DEPARTMENT OF COLLEGIATE AND TECHNICAL...

DEPARTMENT OF COLLEGIATE AND TECHNICAL

EDUCATION

STUDY MATERIAL

ON

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APPLIED THERMAL ENGINEERING

(15ME52T)

UNIT 1: FORMATION OF STEAM

For V semester Diploma in Mechanical Engineering

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Prepared by

AMRUT SELECTION GRADE LECTURER IN MECHANICAL ENGG

GOVERNMENT POLYTECHNIC RAICHUR

2020-21

APPLIED THERMAL ENGINEERING (15ME52T)

2020-21

MECHANICAL ENGINEERING Page 2

Contents:- page no

1.1 Introduction 3

1.2Formation of steam concept, 3

1.3 Definition, Representation of wet stem, dry stem, superheated steam, on T-H diagram 4

1.4 T-h diagram of formation of steam at constant pressure 6

1.5 Properties of steam Concept and determination of dryness fraction-degree of superheat-sensible

heat- latent heat-enthalpy-entropy internal energy-external energy of evaporation and specific

volume of stem- 8

1.6 uses of steam table and mollier diagram- 13

1.7 Mollier chart or enthalpy entropy diagram 14

1.8 Numerical Examples 15

1.9 Steam vapours cycles- 20

1.10Carnot cycle-schematic diagram- representation on pv and T S diagram-rankine cycle-

schematic diagram- representation on PV and TS diagram (no numerical problem on steam vapour

cycles) 20

1.11Rankine cycle 22

1.12 Steam calorimeter 26

1.13Barrel calorimeter, 26

1.14Separating calorimeter, 28

1.15Throttling calorimeter 29

1.16Combined separating & throttling calorimeter- 31

APPLED THERMAL ENGINEERING 15ME52T 2020-21

MECHANICAL ENGINEIRNG Page 3

UNIT I

FORMATION OF STEAM

1.1 Introduction

Steam is the gaseous vapour form of water. It is commonly formed by boiling or evaporating

the water. Steam is an excellent working medium in various thermodynamic systems because of

its following properties:

1) It can carry large quantities of heat

2) It is produced from water which is cheap and readily available

3) It can be used for heating purposes after its duty as working agent is completed.

4) It can be used purely as a heating medium in food processing Industries because of a

fast, easily controllable and hygienic method of heating.

1.2 Formation of steam under constant pressure

Consider 1 kg of water at 00C in a piston - cylinder arrangement as shown in figure 1.1(i). This

piston and cylinder arrangement helps to maintain constant pressure inside the cylinder.

On addition of heat, the temperature of water starts rising until it reaches the boiling temperature

or saturation temperature at constant pressure. During this process, volume of water increases

with increase in temperature as shown in figure 1.1(ii). This heat absorbed by water is called

sensible heat.

Once the boiling temperature is reached, temperature remains constant. Further addition of heat,

makes steam to vaporize. Thus pushes piston upwards at constant pressure thereby increasing the

specific volume of the steam as shown in figure1.1 (iii). This heat absorbed by steam is called

latent heat or hidden heat as shown in figure 1.2 and converts remaining wet steam into dry

saturated steam at same saturation temperature as shown in figure1.1 (iv).



On further heating, the temperature of the dry steam increases beyond saturation temperature and

converts dry saturated steam into superheated steam as shown in figure1.1 (v) . This heat

absorbed is called Amount of superheat.

Fig 1.1: - Formation of steam under constant pressure

Fig1. 2: – Graphical representation of Formation of steam or T-h diagram



1.3 Definition and Representation of Forms/Types of Steam on T-h diagram

Generally, the steam exists in three forms

➢ Wet steam

➢ Dry steam

➢ Superheated steam

➢ The graph is plotted between T and h. The heating of the water from 0 0C to saturation

temperature at constant pressure is represented by the inclined line AB in the graph.

➢ On further heating, the water will gradually starts to evaporate and convert it to steam,

but during this process the temperature remains constant. It is represented by the

horizontal line BC in the graph

➢ On further heating, the temperature of the dry steam increases beyond saturation

temperature and it is represented by inclined line CD in the graph.

1 Wet Steam

The steam that contains water particles in suspension is known as wet steam. It is mixture of

vapour and water particles. The steam at any point of the line BC in the figure 1.2 indicates wet

steam. Wet steam does not obey the gas laws.

2 Dry Steam

The steam that does not contain any water particles is known as dry steam. The point C in the

figure 1.2 indicates dry steam. Steam obeys the gas laws.

3 Superheated Steam

When the dry steam is heated beyond saturation temperature, the steam formed at the

temperature, higher than saturated steam at given constant pressure is known as superheated

steam. The steam at any point of the inclined line CD in the figure 1.2 indicates superheated

steam and this steam obey the gas laws. Hence it is perfect gas.



Advantages of Superheated Steam

1) It has high heat content and so high capacity of doing work. Thus it results in an

economy in steam consumption.

2) The temperature of superheated steam being higher, it gives a high thermal efficiency

in heat engine.

3) The superheated steam can be considerably cooled during expansion in an engine

cylinder, before its temperature falls so low as to cause condensation on cylinder walls

which is a direct heat loss.

4) Super heated steam is obtained from waste furnace gases.

1.4 T-h diagram of formation of steam at constant pressure

Fig 1.3: T-h diagram of formation of steam at constant pressure

➢ The graph is plotted between T and h.

➢ The point A represents the initial condition of water at 00 C and pressure p (in bar) as

shown in Fig

➢ Line ABCD shows the relation between temperature and heat at a specific pressure of p

(in bar).



➢ The heating of water up to boiling temperature at constant pressure is represented by the

line AB in the fig.

➢ The temperature at which water starts boiling is called saturation temperature (Tg).

➢ The heat absorbed by the water is AP, known. as sensible heat or total heat of water

➢ The change of state from liquid to steam is shown by line BC in the fig.

➢ The heat absorbed during this stage is PQ, known as Latent heat of evaporation.

➢ The amount of heat required to evaporate 1kg of water to dry steam at saturation

temperature at a given constant pressure is defined as Latent heat of evaporation.

➢ The superheating process is shown by line CD.

➢ The heat absorbed during this stage is QR, known as degree of superheat.

➢ Line AR represents the total heat of the superheated steam.

Critical pressure and critical temperature

It may be noted from the figure 1.3, that when the pressure and saturation temperature increases,

the latent heat of vaporization decreases and it becomes zero at a point N where liquid and dry

steam lines meet. This point N is known as the critical point and at this point, the liquid and

vapour phases merge, and become identical in every respect.

➢ The temperature corresponding to critical point N is known as critical temperature

➢ The pressure corresponding to critical point N is known as critical pressure.

➢ For steam, the critical temperature is 374.15°C and critical pressure is 221.2 bar.

.



1.5 Properties of Steam

Important properties of the steam are listed below, they are

1 Sensible heat or Liquid heat or Total heat of water (hf)

The amount of heat required to raise the temperature of 1kg of water from 00C to saturation

temperature at a given constant pressure is defined as Sensible heat.

Therefore sensible heat absorbed by 1kg of water is given by

= 𝑚 ∗ 𝑐𝑤 ∗ (𝑇𝑠𝑎𝑡 – Tin) Where

m= mass of water

Cw = specific heat of water = 4.2 KJ/Kg K.

Tin= 00C

2 Latent Heat of Evaporation (hfg).

The amount of heat required to evaporate 1kg of water at its saturation temperature without

change in temperature is defined as Latent heat of evaporation. At atmospheric pressure Latent

heat of evaporation is 2257 KJ/KgK.

3 Dryness Fraction (x)

This term refers to quality of wet steam. It is defined as the ratio of the mass of actual dry steam

to the mass of same quantity of total wet steam. It is denoted by x.

x= mg /mg + mf

Where mg = mass of actual dry steam

mf = mass of water in suspension

m = mg + mf



h = hf + xhfg

hg = hf + hfg

hsup = hg + CP(tsup - tsat)

4 Degree of Superheat.

it is the difference between the temperature of superheated steam and saturation temperature at a

given pressure defined as degree of superheat.

5 Specific Enthalpy of Water (hf)

The amount of heat required to raise the temperature of one kg of water from 0 ℃ to its boiling

temperature or saturation temperature at given constant pressure. It is denoted as hf.

6 Specific Enthalpy of Wet steam It is defined as the quantity of heat required to convert 1 kg of water at 00C into wet steam of a

given quality and at constant pressure. It may be denoted by h.. Therefore

7 Specific Enthalpy of dry Steam

It is defined as the quantity of heat required to convert 1kg of water at 00C into dry saturated

steam at given constant pressure. It may be denoted by hg. Thus

8 Specific Enthalpy of Superheated Steam (hsup)

It is defined as the quantity of heat required to convert 1kg of water at 00C into the superheated

steam at given temperature and pressure. It may be denoted as hsup.

Degree of superheat = tsup - tsat



in

9 Specific Volume of Steam

The volume of a unit mass of steam is known as its specific volume.

10 Specific volume of saturated water (vf)

It is defined as volume of 1kg of water at saturation temperature corresponding to the given

pressure. It is denoted by vf.

11 Specific volume of wet steam of quantity x

It is the volume of 1kg of wet steam and is denoted as

m3/kg

12 Specific volume of dry steam

It is the volume of 1kg of dry steam and is denoted as

in m3/kg [since x=1]

13 Specific volume of Superheated Steam (vsup)

It is the volume of 1kg of superheated steam and can be determined by assuming that the steam

behaves as a perfect gas i.e., obeys the charle’s laws. It is denoted by vsup

Let P = pressure under which steam is superheated.

tsup =temperature of superheated steam

vg = Specific volume of dry saturated steam

tsat = saturation temperature at pressure P.

Since, P = constant, so vsup/tsup=vg/tsat

vsup=vg x tsup/tsat m3/kg

v= vg

v= x.vg



14 External work done during evaporation

When water is evaporated to form saturated steam, its volume increases from vf to vg at a

constant pressure, and thus external work is done by steam due to increase in volume. The

energy for doing the work is obtained during the absorption of latent heat. This work is called

external work of evaporation.

External work of evaporation = P (vg – vf )

W=P*105(vg – vf)

= 100Pvg kJ since, vf is very small and hence neglected

Wet steam with dryness fraction x,

W=100 p.x.vg kJ

Superheated steam

W= 100p.vsup kJ

15 Internal energy of steam (U)

It is defined as the actual energy stored in the steam. (OR) it is the difference between the

enthalpy and external work of evaporation.

For wet steam

U= (hf + xhfg)-(100 p.x.vg )KJ/Kg

For dry steam

U=hf + hfg-100 p.vg

= hg-100 p.vg KJ/Kg

For superheated steam

U= [hf + hfg + cp (tsup – tsat)] – 100p.vsup



16 Entropy

It is defined as a measure of randomness or disorder of a system. That is the measure of a

system's thermal energy per unit temperature that is unavailable for doing useful work.

Specific entropy of saturated water (sf)

The specific entropy of saturated water at a particular pressure P and saturation temperature

Tsat is given as the change in entropy during conversion of one kg of water at 00C into saturated

water at that pressure. It is given as

S = c log 𝑇

f w

273

As the Initial entropy at 273 K is zero, so this change in entropy above 273 K is taken as entropy

of water at temperature T. In case of Saturated Water, T= Tsat.

Change in specific entropy during evaporation (sfg)

During evaporation heat added = hfg = Latent heat of water

Temperature remains constant during evaporation and is equal to saturation Temperature Tsat.

Specific entropy of dry saturated steam (sg)

It is the entropy of one kg of dry saturated steam and is given as the sum of entropy of 1kg of

saturated water and entropy change during evaporation. It is denoted by sg.

Thus sg = sf + sfg

Specific entropy of wet steam

Specific entropy of wet steam is equal to sum of specific entropy of saturated water and change

in specific entropy during evaporation of dry fraction of steam. It is denoted by s

s = sf + x. sfg

Specific entropy of superheated steam (ssup)



It is the sum of specific entropy of dry saturated steam and entropy change during superheating

from saturation temp Ts to superheated temp Tsup.

S = c log Tsup

sup p

Tsat

Where cp = sp. heat of super heated steam

1.6 Steam Tables and their Uses

The properties of dry saturated steam like its temperature like saturation temperature, sensible

heat, latent heat of vaporization, enthalpy or total heat, specific volume, entropy etc. vary with

pressure, and can be found by experiments only. These properties have been carefully

determined, and made available in a tabular form known as steam tables.

There are two important Steam tables

➢ Absolute pressure table

➢ Temperature table

An extract of the properties of dry saturated steam is shown in the following two tables

Table 1: Absolute pressure table



Table 2: Temperature table

1.7 Mollier chart or enthalpy entropy diagram

➢ Graphical presentation of steam table is termed as Mollier chart

➢ The Mollier diagram, also called the enthalpy (h) – entropy (s) chart or h-s chart

➢ In this, x axis represents entropy of 1kg of water and y axis indicates specific enthalpy as

shown in fig.

➢ It is used to solve the problems on isentropic expansion and compression of steam.

➢ The h-s diagram is divided into two portions by a saturation line.

➢ In the lower region the temperature of the steam remains constant

➢ In the upper region the temperature of the steam increases at constant pressure

Fig1.4:- Mollier chart or enthalpy entropy diagram



hsup = 2916.7kJ/kg

1.8 Numerical Problems

1) Steam enters an engine at a pressure of 12 bar with a 67° C of superheat. It is

exhausted at a pressure of 0.15 bar and 0.95 dry. Find the drop in enthalpy of the

steam. (April/May 2018)

Solution.

Given: p 1 = l2 bar

tsup - tsat = 67°C

p2 =0.15bar;

x = 0.95

From steam tables, corresponding to a pressure of 12 bar, we find that

hf= 798.4 Id/kg; hfg = 1984.3 kJ/kg

We know that enthalpy or total heat of 1 kg of superheated steam,

hsup = hg + CP(tsup - tsat) = 798.4+1984.3+2x67

Similarly, from steam tables, corresponding to a pressure p2= 0.15 bar,

hf = 226 kJ/kg; hfg = 2373.2 kJ/kg

We know that enthalpy or total heat of 1kg of wet steam,

h = hf +x hfg = 226+0.95x 1984.3= 2111 kj/kg

Formation of steam 2020-21

= 805.7 kJ/kg

Drop in enthalpy of the steam

= hsup - h = 2916.7-2111

2) Determine the quantity of hear required to produce 1kg of steam at pressure of 6

bar at temperature of 25° C under the following conditions.(April/May 2019 and

Oct/ Nov 2019)

1. When the steam is wet having a dryness fraction 0.9;

2. When the Steam is dry saturated; and

3. when is is superheated at a constant pressure at 250° C assuming the mean specific

heat of superheated steam to be 2.3 kJflg K.

Solution.

Given: p=6 bar

tw= 25°C

x=O.9

tsup, = 250°C

cp= 2.3 kJ/kg K



hf = 670.4 kJ/kg; hfg = 2085 kJ/kg and tsat = 158.8° C

1) When the steam is wet

For 1 kg of wet steam,

h = hf + xhfg

= 670.4+0.9x2085=2546.9 kj/kg



Since the water is at a temperature of 25° C, therefore

Heat already in water = cw * tw = 4.2 x 25 = 105 kJ

Heat actually required = 2546.9- 105 = 2441.9 kJ

When the steam is dry saturated

h = hf + hfg

, = 670.4 + 2085 = 2755.4 kJ


When the steam is superheated


= 2755.4 + 2.3 (250- 158.8) = 2965.16 kJ


3) Determine the condition of steam in the following cases:

1. At a pressure of 10 bar and temperature 2000 C.

2. At a pressure of 10 bar and volume 0.175 inm3/kg.

Solution.

Given: p = 10 bar

t=200°C

v=0.175 m3/kg.

1) Condition of steam at temperature of 200°C

From steam tables, for pressure of 10 bar,

vg= 0.194 m3/kg; hf = 762.6 kJ/kg.; and tsat = 179.9°C



=

Since the given temperature is higher than saturation temperature, the given steam is

superheated.

The degree of superheat= 200 - 179.9 = 20.1°C.

2. Condition of steam at a volume of 0.175 m3/kg

Since the volume of given steam (0.175 m 3/kg) is less than the specific volume of the dry

Saturated steam (0.194 M3 /kg),

Therefore the given steam is wet.

The dryness fraction of steam,

vg 0.194 =

vw 0.175

=0.902.

4) Find the external work done during evaporation per kg of steam at a pressure of 15 bar

when the steam is (a) 90% dry and (b) dry saturated.

Solution.

Given : p= 15 bar


vg= 0.1317m3/kg

(a) Workdone when the steam is 90% dry (i.e. x = 0.9)

W= l00 Pxv = 100x15x0.9x0.1317 = 177.8kJ/kg.

(b) Workdone when the steam is dry saturated

W= 100PV 100xI5x0.1317 = 197.5kJ/kj

5) Find the internal energy of 1 kg of superheated steam at a pressure 10bar and 280° C. If

this steam be expanded to a pressure of 1.6 bar and 0.8 dry. Determine the change in internal

energy. Assume specific heat of superheated steam as 2.1KJ/kg K.

x



Solution.

Given p 1 = 10 bar

t,sup= 280° C

p2= 1.6 bar

x=0.8

cp =2.I kJ/kg K


hf = 762.6 kJ/kg; hfg = 2013.6 kJ/kg; vg = 0.1943 m3 /kg and tsat = 179.9° C

We know that enthalpy of I kg of superheated steam,


= 762.6+2013.6+2.1 (280- 179.9) = 2986.4 kJ/kg

𝑣𝑠𝑢𝑝 = tsup

tsat ∗ vg = 0.1943x

(280+273) = 0.237m3/kg

(179.9+273)

Internal energy of superheated steam,

Usup== hsup – 100Pvsup

= 2986.4- 100 10x0.237 = 2749.4 KJ/kg

From steam tables corresponding to a pressure of 1.6 bar, we find that

hf = 475.4 kilkg; h15 = 2220.9 kJ/kg; and = 1.091 m3/kg

We know that internal energy of the expanded steam,

Ue= hf +x hfg-100 Px vg = 475.4+0.8x2220.9-100x 1.6x0.8x 1.091

= 2252.1-139.7 = 2112.4 kJ/kg

Change in internal energy

= Usup - Ue = 2749.4-2112.4 = 637 kJ/kg





6) Using Mollier chart, find enthalpy drop and final condition of steam when it is expanded

isentropically from an initial pressure of 30 bar

and 350' C to a pressure oil bar.

Solution.

Given p1 = 30 bar

it = 3500 C p2 = I bar

I First of all, on the Mollier chart mark a point A,

where the pressure line passes through p 1 (i.e. 30

bar) and temperature line through (i.e. 350° C) meets as shown in Fig.

Since the steam is expanded isentropic ally (i.e. at a constant entropy), therefore draw a vertical a

line through A to meet the pressure line p2 (i.e. 1 bar) at B.

Enthalpy drop during the process,

From the Mollier chart, we find that enthalpy at A

hA = 3120 kJ/kg

and enthalpy at B, hB = 2450 kJ/kg

Enthalpy drop during the process,

h = hA - hB

3120-2450=670kJlkg



1.9 Vapour power cycles

In a vapour power cycle, the working fluid is water, which undergoes a change of phase and

similar to air standard cycles. They are used in steam power plants. With the help of an engine

the heat energy of the steam is converted into mechanical work .this engines operates with the

following vapour power cycles

➢ Carnot cycle

➢ Rankine cycle

➢ Modified Rankine cycle

1.10 Carnot Cycle

Fig1. 5:– Line diagram of Carnot cycle



Fig1.6 a: – PV diagram Fig1.6 b: – T-S diagram

Figure 12.1 shows a Carnot cycle on T-s and P-V diagrams. It consists of (i) two constant

pressure process (4-1) and (2-3) and (ii) two frictionless adiabatics (1-2) and (3-4) process.

Isothermal expansion (4-1). 1 kg of water at temperature T1 is heated to form wet steam of

dryness fraction x1. Thus during this process, heat is absorbed at constant temperature T1 and

pressure P1.

Heat supplied = T1 (s1 – s4) or T1 (S2 – S3).

Adiabatic expansion (1-2). During this process steam is expanded isentropically to temperature

T2 and pressure P2. The point ‘2’ represents the condition of steam after expansion. Since there is

no exchange of heat during this process, isentropy remains constant.

Isothermal compression (2-3). During this process heat is rejected at constant pressure P2 and

temperature T2. As the steam is exhausted it becomes wet and cooled from 2 to 3.

Heat rejected= T2 (S2 – S3).

Adiabatic compression (3-4). In this process the wet steam at ‘3’ is compressed isentropically

till the steam regains its original state of temperature T1 and pressure P1. Thus cycle is

completed. Since there is no exchange of heat during this process, isentropy remains constant

Net work done = Heat supplied – heat rejected

= T1 (S2 – S3) – T2 (S2 – S3)

= (T1 – T2) (S2 – S3).



η = (T1 – T2)

𝑇1

Carnot cycle η = Work done /Heat supplied

= (T1 – T2) (S2 – S3)

T1 (S2 – S3)

Limitations of Carnot Cycle

Though Carnot cycle has the highest thermal efficiency for given values of T1 and T2, yet it is

extremely difficult to operate in practice because of the following reasons

1. It is difficult to compress a wet vapour isentropic ally to the saturated state.

2. It is difficult to control the quality of the condensate coming out of the condenser so that the

saturation liquid condition reached.

3. since the specific volume of liquid vapour is more. The size of the compressor and work input

has to be large.

4. Higher power is required for compression, thus reduces plant efficiency and work ratio.

5. Superheated steam has to be supplied at constant temperature instead of constant pressure.

Hence the cycle is difficult to operate in practice

1.11 Rankine cycle

Fig1.7: - line diagram of Rankine cycle



Fig1.8a:– PV diagram Fig1.8b:– T-s diagram

➢ The Rankine cycle is an ideal cycle for comparing the performance of steam plants.

➢ It is modified form of Carnot cycle, in which the condensation process (3-4) is continued

until the steam is condensed into water.

➢ Consider 1 kg of saturated water at pressure p 1 and absolute temperature T1

➢ Process 1-2: The saturated water at point 1 is isothermally converted into dry saturated

steam in a boiler, and the heat is absorbed at a Constant temperature T1 and pressure p1 .

The dry state of steam is represented by point 2. This isothermal process is represented by

curve 1-2 on p-v and Ts diagrams

➢ Process 2-3: The isentropic expansion represented by the curve 2-3. The pressure and

temperature falls from p2 to P3 and T2 to 7'3 respectively with a dryness fraction x3.

Since no heat is supplied or rejected during this process, therefore there is no change of

entropy.

➢ Process 3-4: The isothermal compression represented by curve 3-4. The wet steam at

point 3 is now isothermally condensed in a condenser and the heat is rejected at constant

temperature T3 and pressure p3. The heat rejected by steam is its latent heat (equal to x3

hfg3).

➢ Process 4-1: The water at point 4 is now warmed in a boiler at constant volume from

temperature T4 to T1. Its pressure also rises from p4 to p1. The heat absorbed by water

during this operation is equal to the sensible heat.



Let, hf1 = hf3= Sensible heat or enthalpy of water at point 1 corresponding to a pressure of

p1 or p2 (p1 = p2)

hf4=hf3 = Sensible heat or enthalpy of water at point 4 corresponding to a pressure of p4

or p3 ( p4 = p3).

Heat absorbed during warming operation 4-1

hf1 - hf4 = hf2 - hf3

Heat absorbed during the Complete cycle= Heat absorbed during isothermal operation 1-2

+Heat absorbed during warming operation 4-1

hfg2+ ( hf2- hf3) = hf2+ hfg2 - hf3 = h2 - hf3

We know that heat rejected during the cycle

= h3 - hf4 = hf3+x3hfg3 - hf4 = x3hfg3

Workdone during the cycle = Heat absorbed — Heat rejected

(h2- hf3)-x3hfg3

= h2- (hf3+x3hfg3) = h2- h3 (h3=hf3+x3hfg3)

Efficiency

η= Workdone

=( h1-h2)/ (h2-hf3)

Heat absorbed

Modified Rankine Cycle

We have seen in the Rankine cycle, that the Steam is expanded to the extreme toe of the p-v

diagram (at point 3) as shown in Fig. But in actual reciprocating steam engines, it is found to be

too uneconomical (due to larger size of the cylinder) to expand Steam to the full limit. Therefore

the expansion is carried out at the higher pressure than the condensing pressure.

In order to overcome the above mentioned difficulty, the Rankine cycle i s slightly modified.



Fig1.9a:– PV diagram Fig1.9b:– T-s diagram

In a modified Rankine cycle, the expansion stroke of the piston is stopped at point 5 by cutting

the toe portion of the rankine cycle and and the steam is exhausted from the cylinder at a

constant volume. This causes a sudden drop of pressure from p5 to p6. The expansion of steam is,

therefore, completed by a constant volume line 5-6 as shown on p-v and T-s diagram in Fig.

Let p1=p2= Pressure of steam at point 2,

V2= Volume of steam at point 2,

h2 = Enthalpy or total heat of steam at point 2,

U2= Internal energy of steam at point 2,

P3,v3 , h3 , u3 = Corresponding values of steam at point 3.

p4 = Back pressure of steam at point 4, and

hf4 = Sensible heat or enthalpy of water at point 4.

We know that work done during constant pressure process 1-2,

= Area 1-2-6-0 = 100 p2 v2

We also know that work done during isentropic expansion 2-3

= Area 2-3-7-6 = Change in internal energy

=U2-U3



And work done during constant pressure process 4-5

Area 0-5-4-7 = 100 p4 v4 .

Work done during the cycle per kg of steam,

W = Area 1-2-3-4-5

= Area l-2-6-0 + Area 2-3-7-6 - Area 0-5-4-7

= 100 p2 v2 + ( U2-U3) -100 p4 v4

= 100 p2 v2 + (h2 - 100 p2 v2) - (h3 — 100 p3 v3)]— 100 p4 v4

= h2 —h3 + 100(p3 - p4)v3

We know that heat supplied per cycle

= h2 – hf5 =. h2 – hf4 (:hf4 = hf5)

Efficiency of the modified Rankine cycle =-

------- =(h2-h1)+100(p3-p4)v3 / (h2-hf4)

1.12 Steam calorimeter

It is device used to measure dryness fraction of the wet steam.

There are four methods of determining the dryness fraction of steam experimentally. They are

➢ Bucket or barrel calorimeter

➢ Separating calorimeter

➢ Throttling calorimeter

➢ Combined separating and throttling calorimeter.



1.13 Barrel calorimeter

The Bucket/Barrel Calorimeter consists of a copper vessel which contains cold water. Thu upper

vessel is insulated from the surrounding so as to prevent any heat transfer from or to the system.

The top of the vessel is covered with wooden plate containing two holes. In one of the hole a

thermometer is inserted to record the temperature and in another hole, steam supply pipe is inserted

Fig1.10:– line diagram of barrel calorimeter

as shown in the below figure. The whole assembly is placed on the weighing bridge.

In this calorimeter, steam from the main pipe of boiler enters into the calorimeter through

nozzles. The steam coming in contact with water gets condensed and results in an increase in the

mass and temperature of water. Temperatures of water before and after condensation are

recorded. Knowing the quantities of steam and water, we can very easily calculate the dryness

fraction steam.

let, ms= weight of steam condensed

mw = weight of cold water in barrel

Cpw= sp. heat of water

x= dryness fraction of steam

hfg = latent heat of the steam at pressure of steam P,



ms[ x hfg1+ Cpw ( tsat-t2)]=mw Cpw(t2-t1)

t1, =initial temperature of water before mixing steam

t2 = final temperature of steam after mixing of steam

tsat = saturation the temperature of steam at

Therefore, Heat lost by steam = Heat gained by coal water

From this equation, we can find the dryness fraction of steam



1.14 Separating Calorimeter:

Fig1.11:– line diagram of separating calorimeter

It is a simple calorimeter, used for determining the dryness fraction steam by separating the

moisture content from the wet steam.

It consists of two eccentric chambers, inner and outer chambers. They communicate with the

help of opening at the top. The wet steam to be tested enters the inner chamber of separating

calorimeter through a sampling tube controlled by a valve. In the inner chamber of calorimeter,

perforated cups or baffles plates are arranged in such a manner as to separate the moisture from

the steam. Hence, when steam enters the separating calorimeter, it undergoes a sudden reversal

of direction of motion when it strikes the baffles plates. This causes the water particles which

have greater inertia to separate from the wet steam. The water moves downwards and collected at



the bottom portion of the inner chamber. The dry steam moves upwards and this steam is

measured by condensing it.

Let m= Mass of water collected in a particular time.

M= mass of dry steam passing in the same time

x = dryness fraction of wet steam

Therefore, x= mass of dry steam =M/M+m

mass of wet steam

Limitation

This method gives approximate value because of incomplete separation of moisture from wet

steam.

1.15 Throttling calorimeter

In this, the principle of constant enthalpy expansion is adopted for the measurement of moisture

in the wet steam is as shown in the below figure. In this calorimeter, steam from the main pipe

of boiler enters into the calorimeter through controlled valves and moves into the well insulated

expansion chamber in which the temperature is measured.

Fig1.12:– line diagram of throttling calorimeter



During throttling, pressure of the steam decreases and volume increases. Neglecting the heat

losses, total heat of the steam before and after the throttling remains same. Hence Steam

becomes dry; therefore super heated steam is obtained after throttling.

Enthalpy before throttling = (hf1+ x hfg1)

Where, P1 = absolute pressure of steam before throttling

hf1 = sensible heat of steam corresponding to P1

hfg1= latent heat of steam corresponding to P1and

x = dryness fraction of steam (at P1) entering the throttling unit.

Enthalpy after throttling (steam being superheated) = hg2+ Cp (tsup- tsat)

Where, P2 pressure of steam, after throttling

hg2 = enthalpy of 1 kg of dry saturated steam corresponding to P2

tsup = temperature of superheated steam after throttling

tsat= temperature of saturated steam corresponding to pressure P2

Cp = specific heat of superheated steam at constant pressure.

Enthalpy before throttling = Enthalpy after throttling

hf1+ x hfg1= hg2+ Cp (tsup- tsat)

hg2 + Cp tsup − tsat − hf1 𝑥 =

hfg1



x= x1. x2

1.16 Combined Separating and Throttling calorimeter.

Separating calorimeter gives approximate value because of incomplete separation of moisture

from wet steam where as throttling calorimeter can be used to find dryness fraction, only if

superheated is obtained after throttling. Hence to overcome the limitations of these calorimeters

and to obtain accurate results combined separating and throttling calorimeter is used.

When steam sample enters the separating unit, most of the water vapours are separated from

the steam. Then dry steam passes throttling unit where the steam is superheated without change

in enthalpy.

Fig1.13:– line diagram of Combined Separating and Throttling

calorimeter Let x1 = dryness fraction of steam sample in separating unit

x2 = dryness fraction of steam sample in throttling unit

Therefore

Reference

1) A textbook of Thermal engineering by RS Khurmi and J k Gupta

2) Thermal Engineering By R K Rajput

3) Encyclopedia

4) Google hunt



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