Thermo Problems 2

ENGR 224 - Thermodynamics Problem :

7.14 - The Increase of Entropy Principle - 2 pts

Baratuci HW #5 11-May-11

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

The usual problem solving procedure does not really apply th this "problem" because it is really a short-answer question.

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. 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




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

Read :


Answer :

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


hw5-sp11.xlsm, 7.15

5/9/2011




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 ? 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 it is accelerated through the nozzle.


hw5-sp11.xlsm, 7.18

5/9/2011

ENGR 224 - Thermodynamics


7.28 - ΔSSys, ΔSRes, and ΔSUniv, for a H.T. Process - 4 pts

Problem :

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 … a.)

The amount of heat transfer

b.)

The entropy change of the heat sink The total entropy change of the universe for this process.

c.) Read :

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 = 0 and we can evaluate Sfluid using 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 Ssink from the definition of entropy. Finally, we can evaluate Suniv because it is just the sum of Sfluid and Sres.

Given : Find :

Sfluid a.) b.)

Btu/oR

-0.7 QC Ssink

TC

??? ???

Btu Btu/oR

c.)

Suniv

95 554.67 ???

o

F R

o

Btu/oR

Diagram :

TC = 95oF QC S1

S2

HEX

Sfluid = ‐0.7 Btu/oR Assumptions :

123-

The heat transfer process is completely reversible because it is a step in a Carnot Cycle. The reservoir is internally reversible. The working fluid absorbs heat isothermally.

Equations / Data / Solve : Part a.)

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

 S fluid 



Q  S gen,int T

Eqn 1

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

 Sfluid  

Q T

Eqn 2

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 2 can be simplified to :

 Sfluid 

Q TC

Eqn 3

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

Q  TC  Sfluid

Eqn 4

Now, we can plug numbers into Eqn 4 :

Q

-388.27

Btu

The negative value for Q in 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 Q and W values are positive with their direction indicated by an arrow. Therefore :


hw5-sp11.xlsm, 7.28E

QC

388.27

Btu

5/9/2011

Part b.)

Here we take our system to be the reservoir that provides heat to the Carnot Cycle. The reservoir is internally reversible. Therefore, Eqn 1 becomes :

 Sres  

Q T

Eqn 4

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

 S res 

QC TC

Eqn 5

Now, we can plug numbers into Eqn 5 : Part c.)

Sres

0.7000

kJ/-K

Now we can calculate Suniv directly for Sfluid and Sres using :

 Suniv   Sfluid   Sres

Eqn 6

Plugging values into Eqn 6 yields :

Suniv

0.0000

kJ/-K

This makes sense since :

 S univ  S gen,total

Eqn 7

And, for a completely reversible process like this :

S gen,total  0

Eqn 8

Verify :

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

Answers :

a.)

QC

b.)

Sres


388 0.700

Btu kJ/-K

c.)

hw5-sp11.xlsm, 7.28E

Suniv

0

kJ/-K

5/9/2011



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

R-134a enters 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 kJ of heat from the cooled space, which is maintained at -5oC, and leaves the evaporator as a saturated vapor at the same pressure. Determine... a.) b.) c.)

The entropy change of the refrigerant. The entropy change of the refrigerated space. The entropy change of the universe for the process.

Read :

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.

Given :

P x2

Find :

a.) b.) c.)

Diagram :

200 1

QC TC TC

kPa kg vap/kg

SR-134a Sres Suniv

??? ??? ???

1

120 -5.0 268.15

kJ deg C K

kJ/K kJ/K kJ/K 2

Evaporator

P1 = 200 kPa 0 < x1 < 1 T2 = Tsat

P2 = 200 kPa x2 < 1 T2 = Tsat

QC = 120 kJ Cold Reservoir TC = -5 oC

Assumptions : 123-

The pressure everywhere in the evaporator is 200 kPa so that P2 = P1. The temperature of the refrigerated space remains constant throughout the process so that it behaves as a true thermal reservoir. 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

Equations / Data / Solve : Parts a & b) 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.

Q  Q  S       T  Int Re v T

Eqn 1

We can immediately apply Eqn 1 to 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). Sres

-0.4475 kJ/K

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. Tsat Part c.)

-10.076 deg C 263.07 K

SR-134a

0.4561 kJ/K

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

Suniv  Sres  S R 134a

Eqn 2

Verify :


Answers :

a.)

SR-134a

b.)

Sres


0.4561 kJ/K

c.)

Suniv

0.0086 kJ/K

Suniv

0.0086 kJ/K

-0.4475 kJ/K

hw5-sp11.xlsm, 7.29

5/9/2011



Problem : 7.66 - SUniv Upon Quenching an Iron Block - 6 pts

A 12 kg iron block initially at 350oC is quenched in an insulated tank that contains 100 kg of water at 22oC. Assuming the water that vaporizes during the process condenses back into the liquid phase inside the tank, determine the entropy change of the universe for this process. Data :

CP,H2O

kJ/kg-K

Read :

This problem requires the application of the 1st Law to determine T2 and then one of the Gibbs equations to evaluate S.

4.18

CP,Fe

0.45

kJ/kg-K

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

T1,H2O Find :

o

C K o C K

350 623.15 22 295.15

T1,Fe

Suniv

mFe mH2O

12 100

kg kg

kJ/K

???

Diagram :

Iron mFe = 12 kg T1 = 350 oC T2 = ?? oC

Assumptions : 123-

Q

mH2O = 100 kg T1,H2O = 22oC = ?? oC T2

All of the water is in the liquid phase in the final state. The heat capacities of iron and water are constants. 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. Both the water and the iron are perfectly incompressible. The volume of each remains constant throughout this process.

4Equations / Data / Solve :

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

S univ  S Fe  S H 2O

Eqn 1

We can use Gibbs 2nd Equation for incompressible substances with constant heat capacities to evaluate SFe and SH2O.

dS 

 C

 T dT

ˆ  Ln  T2  Sˆ  C    T1 

Eqn 2

Eqn 3

Because the iron and water are incompressible, CP = CV = C. So, all we need to do in order to use Eqns 3 & 1 to 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.

ˆ Q Fe  Wb,Fe  m Fe U Fe

Eqn 4

ˆ Q H 2O  Wb,H 2O  m H 2O U H 2O

Eqn 5

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. Since all of the heat leaving the iron goes into the water:


Q H 2O   QFe

hw5-sp11.xlsm, 7.66

Eqn 6

5/9/2011

Combining Eqns 4 - 6 yields :

ˆ ˆ ˆ ˆ ˆ ˆ U H 2 O  U 2,H 2 O  U1,H 2 O   U Fe  U1,Fe  U 2,Fe

Eqn 7

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 :

ˆ  U

T2

 Cˆ

V

ˆ T  T  dT  C V 2 1

Eqn 8

T1

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







ˆ ˆ m H 2O C H 2 O T2  T1,H 2 O  m Fe CFe T1,Fe  T2



Eqn 9

Notice that at the final state, the water and the iron are at the same temperature, T2. Now, we need to solve Eqn 9 for the unknown T2.

T2 

ˆ ˆ m H 2O C H 2 O T1,H 2 O  m Fe CFe T1,Fe ˆ ˆ m C m C H 2O

H 2O

Fe

Eqn 10

Fe

Plugging values into Eqn 7 yields : mFe*CFe mH2O*CH2O

T2 T2

5.4 kJ/K 418 kJ/K

26.18 oC 299.33 K

Now we can use Eqn 3 for the water and then for the iron to evaluate the entropy change for each. And then plug these values into Eqn 1 to complete the problem. SFe SH2O

-3.9594 kJ/K 5.8829 kJ/K

Suniv

Verify :


Answers :

Suniv


1.9234 kJ/K

1.923 kJ/K

hw5-sp11.xlsm, 7.66

5/9/2011



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

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. Read :

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.

Given :

Q1 T1

Find :

th

750 1500

kJ K

Q2 T2

100 500

kJ K

T3

1000

K

???

Diagram :

Tres,1 Q1 T1

Q2 Tres,2

T2

Wcycle

HER T3

Q3 Tres,3 Assumptions : 123-

The power cycle is internally reversible. Heat transfers can be irreversible, but the temperatures within the system at which the heat trasfers occur must remain constant at the given values. Heat is exchanged only with the three reservoirs mentioned in the problem statement.


hw5-sp11.xlsm, WB-1

5/9/2011

Equations / Data / Solve : Let's begin with the definition of a thermal efficiency of a power cycle :

th 

Wcycle Q in



Wcycle

Eqn 1

Q1

Next, we need to apply the 1st Law to

the cycle to help us determine Wcycle :

Wcycle  Q1  Q 2  Q 3

Eqn 2

Now, all we need to do is determine Q2 and then we can use Eqns 1 & 2 to evaluate th. The key to determining Q2 is 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 :

 Scycle   

Q  S gen,int  0 T

Eqn 3

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

Q1 Q 2 Q 3   0 T1 T2 T3

Eqn 4

Notice that because Q1, Q2 and Q3 are 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 Q2 and Q3 represent heat transfers out of the system, they become negative when we drop the sign convention as we move from Eqn 3 to Eqn 4. Now, we can solve Eqn 4 for Q3 :

Q Q  Q 3  T3  1  2  T2   T1

Eqn 5

Now, plug values into Eqns 5, 2 & 3 :

Q3 Wcycle th

Verify :


Answers :

th

300 350

kJ kJ

0.4667

46.7%


hw5-sp11.xlsm, WB-1

5/9/2011



Problem : WB-2 - Efficiency and Tres for Rev. and Irrev. Cycles - 6 pts Complete the following involving reversible and irreversible cycles. a.)

Reversible and irreversible power cycles each discharge QC to a cold reservoir at TC and receive energy QH from hot reservoirs at TH and T'H, respectively. There are no other heat transfers involved. Show that T'H> TH.

b.)

Reversible and irreversible refrigeration cycles each discharge QH to a hot reservoir at TH and receive energy QC from cold reservoirs at TC and T'C, respectively. There are no other heat transfers involved. Show that T'C> TC.

Read :

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.

Given :

See diagrams, below.

Find :

a.) b.)

Show that T'H > TH. Show that T'C > TC.

Diagram : See below. Assumptions : 1234-

Both heat engines operate on thermodynamic cycles. Changes in kinetic and potential energies are negligible. The mixer is adiabatic. Flow work is the only form of work that crosses the system boundary.


Diagram :

TH

T'H QH

QH

WI

WR HER

HEI

QC

QC TC

The trick to this problem is to VERY CAREFULLY choose your system ! 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. However, in the irreversible cycle, the temperature within the system MAY NOT BE EQUAL to the temperature of the reservoir that is giving 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.


hw5-sp11.xlsm, WB-2

5/9/2011

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

 SR     SI   

Q  S gen,int,R  0 T

Eqn 1

Q  S gen,int,I  0 T

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.

Q H QC  0 TH TC

Eqn 3

Q H QC   S gen,int,I  0 TH TC

Eqn 4

Notice that because QH and QC are 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. For an irreversible process or cycle, the 2nd Law tells us that :

S gen,int,I  0 Solve Eqn 4 for Sgen,int,I :

Eqn 5

S gen,int,I 

QC Q H  0 TC TH

Eqn 6

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

QC Q H  TC TH

Eqn 7

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

1 1  TH TH Part b.)

TH  TH

Eqn 8

Eqn 9

Diagram :

TH QH

WR

RefR

QC TC

QH

WI RefI

QC T'C

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


hw5-sp11.xlsm, WB-2

5/9/2011

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

 SR     SI   

Q  S gen,int,R  0 T

Eqn 10

Q  S gen,int,I  0 T

Eqn 11

The entropy generation for the completely reversible cycle (in Eqn 1) is ZERO. Now, express Eqns 1 & 2 in terms of QH and QC.

QC Q H  0 TC TH

Eqn 12

QC Q H   S gen,int,I  0 TC TH

Eqn 13

Notice that because QH and QC are 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. For an irreversible process or cycle, the 2nd Law tells us that :

S gen,int,I  0

Eqn 14

S gen,int,I 

Solve Eqn 13 for Sgen,int,I :

Q H QC  0 TH TC

Eqn 6

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

QC Q H  TC TH

Eqn 7

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

1 1  TC TC Verify :

TC  TC

Eqn 8

Eqn 9


Answers : a.)

TH  TH


b.)

TC  TC

hw5-sp11.xlsm, WB-2

5/9/2011



WB-3 - Specific Entropy Change Using Tabluar Data - 4 pts

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

c.) d.)

( Use the NIST Webbook ) Water: P1 = 10 MPa, T1 = 400oC and P2 = 10 MPa, T2 = 100oC. o R-134a: H1 = 211.44 kJ/kg, T1 = - 40 C and P2 = 5 bar, x2 = 1.0. (Use the NIST Webbook with the default reference state: U = 0 and S = 0 for saturated liquid at 0.01oC) Air (IG): T1 = 27oC, P1 = 2 bar and T2 = 327oC, P2 = 1 bar. Hydrogen (H2, IG): T1 = 727oC, P1 = 1 bar and T2 = 25oC, P2 = 3 bar.

Read :

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

a.) b.)

Parts (c) and (d) are exercises in the use of the ideal gas property tables and Gibbs' 2nd Equation to evaluate S for ideal gases. Given :

Find :

a.) P1 T1

Water 10000 400

b.) H1 T1

R-134a 211.44 -40

c.) P1 T1 R

Air

d.) P1 T1 R

H2

S

???

kJ/kg-K

1-

For parts (c) and (d), assume that air and H2 behave as ideal gases, as instructed in the problem statement. For parts (c) and (d), assume that air and H2 have variable heat capacities and use the Ideal Gas Property Tables.

200 27 8.314

100 727 8.314

Assumptions :

2Diagram:

o

kPa C

P2 T2

10000 100

o

kJ/kg C

P2 x2

500 1

kPa C J/mole-K

P2 T2 MW

100 327 28.97

kPa C J/mole-K

P2 T2 MW

300 25 2.016

o

o

kPa C

o

kPa kg vap/kg kPa C g/mole

o

kPa C g/mole

o

for each part of the problem.

None needed.


Here we can use the NIST Webbook to evaluate the specific entropy of water and then directly calculate the change in specific entropy. I used the isothermal properties table with the default reference state to obtain : S1 S2

Part b.)

6.2141 kJ/kg-K 1.2996 kJ/kg-K

S

-4.9145 kJ/kg-K

The first step here is to use T1 to obtain data from the Saturation properties — temperature increments option. This will allow us to compare the given value of H to the values of Hsat liq and Hsat vap in order to determine the phase or phases present in the system at equilibrium. Here is the relevant data : Pressure (kPa)

Temp. (oC)

51.209

-40.00

Since :

H (kJ/kg) Sat. Liq 148.14

ˆ ˆ ˆ H sat  H  H sat


liq

vap

Sat. Vap 374.00

S (kJ/kg-K) Sat. Liq 0.79561

Sat. Vap 1.7643

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

hw5-sp11.xlsm, WB-3

5/9/2011

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

x1 

ˆ H ˆ H 1 sat liq ˆ ˆ H H sat vap

Eqn 1

sat liq


0.2803 kg vap/kg

x1

Sˆ sat  x Sˆ sat   1  x  Sˆ sat mix

Eqn 2

liq

vap

1.0671 kJ/kg-K

S1 Because state 2 is a saturated vapor at 500 kPa, we can get S2 directly from the saturation pressure table. S2 Part c.)

S

1.7197 kJ/kg-K

0.6526 kJ/kg-K

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

P  R Ln  2  Sˆ  Sˆ o2  Sˆ 1o  MW  P1 

Eqn 3

We can immediately lookup So2 and So1 in the Ideal Gas Property Table for air because we know T1 and T2. 0.0061681 kJ/kg-K

So1

So2 S

Now, we can plug values into Eqn 3 to get : Part d.)

0.91202 kJ/kg-K

The same procedure is used in part (d) as in part (c). 17.627 kJ/kg-K

So1

So2 S

Now, we can plug values into Eqn 3 to get : Verify :

0.71926 kJ/kg-K

0 kJ/kg-K -22.158 kJ/kg-K

a.)

No assumptions.

c.)

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.

b.)

  RT PV

Eqn 4

  RT V P

V1

12.48 L/mole

V2

No assumptions.

Eqn 5 49.90 L/mole

Since both V1 and V2 are greater than 5 L/mole, the ideal gas assumption is valid. d.)

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. V1

83.15 L/mole

V2

8.26 L/mole

Since both V1 and V2 are greater than 5 L/mole, the ideal gas assumption is valid. Answers :

a.)

S

-4.915 kJ/kg-K

b.)

S

0.653 kJ/kg-K


hw5-sp11.xlsm, WB-3

c.)

S

0.912 kJ/kg-K

d.)

S

-22.158 kJ/kg-K

5/9/2011


WB-4 - "Show That" for a Cycle Interacting with Three Reservoirs - 4 pts


The system shown in the figure undergoes a cycle while receiving energy at the rate Qsurr from the surroundings at temperature Tsurr, receiving QH from a heat source at temperature TH, and rejecting QC to a thermal reservoir at TC. Derive an expression for the maximum theoretical value of QC in terms of QH and T0, TS and TU only. 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, Tsurr and TC are not important in the solution of this problem !

Read :

The key here is to recognize that TH, TC and Tsurr do 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 > 0 we will be able to determine a maximum value for QC as a function of QH and the system temperatures at which heat is transferred.

Given :

TS  TU  T0

Find :

Q C  fxn Q H , TS , TU , T0 

Diagram :

See the problem statement.

Assumptions :

123-

The system operates on a thermodynamic cycle.

TS  TU  T0 The system only exchanges heat with the three reservoirs shown in the diagram.

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

 Scycle   

Q  S gen,int  0 T

Eqn 1

The change in any property, including entropy, for a cycle is zero. We can now expand the cyclic integral in terms of the three heat transfers that actually occur during the cycle.

Q H Qsurr QC    S gen,int  0 TS T0 TU

Eqn 2

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

Qsurr  QC  Q H

Eqn 3

Q H QC  Q H QC    S gen,int  0 TS T0 TU

Eqn 4

Now, we can solve Eqn 4 for QC.

1 1 1  1  QC     QH     S gen,int  T0 TU   T0 TS 


hw5-sp11.xlsm, WB-4

Eqn 5

5/9/2011

Because :

TS  TU  T0

1 1    0  T0 TU 

1 1    0  T0 TS 

Eqn 6

Eqn 7

Therefore, because Sgen > 0, QC is 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. So, for a reversible process, Eqn 5 becomes :

Q C,max

Verify :

Answers :

   QH    

T 1 1    1 0   T0 TS TS   QH  1 1   1  T0   T0 TU  TU 

     

Eqn 8


QC,max

T0   1 T S  QH   1  T0  TU 


     

hw5-sp11.xlsm, WB-4

5/9/2011



WB-5 - Three-Step, Ideal Gas Cycle Analysis - 8 pts

A quantity of air undergoes a thermodynamic cycle consisting of three internally reversible processes 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. Step 1 - 2 : Step 2 - 3 : Step 3 - 1 :

Isothermal expansion at 350 K from 4.75 bar to 1.0 bar. Adiabatic compression to 4.75 bar. Isobaric cooling.

a.) b.) c.) d.)

Sketch the cycle on a PV diagram. Sketch the cycle on a TS diagram. Determine T3 in Kelvin If the cycle is a power cycle, determine its thermal efficiency. If the cycle is a refrigeration cycle, determine its COP.

Read :

In part (c), you can use the 2nd Gibbs Equation and the Ideal Gas Entropy Table for air to determine So3 and then T3. 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 = QC from an entropy balance, taking advantage of the fact that there is no internal entropy generation because the cycle isinternally reversible. We can evaluate W31 from the integra of P dV and then apply the 1st Law to determine Q31 = QH. Finish up by plugging QC and QH into the definition of COPR.

Given :

T1 = T2 P1 P2

350 475 100

K kPa kPa

R MW

Find :

a.) b.)

Diagram :

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

PV diagram TS diagram.

c.) d.)

8.314 28.97 ??? ???

T3 COPR or 

J/mol-K g/mole K

Assumptions : 123-

The air is contained in a closed system. Air is an ideal gas in all three states. This assumption can be verified. All three steps in the cycle are internally reversible.


Part b.)

1

T

3 Isobaric

Isobaric T2

Adiabatic Isothermal

T1 2

1 Isothermal

V


P1

3 Isentropic

P

P2

2 S

hw5-sp11.xlsm, WB-5

5/9/2011

Part c.)

The key to determining T3 is 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 :

P  R Ln  3  Sˆ  Sˆ o3  Sˆ o2  MW  P2 

Eqn 1

So is a function of T only, so we can lookup So1 in the Ideal Gas Property Table for air. S o2 = S o1 o

Since S is a function of T only, we can solve Eqn 1 for determine T2.

So2

1.85708 kJ/kg-K

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

P  R Sˆ o3  Sˆ o2  Ln  3  MW  P2 

Eqn 2 S o3

Plugging values into Eqn 2 yields : T (K) 540 542.73 550

Interpolation on the Ideal Gas Property Table for air :

Part d.)

S

2.30425 kJ/kg-K

o

2.29906 2.30425 2.31809

T3

542.73 K

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 PV and TS diagrams. So, our goal is to determine the value of COPR.

COPR 

We cannot determine COPR using :

1 TH / TC  1

Eqn 3

There are two reasons. 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 T3 to T1. Which T would you use ? If you simply use T3, then the heat transfer in step 3-1 is not reversible and Eqn 3 does not apply. Instead, we can determine COPR from its definition.

COPR 

ˆ Q C ˆ W

cycle



ˆ Q C ˆ Q ˆ Q H C



1 ˆ ˆ 1 QH / Q C

Eqn 4

Note : Because of our use of the sign convention that heat transfer is positive when heat is transferred into the system, QC = Q12 and QH = - Q31. Let's begin by determining Q12. Because this step is internally reversible and isothermal, we can use :

 Sˆ 12   

ˆ ˆ Q Q  Sˆ gen,int  12 T T1

We can solve Eqn 5 for Q12 :


ˆ  T   Sˆ Q 12 1 12

hw5-sp11.xlsm, WB-5

Eqn 5

Eqn 6

5/9/2011

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

Sˆ 12  Sˆ o2  Sˆ 1o 

P  R Ln  2  MW  P1 

Because T1 = T2, So2 = So1 and Eqn 7 simplifies to :

Sˆ 12  

Eqn 7

P  R Ln  2  MW  P1 

Eqn 8 S12 Q12 = QC

Plugging values into Eqns 8 & 6 yields :

0.44717 kJ/kg-K 156.51 kJ/kg

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

ˆ W ˆ  U ˆ Q 31 31 31

Eqn 9

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

W31 

 P dVˆ  P Vˆ    PVˆ 

V1

Eqn 10

V3

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

 

ˆ  U ˆ W ˆ  U ˆ   PV ˆ Q 31 31 31 31

31

ˆ H ˆ H ˆ  H 31 1 3

Eqn 11

Because we know T1 and T3 and both states are assumed to be ideal gas states, we can lookup H1 and H3 in the Ideal Gas Property Table for air. Ho3 requires interpolation. T (K) 540

Ho1

350.49 kJ/kg

Ho3

547.45 kJ/kg

o

H

544.35

Ho3

547.45

550

555.74

Now, we can plug values into Eqn 11 to evaluate Q31 and then plug Q12 and Q31 into Eqn 4 to evaluate COPR. -196.96 kJ/kg

Q31 = -QH Verify :

COP

3.869

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 (diatomic gas).   RT PV

  RT V P

Eqn 12 6.13 L/mole 29.10 L/mole

V1 V2

V3

Eqn 13

9.50 L/mole

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

a.)

See the PV diagram, above.

c.)

T3

b.)

See the TS diagram, above.

d.)

COP


hw5-sp11.xlsm, WB-5

542.7 K 3.87

5/9/2011



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

Steam enters an adiabatic turbine at 800 psia and 900oF and leaves at a pressure of 40 psia. Determine the maximum amount of work that can be delivered by this turbine. Read :

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 P2 and S2 to determine H2. Then, we can apply the 1st Law to evaluate the work output from the isentropic turbine.

Given :

P1 T1

800 900

o

W sh,max

???

Btu/lbm

Find :

P2

psia F

40

psia

Diagram : 1

P1 = 800 psia T1 = 900 oF

W sh = ??? kJ/kg Turbine P2 = 40 psia T2 = ??? oF

2

Assumptions : 12-

The turbine is adiabatic. Changes in kinetic and potentail energies are negligible.

Equations / Data / Solve : Let's begin by applying the 1st Law for open systems to the turbine, assuming changes in kinetic and potential energies are negligible.

ˆ W ˆ  H ˆ Q sh

Eqn 1

Because the turbine is adiabatic, Eqn 1 becomes :

ˆ H ˆ H ˆ W sh 1 2

Eqn 2

Because we know both T1 and P1, we can immediately lookup H1 in the steam tables. At P1 = 800 psia :

Tsat

518.27

o

Since T1 > Tsat, the feed is superheated steam.

H1

1456.0

Btu/lbm

F

It is a bit harder to determine H2 because 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. We can immediately lookup S1 because we know P1 and T1.

S1 = S2

1.6413 Btu/lbm-oR

Now, knowing P2 and S2, we can determine H2. Let's begin by determining what phases are present. The first step here is to use P2 to obtain data from the saturation pressure table. This will allow us to compare the given value of S2 to the values of Ssat liq and Ssat vap in order to determine the phase or phases present in the system at equilibrium. Here is the relevant data : Temp.

Pressure (psia) 40 Since :

(oF) 267.22

H (Btu/lbm) Sat. Liq Sat. Vap 236.15 1169.80

Sˆ sat  Sˆ 2  Sˆ sat liq


vap

S (Btu/lbm-oR) Sat. Liq Sat. Vap 0.39214 1.6766


hw5-sp11.xlsm, WB-6

5/9/2011

Next, we must use S2 to determine the quality of the water in the system. The key equation is:

x2 

Sˆ 2  Sˆ sat liq Sˆ  Sˆ sat vap

Eqn 3

sat liq

x2


0.9725 lbm vap/lbm

Sˆ sat  x Sˆ sat   1  x  Sˆ sat mix

vap

Eqn 4

liq

H2

1144.1 Btu/lbm

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


Answers :

Wsh


312

311.9

Btu/lbm

Btu/lbm

hw5-sp11.xlsm, WB-6

5/9/2011



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

A 0.5 m3 rigid tank contains R-134a initially at 200 kPa and 40% quality. Heat is transferred to the refrigerant from a source at 35oC until the pressure rises to 400 kPa. Determine… a.) b.) c.)

The entropy change of the R-134a. The entropy change of the heat source. The entropy change of the universe for this process.

Read :

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 S can be determined from the definition of entropy).

Given :

V P1 x1

Find :

m3 kPa kg vap/kg

0.5 200 0.4 SR-134a Sres

a.) b.)

??? ???

kJ/K kJ/K

c.)

P2 Tsource

400 35 308.15

Suniv

???

kPa C K

o

kJ/K

Heat Source 35oC

Diagram :

Assumptions : 123-

Q = ?? kJ

1

2

Tank P1 = 200 kPa T1 = Tsat x1 = 0.4 kg vap/kg

Tank P2 = 400 kPa T2 = ?? oC x2 = ?? kg vap/kg

Changes in kinetic and potential energies are negligible. The temperature of the heat source is constant and so it acts as a true thermal reservoir. The tank is a closed system. The mass of R-134a in the tank is constant.


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



S R 134a  m Sˆ 2  Sˆ 1



Eqn 1

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

m

Vtank ˆ V

Eqn 2

1

While we are looking up V1 in the saturation pressure table we might as well lookup all the other relevant properties. Pressure (kPa) 200

Temp. (oC) -10.090

V (m3/kg) Sat. Liq Sat. Vap 0.0007533 0.099867

U (kJ/kg) Sat. Liq Sat. Vap 38.28 224.48

S (kJ/kg-K) Sat. Liq Sat. Vap 0.15457 0.93773

We can now use the quality to determine V1, U1, and S1 using :

ˆ x M ˆ ˆ M 1 1 sat   1  x1  M sat vap

V1 U1

Eqn 3

liq

3 0.04040 m /kg 112.76 kJ/kg


S1

0.4678

kJ/kg-K

m

12.377

kg

Now, we need S2. We can get it because we know P2 and we know that the volume of the tank and the mass of R-134a in the tank are both constant in this process. So: V2 = V1


hw5-sp11.xlsm, WB-7

0.04040

m3/kg

5/11/2011

Next, we determine which phase or phases are present in state 2 by comparing V2 to Vsat liq and Vsat vap at 400 kPa. Pressure (kPa) 400

Since :

Temp. (oC) 8.910

ˆ  V ˆ  V ˆ V sat 2 sat liq

V (m3/kg) Sat. Liq Sat. Vap 0.0007907 0.051201

U (kJ/kg) Sat. Liq Sat. Vap 63.62 235.07

S (kJ/kg-K) Sat. Liq Sat. Vap 0.24761 0.92691


vap

Next, we must use V2 to determine the quality of the water in the system. The key equation is:

x2 

ˆ V ˆ V 2 sat liq ˆ ˆ V V sat vap

Eqn 4

sat liq


x2

0.7857

kg vap/kg

S2

0.7813

kJ/kg-K

SR-134a

3.8802

kJ/k

We can now use the quality to determine U1, and S1 using Eqn 3. 198.33 kJ/kg

U2

Finally, plugging values into Eqn 1 yields : Part b.)

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.

Q  Q  S res      T  T  Int Re v

Eqn 5

I put the minus sign in front of Q in this equation because we will apply the 1st Law to the tank to determine Q. So, Q is the amount of heat transferred out of the source and with our sign convention, -Q is the heat transferred when we apply Eqn 1 to the heat source. Apply the 1st Law for closed systems with negligible changes in kinetic and potential energies to the tank.

ˆ Q  Wb  m U

Eqn 6

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



ˆ U ˆ Qm U 2 1



Eqn 7

Since we calculated U1 and U2 in part (a), all we need to do is plug them into Eqn 7 to determine Q, and put Q back into Eqn 5 to get Sres. Q Part c.)

Ssource

1059.07 kJ

-3.4369

kJ/k

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

Suniv  Sres  S R 134a

Eqn 2

Verify :


Answers :

SR-134a Ssource Suniv


Suniv

0.4433

kJ/K

3.880 kJ/k -3.437 kJ/k 0.4433 kJ/k

hw5-sp11.xlsm, WB-7

5/11/2011

Thermo Problems 2

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