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)...
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...
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...
Ok Morel.
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...
GGW:
This is the question: I would like to use Darcy and Colebrook equations
to compute satured steam flow pipes size, say, 120 psi.
Can you tell me what is this equation of state?
Thanks.
Luiz
For pipe design flowing incompressible fluid, I use:
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
where:
hf = friction loss
f = friction factor
L = length
V = velocity
g = gravity
D = diameter
k = equivalent roughness
Re =...
For pipe design flowing incompressible fluid, I use:
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
where:
hf = friction loss
f = friction factor
L = length
V = velocity
g = gravity
D = diameter
k = equivalent roughness
Re =...