Thermo Problems 2

8/10/2019 Thermo Problems 2

1/24

Baratuci

HW #5

Problem : 7.14 - The Increase of Entropy Principle - 2 pts 11-May-11

Read :

Answers : The entropy of steam will increase as it flows through a real adiabatic turbine.

Real turbines are not reversible. They are irreversible.

Irreversibilities cause entropy generation.

Because the process is adiabatic, there is no way to decrease the entropy.

ENGR 224 - Thermodynamics

Will the entropy of steam increase, decrease or remain the same as it flows through a real adiabatic turbine ?

The usual problem solving procedure does not really apply th this "problem" because it is really a

short-answer question.

Therefore, the entropy generation due to irreversibilities in the real turbine produce an INCREASE

in the entropy of the steam as it passes through the turbine.

Dr. Baratuci - ENGR 224 hw5-sp11.xlsm, 7.14 5/9/2011


2/24

Baratuci

HW #5


Read :

Answer :

Because the cycle is a Carnot Cycle, each step in the cycle is completely reversible.

Nonetheless, adding heat to the working fluid INCREASES its entropy.

Keep in mind that the heat CAME from a thermal reservoir.

The entropy of the working fluid in an ideal Carnot Cycle will increase during the isothermal heat

addition process.

When the heat "left" the thermal reservoir, the entropy of the reservoir decreased by the same

amount that the entropy of the working fluid in the cycle increased.

The result is a zero net change in the entropy of the universe, despite the fact that the entropy of

the working flid INCREASED.


Will the entropy of the working fluid in an ideal Carnot Cycle increase, decrease or remain the same during the isothermal heat addition process ?





3/24

Baratuci

HW #5


Read :

Answer : The entropy of the steam at the nozzle exit will be greater than the entropy at the nozzle inlet.

Real nozzles are not reversible. They are irreversible.

Irreversibilities cause entropy generation.

Because the process is adiabatic, there is no way to decrease the entropy.

Therefore, the entropy generation due to irreversibilities in the real nozzle produce an INCREASE

in the entropy of the steam as i t is accelerated through the nozzle.


Steam is accelerated as it flows through a real, adiabatic nozzle. Will the entropy of the steam at the nozzle exit be greater than, equal to or less

than the entropy at the nozzle inlet ?





4/24

Baratuci

HW #5

Problem : 7.28 - SSys, SRes, and SUniv, for a H.T. Process - 4 pts 11-May-11

a.) The amount of heat transfer

b.) The entropy change of the heat sink

c.) The total entropy change of the universe for this process.

Read :

Finally, we can evaluate Sunivbecause it is just the sum of Sfluidand Sres.

Given : Sfluid -0.7 Btu/oR TC 95

oF

554.67 oR

Find : a.) QC ??? Btu

b.) Ssink ??? Btu/oR c.) Suniv ??? Btu/

oR

Diagram :

Assumptions : 1 -

2 - The reservoir is internally reversible.

3 - The working fluid absorbs heat isothermally.

Equations / Data / Solve :

Part a.)

Eqn 1

Eqn 2

Eqn 3

Eqn 4

Now, we can plug numbers into Eqn 4: Q -388.27 Btu

QC 388.27 Btu


During the isothermal heat rejection process of a Carnot Cycle, the working fluid experiences an entropy change of-0.7 Btu/oR. If the temperature of

the heat sink is 95oF, determine

The heat transfer process is completely reversible because it is a step in a Carnot Cycle.

Let's begin by applying the entropy balance equation to the HEX in which the isothermal heat rejection step in the Carnot Cycle takesplace.

Because this HEX is part of a Carnot Cycle, it is completely reversible and so the entropy generation is ZERO. So, Eqn 1reduces to

the definition of S.

In part (a)the system is the HEX in which heat is rejected to the cold reservoir. Because this step is isothermal and reversible, Sgen=

0and we can evaluate Sfluidusing the definition of entropy.

Part (b)is very similar because the cold reservoir also undergoes a reversible process and remains at a constant temperature. So,

we can also evaluate Ssinkfrom the definition of entropy.

Now, because the heat transfer process is reversible, the temperature of the system at which the heat leaves the HEX must be the

same as the temperature of the reservoir, TC. As a result, Eqn 2can be simplified to :

Next, we can solve Eqn 3for QCand plug in values to complete this part of the problem.

The negative value for Qin association with our usual sign convention, indicates that heat is transferred out of the working fluid.

In terms of cycle or tie-fighter diagrams, all Qand Wvalues are positive

with their direction indicated by an arrow. Therefore :

fluid

QS

T

fluid

C

QS

T

HEX

TC = 95oF

QC

S1 S2

Sfluid=0.7Btu/oR

fluid gen ,int

QS S

T

C fluidQ T S

Dr. Baratuci - ENGR 224 hw5-sp11.xlsm, 7.28E 5/9/2011


5/24

Part b.)

Eqn 4

Eqn 5

Now, we can plug numbers into Eqn 5: Sres 0.7000 kJ/-K

Part c.)

Eqn 6

Plugging values into Eqn 6yields : Suniv 0.0000 kJ/-K

This makes sense since : Eqn 7

And, for a completely reversible process like this : Eqn 8

Verify : None of the assumptions made in this problem solution can be verified.

Answers : a.) QC 388 Btu

b.) Sres 0.700 kJ/-K c.) Suniv 0 kJ/-K

Now we can calculate Sunivdirectly for Sfluidand Sresusing :

Here we take our system to be the reservoir that provides heat to the Carnot Cycle.

The reservoir is internally reversible. Therefore, Eqn 1becomes :

Now, because the heat transfer process is reversible and the temperature of the reservoir, TH, is constant, Eqn 4simplifies to :

res

QS

T

Cres

C

QS

T

univ fluid resS S S

univ gen ,totalS S

gen,totalS 0

Dr. Baratuci - ENGR 224 hw5-sp11.xlsm, 7.28E 5/9/2011


6/24

Baratuci

HW #5

Problem : 7.29 - Entropy Change in the Evaporator of a Refrigerator - 6 pts 11-May-11

a.) The entropy change of the refrigerant.

b.) The entropy change of the refrigerated space.

c.) The entropy change of the universe for the process.

Read :

Given : P 200 kPa QC 120 kJ

x2 1 kg vap/kg TC -5.0 deg C

TC 268.15 K

Find : a.) SR-134a ??? kJ/K

b.) Sres ??? kJ/K

c.) Suniv ??? kJ/K

Diagram :

Assumptions :

1 - The pressure everywhere in the evaporator is 200 kPa so that P2= P1.

2 -

3 -


Parts a & b)

Eqn 1

Sres -0.4475 kJ/K

Tsat -10.076 deg C

263.07 K SR-134a 0.4561 kJ/K

Part c.)

Eqn 2 Suniv 0.0086 kJ/K


Answers : a.) SR-134a 0.4561 kJ/K c.) Suniv 0.0086 kJ/K

b.) Sres -0.4475 kJ/K

The entropy change of the universe is just the sum of the entropy change for the refrigerated space and the R-134a.

The temperature of the R-134a in the evaporator is always equal to the saturation temperature at 200 kPa. As a

result, the R-134a also behaves as a true thermal reservoir

Because both the refrigerated space and the R-134a behave as true thermal reservoirs in this process, we can calculate the

change in entropy for each from the definition of entropy.

We can immediately apply Eqn 1to the refrigerated space because we know its temperature.

Just keep in mind that from the perspective of the refrigerated space Q = -120 kJ because heat is being transferred out of that

system (the refrigerated space).

We can look up the saturation temperature of the R-134a in the saturation pressure table for R-134a and then plug values into

Eqn 1.


R-134aenters the coils of the evaporator of a refrigeration system as a saturated vapor-liquid mixture at a pressure of 200 kPa. The

refrigerant absorbs 120 kJof heat from the cooled space, which is maintained at -5oC, and leaves the evaporator as a saturated vapor at the

same pressure. Determine...

The key here is that the temperature of the R-134a does not change in the evaporator, so it behaves as a true thermal

reservoir and we can determine the change in entropy from the definition of entropy.

The temperature of the refrigerated space remains constant throughout the process so that it behaves as a true

thermal reservoir.

Evaporator

Cold ReservoirTC = -5

oC

QC = 120 kJ

P1 = 200 kPa0 < x1 < 1

T2 = Tsat

P2 = 200 kPax2 < 1

T2 = Tsat

IntRev

Q QST T

univ res R 134aS S S



7/24

Baratuci

HW #5

Problem : 7.66 - SUnivUpon Quenching an Iron Block - 6 pts 11-May-11

Data : CP,H2O 4.18 kJ/kg-K CP,Fe 0.45 kJ/kg-K

Read :

The problem is simplified by the fact that the heat capacites are constant.

Given : T1,Fe 350oC mFe 12 kg

623.15 K mH2O 100 kg

T1,H2O 22oC

295.15 K

Find : Suniv ??? kJ/K

Diagram :

Assumptions :

1 - All of the water is in the liquid phase in the f inal state.

2 - The heat capacities of iron and water are constants.

3 -

4 -


The universe, in this problem, is made up of the water and the iron block. Therefore :

Eqn 1

Eqn 2 Eqn 3

Because the iron and water are incompressible, CP= CV= C.

Eqn 4

Eqn 5

Eqn 6

This problem requires the application of the 1st Law to determine T2and then one of the Gibbs equations to evaluate S.

The process is adiabatic. All the heat leaving the iron enters and stays in the water. No heat or work is

exchanged with the surroundings.


We can use Gibbs 2nd Equation for incompressible substances with constant heat capacities to

evaluate SFeand SH2O.

A 12 kgiron block initially at 350oCis quenched in an insulated tank that contains 100 kgof water at 22

oC. Assuming the water that

vaporizes during the process condenses back into the liquid phase inside the tank, determine the entropy change of the universefor this

process.

Since all of the heat leaving

the iron goes into the water:

Both the water and the iron are perfectly incompressible. The volume of each remains constant throughout this

process.

So, all we need to do in order to use Eqns 3 & 1to complete this problem is to determine T2, the final, equilibrium temperture

fo both the water and the iron.

Let's begin by applying the 1st Law for clases sytems with negligible changes in kinetic and potential energies to the iron and

the water separately.

If we assume that the volume of the water does not change in the process and that the volume of the iron does not change in

the process, then there is no boundary work in the process.

IronmFe = 12 kgT1 = 350

oC

T2 = ??o

C

mH2O = 100 kgT1,H2O = 22

oCT2 = ??

oC

Q

Fe b,Fe Fe FeQ W m U

2 2 2 2H O b,H O H O H OQ W m U

2H O FeQ Q

2univ Fe H OS S S

CdS dT

T

2

1

T S C LnT



8/24

Eqn 7

Eqn 8

Eqn 9

Eqn 10

Plugging values into Eqn 7yields :

mFe*CFe 5.4 kJ/K T2 26.18oC

mH2O*CH2O 418 kJ/K T2 299.33 K

SFe -3.9594 kJ/K

SH2O 5.8829 kJ/K Suniv 1.9234 kJ/K


Answers : Suniv 1.923 kJ/K

Combining Eqns 4 - 6yields :

Now we can use Eqn 3for the water and then for the iron to evaluate the entropy change for each. And then plug these values

into Eqn 1to complete the problem.

Because the heat capacites of both iron and water are assumed to be constant, we can express the change in internal energy

in terms of the heat capacities as :

So, we can replace CVwith Cand combine Eqns 7 & 8to obtain :Because the iron and water are incompressible, CP= CV= C.

Notice that at the final state, the water and the iron are at the same temperature, T2.

Now, we need to solve Eqn 9for the unknown T2.

2 2 2H O 2,H O 1,H O Fe 1,Fe 2,Fe U U U U U U

2

1

T

V V 2 1

T

U C dT C T T

2 2 2H O H O 2 1,H O Fe F e 1,Fe 2 m C T T m C T T

2 2 2

2 2

H O H O 1,H O Fe Fe 1,Fe

2

H O H O Fe Fe

m C T m C TT

m C m C



9/24

Baratuci

HW #5

Problem : WB-1 - Efficiency of an Int. Rev. HE with Multiple Heat Transfers - 4 pts 11-May-11

Read :

Given : Q1 750 kJ Q2 100 kJ

T1 1500 K T2 500 K

T3 1000 K

Find : th ???

Diagram :

Assumptions :

1 - The power cycle is internally reversible.

2 -

3 - Heat is exchanged only with the three reservoirs mentioned in the problem statement.


A system executes a power cycle while receiving 750 kJ by heat transfer at its boundary where the temperature is 1500 K and discharges 100 kJ

by heat transfer at another portion of its boundary where its temperature is 500 K. Another heat transfer from the system occurs at a portion of the

system boundary where the temperature of the system is 1000 K. No other heat transfer crosses the boundary of the system. If no internal

irreversibilities are present, determine the thermal efficiency of the power cycle.

The key to this problem is that the cycle is internally reversible and that the temperatures within the system at which these heat

transfers occur must remain constant. This allows us to use an entropy balance and the 2nd Law to determine the amount of heat

rejected to reservoir 3. The 1st Law then allows to evaluate Wcycle. We can then calculate th from its definition.

Heat transfers can be irreversible, but the temperatures within the system at which the heat trasfers occur must remain

constant at the given values.

HER

Tres,3

Q3

Wcycle

Tres,1

Q1

Tres,2

T3

T2

T1

Q2

Dr. Baratuci - ENGR 224 hw5-sp11.xlsm, WB-1 5/9/2011


10/24


Let's begin with the definition of a thermal efficiency of a power cycle :

Eqn 1

Next, we need to apply the 1st Law to the cycle to help us determine Wcycle:

Eqn 2

Now, all we need to do is determine Q2and then we can use Eqns 1 & 2 to evaluate th .

The key to determining Q2is the fact that the power cycle in INTERNALLY reversible.

This means that the entropy generation within the cycle is ZERO.

Write the entropy balance equation for the cycle :

Eqn 3

Evaluating the cyclic integral in terms of the Q's and T's in this cycle gives us :

Eqn 4

Now, we can solve Eqn 4for Q3:

Eqn 5

Now, plug values into Eqns 5, 2 & 3 : Q3 300 kJ

Wcycle 350 kJ

th 0.4667


Answers : th 46.7%

Notice that because Q1, Q2and Q3are based on the tie-fighter diagram, they are all positive numbers. Their direction is represented

by the arrows in the tie-fighter diagram, above. Because Q2and Q3represent heat transfers out of the system, they become negative

when we drop the sign convention as we move from Eqn 3to Eqn 4.

cycle 1 2 3W Q Q Q

cycle gen,int

QS S 0

T

31 2

1 2 3

QQ Q0

T T T

1 23 3

1 2

Q QQ T

T T

cycle cycle

th

in 1

W W

Q Q



11/24

Baratuci

HW #5

Problem : WB-2 - Efficiency and Tresfor Rev. and Irrev. Cycles - 6 pts 11-May-11

Complete the following involving reversible and irreversible cycles.

a.)

b.)

Read :

Given : See diagrams, below.

Find : a.) Show that T'H > TH.

b.) Show that T'C > TC.

Diagram : See below.

Assumptions :1 - Both heat engines operate on thermodynamic cycles.

2 - Changes in kinetic and potential energies are negligible.

3 - The mixer is adiabatic.

4 - Flow work is the only form of work that crosses the system boundary.


Part a.) Diagram :

The trick to this problem is to VERY CAREFULLY choose your system !

However, in the irreversible cycle, the temperature within the system MAY NOT BE EQUAL to the temperature of the reservoir that isgiving or receiving the heat !

The solution is to choose your systems to be LARGER, so that at the boundary of this larger system, T = Treservoir! See the diagram,

above.

The easiest approach to this problem is to carefully define your systems so that ALL of the irreversibilities lie inside the system

boundary.

We accomplish this by choosing the system boundary to be bigger than the HE's and Ref's. The boundaries are so big, in fact, that

they TOUCH the reservoirs. This permits us to assume that the temperature of this extended system, at which the heat enters or

leaves the extended system, is EQUAL TO THE RESERVOIR T. By defining our systems in this manner, we have made ALL of the

irreversibilities INTERNAL. But the important goal of choosing these larger systems is that, regardless of whether the cycle is

reversible or not, the temperature at which heat transfer occurs is known and constant.

If you choose your system boundary to be snug on each cycle, you run into a problem on the irreversible cycle. Here is why.

The completely reversible cycle is no problem. The heat transfers must be reversible, so the temperatures within the system and the

temperatures of the reservoirs must be equal.


Reversible and irreversible power cycles each discharge QCto a cold reservoir at TC and receive energy QHfrom hot reservoirs at TH

and T'H, respectively. There are no other heat transfers involved. Show that T'H> TH.

Reversible and irreversible refrigeration cycles each discharge QHto a hot reservoir at THand receive energy QCfrom cold reservoirs

at TC and T'C, respectively. There are no other heat transfers involved. Show that T'C> TC.

TC

QC

HEI

WRHER

TH

WI

QH QH

QC

T'H



12/24

Let's apply the entropy balance equation to the larger, extended systems surrounding each HE.

Eqn 1

Eqn 2

The entropy generation for the completely reversible cycle (in Eqn 1) is ZERO.

Now, express Eqns 1 & 2 in terms of QH and QC.

Eqn 3

Eqn 4

For an irreversible process or cycle, the 2nd Law tells us that :

Eqn 5

Solve Eqn 4for Sgen,int,I : Eqn 6

Now, solve Eqn 3for QC/ TC and use the result to eliminate this term from Eqn 6.

Eqn 7

Manipulate Eqn 8to obtain a relationship between THand T'H:

Eqn 8 Eqn 9

Part b.) Diagram :

Notice that because QHand QCare based on the tie-fighter diagrams, they are positive numbers. Their direction is represented by

the arrows in the tie-fighter diagram, above. Because QC represents heat transfer out of the system, it becomes negative when we

drop the sign convention as we move from Eqns 1 & 2 to Eqns 3 & 4.

This part of the problem is very similar to part (a). The key is still to choose your systems wisely.

I gen,int,I

QS S 0

T

R gen ,int,R

QS S 0

T

CH

H C

QQ0

T T

CH

gen,int,I

H C

QQS 0

T T

gen,int,IS 0

C H

gen,int,I

C H

Q QS 0

T T

C H

C H

Q Q

T T

H H

1 1T T

H HT T

TC

QC

RefI

WRRefR

TH

WI

QH QH

QC

T'C



13/24

Eqn 10

Eqn 11

Eqn 12

Eqn 13

For an irreversible process or cycle, the 2nd Law tells us that :

Eqn 14

Solve Eqn 13for Sgen,int,I : Eqn 6

Eqn 7

Manipulate Eqn 8to obtain a relationship between TCand T'C:

Eqn 8 Eqn 9


Answers : a.) b.)

The entropy generation for the completely reversible cycle (in Eqn 1) is ZERO.

Now, express Eqns 1 & 2 in terms of QH and QC.

Notice that because QHand QCare based on the tie-fighter diagrams, they are positive numbers. Their direction is represented by

the arrows in the tie-fighter diagram, above. Because QH represents heat transfer out of the system, it becomes negative when we

drop the sign convention as we move from Eqns 10 & 11 to Eqns 12 & 13.

Now, solve Eqn 12for QC/ TC and use the result to eliminate this term from Eqn 6.

Apply the entropy balance equation to the larger, extended systems surrounding each refrigerator.

I gen,int,I

QS S 0

T

R gen ,int,R

QS S 0

T

C H

C H

Q Q0

T T

C H

gen,int,I

C H

Q QS 0

T T

gen,int,IS 0

CH

gen,int,I

H C

QQS 0

T T

C H

C H

Q Q

T T

C C

1 1

T T

C CT T

H HT T

C CT T



14/24

Baratuci

HW #5

Problem : WB-3 - Specific Entropy Change Using Tabluar Data - 4 pts 11-May-11

a.) Water: P1= 10 MPa, T1= 400oC andP2 = 10 MPa, T2 = 100

oC. ( Use the NIST Webbook)

b.) R-134a: H = 211.44 kJ/kg, T = - 40oC and P = 5 bar , x = 1.0.

c.) Air (IG): T1= 27oC, P1= 2 bar andT2= 327

oC, P2 = 1 bar.

d.) Hydrogen (H2, IG): T1= 727oC, P1= 1 bar and T2 = 25

oC, P2 = 3 bar.

Read :

Given : a.) Water

P1 10000 kPa P2 10000 kPa

T1 400oC T2 100

oC

b.) R-134a

H1 211.44 kJ/kg P2 500 kPa

T1 -40oC x2 1 kg vap/kg

c.) Air

P1 200 kPa P2 100 kPa

T1 27oC T2 327

oC

R 8.314 J/mole-K MW 28.97 g/mole

d.) H2P1 100 kPa P2 300 kPa

T1 727oC T2 25

oC

R 8.314 J/mole-K MW 2.016 g/mole

Find : S ??? kJ/kg-K for each part of the problem.

Assumptions : 1 -

2 -

Diagram: None needed.


Part a.)

I used the isothermal properties table with the default reference state to obtain :

S1 6.2141 kJ/kg-K

S2 1.2996 kJ/kg-K S -4.9145 kJ/kg-K

Part b.)

Here is the relevant data :

Sat. Liq Sat. Vap Sat. Liq Sat. Vap

51.209 -40.00 148.14 374.00 0.79561 1.7643

Since :


(Use the NIST Webbookwi th the default reference state: U = 0 and S = 0 for saturated liqu id at 0.01oC)

Pressure(kPa)

S (kJ/kg-K)

Determine the change in specific entropy in kJ/kg-K for :

H (kJ/kg)Temp.(oC)

Parts (a)and (b) of this problem are exercises in the use of thermodynamic property tables to evaluate S.

Parts (c)and (d) are exercises in the use of the ideal gas property tables and Gibbs' 2nd Equation to evaluate Sfor ideal

gases.

For parts (c)and (d), assume that airand H2behave as ideal gases, as instructed in the problem

statement.

For parts (c)and (d), assume that airand H2have variable heat capacities and use the Ideal Gas

Property Tables.

Here we can use the NIST Webbook to evaluate the specific entropy of water and then directly calculate the change in specific

entropy.

The first step here is to use T1to obtain data from the Saturation properti es temperature increments option.

This will allow us to compare the given value of Hto the values of Hsat liq and Hsat vap in order to determine the phase or phases

present in the system at equilibrium.

the system contains a saturated mixture and the temperature must be equal to the

saturation temperature.sat satliq vap

H H H



15/24

Eqn 1

Plugging values intoEqn 1 yields : x 1 0.2803 kg vap/kg

Eqn 2

S1 1.0671 kJ/kg-K

S2 1.7197 kJ/kg-K 0.6526 kJ/kg-K

Part c.)

Eqn 3

So

10.0061681 kJ/kg-K S

o2

0.71926 kJ/kg-K

Now, we can plug values into Eqn 3to get : S 0.91202 kJ/kg-K

Part d.) The same procedure is used in part (d)as in part (c).S

o1

17.627 kJ/kg-K So

20 kJ/kg-K

Now, we can plug values into Eqn 3to get : -22.158 kJ/kg-K

Verify : a.) No assumptions. b.) No assumptions.

c.)

Eqn 4 Eqn 5

V1 12.48 L/mole V2 49.90 L/mole

d.)

V1 83.15 L/mole V2 8.26 L/mole

Answers : a.) S -4.915 kJ/kg-K c.) S 0.912 kJ/kg-K

b.) 0.653 kJ/kg-K d.) -22.158 kJ/kg-K

Last, we must use the given value of the enthalpy to determine the quality of the water in the system. The key equation is:

We can use the Ideal Gas EOS to verify that the molar volume of the air in both states 1 and 2 is greater than 5L/mole (diatomic gas) using Eqn 5.

Since both V1and V2are greater than 5 L/mole, the ideal gas assumption is valid.

Because state 2 is a saturated vapor at 500 kPa, we can get S2directly from the saturation pressure table.

Gibbs's 2nd equation is most useful in this case because the initial and final pressures are given.

We can immediately lookup So

2and So

1 in the Ideal Gas Property Table for air because we know T1and T2.

We can use the Ideal Gas EOS to verify that the molar volume of the air in both states 1 and 2 is greater than 5

L/mole (diatomic gas) using Eqn 5.

Since both V1and V2are greater than 5 L/mole, the ideal gas assumption is valid.

sat sat satmix vap liq

S x S 1 x S

1 sat liq

1

sat vap sat liq

H Hx

H H

o o 2

2 1

1

PR S S S LnMW P

P V R TR T

VP



16/24

Baratuci

HW #5

Problem : WB-4 - "Show That" for a Cycle Interacting w ith Three Reservoirs - 4 pts 11-May-11

Read :

Given :

Find :

Diagram : See the problem statement.

Assumptions : 1 - The system operates on a thermodynamic cycle.

2 -

3 -


Eqn 1

The change in any property, including entropy, for a cycle is zero.

Eqn 2

We can now use the 1st Law to eliminate Qsurrfrom Eqn 2.

Eqn 3

Eqn 4

Now, we can solve Eqn 4for QC.

Eqn 5

The system only exchanges heat with the three reservoirs shown in the diagram.

Let's begin by appplying the entropy generation form of the 2nd Law to the system shown in the diagram.

We can now expand the cyclic integral in terms of the three heat transfers that actually occur during the cycle.


The system shown in the figure undergoes a cycle while receiving

energy at the rate Qsurrfrom the surroundings at temperature Tsurr,

receiving QHfrom a heat source at temperature TH, and rejecting QCto

a thermal reservoir at TC. Derive an expression for the maximum

theoretical value of QC in terms of QHand T0, TSand TUonly. There is

no work produced by this system. This is an absorption heat pump

system. The heat source might be a propane flame and the heat sink

might be the air inside your camper. So, it is important to note that TS

> TU> Tsurr. Hint: TH, Tsurrand TCare not important in the solution of

this problem !

The key here is to recognize that TH, TC and Tsurrdo not directly effect the solution of this problem and then determine how

entropy generation affects the heat output of this heat pump, QC. Because Sgen> 0we will be able to determine a maximum

value for QC as a function of QHand the system temperatures at which heat is transferred.

cycle gen,intQS S 0T

surr CH

gen,int

S 0 U

Q QQS 0

T T T

S U 0T T T

C H S U 0Q fxn Q , T , T ,T

S U 0T T T

surr C HQ Q Q

C H CH

gen,int

S 0 U

Q Q QQ S 0T T T

C H gen ,int

0 U 0 S

1 1 1 1Q Q S

T T T T



17/24

Because :

Eqn 6 Eqn 7

So, for a reversible process, Eqn 5becomes :

Eqn 8


Answers :

Therefore, because Sgen> 0, QCis maximized when Sgen,int= 0. That is to say this absorption heat pump will deliver the most

heat to the heated space, per Joule of heat supplied from the heat source, if it is reversible. This should come as no surprise.

S U 0T T T

0 U

1 10

T T

0 S

1 10

T T

0

0 S S

C,max H H0

0 U U

T1 11

T T TQ Q Q

1 1 T1

T T T

0

S

C,max H0

U

T1

TQ Q

T1

T



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Baratuci

HW #5

Problem : WB-5 - Three-Step, Ideal Gas Cycle Analysis - 8 pts 11-May-11

Step 1 - 2 : Isothermal expansion at 350 Kfrom 4.75 barto 1.0 bar.

Step 2 - 3 : Adiabatic compression to 4.75 bar.

Step 3 - 1 : Isobaric cooling.

a.) Sketch the cycle on a PVdiagram.b.) Sketch the cycle on a TSdiagram.

c.) Determine T3in Kelvin

d.)

Read : In part (c), you can use the 2nd Gibbs Equation and the Ideal Gas Entropy Table for air to determine So

3and then T3.

Given : T1= T2 350 K R 8.314 J/mol-K

P1

475 kPa MW 28.97 g/mole

P2 100 kPa

Find : a.) PV diagram c.) T3 ??? K

b.) TS diagram. d.) COPRor ???

Diagram : See the solutions to parts (a)and (b)below.

Assumptions :

1 - The air is contained in a closed system.

2 - Air is an ideal gas in all three states. This assumption can be verified.

3 - All three steps in the cycle are internally reversible.


Part a.) Part b.)


A quantity of air undergoes a thermodynamic cycle consisting of three internally reversibleprocesses in series. Assume that the air

behaves as an ideal gas. This may not be a good assumption, but let's work with it here anyway.

If the cycle is a power cycle, determine its thermal efficiency. If the cycle is a refrigeration cycle, determine its COP.

In part (d), we cannot use the reservoir temperatures alone to determine COPR because the cycle is only internally reversible.

It is not completely reversible.

We can determine Q12= QCfrom an entropy balance, taking advantage of the fact that there is no internal entropy generation

because the cycle isinternally reversible.

We can evaluate W31from the integra of P dVand then apply the 1st Law to determine Q31= QH.

Finish up by plugging QCand QHinto the definition of COPR.

P

V

1

2

3

I s o t h e r m a l

Adiabatic

T1

T2

Isobaric

T

S

1 2

3

Isothermal

Isentropic

P1

P2Isobaric



19/24

Part c.)

Eqn 1

So2= S

o1 1.85708 kJ/kg-K

Eqn 2

Plugging values into Eqn 2yields : So3 2.30425 kJ/kg-K

T (K)

540 2.29906

542.73 2.30425

550 2.31809 T3 542.73 K

Part d.)

So, our goal is to determine the value of COPR.

We cannot determine COPRusing : Eqn 3

There are two reasons.

Instead, we can determine COPRfrom its definition.

Eqn 4

Eqn 5

We can solve Eqn 5for Q12: Eqn 6

The key to determining T3is to recognize that since step 2-3 is both adiabatic and internally reversible, it is also isentropic.

The entropy change of an ideal gas with variable heat capacities is given by :

Sois a function of Tonly, so we can lookup S

o1 in the Ideal Gas Property Table for air.

Since Sois a function of Tonly, we can solve Eqn 1for S

o2and then interpolate on the Ideal Gas Property Table for air to

determine T2.

First we must abserve that this cycle is a refrigeration cycle. There are many ways to reach this conclusion, but the easiest is

to observe that the cycle progresses in a counter-clockwise direction on the PVand TSdiagrams.

1- This cycle is only internally reversible, not completely reversible (like a Carnot Cycle).

2- If this cycle were to be completely reversible, what would be the temperature of the hot reservoir ? It would need to change

during step 3-1 from T3to T1. Which Twould you use ? If you simply use T3, then the heat transfer in step 3-1 is not

reversible and Eqn 3does not apply.

Note : Because of our use of the sign convention that heat transfer is positive when heat is transferred into the system,

QC= Q12and QH= - Q31.

Let's begin by determining Q12. Because this step is internally reversible and isothermal, we can use :

Interpolation on the Ideal

Gas Property Table for air :

o o 3

3 2

2

PR S S LnMW P

C CR

cycle H C H C

Q Q 1COP

W Q Q Q / Q 1

R

H C

1COP

T / T 1

12

12 gen ,int

1

QQ S ST T

12 1 12 Q T S

o o 3

3 2

2

PR S S S LnMW P



20/24

Eqn 7

Plugging values into Eqns 8 & 6yields : S12

0.44717 kJ/kg-K

Q12= QC 156.51 kJ/kg

We can determine QHby applying the 1st Law for closed systems.

Eqn 9

Eqn 10

Combining Eqns 9 & 10and making use of the definition of enthalpy yields :

Eqn 11

Ho1

350.49 kJ/kg

T (K) H

540 544.35

Ho

3 547.45

550 555.74 Ho3

547.45 kJ/kg

Q31 = -QH -196.96 kJ/kg COP 3.869

Verify :

Eqn 12 Eqn 13

V1 6.13 L/mole

V2 29.10 L/mole V3 9.50 L/mole

Answers : a.) See the PV diagram, above. c.) T3 542.7 K

b.) See the TS diagram, above. d.) COP 3.87

Eqn 8

Because T1= T2, So

2= So

1and

Eqn 7simplifies to :

Since all three molar volumes are greater than 5 L/mole, the ideal gas assumption is valid to about 2 significant figures.

Now, we can apply Eqn 1to step 1-2 to evaluate S12.

Because we know T1and T3and both states are assumed to be ideal gas states, we can lookup H1and H3in the Ideal Gas

Property Table for air. Ho

3requires interpolation.

Step 3-1 is internally reversible and also isobaric, so the boundary work can be determined as

follows :

Now, we can plug values into Eqn 11to evaluate Q31and then plug Q12and Q31into Eqn 4to evaluate COPR.

We can use the Ideal Gas EOS to verify that the molar volume of the air in all three states is greater than 5 L/mole(diatomicgas).

o o 2

12 2 1

1

PR S S S LnMW P

2

12

1

PRS LnMW P

31 31 31 Q W U

1

3

V

31

V

W P dV P V PV

31 31 31 31 31 1 331 Q U W U PV H H H

P V R TR T

VP



21/24

Baratuci

HW #5

Problem : WB-6 - Maximum Work From an Adiabatic Turbine - 5 pts 11-May-11

Read :

Given :P

1800 psia P

240 psia

T1 900oF

Find : Wsh,max ??? Btu/lbm

Diagram :

Assumptions :

1 - The turbine is adiabatic.

2 - Changes in kinetic and potentai l energies are negligible.


Eqn 1

Because the turbine is adiabatic, Eqn 1becomes : Eqn 2

At P1= 800 psia: Tsat 518.27oF

H1 1456.0 Btu/lbm

S1= S2 1.6413 Btu/lbm-oR

The first step here is to use P2to obtain data from the saturation pressure table.

Here is the relevant data :

Sat. Liq Sat. Vap Sat. Liq Sat. Vap

40 267.22 236.15 1169.80 0.39214 1.6766

Since :

Because we know both T1and P1, we can immediately lookup H1in the steam tables.

Pressure

(psia)

emp.

(oF)

H (Btu/lbm) S (Btu/lbm-oR)


Steam enters an adiabatic turbine at 800 psiaand900oFand leaves at a pressure of 40 psia. Determine the maximum amount of work that

can be delivered by this turbine.

The key to this problem is the fact that the maximum work will be developed by a reversible turbine. A turbine that is both

reversible and adiabatic is isentropic. Because S2= S1, we can use P2and S2to determine H2. Then, we can apply the 1st

Law to evaluate the work output from the isentropic turbine.

Let's begin by applying the 1st Law for open systems to the turbine, assuming changes in kinetic and potential energies are

negligible.

We can immediately lookup S1because we know P1and T1.

Now, knowing P2and S2, we can determine H2. Let's begin by determining what phases are present.

This will allow us to compare the given value of S2to the values of Ssat liqand Ssat vapin order to determine the phase or

phases present in the system at equilibrium.


saturation temperature.

Since T1> Tsat, the feed is superheated steam.

It is a bit harder to determine H2because we only know P2, not T2.

But we do have one more piece of information that we can use.

The maximum shaft work will be produced by an internally reversible turbine.

A device that is both adiabatic and internally reversible is also isentropic: S2= S1.

Turbine

P1 = 800 psiaT1 = 900

oF

P2 = 40 psiaT2 = ???

oF

Wsh = ??? kJ/kg

sh Q W H

sh 1 2 W H H

sat 2 satliq vap

S S S



22/24

Eqn 3

Plugging values intoEqn 3yields : x2 0.9725 lbmvap/lbm

Eqn 4

H2 1144.1 Btu/lbm

Finally, we can plug values back into Eqn 2to evaluate the maximum shaft work for the turbine.

Wsh,max 311.9 Btu/lbm


Answers : Wsh 312 Btu/lbm

Next, we must use S2to determine the quality of the water in the system. The key equation is:

sat sat satmix vap liq

S x S 1 x S

2 sat liq

2

sat vap sat liq

S Sx

S S



23/24

Baratuci

HW #5

Problem : WB-7 - S for Heat Transfer to R-134a in a Rigid Tank - 6 pts 11-May-11

a.) The entropy change of the R-134a.

b.) The entropy change of the heat source.

c.) The entropy change of the universe for this process.

Read :

Given : V 0.5 m P2 400 kPa

P1 200 kPa Tsource 35oC

x1 0.4 kg vap/kg 308.15 K

Find : a.) SR-134a ??? kJ/K

b.) Sres ??? kJ/K c.) Suniv ??? kJ/K

Diagram :

Assumptions :

1 - Changes in kinet ic and potent ial energies are negligible.

2 -

3 -


Part a.)

Eqn 1

Let's begin by determining the mass of R-134a in the tank using :

Eqn 2

Sat. Liq Sat. Vap Sat. Liq Sat. Vap Sat. Liq Sat. Vap

200 -10.090 0.0007533 0.099867 38.28 224.48 0.15457 0.93773

We can now use the quality to determine V1, U1, and S1using :

Eqn 3

V1 0.04040 m /kg

U1 112.76 kJ/kg S1 0.4678 kJ/kg-K

Plugging values into Eqn 2yields : m 12.377 kg

V2= V1 0.04040 m /kg


A 0.5 m3rigid tank contains R-134ainitially at 200 kPaand 40% quality. Heat is transferred to the refrigerant from a source at 35

oC until the

pressure rises to 400 kPa. Determine

Pressure

(kPa)

The keys to this problem are that the specific volume is the same in states 1 and 2 and that the heat source can be treated as

a true thermal reservoir (its Scan be determined from the definition of entropy).

The temperature of the heat source is constant and so it acts as a true thermal reservoir.

We can determine the change in the specific entropy directly by determining the initial and final specific entropy of the R-134a

in the tank.

The tank is a closed system. The mass of R-134ain the tank is constant.

Then, we can multiply by the mass of R-134ain the tank to get SR-134a.

While we are looking up V1in the saturation pressure table we might as well lookup all the other relevant properties.

emp.

(oC)

V (m3/kg) U (kJ/kg) S (kJ/kg-K)

Now, we need S2. We can get it because we know P2and we know that the volume of the tank and the mass of R-134ain the

tank are both constant in this process. So:

TankP1 = 200 kPa

T1 = Tsatx1 = 0.4 kg vap/kg

Q = ?? kJTank

P2 = 400 kPaT2 = ??

oCx2 = ?? kg vap/kg

HeatSource

35oC

tank

1

Vm

V

1 1 sat 1 satvap liq

M x M 1 x M

R 134a 2 1 S m S S



24/24

Sat. Liq Sat. Vap Sat. Liq Sat. Vap Sat. Liq Sat. Vap

400 8.910 0.0007907 0.051201 63.62 235.07 0.24761 0.92691

Since :

Eqn 4

Plugging values intoEqn 4yields : x2 0.7857 kg vap/kg

We can now use the quality to determine U1, and S1using Eqn 3.

U2 198.33 kJ/kg S2 0.7813 kJ/kg-K

Finally, plugging values into Eqn 1yields :SR-134a 3.8802 kJ/k

Part b.)

Eqn 5

Eqn 6

The volume of the tank is constant, so there is no boundary work and Eqn 6becomes:

Eqn 7

Q 1059.07 kJ Ssource -3.4369 kJ/k

Part c.)

Eqn 2 Suniv 0.4433 kJ/K


Answers : SR-134a 3.880 kJ/k

Ssource -3.437 kJ/k

Suniv 0.4433 kJ/k

S (kJ/kg-K)

Next, we must use V2to determine the quality of the water in the system. The key equation is:

Since we calculated U1and U2in part (a), all we need to do is plug them into Eqn 7to determine Q, and put Qback into Eqn 5

to get Sres .

The entropy change of the universe is just the sum of the entropy change for the refrigerated space and the R-134a.

Apply the 1st Law for closed systems with negligible changes in kinetic and potential energies to the tank.

Next, we determine which phase or phases are present in state 2 by comparing V2to Vsat liqand Vsat vapat 400 kPa.


saturation temperature.

We can determine the entropy change of the heat source from the definition of entropy because we assume the source

behaves as a true thermal reservoir.

I put the minus sign in front of Qin this equation because we will apply the 1st Law to the tank to determine Q. So, Qis the

amount of heat transferred out of the source and with our sign convention, -Qis the heat transferred when we apply Eqn 1to

the heat source.

Pressure

(kPa)

emp.

(oC)

V (m3/kg) U (kJ/kg)

sat 2 satliq vap

V V V

2 sat liq

2

sat vap sat liq

V Vx

V V

res

IntRev

Q QS

T T

bQ W m U

2 1 Q m U U

univ res R 134aS S S

Thermo Problems 2

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