TUTORIAL-9 Solution (20/04/2017) Thermodynamics for Aerospace Engineers (AS1300)

Phase change in Pure Substances and Vapour Cycles (steam)

Q1. Determine the phase or phases in a system consisting of water at the following conditions

and sketch p-v or T-v diagrams showing the location of each state with respect to the saturation

lines. (a) 5 bar, 151.9 °C (b) 5 bar, 200 °C (c) 2.5 MPa, 200 °C (d) 4.8 bar, 160 °C.

Solution:

The above mentioned states can be classified into three categories. The first is sub-cooled liquid,

the second is saturated vapour and liquid state, and the final is superheated steam state. Using the

steam tables these information can be obtained.

a.) T=151.9 °C, P=5 bar – It can be observed in the steam tables that this pressure and

temperature lie in the region of saturated liquid and vapour state. The exact location of the

state is unknown as it lies on the horizontal line shown in the Fig. 1.1. Look for state ‘1’ in the P-

v and T-S diagram in the Fig. 1.1. To know the exact location of the state the volume has to be

specified.

b.) T=200 °C, P=5 bar - As the temperature in this case is more than that of option ‘a’ with the

same pressure as option ‘a’ the state lies in the superheated steam regime. Look for state ‘2’ in

the P-v and T-S diagram in the Fig. 1.1.

c.) T=200 °C, P=25 bar - As the saturated pressure for the temperature of 200 °C is 15.54 bar

and the corresponding pressure is more than this the state lies in sub-cooled or compressed

liquid region. Look for state ‘3’ in the P-v and T-S diagram in the Fig. 1.1.

d.) T=160 °C, P=4.8 bar - As the saturated pressure for the temperature of 160 °C is 6.187 bar

and the corresponding pressure is less than this the state lies in superheated steam regime.

Look for state ‘4’ in the P-v and T-S diagram in the Fig. 1.1.

Figure 1.1

Q2. Using steam tables, determine the specified property data from each of the folowing states.

Show the location of each state with respect to the saturation lines. (a) 3 bar, 240 °C, find ‘v’and

‘u’; (b) 3 bar, 0.3 m3/kg, find T and ‘u’; (c) 10 bar, 400 °C find ‘v’ and ‘h’; (d) 320 °C, 0.026

m3/kg find ‘p’ and ‘u’.

Solution

(a) 𝑝 = 3 bar, 𝑇 = 240 °C

𝑇𝑠𝑎𝑡@ 3 bar = 133.6 °C. Therefore the given state is superheated.

Use the superheated table for 𝑝 = 3 bar. Perform a linear interpolation between the values

provided in the table to obtain 𝑣 = 0.78 m3/kg and 𝑢 = 2713.4 kJ/kg.

(b) 𝑝 = 3 bar, 𝑣 = 0.3 m3/kg

𝑣𝑓@ 3 bar = 0.001073 m3/kg and 𝑣𝑔@ 3 bar = 0.606 m3/kg. The given 𝑣 satisfies 𝑣𝑓 < 𝑣 <

𝑣𝑔. Therefore the state is saturated (water steam mixture).

We can find 𝑇 = 133.6 °C (Boiling point at 3 bar).

The dryness factor 𝑥 = (𝑣 − 𝑣𝑓) (𝑣𝑔 − 𝑣𝑓)⁄ = 0.4941

Thus, 𝑢 = 𝑥𝑢𝑔 + (1 − 𝑥)𝑢𝑓 = 1540.77 kJ/kg.

(c) 𝑝 = 10 bar, 𝑇 = 400 °C

𝑇𝑠𝑎𝑡@ 10 bar = 179.9 °C. Therefore the given state is superheated.

Using the superheated table for 𝑝 = 10 bar, we can obtain 𝑣 = 0.307 m3/kg and ℎ = 3264

kJ/kg.

(d) 𝑇 = 320 °C, 𝑣 = 0.026 m3/kg

𝑣𝑔@ 320 °C = 0.015 m3/kg. We can see that 𝑣 > 𝑣𝑔. Therefore, the state is superheated.

We know that 𝑝 must be lower than 𝑝𝑠𝑎𝑡@ 320 °C. In the superheated tables, check for the value

of 𝑝 where 𝑇 = 320 °C and 𝑣 = 0.026 m3/kg. We can find that 𝑝 = 80 bar fits this condition.

By interpolation, we can find 𝑢 = 2645.6 kJ/kg.

Q3. A system consisting of 1 kg of steam/water substance at 150°C is contained in a frictionless

piston-cylinder mechanism. From an initial volume of 0.35 m3, the steam is compressed slowly

until the volume has been reduced to 10% of the initial volume. During the process heat transfer

occurs at such a rate as to keep the temperature constant at 150 °C. Indicate the initial and final

states and the compression process on a p-v diagram clearly, showing the saturation lines.

Calculate the work and the heat interactions for the system.

Solution

Given the mass of the water/steam system (m) is 1 kg. The temperature (T1) is 150 °C. The

substance is in saturated liquid-vapour region.

As the substance is contained in a piston cylinder arrangement, the pressure in the container will

be constant throughout any process undergone by the system.

The initial volume of the water/steam system is (V1) is 0.35 m3. The system is compressed

slowly to the 10% of its volume. So final volume of the system is (V2) is 0.035 m3. The system is

compressed slowly suggests that the process is a reversible process. Look at Fig 1.2

The initial and final specific volume can be calculated by dividing the volume with mass of the

system which is 1 kg.

𝑣1 = 0.35𝑚3

𝑘𝑔

𝑣2 = 0.035𝑚3

𝑘𝑔

Looking at the steam table, the above initial and final specific volume lies in the saturated liquid-

vapour regime with saturated vapour specific volume (𝑣𝑔) of 0.393 m3/kg and saturated liquid

specific volume (𝑣𝑓) of 0.001091 m3/kg

𝑣𝑔 = 0.393𝑚3

𝑘𝑔

𝑣𝑓 = 0.001091𝑚3

𝑘𝑔

So, from the above information the pressure and temperature for this compression process

remains constant as the both initial and final states lie on the saturated liquid-vapour region.

The dryness factor (x) for both the states can be calculated using the volume of the state (𝑣1 or

𝑣2) and the corresponding saturated liquid and vapour specific volume (𝑣𝑓, 𝑣𝑔)

The dryness factor (x) for both the states are calculated below:

𝑥1 =(𝑣1 − 𝑣𝑓)

(𝑣𝑔 − 𝑣𝑓)=

0.35 − 0.001091

0.393 − 0.001091= 0.89

𝑥2 =(𝑣2 − 𝑣𝑓)

(𝑣𝑔 − 𝑣𝑓)=

0.035 − 0.001091

0.393 − 0.001091= 0.086

As this is a constant pressure process, the heat transfer to the system in the above process is

equal to the change in enthalpy of the system. This statement can be proved using first law of

thermodynamics as follows:

𝛿𝑄 = ∆𝑈 + 𝛿𝑊 = 𝐶𝑣∆𝑇 + 𝑃𝑑𝑣 = 𝐶𝑣∆𝑇 + 𝑅𝑔𝑎𝑠∆𝑇 = 𝐶𝑃𝑔𝑎𝑠∆𝑇 = ∆𝐻 (𝑢𝑛𝑖𝑡𝑠 − 𝐽/𝑘𝑔) [3.1]

Using the dryness factor (x) the corresponding enthalpy (H) and internal energy (U) of the

system at initial and final state can be calculated.

ℎ𝑔 = 2746𝑘𝐽

𝑘𝑔

ℎ𝑓 = 632.2𝑘𝐽

𝑘𝑔

𝑢𝑔 = 2559𝑘𝐽

𝑘𝑔

𝑢𝑓 = 631.7𝑘𝐽

𝑘𝑔

Using the above enthalpy and internal energy of the saturated liquid and vapour state, the

enthalpy and internal energy (U) of the system at initial and final state is calculated below.

ℎ1 = ℎ𝑓 + 𝑥1 ∗ (ℎ𝑔 − ℎ𝑓) = 632.2 + 0.89 ∗ (2746 − 632.2) = 2513.482𝑘𝐽

𝑘𝑔

ℎ2 = ℎ𝑓 + 𝑥2 ∗ (ℎ𝑔 − ℎ𝑓) = 632.2 + 0.086 ∗ (2746 − 632.2) = 813.987𝑘𝐽

𝑘𝑔

𝑢1 = 𝑢𝑓 + 𝑥1 ∗ (𝑢𝑔 − 𝑢𝑓) = 631.7 + 0.89 ∗ (2559 − 631.7) = 2346.98𝑘𝐽

𝑘𝑔

𝑢2 = 𝑢𝑓 + 𝑥2 ∗ (𝑢𝑔 − 𝑢𝑓) = 631.7 + 0.086 ∗ (2559 − 631.7) = 797.44𝑘𝐽

𝑘𝑔

The saturated pressure value which remains constant throughout the process obtained from the

steam table is 4.758 bar

𝑇 = 150 ℃

𝑃 = 4.758 ∗ 105 𝑃𝑎

So, work done can be calculated as follows:

𝛿𝑊 = 𝑚 ∗ ∫ 𝑃𝑑𝑣0.035

0.35

= 𝑚 ∗ 𝑃(𝑣2 − 𝑣1) = 1 ∗ 4.758 ∗ 105 ∗ (0.035 − 0.35) = −𝟏𝟒𝟗. 𝟖𝟕𝟕 𝒌𝑱

The change in internal energy can be calculated as follows:

∆𝑈 = 𝑚 ∗ (𝑢2 − 𝑢1) = 1 ∗ (797.44 − 2346.98) = −𝟏𝟓𝟒𝟗. 𝟓𝟒 𝒌𝑱

The heat transfer to the system can be calculated as follows:

𝛿𝑄 = ∆𝐻 = 𝑚 ∗ (ℎ2 − ℎ1) = 1 ∗ (813.987 − 2513.482) = −𝟏𝟔𝟗𝟗. 𝟒𝟗𝟓 𝒌𝑱

The above answer for heat transfer to the system can also be found by adding work done and

internal energy change calculated in the above part.

Figure 1.2

Q4. 1 kg of water at 0.2 MPa is initially enclosed within a volume of 0.10 m3 in a piston-cylinder

arrangement. The piston initially rests on stops and will move when the pressure is 1.0 MPa.

During a certain process, an amount heat equal to 2500 kJ is transferred to the water. Determine

the work done and the final state of the water.

Solution

𝑚 = 1 kg, 𝑃1 = 0.2 MPa, 𝑉1 = 0.1 m3 and 𝑄 = 2500 kJ. We can find 𝑣1 = 0.1 m3/kg.

The piston does not move until the pressure reaches 𝑃2 = 1 MPa. Thus, the volume remains

constant during this process i.e., 𝑣2 = 0.1 m3/kg. We can notice that state 1 is a saturated liquid

vapor mixture since 𝑣1𝑓 < 𝑣1 < 𝑣1𝑔 corresponding to 𝑃1 = 0.2 MPa. The dryness factor 𝑥1 =

(𝑣1 − 𝑣1𝑓) (𝑣1𝑔 − 𝑣1𝑓)⁄ = 0.1118. Thus, the specific internal energy corresponding to state 1 is

given by 𝑢1 = 𝑥1𝑢1𝑔 + (1 − 𝑥1)𝑢1𝑓 = 730.95 kJ/kg. For state 2, we can find that it is saturated

since 𝑣2𝑓 < 𝑣2 < 𝑣2𝑔 corresponding to 𝑃2 = 1 MPa. The dryness factor 𝑥2 =

(𝑣2 − 𝑣2𝑓) (𝑣2𝑔 − 𝑣2𝑓)⁄ = 0.513. Thus, the specific internal energy corresponding to state 2 is

given by 𝑢2 = 𝑥2𝑢2𝑔 + (1 − 𝑥2)𝑢2𝑓 = 1696.54 kJ/kg. For the process 1-2 (isochoric), the first

law becomes 𝑞1−2 = ∆𝑢1−2 = 𝑢2 − 𝑢1 = 965.59 kJ/kg, where 𝑞 is heat supplied per kg.

Once 𝑃2 = 1 MPa is reached, the remaining heat added (𝑞 − 𝑞1−2 = 1534.41 kJ/kg) is used to

lift the piston and take the system to state 3. Thus 𝑞2−3 = 𝑞 − 𝑞1−2 and the process 2-3 is

isobaric (𝑃3 = 𝑃2 = 1 MPa).

Therefore, from first law, 𝑞2−3 = ∆ℎ2−3 = ℎ3 − ℎ2 = 1534.41 kJ/kg. We can find ℎ2 =

𝑥2ℎ2𝑔 + (1 − 𝑥2)ℎ2𝑓 = 1796.6 kJ/kg. Thus, we can evaluate ℎ3 = 3331 kJ/kg. The state 3 is

superheated since ℎ3 > ℎ𝑔@ 1 MPa. Use the superheated table for 1 MPa and interpolate

linearly to find 𝑣3 = 0.322 m3/kg.

Work done by the gas during the process 𝑊 = 𝑚𝑃3(𝑣3 − 𝑣2) = 222 kJ.

Q5. Saturated steam-water mixture of 0.6125 kg mass at 150 kPa, 0.6 m3 is enclosed in a piston

cylinder arrangement. Calculate the work interaction and the change in internal energy for this

system for the following cases: (a) The mixture is heated at constant pressure until its

temperature becomes 200 °C, (b) The mixture follows pV= constant until it becomes a saturated

vapour; (c) The pressure increases linearly with volume until final pressure is 200 kPa and

temperature is 250 °C.

Solution

a.)

Given the mass of the water/steam system (m) is 0.6125 kg. The initial pressure (P1) is 150 kPa

or 1.5 bar. The substance is in saturated liquid-vapour region.

The substance is contained in a piston cylinder arrangement and the initial volume of the

container (V1) is 0.6 m3.

The initial specific volume can be calculated by dividing the volume with mass of the system

which is 0.6125 kg.

𝑣1 =0.6

0.6125= 0.9796

𝑚3

𝑘𝑔

Looking at the steam table, the above initial specific volume lies in the saturated liquid-vapour

regime with saturated vapour specific volume (𝑣𝑔) of 1.159 m3/kg and saturated liquid specific

volume (𝑣𝑓) of 0.001053 m3/kg

𝑣𝑔 = 1.159𝑚3

𝑘𝑔

𝑣𝑓 = 0.001053𝑚3

𝑘𝑔

The temperature of the initial state (T1) which is the saturated temperature for the given pressure

1.5 bar is 111 °C.

As, it is mentioned in the question that the mixture is heated at constant pressure upto a

temperature of 200 °C which is greater than the saturated temperature of 111 °C. So the final

state lies on the superheated steam regime.

The dryness factor (x) for the initial state is as follows:

𝑥1 =(𝑣1 − 𝑣𝑓)

(𝑣𝑔 − 𝑣𝑓)=

0.9796 − 0.001053

1.159 − 0.001053= 0.8451

Using the dryness factor (x) the corresponding enthalpy (H) and internal energy (U) of the

system at initial state can be calculated.

ℎ𝑔 = 2694𝑘𝐽

𝑘𝑔

ℎ𝑓 = 467.1𝑘𝐽

𝑘𝑔

𝑢𝑔 = 2520𝑘𝐽

𝑘𝑔

𝑢𝑓 = 466.9𝑘𝐽

𝑘𝑔

Using the above enthalpy and internal energy of the saturated liquid and vapour state, the

enthalpy and internal energy (U) of the system at initial state is calculated below.

ℎ1 = ℎ𝑓 + 𝑥1 ∗ (ℎ𝑔 − ℎ𝑓) = 467.1 + 0.8451 ∗ (2694 − 467.1) = 2349.05𝑘𝐽

𝑘𝑔

𝑢1 = 𝑢𝑓 + 𝑥1 ∗ (𝑢𝑔 − 𝑢𝑓) = 466.9 + 0.8451 ∗ (2520 − 466.9) = 2201.97𝑘𝐽

𝑘𝑔

The final pressure and temperature of the system is as follows:

𝑇2 = 200 ℃

𝑃2 = 𝑃1 = 1.5 ∗ 105 𝑃𝑎

Looking in the superheated steam table for the pressure of 1.5 bar the following properties at the

final state is obtained.

𝑣2 = 1.626𝑚3

𝑘𝑔

𝑢2 = 2656 𝑘𝐽

𝑘𝑔

ℎ2 = 2873 𝑘𝐽

𝑘𝑔

So, work done can be calculated as follows:

𝛿𝑊 = 𝑚 ∗ ∫ 𝑃𝑑𝑣1.626

0.9796

= 𝑚 ∗ 𝑃(𝑣2 − 𝑣1) = 0.6125 ∗ 1.5 ∗ 105 ∗ (1.626 − 0.9796)

= 𝟓𝟗. 𝟑𝟖𝟖 𝒌𝑱

The change in internal energy can be calculated as follows:

∆𝑈 = 𝑚 ∗ (𝑢2 − 𝑢1) = 0.6125 ∗ (2656 − 2201.97) = 𝟐𝟕𝟖. 𝟎𝟗 𝒌𝑱

The heat transfer to the system can be calculated as follows:

𝛿𝑄 = ∆𝐻 = 𝑚 ∗ (ℎ2 − ℎ1) = 0.6125 ∗ (2873 − 2349.05) = 𝟑𝟑𝟕. 𝟒𝟕𝟖 𝒌𝑱

The above answer for heat transfer to the system can also be found by adding work done and

internal energy change calculated in the above part.

b.)

In this process the product of pressure and the temperature is constant. It is mentioned in the

question that the final volume of the state is in saturated vapour point. As the initial volume is in

the saturated liquid-vapour regime, the final volume will be more than the initial volume. So the

final pressure will be less than the initial pressure.

Calculating the product of pressure and the saturated vapour specific volume matching the initial

product of pressure and specific volume the final state is obtained.

𝑃1 ∗ 𝑣1 = 1.5 ∗ 105 ∗ 0.9796 ∗= 146940 = 0.1 ∗ 105 ∗ 14.68 = 𝑃2 ∗ 𝑣𝑔2= 𝐶

Where C is a constant.

So the final pressure is 0.1 bar and specific volume is 14.68 m3/kg.

𝑃2 = 0.1 ∗ 105 𝑃𝑎

𝑣2 = 14.68 𝑚3

𝑘𝑔

The internal energy at the final state is as follows:

𝑢2 = 2438 𝑘𝐽

𝑘𝑔

So, work done can be calculated as follows:

𝛿𝑊 = 𝑚 ∗ ∫ 𝑃𝑑𝑣0.035

0.35

= 𝑚 ∗ 𝐶 ∗ ∫1

𝑣𝑑𝑣

14.68

0.9796

= 0.6125 ∗ 146950 ∗ ln14.68

0.9796= 𝟐𝟒𝟑. 𝟒𝟎𝟖 𝒌𝑱

The change in internal energy can be calculated as follows:

∆𝑈 = 𝑚 ∗ (𝑢2 − 𝑢1) = 0.6125 ∗ (2438 − 2201.97) = 𝟏𝟒𝟒. 𝟓𝟔 𝒌𝑱

The heat transfer to the system can be calculated as follows:

𝛿𝑄 = ∆𝑈 + 𝛿𝑊 = 𝟑𝟖𝟕. 𝟗𝟔𝟖 𝒌𝑱

c.)

In this case the pressure increases linearly with volume. The final pressure and temperature is

200 kPa and 250 °C.

𝑃2 = 2 ∗ 105 𝑃𝑎

𝑇2 = 250 °C

By looking at the steam table it can be inferred that the above final state lie in the superheated

steam regime. The specific volume and the internal energy is as follows:

𝑣2 = 1.199 𝑚3

𝑘𝑔

𝑢2 = 2731 𝑘𝐽

𝑘𝑔

Let us assume pressure as a linear function of specific volume which can be written as follows:

𝑃 = 𝐴𝑣 + 𝐵

@ 𝑃1 = 1.5 ∗ 105𝑃𝑎 − − − −−→ 𝑣1 = 0.9796 𝑚3

𝑘𝑔

@ 𝑃2 = 2 ∗ 105𝑃𝑎 − − − −−→ 𝑣2 = 1.199 𝑚3

𝑘𝑔

Solving the equation with the above boundary conditions the values of the constants can be

obtained.

𝐴 = 2.2789 ∗ 105𝑘𝑔 ∗ 𝑃𝑎

𝑚3

𝐵 = −0.732 ∗ 105 𝑃𝑎

So, work done can be calculated as follows:

𝛿𝑊 = 𝑚 ∗ ∫ 𝑃𝑑𝑣0.035

0.35

= 𝑚 ∗ 𝐶 ∗ ∫ (𝐴𝑣 + 𝐵)𝑑𝑣1.199

0.9796

= 𝟐𝟑. 𝟓𝟐 𝒌𝑱

The change in internal energy can be calculated as follows:

∆𝑈 = 𝑚 ∗ (𝑢2 − 𝑢1) = 0.6125 ∗ (2731 − 2201.97) = 𝟑𝟐𝟒. 𝟎𝟒 𝒌𝑱

The heat transfer to the system can be calculated as follows:

𝛿𝑄 = ∆𝑈 + 𝛿𝑊 = 𝟑𝟒𝟕. 𝟓𝟔 𝒌𝑱

Q6. Steam at 20 bar, 360°C is expanded in a steam turbine to 0.08 bar. It then enters a

condenser, where it is condensed to saturated liquid water. The pump feeds back the water into

the boiler. (a) Assuming ideal processes, find net work per kg of steam and the cycle efficiency.

(b) If the turbine and the pump each have 80% efficiency, find the percentage reduction in the

net work and cycle efficiency.

Solution

Given 𝑇1 = 360°C and 𝑃1 = 20 bar.

Figure 6.1

(a) Using the steam tables, we can find that 𝑇1 > 𝑇𝑠𝑎𝑡@ 20 bar. Thus state 1 is superheated

steam. Then, ℎ1 = 3159.3 kJ/kg and 𝑠1 = 6.9917 kJ/ kg. K. The steam undergoes an isentropic

expansion to 𝑃2 = 0.08 bar. 𝑠2𝑠 = 𝑠1 = 6.9917 kJ/ kg. K. For state 2, at 𝑃2 = 0.08 bar, 𝑠2𝑓 <

𝑠2𝑠 < 𝑠2𝑔. Therefore, state 2 is saturated liquid vapor mixture. We can calculate the dryness

factor 𝑥2𝑠 = (𝑠2 − 𝑠2𝑓) (𝑠2𝑔 − 𝑠2𝑓)⁄ = 0.838. Thus, ℎ2𝑠 = 𝑥2𝑠ℎ2𝑔 + (1 − 𝑥2𝑠)ℎ2𝑓 = 2187.68

kJ/kg. Use the first law for a control volume to calculate the work done. Since the process is

adiabatic, work extracted by turbine 𝑊𝑇𝑠 = ℎ1 − ℎ2𝑠 = 971.62 kJ/kg.

The condenser takes the system to state 3 (saturated liquid) where 𝑠3 = 𝑠2𝑓, ℎ3 = ℎ2𝑓 and 𝑃3 =

𝑃2 = 0.08 bar. Then, the pump increases the pressure from 𝑃3 = 0.08 bar to 𝑃4 = 20 bar.

During this process, water is in its liquid state. Therefore, the specific volume is almost a

constant i.e., 𝑣4𝑠 = 𝑣3 = 𝑣𝑔@ 0.08 bar. Using the first law 𝑊𝑃𝑠 = ℎ4𝑠 − ℎ3 = 𝑣3(𝑃4 − 𝑃3) =

2.008 kJ/kg.

Net work output 𝑊𝑠,𝑛𝑒𝑡 = 𝑊𝑇𝑠 − 𝑊𝑃𝑠 = 969.612 kJ/kg.

Heat supplied to the system, 𝑄𝐻 = ℎ1 − ℎ4𝑠 = 2983.41 kJ/kg.

Thus, efficiency 𝜂 = 𝑊𝑠,𝑛𝑒𝑡 𝑄𝐻⁄ = 0.325

(b) If 𝜂𝑃 = 0.8 and 𝜂𝑇 = 0.8,

𝑊𝑃 = 𝑊𝑃𝑠 0.8⁄ = 2.51 kJ/kg and 𝑊𝑇 = 0.8 × 𝑊𝑇𝑠 = 777.3 kJ/kg.

Therefore 𝑊𝑛𝑒𝑡 = 𝑊𝑇 − 𝑊𝑃 = 774.8 kJ/kg.

Percentage reduction in work output =𝑊𝑠,𝑛𝑒𝑡−𝑊𝑛𝑒𝑡

𝑊𝑠,𝑛𝑒𝑡× 100 = 20.1%

𝑊𝑃 = ℎ4 − ℎ3, where ℎ3 = ℎ4𝑠 − 𝑊𝑃𝑠.

Therefore, ℎ4 = 𝑊𝑃 − ℎ3 = 176.39 kJ/kg.

Now, 𝑄𝐻′ = ℎ1 − ℎ4 = 2982.91 kJ/kg.

Efficiency 𝜂′ = 𝑊𝑛𝑒𝑡 𝑄𝐻′⁄ = 0.2597

Percentage reduction in efficiency =𝜂−𝜂′

𝜂× 100 = 20.1%