Vtu Btd Notes for Pure Substance Chapter

7/25/2019 Vtu Btd Notes for Pure Substance Chapter

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Pure SubstancesP-T and P-V diagramsTriple point and critical pointSub cooled liquidSaturated liquid, mixture of saturated liquid and vapor

Saturated vapor and superheated vapor states of a pure substance with water asexample.Enthalpy of change of phase, dryness fraction, T-S and H-S diagrams.Representation of various processes on these diagrams.Steam tables and its use.Throttling Calorimeter, Separating Calorimeter.Throttling and Separating Calorimeter.

Introduction:A pure substance is one that has a homogeneous and in variable chemical

composition. It may exist in more than one phase, but the chemical composition isthe same in all phases. Thus liquid water, a mixture of liquid water and water vapor,and a mixture of ice and liquid water are all pure substance: every phase has thesame chemical composition. On the other and mixture of liquid air and gaseous air isnot a pure substance. Because of the composition of the liquid phase is differentfrom that of the vapor phase.

Some times a mixture of gases such as air is considered a pure substance as longas there is no change of phase. Strictly speaking this is not true.

Vapor liquid solid phase equilibrium in a pure substance:

Fig 1: Constant pressure change from liquid to vapor phase for a puresubstance


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Consider a system one kg of water contained in a piston cylinderarrangement as shown in the figure. Suppose that the piston and weight maintain apressure of 0.1 MPa in the cylinder and that initial temperature be 200C. As the heatis transferred to the water the temperature increases appreciably, the specificvolume increases slightly under constant pressure. When the temperature 99.6

0C,

additional heat transfer results in a change of phase. I.e. some of the liquid becomesvapor. During this process both temperature and pressure remain constant where assp. Volume increases considerably. When the last drop of liquid has vaporizedfurther transfer of heat results in an increase in both temperature and sp. volume ofthe vapor.

The term saturation temperature designates the temperature at whichvaporization takes place at a given pressure. Or the pressure is called saturationpressure corresponding to the saturation temperature. Thus for water 99.60C thesaturation pressure is 0.1MPa, and for water at 0.1MPa the saturation temperatureis 99.60C. Thus there is a definite relation between saturation pressure andsaturation temperature.

Pressure Vapor-pressure curve

TemperatureFig 2: Vapor Pressure curve of a pure substance

If the substance exists as liquid at the saturation temperature and pressure itis called saturated liquid. If the temperature of the liquid is lower than the saturationtemperature for the existing pressure it is called either a sub cooled liquid or acompressed liquid.

When the substance exists as a part liquid and part vapor at the saturationtemperature a dryness fraction comes into picture. It is also called as quality and it isdefined as the ratio of mass of vapor to the total mass. It is denoted by the symbolx. Quality has meaning only when the substance is in saturated state. I.e at thesaturation pressure and temperature. The quality x is an intensive property.


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If a substance exists as vapor at the saturation temperature it is calledsaturated vapor or dry saturated vapor with x=1. When the vapor is at a temperaturegreater than the saturation temperature at the saturation pressure, it is said to existas superheated vapor. After that the temperature increases as heat is added atconstant pressure.

Temperature Volume (T-V) diagram for water:

Fig 3: T-V diagram for water

It is clear from the figure that constant pressure lines are ABCD, EFGH, IJKLetc. the peak point of the figure indicated by N is the critical point of water. Thus the

critical pressure 22.089 MPa and corresponding critical temperature is 374.14 0C.

A constant pressure process at a pressure greater than the critical pressure isrepresented by curve PQ. Thus water at 40 MPa, 200C is heated in a constantpressure process, there will be never be two phases present at the state shown.Instead there will be continuously in density at all the times and there will be onlyone phase present. The question arises is: when do we have a liquid and when dowe have a vapor? The answer is that this is not a valid question at super criticalpressures. We simply term the substance as fluid. However rather arbitrarily attemperatures below the critical temperatures we usually refer to it as a compressedliquid and at temperatures above the critical temperatures as superheated vapor. It

should be noted that however at pressures above the critical pressures we neverhave a liquid and vapor phase of pure substance existing in equilibrium.


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Table1: Some Critical point data

CriticalTemperature, 0C

CriticalPressure, MPa

Critical Volume,m3/kg

Water 374.14 22.089 0.003155

Carbon dioxide 031.05 07.39 0.002143

Oxygen -118.35 05.08 0.003438Hydrogen -239.85 01.30 0.032192

Consider another experiment with piston cylinder arrangement. Suppose that thecylinder contains one kg of ice at 200C and one bar. When heat is transferred to theice the pressure remains constant the specific volume increases slightly and thetemperature increases until it reaches 00C, at which point the ice melts andtemperature remains constant. This state is called saturated solid state. For mostsubstances the specific volume increases during this melting process. But for waterspecific volume of the liquid is less than the specific volume of the solid.

Sublimation: If the initial pressure of the ice at 200

C is 0.26 kPa, heat transferredto the ice results in an increase in the temperature to 10

0C. At this point however

the ice passes directly from the solid phase to the vapor phase. This process isknown as sublimation. Further heat transfer results in superheating of the vapor

Triple Point: Consider the ice at 0.6113 kPa and temperature of 200C. Throughheat transfer let the temperature increase until it reaches 0.01C. At this pointhowever further heat transfer may cause some of the ice to become vapor and someto become liquid. At this point it is possible to have three phases in equilibrium. Thispoint is called the triple point. Triple point is defined as the state in which all threephases may be present in equilibrium. The pressure and temperature at the triple

point for a number of substance is given in following table.

Table 2:Triple Point Data

Temperature, 0C Pressure, kPa

Hydrogen -259 7.194

Oxygen -219 0.15

Nitrogen -210 12.53

Carbon Dioxide -56.4 520.8

Mercury -39 0.00000013

Water 0.01 0.6113

Zinc 419 5.066

Silver 961 0.01Copper 1083 6.000079

Consider a solid state as shown in the figure, when the temperature increaseswith constant pressure the substance passes directly from solid to vapor phase.

Along the constant pressure line EF the substance passes from solid to liquid phaseat one temperature and then from liquid to vapor phase at higher temperature.


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Constant pressure line CD passes through the triple point and it is only at thetriple point the three phases exists together in equilibrium. At a pressure abovecritical pressure such as GH line there is no sharp distinction between liquid andvapor phases. The triple point temperature and critical temperature vary greatly fromsubstance to substance. For ex: critical temperature of helium is 5.3K. Therefore

absolute temperature of helium at ambient conditions is over 50 times greater thanthe critical temperature. On the other hand water has a critical temperature 374.140C (647.29K) and at ambient conditions the temperature of water is less than thecritical temperature.

Allotropic transformationIt should be pointed out that a pure substance can exist in a number of different solidphases. A transition from one solid phase to another is called an allotropictransformation. This can be well understood by the following figure.

Temperature

Fig 4: P-T Diagram for water.

Fig 5: P-T Diagram for iron


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Independent properties of a pure substance:One important reason for introducing the concept of pure substance is that

the state of a simple compressible pure substance is defined by two independentproperties. For ex: if the specific volume and temperature of a super heated steamare specified the state of the steam is determined.

To understand the significance of the term independent property, consider thesaturate liquid and vapor state of a pure substance. These two states have the samepressure and temperature but they are definitely not the same state. In a saturationstate therefore pressure and temperature are not independent properties. Twoindependent properties such as pressure -specific volume, pressure and quality,temperature-specific volumes are required to specify a saturation state of a puresubstance.

Equations of state for the vapor phase of a simple compressiblesubstance:

From the experimental observations it has been established that the equationof state under low-density gases is given by PV=RuT. The ideal gas equation of stateand compressibility factor equation are good approximations at low-densityconditions. Therefore ideal gas equation of state is very convenient to use inthermodynamic calculations. The question comes what is low density? Or whatrange of density will the ideal gas equation of state hold with accuracy?

The analysis gives the pressure temperature deviations from ideal gasbehavior. To answer5 the question the concept of compressibility factor Z isintroduced and is defined by the relation Z= PV/RuT. For ideal gas Z=1 and thedeviation of Z from unity is a measure of the deviation of actual relation from theideal gas equation of state.

Fig 6: Compressibility of Nitrogen


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It shows a skeleton compressibility charge for nitrogen. Three observations can bemade from this chart.

All the temperature Z 1 as P 0. i.e. as the pressure approaches zero the P-V-T behavior closely approaches that predicted by the ideal gas equation of state.

Note also that at temperatures of 300K and above i.e. above room temperature thecompressibility factor is near unity up to a pressure of 10MPa. This means that theideal gas equation of state can be used for nitrogen over this range withconsiderable accuracy.

Now suppose we reduce the temperature from 300K but keep the pressure constantat 4MPa, the density will increase and we note a sharp decrease below unity in thevalue of compressibility factor. Values of Z


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It contains values of enthalpy, internal energy, specific volume and entropy. If thevalues of T and P are given other all values can be directly taken from the table.Therefore we should first learn how to use the table. We note that all these areindependent variables.

To finding the correct table other nuisance of everyday use of the table in theinterpolation. I.e. when one of the stated values is not exactly equal to a value listedin the table. Recently computerized tables are in use, which eliminates theseproblems. But the students nevertheless must learn to understand the significance,construction and limitations of the tables.

The actual steam table as presented in various books is a summary table based oncomplicated curve fit to the behavior of water. Here we concentrate three propertiesnamely T,P and v and note that other properties listed namely u, h, s are presented.It is further noted that the separation of phases in terms of values of P and T isactually described by the relation illustrated in the pressure-temperature diagram.

The specific volume of a substance having given quality can be found out by thedefinition of quality. The volume is the sum of volume of liquid and vapor.

V= Vliq + V vap.

In terms of masses: mv = mliq * vf + m vap* vgBy introducing the quality x we have v = (1-x) vf+ x vgWe know that vfg = vg - vf

We can write the specific volume equation for wet steam as v = vf+ x v

fg

In superheated region pressure and temperature are independent properties andtherefore for each pressure a large number of temperature is given and for eachtemperature four thermodynamics properties are listed namely specific volume,enthalpy, entropy and internal energy.

Temperature specific volume diagram :

Figure shows a T-v plot for water with an indication of percent error in

assuming ideal gas behavior along the saturated vapor curve and also in severalarea of superheated region. Generally a slight decrease and only a small error ismade if one same that the volume of a compressed liquid is equal to the specificvolume of the saturated liquid at the same temperature. It is generally acceptedprocedure particularly when compressed liquid data are not available.


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Fig 7: Temperature specific volume diagram for water

In the modern scenario computerized table are available. The main programoperated with a visual interface in the window environment on a PC which isgenerally user friendly. The program operates on DOS environment, which coversnot only the tables of water, but it covers most of the pure substances used inengineering industries. The generalized chart with compressibility factor is alsoincluded so it is possible to get the value of Z a more little accurately than readingthe graph. It is useful in the case of two-phase mixture, like saturated liquid andvapor values are needed.

Thermodynamic Surfaces:

The matter discussed in this chapter can be well summarized by a consideration of apressure, specific volume, temperature surface (PVT surface). Two such surfacesare shown in the figure for a substance such as water in which the specific volumeincreases during freezing and the other in which a substance, its specific decreasesduring freezing.

Fig 8 : PvT Surface -- Water


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Fig 9 : PvT Surface Water

In these diagrams the pressure, specific volume and temperature are plotted on amutually perpendicular coordinates and each possible equilibrium state is thusrepresented by a point on the surface. This follows directly from the fact that a puresubstance has only two independent intensive properties. All points along a quasiequilibrium surface lie on the PvT surface since such a process always passes

through equilibrium states.

The regions of the surfaces that represent a single surface- the solid, liquid andvapor faces are indicated. These surfaces are curved. The two phase regions thesolid-liquid, solid-vapor, liquid-vapor regions are ruled by surfaces. It is understoodthat they are made up of straight lines parallel to the specific volume axis. This ofcourse follows from the fact that in the two-phase region lines of constant pressureare also lines of constant temperature, although the specific volume may change.The triple point actually appears as the triple line on the PVT surface, since thepressure and temperature of the triple point are fixed but the specific volume mayvary depending upon the proportion of each phase.

It is also of interest to note that the pressure temperature and pressure volumeprojections of these surfaces. We have already considered the P-T diagram forwater. It is on this diagram we observe the triple point. Various lines of constanttemperature are shown on the P-v diagram and corresponding constant temperaturesections are identically seen on the P-v-T surfaces. The critical isotherm as a pointof intersection at the critical point.


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Fig 10 : PvT Surface for a Substance which Contracts Upon Freezing

Fig 11 : PvT Surface for a Substance which Expands Upon Freezing


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One notice that for a substance such as water which expands on freezing, thefreezing temperature decreases with an increase in pressure. For a substance thatcontracts on freezing, the freezing temperature increases as the pressure increases.Thus as the pressure of the vapor is increased along the constant temperature line asubstance that expands on freezing first becomes solid and then liquid. For

substance that contract on freezing, the corresponding constant temperature lineindicates that as the pressure of the vapor increases, it first becomes liquid and thensolid.

T-s Diagram for a pure substance:

Consider the heating of 1 kg of ice at 50C to steam at 2500C. The pressure beingmaintained at 1 atm. It is observed that the entropy of steam increases in differentregimes of heating namely

1) Entropy increase of ice to saturated freezing temperature2) Entropy increase of ice as it melts into water.

3) Entropy increase of water as it is heated from 0O

c to 100O

C.4) Entropy increase of water as it is vaporized at 100OC absorbing latent heat of

vaporization.5) Entropy increase of vapor as it is heated from 100OC to 250OC.

Fig 12: T s Diagram for water Constant pressure lines

These entropy changes are shown in T-S graph. It is a constant pressure process. Ifduring heating process the pressure had been maintained constant at 2 atm, asimilar curve would be obtained. If these states for different pressures are joined thephase equilibrium diagram of a pure substance on the T-s coordinate would beobtained as shown below.


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Fig 13: T s Diagram for water

Most often liquid vapor transformation only are of interest, the following figure theliquid, the vapor and the transition zones only at a particular pressure, sf specificentropy of saturated water and sg - specific entropy of saturated vapor. The entropychange of the system during the phase change from a liquid to vapor at thatconstant pressure is sfg. (= sg sf). The value of sfg decreases as pressureincreases. And becomes zero at the critical point.

Fig 14: Phase equilibrium diagram T s Diagram

H-S diagram (Mollier diagram) for a pure substance:From the first and second law of thermodynamics following property relations areobtained.

Tds = dh v dp and ( h / s )p = TThese equations form the basis of h-s diagram of a pure substance. The slope of anisobar on the h-s coordinates is equal to the absolute temperature. If thetemperature remains constant the slope will remain constant. If the temperatureincreases the slope of the isobar will increase.


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Consider once again the heating of ice at 5OC to steam at 250OC the pressurebeing maintained constant at 1 atm. The slope of the isobar of 1 bar on the h-scoordinates first increases as the temperature of the ice increases from 5OC to0OC. the slope then remains constant as ice melts to water at 0 OC. Te slope ofisobar again increases as the temperature of water rises from 0

OC to 100

OC. the

slope again remains constant as water vaporizes at constant temperature. Finallythe slope of the isobar continues to increase as the temperature of steam increasesto 250

OC and beyond (as shown in the figure below.

Fig 15: h s diagram for water - constant pressure lines.

Similarly the isobars of different pressures can be drawn on h-s diagrams as shownin the figure below.

Fig 16: h s for water.


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This figure shows the phase equilibrium diagram of a pure substance on the h-scoordinate indicating the saturated solid line, saturated liquid line saturated vaporline, the various phases and the transition zones.

Fig 17: h-s diagram for water:

This figure is the Mollier diagram indicating only the liquid and vapor phases. As thepressure increases the saturation temperature increases, slope of the isobarincreases. Hence the constant pressure lines diverge from one another and thecritical isobar is at a tangent to the critical point. In the vapor region the states ofequal slopes at various pressures are joined by lines as shown, which are theconstant temperature lines. Here at a particular pressure h f is the specific enthalpy ofsaturated water and hg is specific enthalpy of saturated vapor and h fg (= hg - hf) is

the latent heat of vaporization at that pressure. As the pressure increases hfgdecreases and at the critical pressure hfg becomes zero.

Dryness fraction and various equations:

Dryness fraction is defined as the ratio of mass of dry steam to total mass of steam.It is denoted by x and is also called as quality of steam.

x = mv / (mv +ml ) where mv and ml are the masses of vapor and liquid respectively.Let V be the toal volume of a liquid vapor mixture of quality x in which Vf volume ofsaturated liquid and Vg volume of saturated vapor, the corresponding being m, mf,mg respectively.

We have m = mf+ mg and V= Vf+Vg

Therefore mv = mf* vf+ mg*vg v= (1-x) vf+ x vg.


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Similarly s = (1-x) sf+ x sg h= (1-x) hf+ x hg u= (1-x) uf+ x ug.Same equations are written asv = vf+ x vfg, h = hf+ x hfg u = uf+ x ufg s = sf+ x sfgIf the condition of the steam is superheated then we have Degree of superheat,which is difference between the superheated temperature to the saturation

temperature.

super= TsuperTsaturationThe other properties are calculated as v super = vsat * Tsuper/ Tsat

hsuper= hg + Cp (Tsuper-Tsat)ssuper = sg + Cp log(Tsuper/Tsat)

Measurement of Steam Quality:

The state of a pure substance gets fixed if two independent properties are given.

Thus the pure substance is said to have two degrees of freedom.

Fig 18: T s and h s diagram

Figure shows the values of pressure and temperature would fix up the state. Butwhen the substance is in the saturation state or two phase region the measuredvalues of pressure and temperature could apply equally well to saturated liquid pointf and saturated vapor point g. or two mixtures of any quality points x1, x2 or x3. of the

two properties, P and T only one is independent; the other is a dependent property.If the pressure is given the saturation temperature gets automatically fixed for thesubstance. In order to fix up the state of the mixture one more property such asspecific volume, enthalpy or composition of mixture or internal energy is required tobe known. Since it is relatively difficult measure the specific volume of the mixturedevices such as calorimeters are used for determining the quality or the enthalpy ofthe mixture.


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In the measurement of quality, the object is always to bring the state of thesubstance from the two phase region to the single phase region or superheatedregion where, the pressure and temperature are independent and measured to fixthe state, either by adiabatic throttling or e3lectric heating.

There are four types of calorimeter are common in use namely1) Separating Calorimeter2) Throttling Calorimeter3) Combined separating and throttling calorimeter.4) Electrical calorimeter.

Separating Calorimeter: the steam whose dryness fraction is to be determinedis very wet then separating calorimeter gives the quality of the steam. A knownquantity of steam is passed through a separating calorimeter as shown. The steamis made to change direction suddenly, the water being denser than the dry steam isseparated out. The quantity of water, which is separated out, is measured at the

separator. The dry steam coming out of the separator is sent through a condenserwhere it is condensed separately.The dryness fraction of the steam is calculated by weighing the mass of the waterand mass of dry steam after condensation separately.

Throttling Calorimeter: sample of wet steam of mass m at pressure p1 is takenfrom the steam main through a perforated sampling tube as shown in the figure.

Fig 19: Throttling calorimeter


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Then it is throttled by the partially opened valve to a pressure p 2 measured by amercury manometer and temperature T2 so that after throttling the steam is in thesuperheated region. The process is shown on T-s and H-s diagram.

The steady flow energy equation gives the enthalpy after throttling as equal to

enthalpy before throttling. It is a irreversible process hence joined by a dotted line.Thus the initial state of the steam is p1 , t1 and its dryness fraction is x1 and the finalstate of the superheated steam is p2 x2.

Fig 20: T s and h s diagram for throttling calorimeter

Now h1 =h2hf1+x1 hfg1 = h2

x1 =( h2 hf1 ) / hfg1

With P2 and T2 being known, h2 can be found out from the superheated steam table.The values of hf, hfg are taken from saturated steam table., thus quality of the wetsteam x1 can be calculated.

Combined separating and throttling calorimeter:When the steam is very wet and the pressure after throttling is not low enough totake the steam to the superheated region then a combined separating and throttlingcalorimeter is used for the measurement of quality. Steam from the main is first

passed through a separator as shown in the figure, where some part of the moistureseparates due to sudden change in direction and falls by gravity and partially dry

vapor is then throttled and taken to the superheated region.


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Fig 21: Schematic diagram for separating and throttling calorimeter.

As shown in the figure process 1-2 represents moisture separation from the wetsample of steam at constant pressure P1 and process 2-3 represents throttling topressure P2 withP2 and T3 being measured, h3 can be found out from thesuperheated steam table.

h3= h2 = hf1 + x2 hfg1

Therefore x2, the quality of the steam after partial moisture separation can beevaluated. If m kg of steam is taken through the sampling tube in y seconds, m 1 kg isseparated and m2 kg is throttled and then condensed to water and collected, wehave

m=m1+m2.

The mass of dry vapor will be at state2 is x2m2.Therefore the quality of the sample of the steam at state1 which ids x1 is given byx1= x2m2 / (m1 + m2)

There is one more method of measurement of quality of wet steam by using electriccalorimeter as shown in the figure. The sample of steam is passed in steady flowthrough an electric heater. Electrical energy input Q should be sufficient to taken thesteam to the superheated region where pressure P2 and temperature T2 aremeasured. If I is the current flowing through the heater in amperes and V thevoltage across the coil at steady state Q=VI. If m is the mass of steam taken in tseconds under steady flow condition then the steady flow energy equation for heateris given by m1h1 + Q = m1h2


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Fig 22: Electric calorimeter

Hence h1 + Q / m1 = h2. With h2,Q and m1 =being known h1 can be computed. Thush1= hf1 + x1 hfg1.

Hence x1 can be calculated.

1) A steam boiler initially contains 5 m3 of steam and 5 m3 of water at 1 MPa.Steam is taken out at constant pressure until 4 m3 of water is left. What is theheat transferred during the process?Solution:

Fig 23: Steam drum.

At 1 MPa vf = 0.001127 and vg= 0.1944 m3 / kg.hg=2778.1 kJ/kg uf = 761.68 , ufg = 1822 , ug = 2583.6 kJ/kg

The initial mass of saturated water and steam in the boiler = (Vf /vf) + (Vg / vg) =

[(5/0.001127) + (5/0.1944) ] = [(4.45 * 103) + (25.7) ] kg


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Final mass of saturated water and steam = (4/0.001127) +(6/0.1944) = [ (3.55 *10

3) + 30.8] kg

Mass of steam taken out of the boiler, ms = [4.45 * 103 + 24.7] - [ (3.55 * 103) + 30.8]= 894.9 kg

Making an energy balance we have initial energy stored in saturated water andsteam + heat transfer from external source = final energy stored in saturated waterand steam + energy leaving the steam or

U1 + Q =U1 + ms*hgAssuming that the steam taken out is dry.Hence 4.45 * 10

3* 761.68+27.7*2583.6+ Q = 3.55 * 103 *761.68 +30.8 *2583.6

+894.9 * 2778.1Q = 2425000-685500+13176Q = 1752676 kJ.

2) Steam flows in a pipeline at 1.5Mpa. After expanding to 0.1MPa in athrottling calorimeter, the temperature is found to be 120 o C. Find the qualityof steam in the pipe line. What is the maximum moisture at1.5 MPa that can bedetermined with this set-up if least 5 o C of superheat is required afterthrottling for accurate readings?

Fig 24 : h s Diagram

Solution.At state 2 when p = 0.1 MPa and t = 120 o C by interpolation,h2 = 2716.2 kJ/kg and

p = 1.5 MPa hf = 844.89 and hfg = 1947.3 kJ/kg

and h1 = h2 hf1+ x1 hfg1 = h2


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844.89 + x1 *1947.3 = 2716.2

x1 = 1871.3 / 1947.3 = 0.963 Ans.

When p = 0.1MPa and t = 99.63 + 5 = 104.63 oC,h3=2685.5 kJ/kg

Since h3=h4 2685.5 = 844.89 + x4 *1947.3x4 = 1840.6 / 1947.3 = 0.948

The maximum moisture that can be determined with this set up is only 5.2% Ans

3) The following data were obtained with a separating and throttlingcalorimeter:Pressure in pipeline :1.5 MPaCondition after throttling:0.1 MPa.110o CDuring 5 min moisture collected in the separator:0.150 litre at 70 oC,Steam condensed after throttling during 5 minFind the quality of steam in the pipeline

Fig 25: h s DiagramSolution :

AT 0.1 MPa, 110 oC, h3 = 2696.2 kJ/kgNow h3=h2 = hf2 + x2 hfg22696.2=844.89 + x2 1947.3x2 = 1851.31/1947.3 = 0.955

If m1 = mass of moisture collected in the separator in 5 min and

m2= mass of steam condensed after throttling in 5 min thenx1 = (x2 m2) / (m1 + m2)

at 70oC vf= 0.001023 m3/kg

m1= 0.1462 kg and m2 = 3.24 kg

Hence x1= 0.955*3.24 / (0.1462 +3.24) = 0.915 Ans.


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4) A vessel having a volume of .4 m3 contains 2.0 kg of liquid water and vapormixture in equilibrium at a pressure of 600 kPa. Calculate a) The volume andmass of liquid b) The volume and mass of vapor.

Solution: The specific Volume is calculated first.v =(0.4/2 ) = 0.2 m3/kg

The quality of steam can now be calculated0.2 = 0.001101 + x * 0.3146, x = 0.6322Therefore mass of liquid is 2.0*(1 - 0.6322) = 0.7356 kgMass of vapor is 2.0(0.6322) = 1.2644 kgVolume of liquid is ml vf

= 0.7356(0.001101) =0.0008m3

Volume of vapor is mvvg = 1.2644(0.3157)=0.3992 kg

5) Steam at 1 bar and a dryness fraction of 0.523 is heated in a rigid vesseluntil it becomes saturated vapor. Calculate the heat transferred per kg steam.

Solution:

From the steam table at 1 bar pressure, ts = 99.62oC,

vf=0.001043m3/kg,vfg=1.69296m3/kg,

uf=417.33 kJ/kg, ufg =2088.72kJ/kg,

Volume of one kg of given state of vapor = v f+ x vfg= 0.001043 + 0.523 * 1.69296 = 0.8864 m3/kg

Enthalpy correspond to the state of steam,u= uf+ x ufg = 417.33 + 0.523 * 2088.72 =1509.73kJ/kg

v (0.8864 m3/kg ) correspond to the saturated condition of the steam, from steamtable, which is vg,The pressure found to be 2 bar and ug =2529.49kJ/kg

Heat added = 2529.49 1509.73 = 1019.76 kJ/kg Ans

.6) A rigid vessel contains one kg of mixture of saturated water and saturatedsteam at a pressure of 0.15MPa. When mixture is heated the state passes

trough the critical point. Determine,a) Volume of the vesselb) The mass of the liquid and vaporc) The temperature of the mixture when the pressure rises to 3 MPa.d) The heat transfer required to produce the final state.Solution. Vc: Critical volume = 0.003155 m3/kg (From the steam table.)


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v = vf+ x X vfg0.003155 = 0.001035 + x1 * 1.15828 , x1 = 0.00183

Hence, mass of vapor = x = 0.00183 kg

Mass of liquid = (1-x1) = 0.998 kg.

u1 = uf + x1 X ufg ;u1 = 466.92 + 0.00183 X 2052072 = 470.68 kJ/kg

Saturation temperature correspond to 3 MPa, is 233.9OC, the temperature of themixture

x2 = (0.003155 0.001216)/0.06546 = 0.02962

Heat transfer in constant volume process = u2 u1,

u2 = 1004.76 + 0.02962 X 1599.34 = 1052.13 kJ/kg

Heat transfer = 1052.13 470.68 = 581.45 kJ/kg.

7) Steam initially at 0.3MPa, 250OC is cooled at constant volume. Findi) At what temperature will the steam become saturated vapor?ii) What is the quality at 80OC? What is the heat transferred per kg of steam incooling from 250 OC to 80OC?

Solution:At 300kPa, 250 OC, from the steam table , it is a superheated condition.

v = 0.79636 m3/kg , u = 2728.69 kJ/kg

vg

= 0.79636 m3/kg , Ps

= 225 kPa, ts

= 124 OC

at 80 OC0.79636 = 0.001029 + x 3.45612

x = (0.79636 - 0.001029) /3.45612 = 0.23u2 = 520.45+0.23(2013.1) = 483.463 kJ/kg

Heat transfer = change in internal energy as the process is const. volume,Heat transfer = u1 u2 = 2728.69 983.463 = 1745.227 kJ/kg

8) State whether the following samples of steam are wet, dry or superheated:Justify your answer.

I) Pressure = 1 MPa absolute enthalpy = 2880 kJ/kgII) Pressure=500kPa absolute, volume =0.35m3/kgIII) Temperature = 200oC Pressure = 1.2 MPa.IV) Temperature = 100oC, entropy =6.88kJ/kg K.V) Pressure= 10 kPa, enthalpy = 2584 kJ/kg.

Try as homework.

Date post:	24-Feb-2018
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Vtu Btd Notes for Pure Substance Chapter

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