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

Solution.

VP ¼ 0:4 14462 ¼ 0:93 ft

Suction presssure ¼ 14:6 psia ¼ 14:6 16662 ¼ 33:9 ft

NPSHa ¼ 33:9 3 0:93 þ 10 ¼ 40 ft

9.16

Q:

In the absence of information from the pump supplier, can we estimate NPSHr?

A:

A good estimate of NPSHr can be made from the expression for specific speed S.

S ranges from 7000 to 12,000 for water.

For example, when q ¼ 100 gpm, N ¼ 1770, and assuming that S ¼ 10,000

Even if we took a conservative value of 7000 for S, we would get

NPSHr ¼ 3:43 ft

This information can be used in making preliminary layouts for systems involving pumps.

9.17

Q:

How is NPSHa for a reciprocating pump arrived at?

A:

NPSHa for a reciprocating pump is calculated in the same way as for a centrifugal pump except that the acceleration head Ha is included with the friction losses. This is the head required to accelerate the liquid column on each suction stroke so

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that there will be no separation of this column in the pump suction line or in the pump [1]:

where

L ¼ length of the suction line; ft ðactual length; not developedÞ

V ¼ velocity in the suction line; ft=s

N ¼ pump speed; rpm

C is a constant: 0.066 for triplex pump, 0.04 for quintuplex, and 0.2 for duplex pumps. K is a factor: 2.5 for hot oil, 2.0 for most hydrocarbons, 1.5 for water, and 1.4 for deaerated water. g ¼ 32 ft=s2. Pulsation dampeners are used to reduce L significantly. By proper selection, L can be reduced to nearly zero.

Example

A triplex pump running at 360 rpm and displacing 36 gpm has a 3 in. suction line 8 ft long and a 2 in. line 18 ft long. Estimate the acceleration head required.

Solution. First obtain the velocity of water in each part of the line. In the 3 in. line, which has an inner diameter of 3.068 in.,

In the 2 in. line, which has an inner diameter of 2.067 in.,

36

V ¼ 0:41 ð2:067Þ2 ¼ 3:45 ft=s The acceleration head in the 3 in. line is

0:066

Ha ¼ 8 360 1:57 1:4 32 ¼ 6:7 ft In the 2 in. line,

0:066

Ha ¼ 18 3:45 360 1:4 32 ¼ 32:9 ft The total acceleration head is 32.9 þ 6.7 ¼ 39.6 ft.

9.18

Q:

How can we check the performance of a pump from the motor data?

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A:

A good estimate of the efficiency of a pump or a fan can be obtained from the current reading if we make a few reasonable assumptions. The efficiency of a motor is more predictable than that of a pump owing to its small variations with duty. The pump differential pressure and flow can be obtained rather easily and accurately. By relating the power consumed by the pump with that delivered by the motor, the following can be derived. The pump power consumption, P, in kW from Eq. (8) is

where

q ¼ flow; gpm

DP ¼ differential pressure; psi

E ¼ voltage; V

I ¼ current; A

Zp; Zm ¼ efficiency of pump and motor; fraction cos f ¼ power factor

From Eq. (22) we can solve for pump efficiency given the other variables. Alternatively, we can solve for the flow by making a reasonable estimate of Zp and check whether the flow reading is good. The power factor cos f typically varies between 0.8 and 0.9, and the motor efficiency between 0.90 and 0.95.

Example

A plant engineer observes that at a 90 gpm flow of water and 1000 psi differential, the motor current is 100 A. Assuming that the voltage is 460 V, the power factor is 0.85, and the motor efficiency is 0.90, estimate the pump efficiency.

Solution. Substituting the data into Eq. (22), we obtain

90 1000 ¼ 4 460 100 0:85 0:90 Zp

Solving for Zp, we have Zp ¼ 0.65.

We can use this figure to check whether something is wrong with the system. For instance, if the pump has been operating at this flow for some time but the current drawn is more, one can infer that the machine needs attention. One

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can also check the pump efficiency from its characteristic curve and compare the calculated and predicted efficiencies.

9.19

Q:

Derive an expression similar to (22) relating fan and motor.

A:

Equating the power consumption of a fan with that delivered by its motor,

where

q ¼ flow; acfm

Hw ¼ static head of fan; in: WC

Other terms are as in Q9.18.

If the efficiency of a fan is assumed to be 65% when its differential head is 4 in. WC, the motor voltage is 460, and the current is 7 A, then the power factor is 0.8 and the motor efficiency is 85%. Solving for q, we have

1:17 10 4 q 4 ¼ 0:001732 460 7 0:80 0:85 0:65

or

q ¼ 5267 acfm

One can check from the fan curve whether the flow is reasonable. Alternatively, if the flow is known, one can check the head from Eq. (23) and compare it with the measured value. If the measured head is lower, for example, we can infer that something is wrong with the fan or its drive or that the flow measured is not correct.

9.20

Q:

How is the performance of pumps in series and in parallel evaluated?

A:

For parallel operation of two or more pumps, the combined performance curve (H versus q) is obtained by adding horizontally the capacities of the same heads. For series operation, the combined performance curve is obtained by adding vertically the heads at the same capacities.

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The operating point is the intersection of the combined performance curve with the system resistance curve. Figure 9.6 explains this. Head and flow are shown as percentages [2]. ABC is the H versus q curve for a single pump, DEF is the H versus q curve for two such pumps in series, and AGH is the H versus q curve for two such pumps in parallel. To obtain the curve DEF, we add the heads at a given flow. For example, at q ¼ 100%, H with one pump is 100%, and with two pumps H will be 200%. Similarly, AGH is obtained by adding flows at a given head. At H ¼ 100, q for two pumps will be 200%.

Let the system resistance curve be KBGE. When one pump alone operates, the operating point is B. With two pumps in series, E is the operating point. With two pumps in parallel, G is the operating point.

BHP curves also have been plotted and reveal that with series operation BHP ¼ 250% and with pumps in parallel BHP ¼ 164%, indicating that BHP=q is larger in series operation than in parallel. This varies with pump and system resistance characteristics. NPSHr also increases with pump capacity.

Note that if the full capacity of the plant were handled by two pumps in parallel and one tripped, the operating BHP would not be 50% of that with two pumps, but more, depending on the nature of the H versus q curve and the system resistance curve. In the case above, with KBGE as the system resistance, G is the operating point with two pumps, and if one trips B would be the operating point.

FIGURE 9.6 Series and parallel operations of pumps with flat head capacity curves. (From Ref. 2.)

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BHP at G is 142%, whereas at B it is 100% (see the inset of Fig. 9.6). Hence in sizing drives for pumps in parallel, this fact must be taken into account. It is a good idea to check on whether the pump has an adequately sized drive.

A similar procedure can be adopted for determining the performance of fans in series and in parallel and for sizing drives.

9.21

Q:

Determine the parameters affecting the efficiency of the Brayton cycle [3].

A:

Figure 9.7a shows a simple reversible Brayton cycle used in gas turbine plants. Air is taken at a temperature T1 absolute and compressed, and the temperature after compression is T2. Heat is added in the combustor, raising the gas temperature to T3; the hot gases expand to T4 in the turbine, performing work. Following are some of the terms used to describe the performance.

FIGURE 9.7 (a) Simple and (b) regenerative Brayton cycle.

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Example

A simple cycle takes in air at 80 F and 14.7 psia and compresses it at constant entropy through a pressure ratio of 4. The combustor raises the gas temperature to 1500 F. The heated air expands to 14.7 psia at constant entropy in the turbine. Assume k ¼ 1.3 and Cp ¼ 0.28. Find (1) compression work, Wc; (2) heat input to cycle, Qa; (3) expansion work, Qe; (4) thermal efficiency, TE.

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Solution. From Eq. (28),

T2 ¼ ð80 þ 460Þ 4ð1:3 1Þ=1:3 ¼ 742 R

Note that 4ð1:3 1Þ=1:3 ¼ 1.375. Hence

Wc ¼ Cp ðT2 T1Þ ¼ 0:28 ð742 540Þ

¼ 56:6 Btu=lb

Heat input to cycle ¼ Qa

¼Cp ðT3 T2Þ

¼0:28 ð1500 þ 460 742Þ

¼341 Btu=lb

Expansion work Qe ¼ 0:28 ð1960 1425Þ ¼ 150 Btu=lb

TE ¼ 150 56:6 ¼ 0:273; or 27:3% 341

Using Eq. (30b), TE ¼ 1 7 1=1.375 ¼ 0.273.

It can be seen that as the pressure ratio increases, TE increases. Also, as inlet air temperature decreases, the efficiency increases. That is why some gas turbine suppliers install chillers or air coolers at the compressor inlet so that during summer months the turbine output does not fall off compared to the winter months.

9.22

Q:

How can the efficiency of a simple Brayton cycle be improved?

A:

One of the ways of improving the cycle efficiency is to use the energy in the exhaust gases (Fig. 9.7b) to preheat the air entering the combustor. This is called regeneration.

Assuming 100% regeneration, the exhaust gas at temperature T4 preheats air from T2 to T5 while cooling it to T6. The actual heat rejected corresponds to a temperature drop of T6 T1, while the heat added corresponds to T3 T5, and hence the cycle is more efficient. Assuming constant Cp,

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Now P2 ¼ P5 ¼ P3 and P4 ¼ P6 ¼ P1. Also,

Example

Using the same data as above, compute the following for the ideal regenerative cycle: (1) Work of compression, Wc; (2) heat added to cycle; (3) heat added to regenerator; (4) expansion work in turbine; (5) cycle efficiency.

Solution. For the same inlet temperature and pressure ratio, Wc ¼ 56.6 and T2 ¼ 742 R. Exhaust temperature from above ¼ 1425 R ¼ T5.

Heat added in regenerator ¼ Cp ðT5 T2Þ

¼0:28

ð1425 742Þ ¼ 191:3 Btu=lb

Heat added in combustor ¼ Qa ¼ Cp ðT3 T5Þ

¼0:28 ð1960 1425Þ

¼150 Btu=lb

Heat rejected ¼ Qr ¼ Cp ðT6 T1Þ

¼ 0:28 ð742 540Þ ¼ 56:6 Btu=lb

TE ¼ 1 Qr ¼ 1 56:6 ¼ 0:622; or 62:2% Qa 150

Using (31)

TE ¼ 1 1960540 ¼ 0:621; or 62:1%

It is interesting to note that as the pressure ratio increases, the efficiency decreases. As the combustor temperature increases, the efficiency increases. However, it can be shown that the power output increases with increases in the

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pressure ratio. Hence industrial gas turbines operate at a pressure ratio between 9 and 18 and an inlet gas temperature of 1800–2200 F.

NOMENCLATURE

dTube or pipe diameter, in.; subscript i stands for inner diameter

EVoltage

hEnthalpy, Btu=lb; subscripts f and g stand for saturated liquid and

vapor

HHead developed by pump, ft; subscript a stands for acceleration

ICurrent, A

kRatio of gas specific heats, Cp=Cv

LLength, ft

NSpeed of pump or fan, rpm

PPower, kW

qFlow, gpm or acfm

rPressure ratio

sSpecific gravity

SSpecific speed

TTemperature, R

VVelocity, ft=s

WFlow, lb=h

ZEfficiency, fraction; subscripts f; m; p, and t stand for fan, motor, pump, and turbine

rDensity, lb=cu ft

Copyright © 2003 Marcel Dekker, Inc.