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

Substituting this into Eq. (13) gives us

(m is the viscosity, lb=ft h; V is the velocity, fps.)

7.18b

Q:

Determine the pressure drop per 100 ft in a drawn copper tube of inner diameter 1.0 in. when 250 lb=h of air at a pressure of 30 psig and at 100 F flows through it.

A:

Calculate the density (see Chap. 5):

45

r ¼ 29 492 359 560 15 ¼ 0:213 lb=cu ft

The effect of pressure can be neglected in the estimation of viscosity of gases up to 40 psig. For a detailed computation of viscosity as a function of pressure, readers may refer to Ref. 11. From Table 7.7, m ¼ 0.047 lb=ft h. The velocity is

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The specific gravity is to be found. At 180 F, from Eq. (23) (see Q7.21) it can be shown that the specific volume at 180 F is 0.0176 cu ft=lb. Then

1

s ¼ 0:0176 62:4 ¼ 0:91

Hence cP ¼ 0.91 24.83, where

135

cS ¼ 0:22 118 118 ¼ 24:83

and

m ¼ 2:42 0:91 24:83 ¼ 54:6 lb=ft h

7000

Re ¼ 15:2 0:91 24:83 2:42 ¼ 1948

(2.42 was used to convert cP to lb=ft h.) From Eq. (17a),

7.20a

Q:

For viscous fluids in turbulent flow, how is the pressure drop determined?

A:

For viscous fluids, the following expression can be used for the friction factor:

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7.20b

Q:

A fuel oil system delivers 4500 lb=h of light oil at 70 F in a pipe. What is the flow

that can be delivered at 30 F, assuming that m70=m30 ¼ 0:5; v70=v30 ¼ 0:95, and flow is turbulent?

A:

Using Eq. (22), we have

v1W11:78m01:22 ¼ v2W21:78m02:22

45001:78 0:50:22 0:95 ¼ W21:78

or

W2 ¼ 4013 lb=h

7.21

Q:

What is the flow in gpm if 1000 lb=h of an oil of specific gravity (60=60 F) ¼ 0.91 flows in a pipe at 60 F and at 168 F?

A:

We need to know the density at 60 F and at 168 F.

At 60 F:

Density ¼ r ¼ 0:91 62:4 ¼ 56:78 lb=cu ft

1

v60 ¼ 56:78 ¼ 0:0176 cu ft=lb

Hence at 60 F,

1000

q ¼ 60 56:78 ¼ 0:293 cu ft=min ðcfmÞ ¼ 0:293 7:48 ¼ 2:2 gpm

At 168 F, the specific volume of fuel oils increases with temperature:

where E is the coefficient of expansion as given in Table 7.8 [13]. For this fuel oil, E ¼ 0.0004. Hence,

v168 ¼ 0:0176 ð1 þ 0:0004 108Þ ¼ 0:01836 cu ft=lb

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TABLE 7.8 Expansion Factor for Fuel Oils

Hence

q168 ¼ 1000 0:01836 60

¼0:306 cfm ¼ 0:306 7:48

¼2:29 gpm

7.22

Q:

How is the pressure loss in natural gas lines determined? Determine the line size to limit the gas pressure drop to 20 psi when 20,000 scfh of natural gas of specific gravity 0.7 flows with a source pressure of 80 psig. The length of the pipeline is 150 ft.

A:

The Spitzglass formula is widely used for compressible fluids [13]:

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Hence F ¼ 1.03. From Table 7.9 we see that d should be 114 in. Choosing the next higher standard F or d limits the pressure drop to desired values. Alternatively, if q; d; L, and P1 are given, P2 can be found.

7.23

Q:

Determine the pressure loss in a rectangular duct 2 ft 2.5 ft in cross section if 25,000 lb=h of flue gases at 300 F flow through it. The equivalent length is 1000 ft.

TABLE 7.9 Standard Steel Pipea Data (Black, Galvanized, Welded, and

Seamless)

a ASTM A53-68, standard pipe.

b Schedule numbers are approx. values of 1000 maximum internal service pressure, psig allowable stress in material, psi

p

c F ¼ dð1 þ 0:03d þ 3:6=dÞ for use in Spitzglass formula ¼ 5=23 for gas line pressure loss. Source: Adapted from Ref. 13.

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A:

The equivalent diameter of a rectangular duct is given by

We make use of the equivalent diameter calculated earlier [Eq. (14)] while computing Re:

W

Re ¼ 15:2 dim

From Table 7.7 at 300 F, m ¼ 0.05 lb=ft h.

25,000

Re ¼ 15:2 26:64 0:05 ¼ 285;285

Hence

0:316

f ¼ 285;2850:25 ¼ 0:014

For air or flue gases, pressure loss is generally expressed in inches of water column and not in psi. The following equation gives DPg [11]:

where di is in inches and the specific volume is v ¼ 1=r.

40

r ¼ 460 þ 300 ¼ 0:526 lb=cu ft

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7.24a

Q:

Determine the Reynolds number when 500,000 lb=h of superheated steam at 1600 psig and 750 F flows through a pipe of inner diameter 10 in.

A:

The viscosity of superheated steam does not vary as much with pressure as it does with temperature (see Table 7.5).

m ¼ 0:062 lb=ft h

7.24b

Q:

Determine the Reynolds number when hot air flows over a tube bundle.

Air mass velocity ¼ 7000 lb=ft2 h

Temperature of air film ¼ 800 F

Tube size ¼ 2 in. OD

Transverse pitch ¼ 4.0 in.

A:

The Reynolds number when gas or fluids flow over tube bundles is given by the expression

where

G ¼ fluid mass velocity, lb=ft2 h d ¼ tube outer diameter, in.

m ¼ gas viscosity, lb=ft h

At 800 F, the air viscosity from Table 7.7 is 0.08 lb=ft h; thus

2

Re ¼ 7000 12 0:08 ¼ 14;580

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7.25

Q:

There are three tubes connected between two headers of a super heater, and it is required to determine the flow in each parallel pass. The table gives the details of each pass.

Total steam flow is 15,000 lb=h, and average steam conditions are 800 psia and 750 F.

A:

Because the passes are connected between the same headers, the pressure drop in each will be the same. Also, the total steam flow will be equal to the sum of the flow in each. That is,

DP1 ¼ DP2 ¼ DP3

In other words, using the pressure drop correlation, we have

The effect of variations in steam properties in the various tubes can be neglected, because it will not be very significant.

Substituting the data and using f from Table 7.6, we obtain

W1 þ W2 þ W3 ¼ 15;000

¼ a constant

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