The penstock head loss computation is:
head loss = f * L * 0.08 * Q² / d^5
where:
f = wall friction coefficient of the penstock, taken from the Moody Chart
L = penstock length, m
Q = flow rate in cubic m per second
d = penstock diameter, m
This means that penstock pressure losses highly affected by its length, its
composition, flow rate, and diameter. It is aggravated by the fact that the
frictional coefficient is also inflenced by the flow rate and diameter.
The following equation is reasonably accurate if the flow rate is limited to
about 1-2 m/s and the penstock diameter is from 2" to 6":
head loss = 0.0018 * L * Q² / d^5
But for rough estimation, assuming that the penstock's average slope is greater
than 20° and water velocity does not exceed 2m/s, penstock efficiency should be
around 80-90%.
Example:
pipe nozzle = 1/2" diameter
gross head, X = 20m
penstock diameter = 2"
Assume:
penstock efficiency = 80%
Solution:
pressure head, h = gross head * penstock efficiency = 40m * 80% = 16m
g = gravitational acceleration = 9.81 m / s²
water nozzle velocity = 0.98 * (2*g*h)^.5 = 0.98 * (2 * 9.81 * 32) = 17.36
m/s
nozzle cross-section area, An = pi x D² / 4 = 3.1416 x (0.5 * 0.0254)² /4 =
0.000127m
flow rate = Area x velocity = 0.000127 x 17.36 = 0.0022 m³/s = 2.2 L/s
2.2L/s of water will flow through the penstock
penstock cross-section area, A = pi x D² / 4 = 3.1416 x (2 * 0.0254)² /4 =
0.002m²
penstock water velocity = flow rate / penstock area = 0.0022 / 0.002 = 1.08
m/s
Since the velocity is less than 2m/s, this means that the 2" diameter
penstock is suitable.
If not, computation should be repeated for the next larger penstock.
I can give you specs for the penstock size and length and nozzle diameter of
your 2-4kW system if you can give me the average slope of the site and the
minimum flow rate.
Eilrem
Mihai Radulescu <[EMAIL PROTECTED]> wrote:
I want to build a microhydropower system in the mountains, the aim
would be to generate 4kW, but alos 2kW is acceptable.
I have read and undertood the theories abot head and flow, but I don't
know how to calculate the effective water flow taht would pass through
the penstock/nozzle/turbine; the idea is that the waterflow of the
creek is big enough, and I want to install just a sort of collection
system that would pour water into the penstock, as much as the
penstock/nozzle would run.
How do I calculate, at least approximatively, the waterflow through a
vertical plastic pipe of, let's say, 2" diameter and a head of X
meters.
And more, if the pipi would finish with a 1/2" nozzle, how this would
affect the waterflow ?
Regards,
Mihai
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