Bk 2 Sec 2.2 Basic Concepts and Calculations

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Section 2.2

BASIC CONCEPTS AND CALCULATIONS

CONFIDENTIAL, For Nalco Employee Use Only 2006 Nalco Company

PAC-2 Boiler Water Technical Manual (6-06)

5

In the boiler, water is converted to steam. The steam is used as a medium for transferring energy(in the form of heat) to its point of use.

Many things can happen to the water and steam en route to the final destination. Steam may belost; some is condensed and returned to its liquid state. Some of the highly concentrated boilerwater must be removed to prevent scale deposition and corrosion. Water must be added tocompensate for these losses or to replace water removed deliberately.

For a properly operating system, water entering a boiler equals water and steam exiting the boiler.This section explains the relationships between the various water streams in a boiler system.

STEAM AND ITS USES

Water, when sufficiently heated, is converted toits vapor form, steam. Because of its heat orenergy content and because it is an easilytransportable fluid, steam is an efficient meansof moving energy from one point (the source orboiler) to its intended end use (heating, electricpower generation, etc.). Here, the energy con-tained in the steam is transferred to anotheroperation requiring heat, such as a chemicalreaction or heating of a fluid, or it is convertedinto work, such as in driving a piston or turbine.Steam finds use in a wide variety of applica-

tions: comfort heating in buildings; processheating in many industries such as food,petroleum, chemical processing, and pulpand paper; and electric power generation.

STEAM GENERATION

Steam is produced in a boiler, which is usuallyheated by the burning of coal, oil, gas, woodwaste, or organic by-products of a plant process.Steam is also produced at some electric utilityplants by nuclear reaction. Water is boiled, and

the vapor is discharged at a controlled tempera-ture and pressure. As vapor leaves the boilingwater, the dissolved solids originally in thewater fed to the boiler are left behind. A simpleconceptual diagram of the steam generationsection of a boiler system is shown inFigure 2.2.1.

The boiler water left behind becomes increas-

ingly concentrated, and eventually reaches alevel where further concentration could causescale or deposits to form or result in problemswith steam quality or steam purity. This highlyconcentrated water must be removed to preventscale deposition or corrosion. This removalprocess is known as blowdown. After blow-down, additional water (makeup) must beadded to the system to replace any steam orcondensate not returned to the boiler as wellas the concentrated water drawn from thesystem as blowdown.

In the steam generator, water circulates in apattern produced by the path of heat from thefuel through a maze of boiler tubes, as illustratedin Figures 2.2.2, 2.2.3, and 2.2.4.

Condensate

When the steam delivers its heat to the pointof use, it gives up, depending upon pressure,between 210-1100 Btu/lb (488-2560 kJ/kg) as

it condenses at the prevailing temperature andpressure and returns to the lower energy levelstate of a liquid. Condensate is usually free ofmineral contamination and contains heat, soit is a very valuable source of water. As much


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Figure 2.2.1 Simple material balance in a boiler

Figure 2.2.2 Water circulation in a firetube

boiler

Figure 2.2.3 Circulation in a two-drum

watertube boiler


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condensate as possible is collected and returnedto the boiler because of its heat value and purity.

As condensate forms, it must be drained to areservoir (receiver) for return to the feedwaterstream, usually at the deaerator. The drainagesystem, which is usually separate for each majorsteam user and condensate producer, consists ofthe following components:

Steam traps to maintain backpressure and topermit condensate flow but not steam flow

Collection tanks, which are usually vented to alow-pressure steam line or to the atmosphere

Pumps to deliver the condensate to thedeaerator or another point in the system

A typical drainage system is shown in Figure2.2.5, and a schematic with extended possibleroutes is shown in Figure 2.2.6.

Losses of Steam or Condensate

Very little steam or condensate is lost in a largeutility station, except in a few operations that areusually intermittent, such as soot blowing.

However, in an industrial plant, it is quitecommon to lose 30-80% of steam, condensate,or both. Major causes of steam and condensateloss include consumption in the process, con-tamination of condensate through productinleakage, and pipe leaks. Other losses mightresult from erratic demands for steam by differ-ent operating departments, which can upset theheat balance and cause a loss of low-pressuresteam because of a momentary or seasonalinability to use it.

Economics also play a role; the cost of installingcondensate collection systems with piping,especially for minor use, is often not justified bythe value of recovered condensate, particularlyin older plants where piping installation wouldbe a problem or where fuel costs are very low.

Makeup

Fresh water of acceptable quality must be addedto compensate for these losses. Hence, this freshwater is called makeup. Scale-forming mineralsare generally removed from makeup by precipi-tation softening, ion exchange, or reverseosmosis before it is added to the boiler system.Makeup may also be heated by waste heatstreams such as blowdown before it is added tothe boiler system. A simplified but typicalmakeup scheme is shown in Figure 2.2.7.

Concentration by Evaporation

Even after primary treatment, makeup stillcontains dissolved solids, such as traces ofhardness, and variable amounts of sodium,

alkalinity and silica. Makeup must be treated toreduce this dissolved matter to acceptable levels(depending on the pressure of the boiler and theuses of the steam); the choice of pretreatmenttechnology is based principally on economics.

Figure 2.2.4 Pumps assist in circulating water

through a boiler when natural circulation is

restricted by limited density difference because

of pressure or limited elevation


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Figure 2.2.5 Typical condensate drainage system


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Figure 2.2.6 Extended condensate system

Figure 2.2.7 Typical boiler makeup water system


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As relatively pure steam is withdrawn from theboiler, the solids left behind concentrate in theboiler water. Based upon pressure and steamquality, there is a limit to the maximum concen-

tration levels the boiler can tolerate. Exact limitsare impossible to determine; however, severalconsensus guidelines have been gathered basedupon operating experience, laboratory data,etc., and have been published by variousorganizations including the American Societyof Mechanical Engineers (ASME), the AmericanBoiler Manufacturing Association (ABMA),and Nalco. In addition to the above references,there are other references such as the VGB(a German organization of large power stationoperators), the VdTV (Association of German

Technical Inspectorates), the BS (BritishStandards), and the JIS (Japanese IndustrialStandard). Each of these may approach theproblem of defining feedwater and boiler waterconcentrations from a different perspective;therefore, the recommended values may differ,depending on the source.

Remember that the values provided are sug-gested guidelines and that actual boiler specifi-cation limits must be established on an indi-vidual basis, depending on steam purity and end

use requirements. Typical limits for non-utilityboiler applications as established by the ASMEare shown in Tables 2.13.1, 2.13.2, and 2.13.3in Section 2.13 - Blowdown Control and HeatRecovery. Additional recommended boilercontrol limits are presented in Section 2.16 -Boiler Scale and Deposits.

Blowdown

Blowdown is a deliberate loss of boiler waterintended to keep the critical components dis-solved in the boiler water from over-con-centrating. Blowdown must also be replacedby makeup, so that the following relationshipapplies.

TM SL CL BL= + + (1)

Where:TM = Total makeupSL = steam loss including soot blowing

and leaks

CL = condensate loss including sewered and leaksBL = blowdown loss, deliberate from mud

drum and steam drum

In low-pressure units or in boiler systems that donot produce large quantities of steam, blowdownis often sewered. In larger systems, considerableamounts of blowdown contain valuable heat,which can be recovered in flash tanks and heatexchangers. Flash tanks extract low-pressuresteam and send it to various users, which may

include the deaerator or lower pressure pro-cesses. Water from the flash tank still hasconsiderable heat value and may be used furtherto preheat incoming makeup water. Figure 2.2.8shows a blowdown system and possibleschemes. Calculations showing the relationshipbetween blowdown and feedwater are presentedlater in this section.

Feedwater

Makeup and return condensate combine tobecome feedwater. A simple diagram showing

a typical feedwater system is presented inFigure 2.2.9. Guidelines for feedwater qualityhave been established by ASME and are pre-sented in Tables 2.13.1, 2.13.2, and 2.13.3 inSection 2.13 - Blowdown Control and HeatRecovery.

MATEMATICAL RELATIONSHIPS

All of the flows, input and output, in a typical

industrial boiler system are shown schematicallyin Figure 2.1.1 in Section 2.1 - Introduction. Inthe system illustrated, steam discharges from aboiler to a turbine and other users. All or part ofthe steam may be condensed in the condenser;some may be extracted for process use. Othersteam may go directly to a process where it may


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Figure 2.2.8 Typical boiler blowdown system

Figure 2.2.9 Typical boiler feedwater system


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be lost or may become contaminated and mustbe dumped. Still other steam may go directly toa process and then be condensed and returned tothe boiler. The relationships among blowdown,

makeup, and feedwater are discussed in detailhere and are illustrated in the materials balancediagram in Figure 2.2.1.

BLOWDOWN AND MAKEUP

Blowdown and makeup are usually expressedas a percentage of feedwater. (They can alsobe calculated as a percentage of steam flow;however, this text will use the feedwatercalculation convention.) When the flow ratesof these streams are known, the following

equations can be used to describe theirrelationship.

%%

BD Quantity of BD

Quantity of FW=

100 (2)

%%

MU Quantity of MU

Quantity of FW=

100 (3)

Where:BD = blowdown mass flowMU = makeup mass flowFW = feedwater mass flow

Many times, actual quantities of blowdown,feedwater, and makeup are not known. In suchcases, percentages of blowdown and makeup canbe calculated based upon the concentration of aparticular dissolved solid or total dissolvedsolids (TDS) in the various streams. Thismethod is shown in Equations 4 and 5.

%%

BD TDS in FW

TDS in BD

=100

(4)

%%

MU TDS in FW

TDS in MU=

100 (5)

Where:TDS = total dissolved solids

It is very important that when dissolved solidsare used to measure the percent blowdown, thefeedwater sample must be taken after all chemi-cals have been added to the system. If conduc-

tivity is used for the measurement of dissolvedsolids, the boiler water sample must first beneutralized before measuring the conductivitybecause any hydrate (O) alkalinity will greatlyincrease the conductivity of the sample, leadingto a low percent blowdown.

The percent makeup flow can be calculated fromthe TDS of the makeup water, the feedwater, andthe condensate as follows:

%( )( )

( )MU

TDS TDS

TDS TDS

FW C

MU C

=

100 (6)

Where:TDS

FW= TDS of the feedwater

TDSC

= TDS of the condensateTDS

MU= TDS of the makeup water

CYCLES OF CONCENTRATION (COC)

As discussed earlier, when steam (devoid ofsolids) is generated, dissolved solids remainbehind and concentrate in the boiler water. The

ratio of the quantity of dissolved solids in theboiler water to those in the feedwater is definedas the concentration ratio (CR) or cycles ofconcentration (COC). (See Figure 2.2.1.)Expressed another way, the number of cyclesof concentration in a boiler system is simplythe reciprocal of the percent blowdown asshown in Equation 7.

COCBD

=100

% (7)

There are times, such as when the makeup wateris demineralized and contains little to no dis-solved solids, when conductivity cannot be usedto measure the cycles of concentration. In thesecases, some other methods must be used tomeasure the cycles of concentration. This can be


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Section 2.2




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done by injecting into the feedwater somethingthat is non-volatile, non-reacting, and non-precipitating such as TRASAR. By measuringthe concentration of TRASAR in the feedwater

and in the boiler water, the ratio of the boilerwater concentration of TRASAR to the concen-tration of TRASAR in the feedwater representsthe cycles of concentration. This method isshown in Equation 7a.

COC BW concentration

FW concentration=

(7a)

If the boiler continuous blowdown and the steamflows are measured, the cycles of concentrationcan be calculated using Equation 7b.

COC Steam flow BD flow

BD flow=

+( )(7b)

Once the COC is known, the blowdown flowcan be calculated from the following:

BD flow Steam flow

COC =

( )1 (7c)

Again, once the COC is known, the feedwater

flow can be calculated from Equations 7dand 7e.

FW flow Steam flow BD flow= + (7d)

FW flow Steam flow COC

COC =

( )( )

( - )1 (7e)

Using Equation 7f, the blowdown flow can alsobe calculated from measurements of the steamflow and the feedwater flow, although thiscalculation will result in a higher error becausea relatively small number is calculated from thedifference between two larger numbers, andthese numbers are taken from two separatemeters, both of which may be out of calibration.

BD flow FW flow Steam flow= (7f)

MASS BALANCE EQUATIONS

Figures 2.2.6, 2.2.7, 2.2.8, and 2.2.9 show theboiler plus the three support systems: pretreat-ment, feedwater, and condensate systems. (see

also Figure 2.2.1). Simple mass balances de-scribing these systems and their relationships aredefined by the following equations.

FW flow MU flow RC flow DA V= + + (8)

FW flow Steam flow BD flow= + (9)

Steam Flow RC flow NRC flow= + (10)

Where:DA = Deaerator steam flowV = Deaerator vent flowRC = returned condensateNRC = non-returned condensate

An accurate water analysis plus one or twomeasured flow rates can be used to determinemost other system flow rates. (See Example2.2.1.) These, in turn, can be used to calculatechemical dosages, size equipment properly,and estimate the heat and energy content ofvarious streams.

Many times, feedwater and makeup flowrates are not metered but steaming rate is.Knowing boiler water and feedwater chemicalcompositions will allow feedwater, makeup,and blowdown flow rates to be calculated.(See Example 2.2.2.)

OPTIMIZING THE WATER BALANCE

Because boiler systems are energy intensive andthe fuel needed to produce energy is expensive,it is in a plants best interest to maximize steamproduction and minimize fuel consumption.

The first step in optimizing the system is to takeevery practical measure to eliminate steam andcondensate losses, working within the economic


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EXAMPLE 2.2.1 MASS BALANCE CALCULATIONS

Given: Makeup water flow = 115000 gpd (435.3 m3/d)

Water Analyses

Constituent Makeup, ppm Feedwater, ppm Boiler, ppmNaCl 32 24 192SiO

28 6 48

TDS 220 165 1320

Find: Makeup flow (MU), lb/d (kg/d) Percent blowdown, %Feedwater flow (FW), lb/d (kg/d) Cycles of concentration (COC)Returned condensate (RC), lb/d (kg/d) Steam flow (S), lb/d (kg/d)Blowdown flow (BD), lb/d (kg/d) Condensate losses (NRC), lb/d (kg/d)

Solution: U.S. Units Metric UnitsMU = (115000 gpd)(8.34 lb/gal) = 959100 lb/d MU = (435.3 m3/d)(1000 kg/m3) = 435300 kg/d

Using Equation 5: Using Equation 5:

%MU = (100)(FW TDS)/MU TDS %MU = (100)(FW TDS)/MU TDS%MU = (100)(165)/220 = 75% %MU = (100)(165)/220 = 75%


%MU = (100)(MU)/FW %MU = (100)(MU)/FW

Rearranging: Rearranging:

FW = (100)(MU)/%MU FW = (100)(MU)/%MUFW = (100)(959100)/75 = 1278800 lb/d FW = (100)(435300)/75 = 580400 kg/d


FW = MU + RC + DA - V FW = MU + RC + DA - VAssume Deaerator steam (DA) included in RC Assume Deaerator steam (DA) included in RCAssume Deaerator vent (V) negligible Assume Deaerator vent (V) negligible


RC = FW - MU = 1278800 - 959100 = 319700 lb/d RC = FW - MU = 580400 - 435300 = 145100 kg/dUsing Equation 4: Using Equation 4:

%BD = (100)(FW TDS)/(BD TDS) %BD = (100)(FW TDS)/(BD SiO2)

%BD = (100)(165)/1320 = 12.5% %BD = (100)(165)/1320 = 12.5%


COC = 100/(%BD) = 100/12.5 = 8 COC = 100/(%BD) = 100/12.5 = 8


%BD = (100)(BD)/FW %BD = (100)(BD)/FW


BD = (FW)(%BD)/100 BD = (FW)(%BD)/100BD = (1278800)(12.5)/100 = 159850 lb/d BD = (580400)(12.5)/100 = 72550 kg/d


FW = S + BD FW = S + BDRearranging: Rearranging:

S = FW - BD = 1278800 - 159850 = 1118950 lb/d S = FW - BD = 580400 - 72550 = 507850 kg/d


S = RC + NRC S = RC + NRC


NRC = S - RC = 1118950 - 319700 = 799250 lb/d NRC = S - RC = 507850 - 145100 = 362750 kg/d


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EXAMPLE 2.2.2 MASS BALANCE CALCULATIONS

Given: Chemical Analysis

Blowdown SiO2 48 ppmMakeup SiO

28 ppm

Feedwater SiO2

6 ppmSteam flow rate 1000000 lb/h (453590 kg/h)

Find: Feedwater (FW) flow, lb/h (kg/h)Makeup (MU) flow, lb/h (kg/h)Blowdown (BD) flow, lb/h (kg/h)

Solution:

U.S. Units Metric Units


%BD = (100)(FW SiO2)/(BD SiO

2) %BD = (100)(FW SiO

2)/BD SiO

2

%BD = (100)(6)/48 = 12.5% %BD = (100)(6)/48 = 12.5%


%BD = (100)(BD)/FW %BD = (100)(BD)/FW

Rearranging and substituting: Rearranging and substituting:

BD = (0.125) (FW) (A) BD = (0.125) (FW) (A)


FW = S + BD (B) FW = S + BD (B)

Solving Equations A and B simultaneously: Solving Equations A and B simultaneously:

FW = S + (0.125) (FW) FW = S + (0.125) (FW)


FW = 1000000/(1 - 0.125) = 1142857 lb/h FW = 453590/(1 - 0.125) = 518389 kg/h


%MU = (100)(FW SiO2)/(MU SiO

2) %MU = (100)(FW SiO

2)/(MU SiO

2)

%MU = (100)(6)/8 = 75% %MU = (100)(6)/8 = 75%


%MU = (100)(MU)/FW %MU = (100)(MU)/FW


MU = (%MU)(FW)/100 = (75)(1142857)/100 MU = (%MU)(FW)/100 = (75)(518389)/100

= 857143 lb/h = 388792 kg/h


FW = S + BD FW = S + BD


BD = FW - S = 1142857 - 1000000 = 142857 lb/h BD = FW - S = 518389 - 453590 = 64799 kg/h


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constraints of the cost of revisions, new equip-ment, piping, and installation. For example,it might be practical to install piping to returncondensate from a tank farm in a refinery or

petrochemical plant, but if the condensate isbadly contaminated with oil, the extra cost ofinstalling and operating condensate filters maynot be justified.

The next major step is to minimize blowdown,or to get the most mileage from the makeup forthe control limits required by the boiler. Becausethe makeup treatment system is in place, a plantusually must do its best with the system at hand;in some cases, however, a new system may berequired to optimize makeup use. Major unit

operations of makeup treatment are discussedin Sections 2.3, 2.4, and 2.5 of this manual.

As shown in the ASME guidelines, the limitson such controls as total dissolved solids, silica,and alkalinity are related to the amount of thesematerials coming in with the makeup water.These concentrations are usually adjusted byblowdown, but some manipulation of themakeup treatment system might be possible,depending on system flexibility. Other constitu-ents such as phosphate, organics, and sulfite

are introduced as internal treatment chemicals,and their concentration can be adjusted bychanging their rate of application.

Table 2.2.1 illustrates how minimizing boilerblowdown can optimize a boiler system. Con-sider a 600 psig (4.1 MPag) boiler system in

a paper mill. The makeup is treated by hot limesoftening followed by ion exchange softening,and after treatment has a total dissolved solids(TDS) concentration of 150 ppm and a silica

concentration of 3 ppm.

Using 50% condensate return for this examplewill yield a feedwater TDS and silica of 75 ppmand 1.5 ppm, respectively. With an allowablelimit of only 30 ppm silica in the boiler, silicais the controlling factor that sets the maximumconcentration ratio at 20 (5% blowdown).However, the water could be concentrated by afactor of 26.7 based on total dissolved solids, sothere is incentive for additional silica reductionin the treatment unit. The addition of dolomitic

lime (rather than hydrated lime) or the additionof magnesium oxide might permit a reductionfrom 3 ppm to less than 2 ppm silica in themakeup. With lower silica in the makeup, silicain the feedwater will be reduced accordingly.

Reducing silica in the makeup from 3 ppm to2 ppm will reduce it equivalently in thefeedwater. Therefore, the feedwater would thencontain only 1.0 ppm silica, and blowdownbased on silica could potentially be reduced to1.0/30 or 3.3%. Now the limiting factor would

no longer be silica but total dissolved solids.The boiler may now be cycled up to its maxi-mum for conductivity at 26.7 cycles or 3.75%blowdown. Depending upon steam production,the availability of blowdown heat recoveryequipment, and fuel costs, the additional 25%blowdown reduction (5% blowdown to 3.75%

Maximum

Boiler Concentration

Factor Makeup Feedwater Limit Ratio

TDS, ppm 150 75 2000 26.7

SiO2, ppm 3 1.5 30 20.0

SiO2, ppm 2 1.0 30 30

(after reduction)

Table 2.2.1 Maximum concentration ratio


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blowdown) could result in sizeable fuel savings.There will also be additional savings in reducedmakeup water costs and blowdown sewer costs.

This example shows that the concentration ratiois determined easily by chemical analysis.Blowdown rate is rarely metered, although mostplants meter makeup, and some meter feedwateras well. Steam flow is usually metered in mostplants, and the feedwater flow can be calculatedbased on the concentration ratio and the steamproduction by using the mathematical relation-ships discussed previously.

EXAMPLE 2.2.3 DETERMINING HEAT CONTENT (U.S UNITS)

Referring to Table 2.2.2, what is the available heat when steam is condensed at 300F?

The heat content of saturated steam at 300F is 1179.7 Btu/lb. The heat content of water at300F is 269.7 Btu/lb. The difference between the heat contents of saturated steam and waterrepresents the amount of heat or energy available when the steam condenses at 300F. In thiscase, 1179.7 - 269.7 = 910 Btu/lb.

EXAMPLE 2.2.4 DETERMINING HEAT CONTENT (METRIC UNITS)

Referring to Table 2.2.3, what is the available heat when steam is condensed at 150C?

The heat content of saturated steam at 150C is 2746.5 kJ/kg. The heat content of water at150C is 632.2 kJ/kg. The difference between the heat contents of saturated steam and waterrepresents the amount of heat or energy available when the steam condenses at 150C. In thiscase, 2746.5 - 632.2 = 2114.3 kJ/kg.

THERMODYNAMIC PROPERTIESOF STEAM

Steam and water at different pressures andtemperatures contain different amounts of

energy. The difference between the energycontent of steam and water at the same pressureand temperature is significant and represents thedriving force of the steam. The thermodynamicproperties of saturated steam and water areshown in Tables 2.2.2 and 2.2.3. Examples 2.2.3and 2.2.4 illustrate how this table can be used todetermine available heat content. Examples2.2.5 and 2.2.6 show how to calculate the heatvalue of recovered steam condensate.


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EXAMPLE 2.2.5 DETERMINING HEAT VALUE OF CONDENSATE(U.S. UNITS)

Given that the temperature of the condensate is 150F and the temperature of the makeup water is60F, what is the recoverable heat from the condensate?

When the water temperature is less than 212F, the enthalpy of the water, in Btu/lb, can becalculated by subtracting 32 from the water temperature. Therefore, the enthalpy of thecondensate is 150-32 or 118 Btu/lb; the enthalpy of the makeup water is 60-32 or 28 Btu/lb.The recoverable heat value of the condensate is then the difference between the two enthalpiesor 118-28 = 90 Btu/lb.

Note, in a boiler system, the base heat content (enthalpy) is that of the makeup water. Anyrecovered liquid, such as steam condensate, will replace an equivalent amount of makeup water.Since the makeup water contains some heat (enthalpy greater than zero), the recoverable heat is

the difference in enthalpies between the recovered liquid and the makeup water.

EXAMPLE 2.2.6 DETERMINING HEAT VALUE OF CONDENSATE

(METRIC UNITS)

Given that the temperature of the condensate is 65C and the temperature of the makeup water is15C, what is the recoverable heat from the condensate?

When the water temperature is less than 100C, the enthalpy of the water, in kJ/kg, can becalculated by multiplying the temperature of the water in degrees Celsius by 4.1868. Therefore,the enthalpy of the condensate is (65)(4.1868) or 272 kJ/kg; the enthalpy of the makeup water is(15)(4.1868) or 62.8 kJ/kg. The recoverable heat value of the condensate is then the differencebetween the two enthalpies or 272-62.8 = 209.2 kJ/kg.

Note, in a boiler system, the base heat content (enthalpy) is that of the makeup water. Anyrecovered liquid, such as steam condensate, will replace an equivalent amount of makeup water.Since the makeup water contains some heat (enthalpy greater than zero), the recoverable heat isthe difference in enthalpies between the recovered liquid and the makeup water.


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Table 2.2.2 Saturated steam table (U.S. units)


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Table 2.2.3 Saturated steam table (metric units)


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TERMINOLOGY

The concepts of work and energy are important to the understanding of how steam can be utilizedin a wide variety of applications. Following are terms commonly used for defining energy, work,and power.

Work The product of a force (F) applied uniformly along or through a distance (d) is defined as

work. For example, the work expended in lifting a weight requiring an applied force (F) where thedistance (d) in transit is upward is calculated as follows.

work = (F)(d)Work is expressed in ft-lb (kg-m)

Heat The amount of energy given up or absorbed by a substance is equal to the product of themass of a substance (m) heated uniformly, its temperature change (T), and its specific heat (C

p),

which is a function of the nature of the substance heated (for water, Cp= 1 Btu/lbF, 1 kcal/kgC,

or 4.1868 kJ/kgC).

heat = mCp

THeat is expressed in Btu (kcal or kJ)

Heat equivalent of work A Btu is defined as the amount of heat necessary to raise the temperatureof one pound of water one degree Fahrenheit. Also, 1 Btu = 778 ft-lb.

A kilocalorie is defined as the amount of heat necessary to raise the temperature of one kilogramof water one degree Celsius, and one kilocalorie equals 4.1868 kilojoules. Also, 1 joule =0.1020 kg-m.

Electrical equivalent of work The unit of electrical energy is the kilowatt-hour (kWh). Themechanical energy of an engine driving an electrical generator is converted to electrical energy.

Common units of energy include Btu, ft-lb, kWh, kcal, and kJ.

1 kWh =3412 Btu = 2655000 ft-lb1 kWh = 860 kcal = 3600 kJ = 367100 kg-m

This is the most commonly misunderstood energy relationship, because kilowatts are oftenerroneously considered an energy term rather than a power term.

Power This is the rate or speed of doing work, generating heat, or producing electricity. Forexample, a person weighing 150 pounds (68 kg) and climbing a 100 foot (30.5 m) hill wouldexpend 15000 ft-lb (2074 kg-m) of energy. If this climb requires 10 minutes (600 seconds), theenergy expended would be 25 ft-lb/s (3.46 kg-m/s). The power applied is defined as:

U.S. Units: Metric Units:

25 ft-lb/s/(550 ft-lb/s/hp) = 0.045 hp 3.46 kg-m/s/(76 kg-m/s/hp) = 0.046 hp

If the hill is climbed in 5 minutes, 15000 ft-lb (2074 kg-m) of energy are still used, but the powerexerted is doubled.

U.S. Units: Metric Units:

15000 ft-lb/300 s/(550 ft-lb/s/hp) = 0.09 hp 2074 kg-m/300 s/(76 kg-m/sec/hp) = 0.09 hp

Common units of power include horsepower, kilowatt, Btu/h, kcal/h, and kJ/h.1 hp = 550 ft-lb/s = 0.7457 kW = 2544 Btu/h1 hp = 76 kg-m/s = 642 kcal/h = 2685 kJ/h


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Bk 2 Sec 2.2 Basic Concepts and Calculations