In steam power plant Rankine cycle is used. Fig. 11.8 shows schematic diagram for steam power plant. It consists of four steady state steady flow processes

Process Description

1-2 Reversible adiabatic pumping

2-3 Heat addition at constant pressure

3-4 Reversible adiabatic expansion in turbine

4-1 Heat rejection at constant pressure in the condenser

**11.9.1 T-S Diagram for Rankine Cycle**

The process is 1-2- 2′-3-4-1 cyclic.

In case the boiler supplies superheated steam, the cycle will be 1- 2-2′ -3-3′-4′-1.

**11.9.2 P-v Diagram for Rankine Cycle**

Here 1-2 is not vertical line it is curve. Similarly 3 to 4 and 3′ to 4′.

2-2′-3-3′ is constant pressure one line. Also 1-4-4′ constant pressure one line.

**1.9.3 h-s Diagram for Rankine Cycle**

1-2 vertical line, 3-4 and 3′-4 vertical lines, 2-2′-3-3′ on curve line also 4′-4-1 on curve line

**11.9.4 Thermal Efficiency Calculations**

For Process 1- 2 W _{p}= l_{2} – h_{1}

For Process 2- 3 q, = h_{3} – h_{2},

For Process 3-4 W _{t} = h_{3} -h_{4},

For Process 4-1 q _{R} = h_{4} – h_{1}

For Network = W _{r}-W _{p}

= (h_{3}– h_{4}) – (h_{2}– h_{l})

Heat supply = q s = h_{3} – h_{2}

ɳ Thermal = Net work / heat supply = ( h_{2} – h_{3} ) – (h_{2} – h_{1}) / h_{2} – h_{2}

= (h_{3} – h_{2}) – (h_{4} – h_{1}) / h_{3} – h_{2}

= 1 – h_{4} – h_{1} / h_{3} – h_{2} = 1 – q R / q s (obvious)

If w p is neglected then ,

ɳ _{Thermol }= h_{3} – h_{4} / h_{3} – h_{2}

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**11.9.5 Effect of Superheat**

1-2-2′-3-4–1 ~ dry saturated steam is used.

1-2-2′- 3- 3′-4′-1 = super heated steam used.

There is more work delivered by turbine= 3–3′-4′-4. Excess heat supply= 3- 3′- B- A.

As superheating increases the average temperature of heat addition to the cycle, the efficiency increases.

**11.9.6 Effect of Boiler Pressure**

P_{2} > P_{1} =>Increase in boiler pressure to P2, condenser pressure same at w1 = w2 (area of

the shaded portion). But heat rejection is decreased by 4-B-A-4′. Thus efficiency increases.

**11.9.7 Effect of Condenser Pressure**

P1 < P2 => Decrease in condenser pressure.

Increase in work ==> 1-4-4′-1′-2′-2-1.

Increase in heat supply to steam==> A-2′-2-B.lncrease in work> Increase in heat supply

Thus 11 increases.

**11.9.8 Comparison between Carnot Cycle and Rankine Cycle**

Carnot cycle ·

=> 1′- 2′- 3-4-1′

Rankine cycle

=> 1-2-2′-3-3′-4′-1

Rankine cycle is used for steam, Carnot cycle is not used for steam – this may be

explained by following.

(a) **Pumping process**: State 1′ is a mixture of liquid and vapour. We face great difficulty to encounter in manufacturing a pump that will handle the mixture of liquid +vapour at 1′ and deliver saturated liquid at 2′. It is easier to condense the vapour completely and handle only liquid in the pump. The Rankine cycle is based on this fact.

(b) **Superheating the vapour**: 3-3′ is a constant pressure process where the vapour superheated. In Carnot the operation of heat transfer is carried out at temperature, and therefore the vapour is superheated in process 3-3″. However pressure drops during the process which results in heat transfer to the vapour as it undergoes an expansion process in which work is done. There is a practical difficult) to transfer heat to achieve this. Thus Rankine cycle is the ideal cycle that can be approximated in practice for steam power plant.

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**SOLVED EXAMPLES**

**Example 11.1.** Steam at 50 bar pressure and 500°C expands adiabatically to a pressure of 1 bar in a steam turbine. Determine the work done in the turbine. The rate of steam flow through the turbine is 200 kg per minute.

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**Solution. **Given P1 = 50 bar T sup = 500°C, P2 = 1 bar.

Assuming the steam to be dry saturated at the end of expansion. From steam tables for

superheated steam corresponding to a pressure of 50 bar and 500°C, we find that

u_{1} = 3091 KJ /kg

and from steam tables for saturated steam corresponding to a pressure of 1 bar, we find that

u_{2 }= 2506.1 KJ/ kg

Change of internal energy= u_{1} – u_{2} = 584.9 KJ / kg.

Applying 1st law of thermodynamics

q= w+(u_{2} -u_{1})

Since heat transferred is zero during adiabatic process,

q= 0

Since w = u_{1} – u_{2} = 584.9 KJ /kg.

m = 200 kg / min = 200 60 kg / sec

(power) work done in the turbines/ sec. = 200 / 6 x 584.9 x KJ / kg

= 1949.66 KW.

**Example 11.2.** Steam at 5 bar, 250°C expands isentropically to a pressure of 0.7 bar. Determine the final condition of steam.

**Solution.** Let x2 = Final dryness fraction of steam. The problem may be solved either by

steam tables or by Mollier diagram.

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**1. By Steam Tables** : In an isentropic process, entropy remains constant.

From steam tables of superheated steam, corresponding to a pressure of 5 bar and

250°C, we find that

S_{SUP} = 7.271 KJ / kg K

From dry and saturated steam table at 0.7 bar pressure .

S _{f}= 1.192KJ/kg K

S _{fg} = 6.288 KJ /kg K

S _{SUP} = s _{f}+ x_{2} s _{fg}

7.271 = 1.192 + x_{2} (6.288)

x_{2 }= 0.966 Ans.

**2. By Mollier diagram:** The final condition of steam (x2) may be obtained by using Mollier

diagram as shown in Fig. 11.16. A vertical line is drawn from point 1 at an initial condition

of steam i.e., pressure 5 bar and temperature 250°C, upto the pressure line of 0.7 bar. The

dryness fraction is read at the point of intersection (i.e., 2) from which x_{2} = 0.966.

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**Example 11.3**. A cylinder contains steam at 1 bar and temperature 150°C. The steam is

compressed reversibly and isothermally (T = C) to a state where the specific volume is 0.28 m3/kg.

Find the change of internal energ1; , change of entropy and work done per kg of steam.

**Solution.** Given data, isothermal process

P_{1} = 1 bar= 100 KN/m^{2} T

t_{1} = t_{2} = 150°C

superheated steam table at

P_{1} = 1 bar and 150 0C we get,

vg_{1} = 1.936 m^{3}/ kg

hg_{1} = 2776.4KJ/kg

sg_{1} = 7.613KJ/kg K

u_{1} = ug_{1} = 2582.8 KJ /kg or

hg_{1} = ug_{1} + P_{l}vg_{1}

2776.4 = u g _{l} + 100 x 1.936

Since the process is isothermal, the saturation pressure, corresponding to 150°C is 4.8 bar

(P_{2}) in steam table.

Hence P_{2} = 4.8 bar h f_{2 }= 633 KJ / kg

vg_{2} = 0.39 h fg_{2}

v_{2} = x_{2} vg_{2 }s f_{2} = 1.845 KJ / kg

0.28 = x_{2} x 0.39 s fg_{2} = 4.990 KJ / kg

x2 = 0.718

h_{2} = hf_{2} + x_{2} hfg_{2}

= 633.6+ 0.718 X 2113.2

= 2150.8 KJ/kg.

u_{2} = h_{2}– p_{2} v_{2}

= 2150.8 – 4.8 X 100 X 0.28

= 2016.4 KJ /kg

s_{2} = sf_{2} + x_{2}sfg_{2}

= 1.845 + 0.718 X 4.990

= 5.428KJ/kg K

Change of entropy = s_{2} –s t

= 5.428 – 7.613

= – 2.185KJ/kg K. Ans.

Change of internal energy

= u_{2} – u_{1}

= 2016.4- 2582.8 KJ /kg

= -566.4KJ/kg

Heat transferred

q = T (s_{2} – s_{1})

= (273 + 150) (- 2.185)

= – 924.3 KJ /kg

Work done

=q-(u_{2}-u_{1})

= – 924.3- (- 566.4)

= – 924.3 + 566.3

= – 358KJ/kg. Ans.

**Example 11.4**. A spherical vessel having a capacity of0.85 m3 contains steam at 12 bar and 0.95 dryness. Steam is blown off until the pressure drops to 6 bar. The valve is then closed and the steam is allowed to cool until the pressure falls to 4 bar. Assuming that the enthalpy per kg of steam in the vessel remains constant during blowing off period, determine,

(i) The quantity of steam blown off

(ii) The quality of steam in the vessel after cooling.

(iii) The heat lost by steam per kg during cooling.

**Solution.** Given data

P_{1} = 12 bar

P_{2} = 6 bar

V= 0.85 m^{3}

X_{1} = 0.95

P_{3} = 4bar

Process 1- 2. Constant enthalpy (adiabatic)

h_{l} = /h_{2}

From dry and saturated steam table at 12 bar pressure

h f_{1} = 798.6 KJ / kg

h fg_{1} = 1986.2 KJ /kg

vg_{1} = 0.163 m3/kg

Capacity of vessel V= m_{1} x_{1} Vg_{1}

0.85= m_{1} x 0.95 X 0.163

or m_{1} = 5.489 kg

h_{1} = h_{f1} + x_{1} h_{gf1}

= 798.6 + 0.95 x 1986.2

= 2685.49 KJ /kg

h_{2} = h_{1} = 2685.49 KJ /kg.

From dry and saturated steam table at 6 bar pressure

h _{f2} = 670.6 KJ/kg

h_{fg2} = 2086.3 KJ / kg

v_{g2} = 0.316 m3/ kg

h_{2} = h_{f2 }+ x_{2} x h_{1}g_{2}

2685.49 = 670.6 + x2 x 2086.3

x_{2} = 0.965

V_{2} = X_{2} vg_{2} = 0.965 X 0.316

= 0.305 m3/kg.

V= m_{2} v_{2}

0.85 = m_{2} x 0.305

m_{2} = 2.785 kg

Quantity of steam blown off= m_{1} – m_{2}

= 5.489 – 2.785

= 2.704 kg. Ans.

Process 2- 3. Constant volume (v2 = v3 = 0.305 m3/ kg)

From dry and saturated steam table at 4 bar pressure

h _{f3} = 604.7 KJ /kg u_{3} = x_{2} ug_{3}

h _{fg3} = 2133.8 KJ / kg 0.305 = x_{3} x 0.463

v_{g3 }= 0.463 m3 /kg x_{3} = 0.658

Quality of steam in the vessel after cooling

= 0.658

= h_{f3} + x 3 h _{fg3}

= 604.7 + 0.658 X 2133.8

= 2008.74KJ/kg

Heat lost by steam during cooling

= (h_{2}– h_{3})- v_{3} (P_{2}– P_{3})

= (2685.49- 2008.74) – 0.305 (6- 4}100

= 615.75 KJ /kg Ans.

**Example 11.5.** 0.05 kg of steam at 15 bar is contained in a rigid vessel of volume 0.0076 m^{3 }What is the temperature of the steam? If the vessel is cooled, at what temperature will the steam be just dry saturated? Cooling is continued 11ntil the pressure in the vessel is 11 bar. Calculate the final dryness fraction of the steam.

**Solution.** Given data m = 0.05 kg

P_{1} = 15 bar

V= 0.0076 m^{3}

P_{2} = 11 bar

The steam is superheated

V= mv_{1}

0.0076 = 0.05 v_{1}

v_{1} = 0.152m^{3 }/ kg.

From superheated steam table, corresponding to a steam press of 15 bar and volume 0.152 m3 / kg, the temperature of steam is 250°C. [vg = 0.152 at 15 bar].

Process 1-2. Constant volume cooling (v1 = v2) from dry and saturated steam table,

corresponding to the pressure of 15 bar the saturation temperature of steam is 191.6°C.

From dry and saturated steam table corresponding to the pressure of 11 bar we get.

v_{g2} = 0.178 m3 / kg

v_{1 }= v_{2} = x_{2} x v_{g2}

0.152 = x2 x 0.178

x_{2 }= 0.854.

Thus the steam will be just dry saturated at the temperature of 191.6°C. Final dryness

fraction of steam = 0.854. Ans.

**Example 11.6.** Find the internal energy of 1 kg of steam at a pressure of 10 bar when the

condition of steam is ·

1. Wet and dry11ess fraction 0.85

2. Dry and saturated

3. Super heated, the degree of superheat being 50°C.

Tile specific heat of superheated steam at constant pressure is 2.01 KJ/kg K. You can 11se steam table or Moller’s chart

**Solution.** Given data

P= l0bar

X= 0.85

From dry and saturated steam table corresponding to a pressure of 10 bar we get

u _{f }= 761.7 KJ /kg

u _{f }= 2583.6 KJ /kg

= u _{fg} – u _{f}

= 2583.6 – 761.7

= 1821.9 KJ /kg

u = u _{f} +x u _{fg}

= 761.7 + 0.85 X 1821.9

= 2310.3 KJ / kg.

(2) Dry and saturated steam

u _{g} = 2583.6 KJ /kg

(3) Superheated steam

Degree of superheat= 50°C

From steam table, saturation temperature of steam at 10 bar is

t _{sat}= 180°C

Temperature of superheated steam

t _{sup}‘ = t _{sat} + 50 = 180 + 50

= 230°C.

From superheated steam table corresponding to a pressure of 10 bar 230°C temperature,

u _{g} = 2674.7KJ/ kg. Ans.

**Example 11.7.** 1 kg of steam at 120 bar and 380°C expands isentropically to a pressure of 38

bar. Calculate the work done by steam using (i) steam table (ii) Mollier clwrt.

**Solution. Process 1- 2.** Isentropic expansion (s1 = s2, q = 0)

m= 1kg

P_{1} = 120 bar

t_{1} = 380°C

P_{2} = 38 bar

**Solution.**

** (1). By using steam table**

From steam table corresponding to 120 bar press, the saturation temperature of steam is

T _{sat} = 324.8°C.

Since the temperature is more than saturation temperature, it is superheated steam. From

superheated steam table corresponding to a press of 120 bar and 380°C temperature.

S_{1} = 5.963 KJ /kg^{o} K

h_{1} = 2977.4KJ/kg

v_{1 }= 0.01965 m3 /kg

u_{1} = h_{1}– P_{l }v _{l}

= 2977.4- 120 x 100 x 0.01965

= 2741.6 KJ /kg.

From dry and saturated steam table corresponding to a pressure of 38 bar, we get,

s_{f2} = 2.769 KJ/kg

s_{fg2} = 3.323 KJ /kg K

u_{f2} = 1068 KJ/kg

U_{g2} = 2603 KJ /kg

u_{fg2} = ug_{2}– uf_{2}

= 2603-1068

= 1535KJ/kg

For isentropic process

s_{1} = s_{2} =5.963KJ/kg K

s_{2} = s_{f2} + x_{2} s_{fg2}

5.963 = 2.769 + x2 x 3.323

x_{2} = 0.96

u_{2} = u_{f2} + x_{2} u_{fg2}

= 1068 + 0.96 X 1535

= 2541.6 KJ/kg.

From first law of thermodynamics

q= W+u_{2}-u_{1}

o = W+U_{2}-u_{1}

or w = u_{1} – u_{2} = 2741.6-2541.6

= 200 KJ I kg Ans.

**(2) By Mollier chart**

Locate point 1 at the intersection of 120 bar press line and 380°C temperature line. Read the value of h_{1} and v_{1}

h_{1} = 2980 KJ /kg

v_{1} = 0.02 m3 / kg

From point 1, draw a vertical line and mark point 2 where this line cuts the constant pressure line of 38 bar. Read the value of h_{2} and v_{2}

h_{2} = 2:730 KJ /kg

v_{2} = 0.05 m3 /kg

Internal energy is given by

u_{1} = h_{1} – P_{1}v_{1}

= – 2980- 120 x 100 x 0.02

= 2740 KJ/ kg

u_{2 }= h_{2}– P_{2}v_{2}

= 2730- 38 x 100 X 0.05

= 25401<-.1 / kg

work done w = u_{1} – u_{2}

= 2740- 2540 = 200 KJ /kg Ans.

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**Example 11.8.** 10 kg/s of superheated steam at 5 bar and 300°C expands through a turbine to

final pressure of 0.1 bar and dryness fraction 0.95. find out the work generated by the turbine.

Is the flow reversible? Explain.

**Solution.** Given data

m= 10kg/s

P_{1} =5bar

t_{1} = 3oooc

P_{2} = 0.1 bar

x_{2 }= 0.95

From superheated steam table corresponding to a pressure of 5 bar and temperature 300°C,

h_{1} = 3064.2 KJ /kg

From dry and saturated steam table corresponding to press 0.1 bar. hp. = 191.8 KJ/kg

hfg_{2} = 2392.8 KJ / kg

h_{2} = hp. + x_{2} h_{1}g_{2}

= 191.8 + 0.95 x 2392.8

= 2465 KJ/kg

Work generated (power) by the turbine

w= m (h_{1}-h_{2})

= 10 (3064.2- 2465)

= 5992kW. Ans.

The flow through the turbine is not reversible due to the following reasons.

1. There is loss of heat due to friction of steam with blades and casting of the turbine.

The entropy of steam increases due to friction.

2. There is loss of heat due to radiations.

3. There is leakage of steam to atmosphere.

4. The superheated steam becomes wet at the end of expansion.