01 POWER ISLAND / 01 CCPP / V. Ganapathy-Industrial Boilers and HRSG-Design (2003)

TABLE 6.1 Typical Heat Contents of Various Oils

Copyright © 2003 Marcel Dekker, Inc.

6.04

Q:

Estimate the annual fuel cost for a 300 MW coal-fired power plant if the overall efficiency is 40% and the fuel cost is $1.1=MM Btu. The plant operates for 6000 h=yr.

A:

Power plants have efficiencies in the range of 35–42%. Another way of expressing this is to use the term heat rate, defined as

3413

Heat rate ¼ efficiency Btu=kWh

In this case it is 3413=0.4 ¼ 8530 Btu=kWh.

Annual fuel cost ¼ 1000 megawatt heat rate ðh=yrÞ cost of fuel

in $=MM Btu

1:1 ¼ 1000 300 8530 6000 106

¼ $16:9 106

The fuel cost for any other type of power plant could be found in a similar fashion. Heat rates are provided by power plant suppliers.

6.05

Q:

A 20 MM Btu=h burner was firing natural gas of HHV ¼ 1050 Btu=scf with a specific gravity of 0.6. If it is now required to burn propane having HHV ¼ 2300 Btu=scf with a specific gravity of 1.5, and if the gas pressure to the burner was set at 4 psig earlier for the same duty, estimate the new gas pressure. Assume that the gas temperature in both cases is 60 F.

A:

The heat input to the burner is specified on HHV basis. The fuel flow rate would be Q=HHV, where Q is the duty in Btu=h. The gas pressure differential between the gas pressure regulator and the furnace is used to overcome the flow resistance according to the equation

where

DP ¼ pressure differential, psi

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K ¼ a constant

r ¼ gas density ¼ 0.075s (s is the gas specific gravity; s ¼ 1 for air) Wf ¼ fuel flow rate in lb=h ¼ flow in scfh 0.075s

Let the subscripts 1 and 2 denote natural gas and propane, respectively.

DP1 ¼ 4; r1 ¼ 0:075 0:6, and r2 ¼ 0:075 1:5. Hence, from Eq. (7),

or

DP2 ¼ 2:08 psig

Hence, if the gas pressure is set at about 2 psig, we can obtain the same duty. The calculation assumes that the backpressure has not changed.

6.06

Q:

Gas flow measurement using displacement meters indicates actual cubic feet of gas consumed. However, gas is billed, generally, at reference conditions of 60 F and 14.65 psia (4 oz). Hence gas flow has to be corrected for actual pressure and temperature. Plant engineers should be aware of this conversion.

In a gas-fired boiler plant, 1000 cu ft of gas per hour was measured, gas conditions being 60 psig and 80 F. If the gas has a higher calorific value of 1050 Btu=scf, what is the cost of fuel consumed if energy costs $4=MM Btu?

A:

The fuel consumption at standard conditions is found as follows.

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where

Vs; Va ¼ fuel consumption, standard and actual, cu ft/h

Ts ¼ reference temperature of 520 R

Ta ¼ actual temperature, R

Ps; Pa ¼ standard and actual pressures, psia

520

Vs ¼ 100 ð30 þ 14:22Þ 14:65 540 ¼ 2900 scfh

Hence

Energy used ¼ 2900 1050 ¼ 3.05 MM Btu=h

Cost of fuel ¼ 3.05 4 ¼ $12.2=h.

If pressure and temperature corrections are not used, the displacement meter reading can lead to incorrect fuel consumption data.

6.07

Q:

Estimate the energy in Btu=h and in kilowatts (kW) for heating 75,000 lb=h of air from 90 F to 225 F. What is the steam quantity required if 200 psia saturated steam is used to accomplish the duty noted above? What size of electric heater would be used?

A:

The energy required to heat the air can be expressed as

Q ¼ WaCpDT ð9Þ

where

Q ¼ duty, Btu/h

Wa ¼ air flow, lb/h

Cp ¼ specific heat of air, Btu/lb F

DT ¼ temperature rise, F

Cp may be taken as 0.25 for the specified temperature range.

Q ¼ 75,000 0:25 ð225 90Þ ¼ 2:53 106 Btu=h

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Using the conversion factor 3413 Btu ¼ 1 kWh, we have

Q ¼ 2:53 106 ¼ 741 kW 3413

A 750 kW heater or the next higher size could be chosen.

If steam is used, the quantity can be estimated by dividing Q in Btu=h by the latent heat obtained from the steam tables (see the Appendix). At 200 psia, the latent heat is 843 Btu=lb. Hence

Steam required ¼ 2:55 106 ¼ 3046 lb=h 843

6.08

Q:

Estimate the steam required at 25 psig to heat 20 gpm of 15 API fuel oil from 40 F to 180 F. If an electric heater is used, what should be its capacity?

A:

Table 6.2 gives the heat content of fuel oils in Btu=gal [2]. At 180 F, enthalpy is 529 Btu=gal, and at 40 F it is 26 Btu=gal. Hence the energy absorbed by the fuel oil is

Q ¼ 20 60 ð529 26Þ ¼ 0:6 106 Btu=h

¼ 0:6 106 ¼ 175 kW 3413

The latent heat of steam (from the steam tables) is 934 Btu=lb at 25 psig or 40 psia. Hence

Steam required ¼ 0:6 106 ¼ 646 lb=h 934

If an electric heater is used, its capacity will be a minimum of 175 kW. Allowing for radiation losses, we may choose a 200 kW heater.

In the absence of information on fuel oil enthalpy, use a specific gravity of 0.9 and a specific heat of 0.5 Btu=lb F. Hence the duty will be

Q¼ 20 60 62:40 70::489 0:5 ð180 40Þ ¼ 0:63 106 Btu=h

(7.48 is the conversion factor from cubic feet to gallons.)

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TABLE 6.2 Heat Content (Btu=gal) of Various Oilsa

aValues in regular type are for liquid; bold values are for vapor.

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TABLE 6.3 Combustion Constants

All gas volumes corrected to 60 F and 30 in. Hg dry. For gases saturated with water at 60 F, 1.73% of the Btu value must be deducted.

aCalculated from atomic weights given in Journal of the American Chemical Society, February 1937.

bDensities calculated from values given in gL at 0 C and 760 mmH in the International Critical Tables allowing for the known deviations from the gas laws. Where the coefficient of expansion was not available, the assumed value was taken as 0.0037 per C. Compare this with 0.003662, which is the coefficient for a perfect gas. Where no densities were available, the volume of the mole was taken as 22.4115 L.

cConverted to mean Btu per lb (1=180 of the heat per lb of water from 32 to 212 F) from data by Frederick D. Rossini, National Bureau of Standards, letter of April 10, 1937, except as noted.

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dDeduction from gross to net heating value determined by deducting 18,919 Btu=lb mol water in the products of combustion. Osborne, Stimson and Ginnings, Mechanical Engineering, p. 163, March 1935, and Osborne, Stimson, and Flock, National Bureau of Standards Research Paper 209.

eDenotes that either the density or the coefficient of expansion has been assumed. Some of the materials cannot exist as gases at 60 F and 30 in.Hg pressure, in which case the values are theoretical ones given for ease of calculation of gas problems. Under the actual concentrations in which these materials are present their partial pressure is low enough to keep them as gases.

fFrom third edition of Combustion. Adapted from Ref. 8.

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6.09a

Q:

Natural gas having CH4 ¼ 83:4%; C2H6 ¼ 15:8%, and N2 ¼ 0:8% by volume is fired in a boiler. Assuming 15% excess air, 70 F ambient temperature, and 80% relative humidity, perform detailed combustion calculations and determine flue gas analysis.

A:

From Chapter 5 we know that air at 70 F and 80% RH has a moisture content of 0.012 lb=lb dry air. Table 6.3 can be used to figure air requirements of various fuels. For example, we see that CH4 requires 9.53 mol of air per mole of CH4, and C2H6 requires 16.68 mol.

Let us base our calculations on 100 mol of fuel. The theoretical dry air required will be

83:4 9:53 þ 16:68 15:8 ¼ 1058:3 mol

Considering 15% excess,

Actual dry air ¼ 1.15 1058.3 ¼ 1217 mol Excess air ¼ 0.15 1058.3 ¼ 158.7 mol Excess O2 ¼ 158.7 0.21 ¼ 33.3 mol Excess N2 ¼ 1217 0.79 ¼ 961 mol

(Air contains 21% by volume O2, and the rest is N2.)

(We multiplied moles of air by 29 to get its weight, and then the water quantity was divided by 18 to get moles of water.)

Table 6.3 can also be used to get the moles of CO2; H2O, N2 and O2 [3].

CO2 ¼ 1 83:4 þ 2 15:8 ¼ 115 mol

H2O ¼ 2 83:4 þ 3 15:8 þ 23:5 ¼ 237:7 mol

O2 ¼ 33:3 mol

N2 ¼ 961 þ 0:8 ¼ 961:8 mol

The total moles of flue gas produced is 115 þ 237.7 þ 33.3 þ 961.8 ¼ 1347.8. Hence

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The analysis above is on a wet basis. On a dry flue gas basis,

100

%CO2 ¼ 8:5 100 17:7 ¼ 10:3%

Similarly,

%O2 ¼ 3:0%; %N2 ¼ 86:7%

To obtain wda,wwa,wdg, and wwg, we need the density of the fuel or the molecular weight, which is

1001 ð83:4 16 þ 15:8 30 þ 0:8 28Þ ¼ 18:30

29

wda ¼ 1217 100 18:3 ¼ 19:29 lb dry air=lb fuel

23:5 18

wwa ¼ 19:29 þ 18:3 100 ¼ 19:52 lb wet air=lb fuel

wdg ¼ 115 44 þ 33:3 32 þ 961 28 1830

¼ 18 lb dry gas=lb fuel

wwg ¼ 115 44 þ 33:3 32 þ 237:7 18 þ 961:8 28 1830

¼ 20:40 lb wet gas=lb fuel

This procedure can be used when the fuel analysis is given. More often, plant engineers will be required to estimate the air needed for combustion without a fuel analysis. In such situations, the MM Btu basis of combustion and calculations will come in handy. This is discussed in Q6.10a.

6.09b

Q:

For the case stated in Q6.09a, estimate the partial pressure of water vapor, pw, and of carbon dioxide, pc, in the flue gas. Also estimate the density of flue gas at 300 F.

A:

The partial pressures of water vapor and carbon dioxide are important in the determination of nonluminous heat transfer coefficients.

volume of water vapor

pw ¼ total flue gas volume ¼ 0:177 atm ¼ 2:6 psia

pc ¼ volume of carbon dioxide ¼ 0:085 atm ¼ 1:27 psia total flue gas volume

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