Pipe flowing compressible fluid 2

[Luiz]

For satured steam in steel pipe under 8 Kgf/cm2 (113.7 psi)
and
D=0.10226 m (4" schedule 40)
k=0.0001m (0.0000305 ft)
V=30 m/s (9.14 ft/s)
v=0.0000035 m2/s (0.0000011 ft2/s)
L=500 m (152.4 ft),

by Colebrook and Darcy formulas, we have:
hf=4,442.73 m (1,354.14 ft)

But, by Spirax Sarco formula:
hf=26.4 m (8.05 ft)


Why the difference? Anything is wrong.
Can anyone help me?

Regards
Luiz
(also posted in sep 1, 2001)

 

[TBP]

If nothing else, you have some unit conversion issues. 500 meters will be a little over 1,500 feet, etc. Looks like your decimal point is misplaced.

I don't know how precise you have to be with your calculations, but the Spirax Sarco literature such as "Hook-Ups" has never failed me in field work.

 

[jproj]

Luiz:

Your units conversion is not correct. See below for correct conversions:

D=0.10226 m (4" schedule 40 - 4.026 in)- ok

k=0.0001m (0.0000305 ft)- backwards ->
0.0001 ft = 0.0000305 m
0.0001 m = 0.000328 ft

V=30 m/s (9.14 ft/s) - backwards ->
30 ft/s = 9.14 m/s
30 m/s = 98.425 ft/s

v=0.0000035 m2/s (0.0000011 ft2/s)- not correct ->
0.0000035 m2 = 0.00003767 ft2
0.0000035 ft2 = 0.000000325 m2

L=500 m (152.4 ft) - backwards ->
500 ft = 152.4 m
500 m = 1640.42 ft

for general reference:

1 m = 3.2808 ft 1 ft = .3048 m
1 m2 = 10.7639 ft2 1 ft2 = .0929 m2

Try your calculations again. Hope this helps!

Jproj

 

[Luiz]

jproj wrote:
> Your units conversion is not correct...

Yes. I'm sorry.
But, forget the english units. It is not correct, but the SI units is correct. Then

by the Colebrook/Darcy formula, we have:
hf=4,442.73 m

but, by the Spirax Sarco formula:
hf=26.4 m

Why this huge difference? Is anything wrong?
Can anyone tell me?

Regards
Luiz

 

[TBP]

I could be missing something, but the only reference I can find for "hf" in Spirax Sarco's "Hook-Ups" is for friction loss in suction piping when calculating NPSH for pumps. I don't see "hf" in Crane's Technical Paper #410 either.

What are you trying to calculate?

 

[Luiz]

TBP wrote:
> What are you trying to calculate?

Darcy equation hf = f*(LV^2)/(2gD) and
Colebrook equation f = 0.25*[-log(0.27*k/D + 2.51/(Re*f^0.5))]^-2
(with SI units)

Sarco equation:
J = 3650.88 * Q^1.95 * v^0.95 * D^-5.1
(with: [D]=mm, [v]=m3/Kg, [Q]=Kg/h and [J]=Kgf/cm2 per 100m)

where:
hf = total friction head loss
f = friction factor
L = length
V = velocity
g = gravity
D = diameter
k = equivalent roughness
Re = Reynolds number
J = friction head loss per 100m

 

[Luiz]

Ah, I forgot:

v = especific volume

Regards.
Luiz

 

[jproj]

Luiz:

Both equations yield approx. the same answers if you keep the units straight. I calculated the friction factor (f)(using the Colebrook equation) to be 0.00497. This compared nicely with the friction factor value of 0.005 from the Moody diagram. Next, I determined the friction term (F) from the following equation:

F(mech/chem) = [(2*f*L*V^2)/(D)]

This is the correct equation to use with friction factor as found with the Colebrook equation. The friction factors between civil and mechanical/chemical engineering varies by a factor of 4:

f(civ) = 4*f(mech/chem)

thus

F(civ) = 4*F(mech/chem) = [(f*L*V^2)/(2*D)]

this calculation yields F(mech)=43775.53 m^2/s

To find the pressure drop, you must use Bernoulli's equation. Ignoring kinetic and potential energy changes and assuming no work is being done, the equation simplifies to the following:

Delta [Pressure/density] = F

For time, assume density changes are negligible, thus:

Pressure drop = F*density - in units of kgf/cm^2

(you may have to divide by gc to get the units to work out)
I did this in english units (gc = 32.2 lbm*ft/lbf*s^2)

As far as the Sarco equation goes... you must first calculate the steam flowrate (Q). This is accomplished by:

Q= (V*PI*(D/2)^2)/density - make sure it's in (kg/hr)

then just plug it in.

I calculated the following values:

By Colebrook and Darcy formulas, we have:

Pressure drop = 25.87 psi (1.819 kgf/cm^2) or 59.68 ft (18.19 m)

By Spirax Sarco formula:

Pressure drop = 33.41 psi (2.35 kgf/cm^2) or 77.06 ft (23.49 m)

Considering the Sarco formula is an estimate, I would expect it to give a bigger pressure drop (conservative). These numbers are not that far off, though. If I were you, I would make sure your units are correct. I would also check the physical properties you are using for the 8 Kgf/cm2 (113.7 psi) sat. steam. There is a lot of room for error when using the Colebrook and Darcy formulas.

Be diligent and keep those units straight!

Good luck

Jproj