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

Example

How many lb=h of superheated steam at 1000 psia, 900 F, is required to generate 7500 kW in a steam turbine if the backpressure is 200 psia and the overall efficiency of the turbine generator system is 70%?

Solution. From the steam tables, at 1000 psia, 900 F, h1 ¼ 1448.2 Btu=lb and entropy s1 ¼ 1.6121 Btu=lb F. At 200 psia, corresponding to the same entropy, we must calculate h2s by interpolation. We can note that steam is in superheated condition. h2s ¼ 1257.7 Btu=lb. Then

3413

ASR ¼ 0:70 ð1448 1257:7Þ ¼ 25:6 lb=kWh

Hence, to generate 7500 kW, the steam flow required is

Ws ¼ 25:6 7500 ¼ 192;000 lb=h

9.02a

Q:

What is cogeneration? How does it improve the efficiency of the plant?

A:

Cogeneration is the term used for simultaneous generation of power and process steam from a single full source, as in a system of gas turbine and process waste heat boiler, wherein the gas turbine generates electricity and the boiler generates steam for process (see Fig. 9.1).

In a typical power plant that operates at 35–43% overall efficiency, the steam pressure in the condenser is about 2–4 in.Hg. A lot of energy is wasted in the cooling water, which condenses the steam in the condenser.

If, instead, the steam is generated at a high pressure and expanded in a steam turbine to the process steam pressure, we can use the steam for process, and electricity is also generated. A full credit for the process steam can be given if the steam is used—hence the improvement in overall energy utilization. Q9.02b explains this in detail.

9.02b

Q:

50,000 lb=h of superheated steam at 1000 psia and 900 F is available in a process plant. One alternative is to expand this is a steam turbine to 200 psia and use the 200 psia steam for process (cogeneration). Another alternative is to expand the

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FIGURE 9.1 Cogeneration produces power and steam from the same fuel source by converting the turbine exhaust heat in a boiler, which produces steam for process.

superheated steam in a steam turbine to 1 psia, generating electricity alone in a power plant. Evaluate each scheme.

A:

Scheme 1. The steam conditions are as in Q9.01, so let us use the data on enthalpy. Assume that the turbine efficiency is 70%. The electricity produced can be written as follows using Eq. (1):

P ¼ 50;000 ð1448 1257:7Þ 34130:70 ¼ 1954 kW

Now let us calculate the final enthalpy at condition 2, h2. Using the equation

we obtain

0:70 ð1448 1257:7Þ ¼ 1448 h2

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or

h2 ¼ 1315 Btu=lb

This enthalpy is available for process in the cogeneration mode. The energy Q available in the cogeneration mode is the sum of the electricity produced and the energy to process, all in Btu=h. Hence the total energy is

Q ¼ 1954 3413 þ 50;000 1315 ¼ 72:4 106 Btu=h

Scheme 2. Let us take the case when electricity alone is generated. Let us calculate the final steam conditions at a pressure of 1 psia. s1 ¼ 1.6121 ¼ s2s. At

us estimate the quality x. From basics, 0:1326ð1 xÞ þ 1:9782x ¼ 1:6121

Hence

x ¼ 0:80

The enthalpy corresponding to this condition is h ¼ ð1 xÞhf þ xhg

or

h2s ¼ 0:80 1106 þ 0:2 70 ¼ 900 Btu=lb

(hf and hg are 70 and 1106 at 1 psia.) Using a turbine efficiency of 75%, from Eq. (3) we have

P¼ 50;000 ð1448 900Þ 34130:75 ¼ 6023 kW ¼ 20:55 106 Btu=h

Hence we note that there is a lot of difference between the energy patterns of the two cases, with the cogeneration scheme using much more energy than that used in Scheme 2.

Even if the steam in Scheme 1 were used for oil heating, the latent heat of 834 Btu=lb at 200 psia could be used.

Total output ¼ 1954 3413 þ 50;000 834

¼ 48:3 106 Btu=h

This is still more than the output in the case of power generation alone.

Note, however, that if the plant electricity requirement were more than 2000 kW, Scheme 1 should have more steam available, which means that a bigger boiler should be available. Evaluation of capital investment is necessary before a

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particular scheme is chosen. However, it is clear that in cogeneration the utilization of energy is better.

9.03

Q:

Which is a better location for tapping steam for deaeration in a cogeneration plant with an extraction turbine, the HRSG or the steam turbine?

A:

When steam is taken for deaeration from the HRSG and not from an extraction point in a steam turbine, there is a net loss to the system power output because the steam is throttled and not expanded to the lower deaerator pressure. Throttling is a mere waste of energy, whereas steam generates power while it expands to a lower pressure. To illustrate, consider the following example.

Example

An HRSG generates 80,000 lb=h of steam at 620 psig and 650 F from 550,000 lb=h of turbine exhaust gases at 975 F. The steam is expanded in an extraction-condensing steam turbine. Figure 9.2 shows the two schemes. The

FIGURE 9.2 Options for taking steam for deaeration.

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condenser operates at 2.5 in.Hg abs. The deaerator is at 10 psig. Blowdown losses ¼ 2%. Neglecting flash steam and vent flow, we can show that when steam is taken for deaeration from the HRSG,

81;700 208 ¼ 1700 28 þ ð80;000 X Þ 76 þ 1319X

where 208, 28, 76, and 1319 are enthalpies of feedwater at 240 F, makeup water at 60 F, condensate at 108 F, and steam at 620 psig, 650 F.

The deaeration steam X ¼ 8741 lb=h; use 8785 to account for losses. Now compute the actual steam rate (ASR) in the steam turbine (see Q9.01). It can be

assuming 4% loss in the generator.

Similarly, when steam is taken at 30 psia from the extraction point in the steam turbine, the enthalpy of steam for deaeration is 1140.6 Btu=lb. An energy balance around the deaerator shows

81;700 208 ¼ 1140:6X þ ð80;000 X Þ 76 þ 1700 28

Thus a significant difference in power output can be seen. However, one has to review the cost of extraction machine versus the straight condensing type and associated piping, valves, etc.

9.04

Q:

A fan develops an 18 in. WC static head when the flow is 18,000 acfm and static efficiency of the fan is 75%. Determine the brake horsepower required, the horsepower consumed when the motor has an efficiency of 90%, and the annual cost of operation if electricity costs 5 cents=kWh and the annual period of operation is 7500 h.

A:

The power required when the flow is q acfm and the head is Hw in. WC is

where Zf is the efficiency of the fan, fraction; in this case, Zf ¼ 0.75.

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where Zm is the motor efficiency, fraction. Substituting the data, we have

18

BHP ¼ 18;000 0:75 6356 ¼ 68 hp

and

68

HP ¼ 0:9 ¼ 76 hp

The annual cost of operation will be

76 0:74 0:05 7500 ¼ $21;261

(0.74 is the conversion factor from hp to kW.)

9.05

Q:

A fan develops 18,000 acfm at 18 in. WC when the ambient conditions are 80 F and the elevation is 1000 ft (case 1). What are the flow and the head developed by the fan when the temperature is 60 F and the elevation is 5000 ft (case 2)?

A:

The head developed by a fan would vary with density as follows:

where r is the density, lb=cu ft, and the subscripts 1 and 2 refer to any two ambient conditions.

The flow q in acfm developed by a fan would remain the same for different ambient conditions; however, the flow in lb=h would vary as the density changes.

Let us use Table 9.2 for quick estimation of density as a function of elevation and temperature. r ¼ 0.075=factor from Table 9.2. At 80 F and 1000 ft elevation,

r1 ¼ 01::07506 ¼ 0:0707 lb=cu ft

At 60 F and 5000 ft,

r2 ¼ 01::07518 ¼ 0:0636 lb=cu ft

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TABLE 9.2 Temperature and Elevation Factors

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Substitution into Eq. (7) yields

18 ¼ Hw2

0:0707 0:0636

Hw2 ¼ 16:1 in. WC

In case 1 flow will be

18;000 0:0707 60 ¼ 76;356 lb=h

and in case 2 the flow will be

18;000 0:0636 60 ¼ 68;638 lb=h

The exact operating point of the fan can be obtained after plotting the new Hw versus q characteristic and noting the point of intersection of the new curve with the system resistance curve.

9.06a

Q:

Why should the capacity of forced draft fans for boilers be reviewed at the lowest density condition?

A:

For the same heat input to boilers, the air quantity required in mass flow units (lb=h) remains the same irrespective of the ambient conditions.

W ¼ 60rq

where

W ¼ mass flow; lb=h r ¼ density; lb=cu ft

q ¼ volumetric flow; acfm

Fans discharge constant volumetric flow at any density. Hence if the fan is sized to give a particular volumetric flow at the high density condition, the mass flow would decrease when density decreases as can be seen in the equation above. Hence the fan must be sized to deliver the volumetric flow at the lowest density condition, in which case the output in lb=h will be higher at the higher density condition, which can be then controlled.

Also, the gas pressure drop DP in in. WC across the wind-box is proportional to W 2=r. If the air density decreases as at high temperature conditions, the pressure drop increases, because W remains unchanged for a given heat input. Considering the fact that H=r is a constant for a given fan,

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where H is the static head in in. WC, using the lowest r ensures that the head available at higher density will be larger.

9.06b

Q:

How does the horsepower of a forced draft fan for boilers or heaters change with density?

A:

Equation (5) gives the fan horsepower:

BHP ¼ qHw 6356Zf

Using the relation W ¼ 60qr, we can rewrite the above as

For a boiler at a given duty, the air flow in lb=h and the head in in. WC, Hw, remain unchanged; hence as the density decreases, the horsepower increases. This is yet another reason to check the fan power at the lowest density condition. However, if the application involves an uncontrolled fan that delivers a given volume of air at all densities, then the horsepower should be evaluated at the highest density case because the mass flow would be higher as well as the gas pressure drop.

9.07

Q:

A triplex reciprocating pump is used for pumping 40 gpm (gallons per minute) of water at 100 F. The suction pressure is 4 psig and the discharge pressure is 1000 psig. Determine the BHP required.

A:

Use the expression

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where

q ¼ flow; gpm

DP ¼ differential pressure; psi

Zp ¼ pump efficiency; fraction

In the absence of data on pumps, use 0.9 for triplex and 0.92 for quintuplex pumps.

A 30 hp motor can be used.

The same expression can be used for centrifugal pumps. The efficiency can be obtained from the pump characteristic curve at the desired operating point.

9.08

Q:

A pump is required to develop 230 gpm of water at 60 F at a head of 970 ft. Its efficiency is 70%. There are two options for the drive: an electric motor with an efficiency of 90% or a steam turbine drive with a mechanical efficiency of 95%. Assume that the exhaust is used for process and not wasted.

If electricity costs 50 mills=kWh, steam for the turbine is generated in a boiler with an efficiency of 85% (HHV basis), and fuel costs $3=MM Btu (HHV basis), determine the annual cost of operation of each drive if the plant operates for 6000 h=year.

A:

Another form of Eq. (8) is

W ¼ flow; lb=h

H ¼ head developed by the pump; ft of liquid

For relating head in ft with differential pressure in psi or flow in lb=h with gpm, refer to Q5.01. Substituting into Eq. (9) and assuming that s ¼ 1, W ¼ 230500 lb=h,

970

BHP ¼ 230 500 0:70 1;980;000 ¼ 81 hp

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