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

ENGR 224 - Thermodynamics

Problem : 7.14 - The Increase of Entropy Principle - 2 pts Will the entropy of steam increase, decrease or remain the same as it flows through a real adiabatic turbine ?

Read : Answers :

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

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.

Baratuci HW #5

11-May-11

Dr. Baratuci - ENGR 224

hw5-sp11.xlsm, 7.14

5/9/2011

ENGR 224 - Thermodynamics Problem : 7.15 - The Increase of Entropy Principle - 2 pts

Baratuci HW #5

11-May-11

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

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.

Dr. Baratuci - ENGR 224

hw5-sp11.xlsm, 7.15

5/9/2011

ENGR 224 - Thermodynamics Problem : 7.18 - The Increase of Entropy Principle - 2 pts

Baratuci HW #5

11-May-11

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

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.

Dr. Baratuci - ENGR 224

hw5-sp11.xlsm, 7.18

5/9/2011

ENGR 224 - Thermodynamics

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

Baratuci HW #5

11-May-11

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

c.)

The total entropy change of the universe for this process.

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 : Diagram :

Sfluid

a.) b.)

S1

-0.7

Btu/oR

QC Ssink

???

Btu

???

Btu/oR

TC = 95oF

QC

HEX

S2

TC

95

oF

554.67 oR

c.)

Suniv

???

Btu/oR

Sfluid = -0.7 Btu/oR

Assumptions :

1 -

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

2 -

The reservoir is internally reversible.

3 -

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.

Sfluid

Q T

Sgen,int

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

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

Eqn 3

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 :

QC

388.27 Btu

Dr. Baratuci - ENGR 224

hw5-sp11.xlsm, 7.28E

5/9/2011

Part b.) Part c.)

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 :

Sres

QC TC

Now, we can plug numbers into Eqn 5 :

Sres

Eqn 5 0.7000 kJ/-K

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

Suniv Sfluid Sres

Plugging values into Eqn 6 yields :

This makes sense since :

Suniv Sgen,total

Suniv

0.0000 kJ/-K

Eqn 6 Eqn 7

And, for a completely reversible process like this :

Sgen,total 0

Eqn 8

Verify : Answers :

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

a.)

QC

388

Btu

b.)

Sres

0.700 kJ/-K

c.)

Suniv

0

kJ/-K

Dr. Baratuci - ENGR 224

hw5-sp11.xlsm, 7.28E

5/9/2011

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