tutorial-8 solution (4 april 2017) thermodynamics for...

TUTORIAL-8 Solution (4th April 2017) Thermodynamics for Aerospace Engineers (AS1300)

Gas Power Cycles

Q1. A gas turbine power plant operating on an ideal Brayton cycle has a pressure ratio of 8. The

gas temperature is 300 K at the compressor inlet and 1300 K at the turbine inlet. Utilising the air

standard assumptions, determine (a) the gas temperature at the exits of the compressor and the

turbine, (b) what percentage of turbine power is used for running the compressor and (c) thermal

efficiency.

Solution:

Fig. 1.1 Reversible and irreversible Brayton cycle

Given: The pressure ratio across the isentropic compression process is 8. So, 𝑃2 𝑃1⁄ = 8. The

gas temperature at the inlet of the compressor (T1) is 300 K & at the inlet of the turbine (T3) is

1300 K.

𝑃2

𝑃1= 8, 𝑇1 = 300 𝐾, 𝑇3 = 1300 𝐾

The isentropic relationship between the various state properties can be written as follows:

𝑃 ∗ 𝑉𝛾 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 [1.1]

𝑇 ∗ 𝑉𝛾−1 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 [1.2]

𝑇𝛾

𝑃𝛾−1 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 [1.3]

Using equation [1.3] and pressure ratio value across the compressor, the temperature at the end

of the compressor can be found out. The calculation is as follows:

𝑇1

𝛾

𝑃1𝛾−1 =

𝑇2𝛾

𝑃2𝛾−1 [1.4]

𝑇2

𝑇1=

𝑃2

𝑃1

𝛾−1

𝛾 [1.5]

𝑇2 = 𝑇1𝑃2

𝑃1

𝛾−1

𝛾 = 300 ∗ 80.4

1.4 = 𝟓𝟒𝟑. 𝟒𝟑 𝑲 Ans

Similarly the temperature at the outlet of the turbine is as follows:

𝑇4 = 𝑇3𝑃4

𝑃3

𝛾−1

𝛾 = 1300 ∗1

8

0.4

1.4 = 𝟕𝟏𝟕. 𝟔𝟔 𝑲 Ans

Note: 𝑃2 = 𝑃3 & 𝑃1 = 𝑃4 as they lie on a constant pressure line. So, 𝑃2

𝑃1=

𝑃3

𝑃4= 8 .

Work done by the turbine can be calculated as follows:

𝑊𝑇 = 𝐶𝑝 ∗ (𝑇3 − 𝑇4) = 1005 ∗ (1300 − 717.66) = 585.252 𝑘𝐽/𝑘𝑔

Similarly, work done by the compressor is calculated below:

𝑊𝑐 = 𝐶𝑝 ∗ (𝑇1 − 𝑇2) = 1005 ∗ (300 − 543.43) = −244.647 𝑘𝐽/𝑘𝑔

Note : The negative sign conveys that work is done on compressor. In an isentropic process,

𝛿𝑄 = 𝛿𝑊 + ∆𝑈 = 0 which suggests 𝛿𝑊 = −∆𝑈. But in the case of Brayton cycle (open cycle),

we consider 𝛿𝑊 = −∆ℎ as flow work is also present here.

The net work produced by the Brayton cycle is as follows:

𝑊𝑛𝑒𝑡 = 𝑊𝑇 + 𝑊𝐶 = |𝑊𝑇| − |𝑊𝐶| = 585.252 − 244.647 = 340.605 𝑘𝐽/𝑘𝑔

Heat supplied in the constant pressure combustion is as follows:

𝑄𝑆 = 𝐶𝑝 ∗ (𝑇3 − 𝑇2) = 1005 ∗ (1300 − 543.43) = 760.352 𝑘𝐽/𝑘𝑔

The percentage of the work produced by the turbine used by the compressor is as follows:

% 𝑊𝑜𝑟𝑘 𝑢𝑠𝑒𝑑 =|𝑊𝑐|

|𝑊𝑇|∗ 100 =

(𝑇2−𝑇1)

(𝑇3−𝑇4)∗ 100 =

543.43−300

1300−717.66∗ 100 = 𝟒𝟏. 𝟒𝟐% Ans

Thermal efficiency of the cycle is defined as follows:

ŋ =|𝑊𝑇|−|𝑊𝑐|

|𝑄𝑆|=

(𝑇3−𝑇4)−(𝑇2−𝑇1)

(𝑇3−𝑇2)=

(𝑇3−𝑇2)−(𝑇4−𝑇1)

(𝑇3−𝑇2)= 1 −

(𝑇4−𝑇1)

(𝑇3−𝑇2) [1.6]

As, 𝑃2

𝑃1=

𝑃3

𝑃4 which suggests that

𝑇2

𝑇1=

𝑇3

𝑇4 . This expression can be obtained using isentropic

relationship in equation [1.3]. So, 𝑇4

𝑇1=

𝑇3

𝑇2 is also a valid relationship.

ŋ = 1 −(𝑇4−𝑇1)

(𝑇3−𝑇2)= 1 −

𝑇1

𝑇2

(𝑇4𝑇1

−1)

(𝑇3𝑇2

−1)= 1 −

𝑇1

𝑇2= 1 −

300

543.43= 𝟎. 𝟒𝟒𝟕𝟗 Ans

% ŋ = 0.4479 ∗ 100 = 𝟒𝟒. 𝟕𝟗 % Ans

Note: In the irreversible process the isentropic relationships cannot be used. So, the relationships

like 𝑇4

𝑇1=

𝑇3

𝑇2 will not hold in that case. So, the general formula for the thermal efficiency (equation

[1.7]) should be used in this case.

ŋ𝑖𝑟𝑟 = 1 −(𝑇4−𝑇1)

(𝑇3−𝑇2) [1.7]

Q2. Assuming a compressor efficiency of 80 % and turbine efficiency of 85 %, determine (a)

what percentage of turbine power is used for running the compressor, (b) the thermal efficiency

and (c) the turbine exit temperature. Take the gas turbine cycle described in the preceding

problem.

Solution :

This problem is a continuation of the question no. 1. In this question the process is irreversible as

opposed to the question no. 1 where reversible Brayton cycle is considered.

Given: The turbine efficiency is 85 % while the compressor efficiency is 80 %.

Note: We know that maximum work is obtained from a heat engine in a reversible process. In

this case, maximum work will get produced by the turbine while operating in a reversible

process. So, in an irreversible process the turbine work will get reduced.

Similarly, we also know that minimum work has to be given to a refrigerator or a heat pump

while operating in a reversible process. So, in this case work done needed by the turbine will be

minimum in a reversible process and will be greater than this value while operating in an

irreversible process.

So, you will expect the turbine work to get reduced and the work needed by the compressor to

get increased. So, the percentage of turbine work used by the compressor will be higher and the

net work done will get reduced. In overall, the thermal efficiency of the Brayton cycle will get

reduced because of the reduced net work done.

The definition of the turbine efficiency is as follows:

ŋ𝑇 =ℎ3−ℎ4

ℎ3−ℎ4𝑠=

𝑇3−𝑇4

𝑇3−𝑇4𝑠 [2.1]

Please refer the Fig. 1.1 for the meaning of various terms. The state ‘4s’ is the ideal state occur in

the case of a reversible process. The properties at state ‘2s’ and ‘4s’ is calculated in the question

1. and can be used here whenever required.

So, the actual temperature T4 and T2 can be found out by applying turbine efficiency and

compressor efficiency formula.

𝑇4 = 𝑇3 − ŋ𝑇 ∗ (𝑇3 − 𝑇4𝑠) = 1300 − 0.85(1300 − 717.66) = 𝟖𝟎𝟓. 𝟎𝟏 𝑲 Ans

The definition of the compressor efficiency is as follows:

ŋ𝐶 =ℎ2𝑠−ℎ1

ℎ2−ℎ1=

𝑇2𝑠−𝑇1

𝑇2−𝑇1 [2.2]

So the temperature T2 in the irreversible Brayton cycle can be calculated as follows:

𝑇2 = 𝑇1 + (𝑇2𝑠−𝑇1)

ŋ𝐶= 300 +

(543.43−300)

0.8= 𝟔𝟎𝟒. 𝟐𝟗 𝑲 Ans

Work done by the turbine can be calculated as follows:

𝑊𝑇 = 𝐶𝑝 ∗ (𝑇3 − 𝑇4) = 1005 ∗ (1300 − 805.01) = 497.464 𝑘𝐽/𝑘𝑔

Similarly, work done by the compressor is calculated below:

𝑊𝑐 = 𝐶𝑝 ∗ (𝑇1 − 𝑇2) = 1005 ∗ (300 − 604.29) = −305.811 𝑘𝐽/𝑘𝑔



Heat supplied in the constant pressure combustion is as follows:

𝑄𝐻 = 𝐶𝑝 ∗ (𝑇3 − 𝑇2) = 1005 ∗ (1300 − 604.29) = 699.188 𝑘𝐽/𝑘𝑔

The percentage of the work produced by the turbine used by the compressor is as follows:

% 𝑊𝑜𝑟𝑘 𝑢𝑠𝑒𝑑 =|𝑊𝑐|

|𝑊𝑇|∗ 100 =

(𝑇2−𝑇1)

(𝑇3−𝑇4)∗ 100 =

604.29−300

1300−805.01∗ 100 = 𝟔𝟏. 𝟒𝟕 % Ans



|𝑄𝐻|=

(𝑇3−𝑇4)−(𝑇2−𝑇1)

(𝑇3−𝑇2)=

(𝑇3−𝑇2)−(𝑇4−𝑇1)

(𝑇3−𝑇2)= 1 −

(𝑇4−𝑇1)

(𝑇3−𝑇2) [2.3]

As mentioned in the note section of the above question that the equation [2.3] cannot be further

simplified as isentropic relationship doesn’t hold true for an irreversible process.

So, the thermal efficiency can be calculated as follows:

% ŋ = (1 −(𝑇4−𝑇1)

(𝑇3−𝑇2)) ∗ 100 = (1 −

805.01−300

1300−604.29) ∗ 100 = 𝟐𝟕. 𝟒𝟏 % Ans

Q3. An ideal Otto cycle has compression ratio of 8. At the beginning of the compression process,

air is at 100 kPa and 17°C, and 800 kJ/kg of heat is transferred to air the constant volume heat

addition process. Assuming cold air standard assumption, determine (a) the maximum pressure

and temperature that occur during the cycle, (b) the net work done, (c) thermal efficiency and (d)

mean effective pressure for the cycle.

Solution:

Fig. 3.1 Reversible Otto cycle

Given: The compression ratio of Otto cycle is 8. The pressure and temperature at the beginning

of the compression process 100 kPa and 17°C respectively.

𝑣1

𝑣2= 8, 𝑇1 = 17°C = 290 𝐾, 𝑃1 = 100 𝑘𝑃𝑎

where 𝑣 is the specific volume (units – m3/kg)

The process 1-2 shown in the Fig. 3.1 is an isentropic compression process. The isentropic

relationship between various state properties can be written as follows:

𝑃 ∗ 𝑣𝛾 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 [3.1]

𝑇 ∗ 𝑣𝛾−1 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 [3.2]

𝑇𝛾

𝑃𝛾−1= 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 [3.3]

So, using the compression ratio (𝑣1

𝑣2⁄ ) the pressure and temperature at the state 2 can be found

out.

𝑃2 = 𝑃1 ∗𝑣1

𝑣2

𝛾

= 100 ∗ 81.4 = 1837.92 𝑘𝑃𝑎

𝑇2 = 𝑇1 ∗𝑣1

𝑣2

𝛾−1

= 290 ∗ 80.4 = 666.25 𝐾

The process 2-3 shown in Otto cycle shown in the Fig 3.1 is a constant volume or isochoric

process. It has been mentioned in the question that 800 kJ/kg amount of heat has been transferred

to the system in this process. As the process is a constant volume process, all the heat will be

used for changing internal energy of the system without any work done.

Using first law of Thermodynamics:

𝛿𝑄 = 𝑄𝐻 = ∆𝑈 = 𝐶𝑣 ∗ (𝑇3 − 𝑇2) = 800 𝑘𝐽/𝑘𝑔 [3.4]

Temperature at state 3 (maximum temperature) can be obtained from equation [3.4] as follows:

𝑇3 = 𝑇2 +𝑄𝐻

𝐶𝑣= 666.25 +

800∗1000

718= 𝟏𝟕𝟖𝟎. 𝟒𝟔 𝑲 Ans

In an isochoric process, temperature is directly proportional to pressure as volume remains

constant. So, the maximum pressure (P3) reached at the end of the isochoric pressure can be

calculated as follows:

𝑃3

𝑃2=

𝑇3

𝑇2 [3.5]

𝑃3 = 𝑃2 ∗𝑇3

𝑇2= 1837.92 ∗

1780.46

666.25= 𝟒𝟗𝟏𝟏. 𝟓𝟖 𝒌𝑷𝒂 Ans

Because of the constant volume heat addition and rejection 𝑣2 = 𝑣3 & 𝑣1 = 𝑣4.

𝑣1

𝑣2=

𝑣4

𝑣3= 8 [3.6]

The process 3-4 is the isentropic expansion process. So, the isentropic relationship in equation

[3.1] to [3.3] can be used to get temperature and pressure at the 4th state.

𝑃4 = 𝑃3 ∗𝑣3

𝑣4

𝛾

= 4911.69 ∗1

8

1.4

= 267.24 𝑘𝑃𝑎

𝑇4 = 𝑇3 ∗𝑣3

𝑣4

𝛾−1

= 1780.46 ∗1

8

0.4

= 774.99 𝐾

Note: The Otto cycle is a closed cycle. So, the system in this case is considered to be a control

mass or closed system. In the isentropic process, 𝛿𝑄 = 𝛿𝑊 + ∆𝑈 = 0 which suggests 𝛿𝑊 =

−∆𝑈. This is different from the work done expression in the Brayton cycle (Open cycle). In the

Brayton cycle, because of the consideration of flow work the expression for work done becomes

as 𝛿𝑊 = −∆ℎ.

Work done in the power stroke of Otto cycle can be calculated as follows:

𝑊𝑇 = 𝐶𝑣 ∗ (𝑇3 − 𝑇4) = 718 ∗ (1780.46 − 774.99) = 721.927 𝑘𝐽/𝑘𝑔

Similarly, work done in the compression process is calculated below:

𝑊𝑐 = 𝐶𝑣 ∗ (𝑇1 − 𝑇2) = 718 ∗ (290 − 666.25) = −270.147 𝑘𝐽/𝑘𝑔



Heat supplied in the constant volume combustion is as follows:

𝑄𝐻 = ∆𝑈 = 𝐶𝑣 ∗ (𝑇3 − 𝑇2) = 718 ∗ (1780.46 − 666.25) = 800.002 𝑘𝐽/𝑘𝑔



|𝑄𝐻|∗ 100 =

(𝑇3−𝑇4)−(𝑇2−𝑇1)

(𝑇3−𝑇2)=

(𝑇3−𝑇2)−(𝑇4−𝑇1)

(𝑇3−𝑇2)= 1 −

(𝑇4−𝑇1)

(𝑇3−𝑇2) [3.6]

Isentropic relationship is used to obtain following expressions.

𝑣1

𝑣2=

𝑣4

𝑣3 &

𝑇1

𝑇2=

𝑇4

𝑇3 &

𝑃1

𝑃2=

𝑃4

𝑃3 [3.7]

Using the above expressions the formula for the thermal efficiency is derived below.

ŋ = 1 −(𝑇4−𝑇1)

(𝑇3−𝑇2)= 1 −

𝑇1

𝑇2

(𝑇4𝑇1

−1)

(𝑇3𝑇2

−1)= 1 −

𝑇1

𝑇2= 1 −

300

543.43= 1 −

𝑣2

𝑣1

𝛾−1= 1 −

1

8

𝛾−1= 𝟎. 𝟓𝟔𝟒𝟕 Ans

% ŋ = 0.4479 ∗ 100 = 𝟓𝟔. 𝟒𝟕 % Ans

The specific volume at state 1 is as follows:

𝑣1 =𝑅𝑎𝑖𝑟 ∗ 𝑇1

𝑃1=

287 ∗ 290

100000= 0.8323

𝑚3

𝑘𝑔

𝑣2 =𝑣1

8=

0.8323

8= 0.104

𝑚3

𝑘𝑔

The definition of the mean effective pressure is as follows:

𝑃𝑚𝑒𝑎𝑛 =𝑊𝑛𝑒𝑡

𝑣𝑚𝑎𝑥−𝑣𝑚𝑖𝑛=

𝑊𝑛𝑒𝑡

𝑣1−𝑣2=

451.779

0.8323−0.104= 𝟔𝟐𝟎. 𝟑𝟐 𝒌𝑷𝒂 Ans

Where units of the 𝑊𝑛𝑒𝑡 is in J/kg and units of the specific volume 𝑣 is in m3/kg.

The mean effective pressure can also be written as follows:

𝑃𝑚𝑒𝑎𝑛 =𝑊𝑛𝑒𝑡

𝑉𝑚𝑎𝑥−𝑉𝑚𝑖𝑛=

𝑊𝑛𝑒𝑡

𝑉1−𝑉2 [3.8]

Where units of the 𝑊𝑛𝑒𝑡 is in J and units of the volume 𝑉 is in m3.

Q4. An ideal Diesel cycle with air as the working fluid ha a compression ratio of 18 and a cut-off

ratio of ’2’. At the beginning of the compression process, the working fluid is at 100 kPa, 27°C,

and 1917 cm3. Utilising the cold standard assumptions, determine (a) the temperature and

pressure of air at the end of each process, (b) net work output, (c) thermal efficiency and (d) the

mean effective pressure.

Solution

Given 𝑉1 𝑉2⁄ = 18, cut off ratio 𝑉3 𝑉2⁄ = 2, 𝑇1 = 300 K, 𝑃1 = 100 kPa and 𝑉1 = 1917 cm3.

Figure 4.1

The mass of the air can be found out using the perfect gas law

𝑚 = 𝑃1𝑉1 𝑅𝑇1⁄ = 2.2265 × 10−3 kg.

Note that 𝛾 = 1.4 for air.

(a) The process 1 − 2 is isentropic. Therefore, 𝑃2 = 𝑃1(𝑉1 𝑉2⁄ )𝛾 = 5719.81 kPa and 𝑇2 =

𝑇1(𝑉1 𝑉2⁄ )(𝛾−1) = 𝟗𝟓𝟑. 𝟑𝟎 K.

The process 2 − 3 is isobaric. Thus, 𝑃3 = 𝑃2 = 𝟓𝟕𝟏𝟗. 𝟖𝟏 kPa.

Also, 𝑉3 = 2𝑉2 = 𝑉1 9⁄ = 213 cm3.

Thus, 𝑇3 = 𝑃3𝑉3 𝑚𝑅⁄ = 𝟏𝟗𝟎𝟔. 𝟓𝟗 K.

The process 3 − 4 is isentropic and 𝑉4 = 𝑉1. Thus, using isentropic relations, we get 𝑃4 =

𝑃3(𝑉3 𝑉4⁄ )𝛾 = 263.9 kPa and 𝑇4 = 𝑇3(𝑉3 𝑉4⁄ )(𝛾−1) = 𝟕𝟗𝟏. 𝟕 K.

(b) Net work output 𝑊𝑛𝑒𝑡 = 𝑄𝐻 − 𝑄𝐶 = 𝑚𝐶𝑃(𝑇3 − 𝑇2) − 𝑚𝐶𝑉(𝑇4 − 𝑇1) = 𝟏𝟑𝟒𝟕. 𝟒𝟗 J.

In the above equation, 𝑄𝐻 is the heat supplied and 𝑄𝐶 is the heat rejected.

(c) Thermal efficiency 𝜂 = 𝑊𝑛𝑒𝑡 𝑄𝐻⁄ = 𝟎. 𝟔𝟑𝟏𝟔

(d) Mean effective pressure = 𝑊𝑛𝑒𝑡 (𝑉𝑚𝑎𝑥 − 𝑉𝑚𝑖𝑛)⁄ = 𝑊𝑛𝑒𝑡 (𝑉1 − 𝑉2)⁄ = 𝟕𝟒𝟒. 𝟐𝟔𝟒 kPa.

Q5. An air standard dual cycle has a compression ratio of 16, and compression begins at 1 bar,

50°C. The maximum pressure is 70 bar. The heat transferred to air at constant pressure is equal

to that at constant volume. Estimate (a) the pressures and temperatures at the cardinal points of

the cycle, (b) the cycle efficiency, and (c) the MEP of the cycle. Take 𝐶𝑉 = 0.718 kJ/kg K and

𝐶𝑃 = 1.005 kJ/kg K

Solution

Given 𝑇1 = 323 K, 𝑃1 = 1 bar, 𝑃3 = 𝑃4 = 70 bar, 𝑣1 𝑣2⁄ = 16. Also, 𝑄2−3 = 𝑄3−4.

Figure 5.1

(a) Using perfect gas law, 𝑣1 = 𝑅𝑇1 𝑃1⁄ = 0.9149 m3/kg.

From isentropic relation,

𝑇2

𝑇1= (

𝑣1

𝑣2)

(𝛾−1)

= 160.4

Therefore, 𝑇2 = 𝟗𝟕𝟗.15 K.

Also, 𝑃2

𝑃1= (

𝑣1

𝑣2)

𝛾

= 161.4

Therefore, 𝑃2 = 𝟒𝟖. 𝟓 bar.

The process 2 − 3 is isochoric and 𝑃3 = 70 bar

Thus, 𝑃3 𝑇3⁄ = 𝑃2 𝑇2⁄ . Substitute the values to get 𝑇3 = 𝟏𝟒𝟏𝟑. 𝟐𝟏 K.

𝑄2−3 = 𝑄3−4, which can be written as 𝑚𝐶𝑉(𝑇3 − 𝑇2) = 𝑚𝐶𝑃(𝑇4 − 𝑇3).

From the above relation, we can get 𝑇4 = 𝟏𝟕𝟐𝟑. 𝟐𝟓 K.

Now, 𝑣2 = 𝑣1 16⁄ = 0.05718 m3/kg.

Also, 𝑣3 = 𝑣2 = 0.05718 m3/kg.

Since the process 3 − 4 is isobaric, i.e., 𝑣4 = 𝑣3(𝑇4 𝑇3⁄ ) = 0.06973 m3/kg.

Note that 𝑣5 = 𝑣1 = 0.9149 m3/kg.

Using isentropic relations for the process 4 − 5, we get

𝑃5 = 𝑃4(𝑣4 𝑣5⁄ )𝛾 = 𝟏. 𝟗𝟎𝟓 bar

𝑇5 = 𝑇4(𝑣4 𝑣5⁄ )(𝛾−1) = 𝟔𝟏𝟓. 𝟒𝟏 K

(b) Heat supplied 𝑄𝐻 = 𝑄2−3 + 𝑄3−4 = 2𝐶𝑉(𝑇3 − 𝑇2) = 623.31 kJ/kg

Heat rejected 𝑄𝐶 = 𝐶𝑉(𝑇5 − 𝑇1) = 209.95 kJ/kg

Therefore, work done by cycle (from first law) 𝑊𝑛𝑒𝑡 = 𝑄𝐻 − 𝑄𝐶 = 413.36 kJ/kg

Thus, cycle efficiency 𝜂 = 𝑊𝑛𝑒𝑡 𝑄𝐻⁄ = 𝟎. 𝟔𝟔𝟑𝟐

(c) MEP = 𝑊𝑛𝑒𝑡 (𝑣𝑚𝑎𝑥 − 𝑣𝑚𝑖𝑛)⁄ = 𝑊𝑛𝑒𝑡 (𝑣1 − 𝑣2)⁄ = 𝟒𝟖𝟏. 𝟗𝟑 kPa

tutorial-8 solution (4 april 2017) thermodynamics for...

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