressor inlet (pt 1)is at 300K (room temperature).

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 efficiencyFind the thermal efficiency.

ThermodynamicsUsing the cold air standard assumptions and assuming negligible changes in kinetic and potential energy:

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kinetic and potential energy:

R air = 0.3704 psia ft3/lbm R

Cp = 0.24 Btu/lbm R

K = 1.4

ηth,Brayton = 1 – 1/r (k-1)/k

η = 1 1/(8) (1 4-1)/1 4ηth,Brayton = 1 – 1/(8) (1.4 1)/1.4

ηth,Brayton = 0.448 or 44.8%

T2/T1 = (8) (1.4-1)/1.4= 1.8112 1

T2 = (300)( 1.811 ) = 543.4K

And

T4/T3 = (1/8) (1.4-1)/1.4 = .552

T4 = (1300)(0.522) = 717.6K

Power Cycles Review

qin = h3 – h2 = Cp (T3 – T2)

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qin = ( 1.005 kJ/kg K)(1300-543.4)

qout = Cp(T4 – T1) = (1.005 kJ/kg K)(717.6-300)qout Cp(T4 T1) (1.005 kJ/kg K)(717.6 300)

ηth,Brayton = (qin-qout)/qin =

0.448 or 44.8%

Back work ratio = Cp(T2-T1)/Cp(T3-T4) =

(543.5-300)/(1300-717.6) =

0 420.42

Power Cycles ReviewUsing air standard assumptions (see text):

Back work ratio = 0 402

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Back work ratio = 0.402

Thermal efficiency = 42.6%

T2 = 540K

T4 = 770 K

Our cold air standard assumptions worked wellwell.

Power Cycles ReviewThe Rankine or Vapor Power Cycle• Used to steam power plant operations

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Used to steam power plant operations

Power Cycles ReviewSimple Ideal Rankine Cycle

1-2 isentropic compression with a pump

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3-4 constant pressure heat addition in a boiler

5-6 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)Losses (thermal and pressure)

Power Cycles ReviewWe will consider water(steam) as our motive fluid.

The steam leaving the boiler (pts. 4 and 5) is usually superheated steam

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usually superheated steam.

The steam leaving the turbine (pt 6) is

usually high quality steam. Remember

that s5 = s6

The water leaving the condenser is either

d li id b l d li idsaturated liquid or subcooled liquid water.

We usually assume that h2 = h1.

Or we can estimate h2 = h1 + v(P2-P1).Or we can estimate h2 h1 v(P2 P1).

Let’s work an example problem to

illustrate how this works.

Power Cycles Review

P

Consider a steam power plant and assume an ideal Rankine cycle to model the system.

Th t t th t bi t 3MP d 350 C

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The steam enters the turbine at 3MPa and 350 C

Condenser pressure is at 75 kPa.

What is the thermal efficiency for this cycle?

14

y y

ηth,Rankine = wnet/qin= (qin – qout)/qin

qin = h1 – h4 23

h

qout = h3-h2

h1 = h of superheated steam = 3115.3 kJ/kg

s = s = 6 7428 kJ/kg Ks1 = 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

Power Cycles Review

P

h3 = hf75kPa = 384.39 kJ/kg

h4 = h3 approx.

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qin = h1 – h4= 3115.3 – 384.39 kJ/kg = 2730.9 kJ/kg

qout = h3-h2 = 384.39 - 2402.6 = -2018.2 kJ/kg

(2730 9 2018 2)/2730 9

14

ηth,Rankine = (2730.9-2018.2)/2730.9 =

26%

Note that I calculated the same efficiency 23

h

yneglecting any change in h across the pump. I assumed the pump was isenthalpic instead of isentropic.

If we do consider the pump ∆h = 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.