1-9 Carnot CycleA Carnot Cycle is a cycle involving two reversible isothermal transitions and two reversible adiabatic transitions. If the working substance is an ideal gas, the 𝑃 − 𝑉diagram of the cycle and the 𝑇 − 𝑆 diagram, looks like Fig. 1-9, The processes:

Fig 1-9: Carnot cycle on the 𝑃 − 𝑉 and 𝑇 − 𝑠 diagrams.

1-2: reversible adiabatic: compression2-3: reversible constant temperature: heat addition3-4: reversible adiabatic: expansion4-1: reversible constant temperature : heat rejection

Between 2 and 3, temperature is always 𝑇𝐻

∆𝑈2−3 = 0𝑄2−3 = 𝑊2−3 > 0

𝑄2−3 = 𝑊2−3 = 𝑅𝑇𝐻𝑙𝑛𝑉3𝑉2

> 0, 𝑃𝑉 = 𝑅𝑇

∆𝑈4−1 = 0

𝑄4−1 = 𝑊4−1 = 𝑅𝑇𝐶𝑙𝑛𝑉4𝑉1< 0

𝑇𝐻𝑉2𝛾−1

= 𝑇𝐻𝑉1𝛾−1

, 𝑇𝐶𝑉3𝛾−1

= 𝑇𝐶𝑉4𝛾−1

𝑇𝐻𝑇𝐶

=𝑉1𝑉2

𝛾−1

=𝑉4𝑉3

𝛾−1

𝑉1𝑉2

=𝑉4𝑉3,𝑉3𝑉2=𝑉4𝑉1

𝑄𝐻𝑄𝐶

=𝑄2−3𝑄4−1

= −𝑇𝐻𝑇𝐿

Carnot heat engine efficiency:

𝜂𝑐 =𝑊

𝑄𝐻=

𝑄𝐻−𝑄𝐶

𝑄𝐻= 1 −

𝑄𝐶

𝑄𝐻= 1 −

𝑇𝐶

𝑇𝐻(1-32)

Carnot’s cycle is completely reversible. Run backwards, it is a Carnot refrigerator:

𝜂𝑃 =𝑄𝐶

𝑊=

𝑄𝐶

𝑄𝐻 − 𝑄𝐶=

𝑇𝐶

𝑇𝐻−𝑇𝐶(1-33)

For a Carnot machine functioning as a refrigerator, the "effectiveness" is the ratio of the energy removed from the low-temperature reservoir to the work required to force the machine around its cycle (the energy consumed and paid for)

THE QUESTION IS: CAN CARNOT CYCLE OPERATE AT Absolute zero???

The best refrigerator you can get (Carnot) has performance:

𝜂𝑃 =𝑇𝐶

𝑇𝐻 − 𝑇𝐶The colder you try to go, the less efficient the refrigerator GETS:

𝜂𝑃 =𝑇𝐶

𝑇𝐻 − 𝑇𝐶, 𝑇𝐶 → 0

Since heat leaks will not disappear as the object is cooled, you need more cooling power the colder it gets. The integral of the power required diverges as: 𝑇𝐶 → 0

Therefore WE cannot cool a system to absolute zero.

Noting that, the change in entropy during heat addition and rejection are equal in magnitude thus:𝑄𝐴 = 𝑇𝐻 𝑆3 − 𝑆2 , (1-34)

𝑄𝑅 = 𝑇𝐿 𝑆1 − 𝑆4 (1-35)Or 𝑄𝑅 = 𝑇𝐿 𝑆4 − 𝑆1 = 𝑇𝐿 𝑆3 − 𝑆2Here 𝑇𝐻 = 𝑇2 and 𝑇𝐿 = 𝑇1The network and cycle efficiency:

∆𝑊𝑁𝐸𝑇 = 𝑄𝐴 − 𝑄𝑅 (1-36)

And 𝜂𝐶 =∆𝑊𝑁𝐸𝑇

𝑄𝐴(1-37)

Thus the thermal efficiency of the Carnot cycle 𝜂𝐶 is given by:

𝜂𝐶 =𝑇𝐻 − 𝑇𝐿𝑇𝐻

Example 1-1The inside of a refrigerator is held at 𝑇𝐶 = 3℃. The surroundings are at 𝑇𝐻 = 27 ℃. We must remove 1.25 𝑘𝑊 of thermal energy from the inside of the refrigerator to balance the thermal energy entering from the environment. Find the best possible coefficient of performance? and the minimum power requirement to operate the refrigerator?.

Answer:First, we need to convert to an absolute scale. So, 𝑇𝐶 = 3 + 273.15 = 276.15 𝐾, and 𝑇𝐻 = 300.15 𝐾.

𝜂𝑃 =𝑇𝐶

𝑇𝐻 − 𝑇𝐶=

276.15

300.15 − 276.15= 11.5

Now,

𝜂𝑃 =𝑄𝐶𝑊𝐶

,𝑊𝐶 =1.25

11.5= 0.10836𝑘𝑊

Note the rules of heat transfer determine the energy load needed to keep the temperature constant. We have not considered those here. This is the smallest possible engine that would be needed. Inefficiencies would cause the actual sized needed to be higher.

Figure 1-10: Sketch of a Carnot cycle for water represented in the 𝑃 – 𝑉 and 𝑇 − 𝑠 planes.

1

2 3

4

𝑇𝐻

𝑇𝐿

𝑃

𝑉

1

2 3

4

𝑇

𝑆

Isothermal

Isothermal

Isotropic Isotropic

Example 1-2.A mass of 𝑚 = 1 𝑘𝑔 of water executes a Carnot cycle. The high temperature isothermal expansion is from 𝑃2 = 15 𝑏𝑎𝑟, 𝑥2 = 0.25 to the saturated vapor state. The adiabatic expansion is to 𝑃4= 1 𝑏𝑎𝑟. Analyse the system?Answer:The system is sketched in Fig. 1-10. We have 𝑃2 = 1500 𝑘𝑃𝑎, 𝑥2 = 0.25. Find the following

𝑇2 = 198.32 ℃ = 471.47𝐾

𝑣𝑓2 = 0.001154𝑚3

𝑘𝑔, 𝑣𝑓𝑔2 = 0.13062

𝑚3

𝑘𝑔

𝑢𝑓2 = 843.14𝑘𝐽

𝑘𝑔, 𝑢𝑓𝑔2= 1751.3

𝑘𝐽

𝑘𝑔

𝑠𝑓2 = 2.315𝑘𝐽

𝑘𝑔. 𝐾, 𝑆𝑓𝑔2 = 4.129

𝑘𝐽

𝑘𝑔. 𝐾So we get the properties of state 2 via

𝑣2 = 𝑣𝑓2 + 𝑥2𝑣𝑓𝑔2 = 0.0338𝑚3

𝑘𝑔

𝑢2 = 𝑢𝑓2 + 𝑥2𝑢𝑓𝑔2 = 1280.97𝑘𝐽

𝑘𝑔

𝑠2 = 𝑠𝑓2 + 𝑥2𝑠𝑓𝑔2 = 3.347𝑘𝐽

𝑘𝑔.𝐾At state 3, we know𝑃3 = 𝑃2 = 1500𝑘𝑃𝑎, and 𝑥3 = 1, so the properties can be read from the saturated water tables for 𝑥3 = 1. We find

𝑣3=𝑔 = 0.13177𝑚3

𝑘𝑔

𝑢3=𝑔 = 2594.5𝑘𝐽

𝑘𝑔

𝑠3=𝑔 = 6.448𝑘𝐽

𝑘𝑔. 𝐾At state 4, we know 𝑃4 = 100𝑘𝑃𝑎. We also know that 𝑠4 = 𝑠3 = 6.4458𝑘𝐽/𝑘𝑔/𝐾. So the state is fixed. At state 4, the tables give

𝑇4 = 99.62℃ = 372.77𝐾

𝑣𝑓4 = 0.001154𝑚3

𝑘𝑔, 𝑣𝑓𝑔4= 0.13062

𝑚3

𝑘𝑔

𝑢𝑓4 = 843.14𝑘𝐽

𝑘𝑔, 𝑢𝑓𝑔4 = 1751.3

𝑘𝐽

𝑘𝑔

𝑠𝑓4 = 2.315𝑘𝐽

𝑘𝑔. 𝐾, 𝑠𝑓𝑔4 = 4.1298

𝑘𝐽

𝑘𝑔. 𝐾

Knowing 𝑠4, we can find the quality 𝑥4 via

𝑥4 =𝑠4 − 𝑠𝑓4

𝑠𝑓𝑔4=

6.4448𝑘𝐽

𝑘𝑔. 𝐾− 2.315

𝑚3

𝑘𝑔

4.1298𝑘𝐽

𝑘𝑔. 𝐾

= 0.849

We then find

𝑣4 = 𝑣𝑓4 + 𝑥4𝑣𝑓𝑔4 = 1.4383𝑚3

𝑘𝑔

𝑢4 = 𝑢𝑓4 + 𝑥4𝑢𝑓𝑔4 = 2190.68𝑘𝐽

𝑘𝑔

Now, we know that for state 1, 𝑠1 = 𝑠2 = 3.34745𝑘𝐽/𝑘𝑔/𝐾, and 𝑠𝑓1 = 𝑠𝑓4, 𝑠𝑓𝑔1= 𝑠𝑓𝑔4. So

𝑥1 =𝑠1 − 𝑠𝑓1

𝑠𝑓𝑔1= 0.3376

We also know that 𝑇1 = 𝑇4 = 372.77𝐾. We also know that

𝑣𝑓1 = 𝑣𝑓4, 𝑣𝑓𝑔1 = 𝑣𝑓𝑔4, 𝑢𝑓1 = 𝑢𝑓4, 𝑢𝑓𝑔1= 𝑢𝑓𝑔4So

𝑣1 = 𝑣𝑓1 + 𝑥1𝑣𝑓𝑔1 = 0.5726𝑚3

𝑘𝑔

𝑢1 = 𝑢𝑓1 + 𝑥1𝑢𝑓𝑔1 = 1122.54𝑘𝐽

𝑘𝑔

Now, let us consider the first law for each process.

• 1 → 2: The first law gives

𝑢2 − 𝑢1 = 12𝑞 − 1

2𝑤

12𝑞 = 0

𝑢2 − 𝑢1 = −12𝑤

12𝑤 = 𝑢1 − 𝑢2

12𝑤 = 1122.54

𝑘𝐽

𝑘𝑔− 1280.97

𝑘𝐽

𝑘𝑔

= −158𝑘𝐽

𝑘𝑔• 2 → 3: The first law gives (constant pressure)

𝑢3 − 𝑢2 = 23𝑞 − 2

3𝑤Now

23𝑤 =

2

3

𝑃𝑑𝑣

= 𝑃3(𝑣3 − 𝑣2)

= (1500𝑘𝑃𝑎) 0.13177𝑚3

𝑘𝑔− 0.0338

𝑚3

𝑘𝑔

= 146.942𝑘𝐽

𝑘𝑔So

23𝑞 = 𝑢3 − 𝑢2 + 2

3𝑤

= 2594.5𝑘𝐽

𝑘𝑔− 1280

𝑘𝐽

𝑘𝑔+ 146.942

𝑘𝐽

𝑘𝑔

= 1460.48𝑘𝐽

𝑘𝑔• 3 → 4: The first law gives

𝑢4 − 𝑢3 = 34𝑞 − 3

4𝑤

34𝑞 = 0, adiabatic𝑢4 − 𝑢3 = −3

4𝑤

2190.68𝑘𝐽

𝑘𝑔− 2594.5

𝑘𝐽

𝑘𝑔= −3

4𝑤

34𝑤 = 403.82

𝑘𝐽

𝑘𝑔𝑢1 − 𝑢4 = 4

1𝑞 − 41𝑤

Now

41𝑤 =

4

1

𝑃𝑑𝑣

= 𝑃1(𝑣1 − 𝑣2)

= 100𝑘𝑃𝑎 0.5726𝑚3

𝑘𝑔− 1.4383

𝑚3

𝑘𝑔

= −86.575𝑘𝐽

𝑘𝑔So

41𝑞 = 𝑢1 − 𝑢4 + 4

1𝑤

= 1122.54𝑘𝐽

𝑘𝑔− 2190.68

𝑘𝐽

𝑘𝑔+ −86.5752

𝑘𝐽

𝑘𝑔

= −1154.71𝑘𝐽

𝑘𝑔The network for the cycle is

𝑤𝑐𝑦𝑐𝑙𝑒 = 12𝑤 + 2

3𝑤 + 34𝑤 + 4

1𝑤

= −158.423𝑘𝐽

𝑘𝑔+ 146.942

𝑘𝐽

𝑘𝑔+ 403.82

𝑘𝐽

𝑘𝑔+ −86.575

𝑘𝐽

𝑘𝑔

= 305.763𝑘𝐽

𝑘𝑔The net heat transfer for the cycle is

𝑞𝑐𝑦𝑐𝑙𝑒 = 23𝑞 + 4

1𝑞

= 1460.48𝑘𝐽

𝑘𝑔+ −1154.71

𝑘𝐽

𝑘𝑔

= 305.763𝑘𝐽

𝑘𝑔Note that, as expected 𝑤𝑐𝑦𝑐𝑙𝑒 = 𝑞𝑐𝑦𝑐𝑙𝑒

Now, let us calculate the thermal efficiency.

𝜂 =𝑤𝑐𝑦𝑐𝑙𝑒

𝑞𝑖𝑛=𝑤𝑐𝑦𝑐𝑙𝑒

23𝑞

=305.763

1460.48= 0.209

This should be the same as the Carnot theory’s prediction

𝜂 = 1 −𝑇𝐶𝑇𝐻

= 1 −372.77𝐾

471.47𝐾= 0.209

We summarize the first law statements in table.

Process∆𝑢

𝑘𝐽

𝑘𝑔𝑞

𝑘𝐽

𝑘𝑔𝑤

𝑘𝐽

𝑘𝑔

1-2

2-3

3-4

4-1

158.423

1313.54

-403.82

-1068.14

0

1460.48

0

-1154.71

-158.423

146.942

403.82

-86.5752

Total 0 305.763 305.763

Example 1-3Consider an ideal gas Carnot cycle with air in a piston cylinder with a high temperature of 1200𝐾 and a

heat rejection at 400𝐾. During the heat addition, the volume triples. The gas is at 1.0𝑚3

𝑘𝑔before the

isentropic compression. Analyse the cycle?Answer:

Take state 1 to be the state before the compression. Then 𝑇1 = 400𝐾 , 𝑣1 = 1.0𝑚3

𝑘𝑔

By the ideal gas law

𝑃1 =𝑅𝑇1𝑣1

=0.287

𝑘𝐽𝑘𝑔. 𝐾

400𝐾

1.0𝑚3

𝑘𝑔

= 1.148 × 102𝑘𝑃𝑎

Now isentropically compress to state 2. By the standard relations for a perfect ideal gas, one finds

𝑇2𝑇1=

𝑣1𝑣2

𝛾−1

=𝑃2𝑃1

𝛾−1𝛾

So

𝑣2 = 𝑣1𝑇1𝑇2

1𝛾−1

= 1𝑚3

𝑘𝑔

400𝐾

1200𝐾

11.4−1

= 0.064𝑚3

𝑘𝑔

Note that 𝑣2 < 𝑣1as is typical in a compression. The isentropic relation between pressure and volume can be rearranged to give the standard

𝑃2𝑣2𝛾= 𝑃1𝑣1

𝛾

Thus, one finds that

𝑃2 = 𝑃1𝑣1𝑣2

𝛾

= 1.148 × 102𝑘𝑃𝑎1.0

𝑚3

𝑘𝑔

0.064𝑚3

𝑘𝑔

1.4

= 5.368 × 103𝑘𝑃𝑎

Note the pressure has increased in the isentropic compression. Check to see if the ideal gas law is satisfied at state 2:

𝑃2 =𝑅𝑇2𝑣2

=0.287

𝑘𝐽𝑘𝑔

(1200𝐾)

0.064𝑚3

𝑘𝑔

= 5.368 × 103𝑘𝑃𝑎

This matches. Now the expansion from state 2 to 3 is isothermal. This is the heat addition step in which the volume triples. So one gets

𝑣3 = 3𝑣2 = 3 0.064𝑚3

𝑘𝑔= 0.192

𝑚3

𝑘𝑔

𝑇2 = 𝑇3 = 1200𝐾The ideal gas law then gives

𝑃3 =𝑅𝑇3𝑣3

=0.287

𝑘𝐽𝑘𝑔. 𝐾

1200𝐾

0.192𝑚3

𝑘𝑔

= 1.789 × 103𝑘𝑃𝑎

Process 3 to 4 is an isentropic expansion back to 400 𝐾. Using the isentropic relations for the perfect ideal gas, one gets

𝑣4 = 𝑣3𝑇3𝑇4

1𝛾−1

= 0.192𝑚3

𝑘𝑔

1200𝐾

400𝐾

11.4−1

= 3.0𝑚3

𝑘𝑔

𝑃4 = 𝑃3𝑣3𝑣4

𝛾

= 1.789 × 103𝑘𝑃𝑎0.192

𝑚3

𝑘𝑔

3.0𝑚3

𝑘𝑔

1.4

= 3.82 × 101𝑘𝑃𝑎

Check:

𝑃4 =𝑅𝑇4𝑣4

=0.287

𝑘𝐽𝑘𝑔. 𝐾

400𝐾

3.0𝑚3

𝑘𝑔

= 3.82 × 101𝑘𝑃𝑎

Now calculate the work, heat transfer and efficiency. Take the adiabatic exponent for air to be = 1.4. Now since

𝛾 =𝑐𝑃𝑐𝑣, 𝑐𝑃 − 𝑐𝑣 = 𝑅

One get

𝛾 =𝑅 + 𝑐𝑣𝑐𝑣

𝛾𝑐𝑣 = 𝑅 + 𝑐𝑣𝑐𝑣 𝛾 − 1 = 𝑅

𝑐𝑣 =𝑅

𝛾 − 1=0.287

𝑘𝐽𝑘𝑔. 𝐾

1.4 − 1= 0.717

𝑘𝐽

𝑘𝑔. 𝐾Recall the first law:

𝑢2 − 𝑢1 = 12𝑞 − 1

2𝑤Recall also the caloric equation of state for a perfect gas

𝑢2 − 𝑢1 = 𝑐𝑣 𝑇2 − 𝑇1Now process 1 → 2 is isentropic, so it is also adiabatic, hence 1

2𝑞 = 0, so one has𝑢2 − 𝑢1 = −1

2𝑤𝑐𝑣 𝑇2 − 𝑇1 = −1

2𝑤

0.717𝑘𝐽

𝑘𝑔. 𝐾1200𝐾 − 400𝐾 = −1

2𝑤

12𝑤 = −5.74 × 102

𝑘𝐽

𝑘𝑔The work is negative as work is being done on the system in the compression process.Process 2 → 3 is isothermal, so there is no internal energy change. The first law gives

𝑢3 − 𝑢2 = 23𝑞 − 2

3𝑤𝑐𝑣 𝑇3 − 𝑇2 = 2

3𝑞 − 23𝑤

Hence𝑐𝑣 𝑇3 − 𝑇2 = 0

Then

23𝑞 − 2

3𝑤 = 0

23𝑞 = 2

3𝑤 = 𝑣2

𝑣3

𝑃𝑑𝑣

= 𝑣2

𝑣3 𝑅𝑇

𝑣𝑑𝑣

= 𝑅𝑇2 𝑣2

𝑣3 1

𝑣𝑑𝑣

= 𝑅𝑇2𝑙𝑛𝑣3𝑣2

= 0.287𝑘𝐽

𝑘𝑔. 𝐾1200𝐾 𝑙𝑛

0.1924𝑚3

𝑘𝑔

0.0641𝑚3

𝑘𝑔

= 3.783 × 103𝑘𝐽

𝑘𝑔

The work is positive, which is characteristic of the expansion process.Process 3 → 4 is adiabatic so 3

4𝑞 = 0. The first law analysis gives then𝑢4 − 𝑢3 = 3

4𝑞 − 34𝑤

34𝑞 = 0

𝑢4 − 𝑢3 = −34𝑤

𝑐𝑣 𝑇4 − 𝑇3 = −34𝑤

0.717𝑘𝐽

𝑘𝑔. 𝐾400𝐾 − 1200𝐾 = −3

4𝑤

34𝑤 = 5.74 × 103

𝑘𝐽

𝑘𝑔Process 4 → 1 is isothermal. Similar to the other isothermal process, one finds

𝑢1 − 𝑢4 = 41𝑞 − 4

1𝑤𝑐𝑣 𝑇1 − 𝑇4 = 4

1𝑞 − 41𝑤

Hence

𝑐𝑣 𝑇1 − 𝑇4 = 0Then

41𝑞 − 4

1𝑤 = 0

41𝑞 − 4

1𝑤 = 𝑣4

𝑣1

𝑃𝑑𝑣

= 𝑣4

𝑣1 𝑅𝑇

𝑣𝑑𝑣

= 𝑅𝑇4 𝑣4

𝑣1 1

𝑣𝑑𝑣

= 𝑅𝑇4𝑙𝑛𝑣1𝑣4

= 0.287𝑘𝐽

𝑘𝑔. 𝐾400𝐾 𝑙𝑛

1.0𝑚3

𝑘𝑔

3.0𝑚3

𝑘𝑔

= −1.26 × 102𝑘𝐽

𝑘𝑔

The cycle work is found by adding the work of each individual process:

𝑤𝑐𝑦𝑐𝑙𝑒 = 12𝑤 + 2

3𝑤 + 34𝑤 + 4

1𝑤

= −5.74 + 3.78 + 5.74 − 1.261 × 102 = 2.522 × 102𝑘𝐽

𝑘𝑔The cycle heat transfer is

𝑞𝑐𝑦𝑐𝑙𝑒 = 12𝑞 + 2

3𝑞 + 34𝑞 + 4

1𝑞

= 0 + 3.78 + 0 − 1.261 × 102 = 2.522 × 102𝑘𝐽

𝑘𝑔Note that

𝑤𝑐𝑦𝑐𝑙𝑒 = 𝑞𝑐𝑦𝑐𝑙𝑒Check now for the cycle efficiency. Recall that the thermal efficiency is

𝜂 =𝑤𝑐𝑦𝑐𝑙𝑒

𝑞𝑖𝑛=2.52 × 102

𝑘𝐽𝑘𝑔

3.78 × 102𝑘𝐽𝑘𝑔

= 0.67

So general Carnot theory holds that the efficiency should be

𝜂 = 1 −400𝐾

1200𝐾=2

3Recall further that for a Carnot cycle, one has

𝑞𝑙𝑜𝑤𝑞ℎ𝑖𝑔ℎ

=𝑇𝑙𝑜𝑤𝑇ℎ𝑖𝑔ℎ

For this problem then one has

−41𝑞

23𝑞

=𝑇𝑙𝑜𝑤𝑇ℎ𝑖𝑔ℎ

1.261 × 102𝑘𝐽𝑘𝑔

3.78 × 102𝑘𝐽𝑘𝑔

=400𝐾

1200𝐾

1

3=1

3

Table 1-3 summarizes the first law parameters for Carnot cycle.

Process∆𝑢

𝑘𝐽

𝑘𝑔𝑞

𝑘𝐽

𝑘𝑔𝑤

𝑘𝐽

𝑘𝑔

1-2

2-3

3-4

4-1

5.74× 102

0

-5.74× 102

0

0

3.783× 102

0

-1.2612× 102

-5.74× 102

3.783× 102

5.74× 102

-1.2612× 102

Total 0 2.5224× 102 2.5224× 102

Example 1-4

Given a steam turbine with 𝑚 = 1.5𝑘𝑔

𝑠, 𝑄𝐶𝑉 = −8.5𝑘𝑊 with the following inlet and exit

Conditions:

𝑃𝑖 = 2𝑀𝑃𝑎, 𝑇𝑖 = 350℃, 𝑣𝑖 = 50𝑚

𝑠, 𝑧𝑖 = 6𝑚, 𝑃𝑒 = 0.1𝑀𝑃𝑎, 𝑥𝑒 = 1, 𝑣𝑒 = 200

𝑚

𝑠, 𝑧𝑒 = 3𝑚

Find the power output. The same simple sketch of Fig. 1-11 applies.

Fig. 1-11: Sketch of turbine problem

Answer:

The first law states:

𝑚(ℎ𝑖 +𝑣𝑖2

2+ 𝑔𝑧𝑖 + 𝑞𝑖) = 𝑚(ℎ𝑒 +

𝑣𝑒2

2+ 𝑔𝑧𝑒 + 𝑤𝑒)

𝑚𝑤𝑒 = 𝑄𝐶𝑉 + 𝑚 ℎ𝑖 − ℎ𝑒 +1

2(𝑣𝑖

2 − 𝑣𝑒2) + 𝑔 𝑧𝑖 − 𝑧𝑒

𝑄𝐶𝑉 = 𝑚𝑞𝑖 , 𝑊𝑒= 𝑚𝑤𝑒From the steam tables, we learn that 2𝑀𝑃𝑎 saturated below 350℃ so that the vapour under super heat region. Under 2𝑀𝑃𝑎 and 350℃ from super heat tables:

ℎ𝑖 = 3137𝑘𝐽

𝑘𝑔, ℎ𝑒 = 2675.5

𝑘𝐽

𝑘𝑔So

𝑊𝐶𝑉

= −8.5𝑘𝑊

+ (1.5𝑘𝑔

𝑠) 3137

𝑘𝐽

𝑘𝑔− 2675.5

𝑘𝐽

𝑘𝑔+1

2

𝑘𝐽

1000𝐽50

𝑚

𝑠

2

− 200𝑚

𝑠

2

Where

461.6𝑘𝐽

𝑘𝑔= ∆ℎ, 18.75

𝑘𝐽

𝑘𝑔= ∆𝐾𝐸, 0.0294

𝑘𝐽

𝑘𝑔= ∆𝑃𝐸

So that= 655.7 𝑘𝑊

Note

• The dominant term is the ∆ℎ term. Kinetic and potential energy changes, as well as heat transfer effects are small in comparison. This is typical for turbines.

Final notes:

For heat exchangers shown in Fig 1-12, we typically neglect all work, as well as changes in kinetic and potential energy. Also,

• There will be exchange of thermal energy between individual flow streams, but

• Globally for the entire device, there will be no heat transfer with the environment.

Fig. 1-14: Sketch of counter flow heat exchanger

Let us consider a counter flow heat exchanger. The mass balance for steady flow is trivial.The energy balance, neglecting changes in 𝐾𝐸 and 𝑃𝐸 states

𝑚𝑖ℎ𝑖 +𝑄𝐶𝑉 = 𝑊𝐶𝑉 + 𝑚𝑒ℎ𝑒

Where𝑊𝐶𝑉 = 0, 𝑄𝐶𝑉 = 0

Applying this to the counter flow heat exchanger gives 𝑚1ℎ1,ℎ𝑜𝑡 + 𝑚2ℎ2,𝑐𝑜𝑙𝑑 = 𝑚1ℎ1,𝑐𝑜𝑙𝑑 + 𝑚2ℎ2,ℎ𝑜𝑡

𝑚1(ℎ1,ℎ𝑜𝑡−ℎ1,𝑐𝑜𝑙𝑑) = 𝑚2(ℎ2,ℎ𝑜𝑡 − ℎ2,𝑐𝑜𝑙𝑑)