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How to calculate the size of the pipe?
- Thread starter kanna
- Start date
owlscreech
Mechanical
The initial calculation can be found in a basic hydraulics text.
t = 2v/q where the dimensions are demonstrated below.
For your problem you calculate initial flow as follows:
q = (2 x 10)/60 = 0.00555 m^3/sec = 0.196 ft^3/sec [Shifting to English units in order to use available equations]
If we assume the initial level in the tank is 6 feet, then
q = (0.0438)(d^2)(h/K)^0.5 [Crane TP410]
where q =ft^3/sec, d=diameter of pipe inches, h=height ft, K=Crane resistance factor = fL/d, f = Moody friction factor, L = pipe length, d = pipe diameter (same units as pipe length).
Solving for the above, we have (d^4)/K = (522)( q^2)/h = 3.33
From here on you will need to use trial and error based on the physical situation you have (length of pipe, fittings required, etc) and find a configuration that meets the ratio (d^4)/K. Problems that will help are found in TP410.
For example a lower bound on the diameter can be found quickly by determining the diameter of a pipe with just an entrance and exit loss [K= 0.5 + 1.0 = 1.5]. This calculates to be d > 1.49 inches. Then, to start, if you assume a 2”/sch40 pipe [d=2.067”], then K max = (2.067)^4)/(3.33) = 5.5. If we subtract the entrance and exit losses, Kpipe = 4.
From a Moody chart for 2” pipe at velocity of 8.4 ft/sec , f = .022 and L = dK/f, or
L = (2.067)(4)/(.022) = 375 inches = 31+ feet.
Therefore a 2”/sch40 pipe, 33 feet long, without any elbows, etc., will drain a tank containing 10 m^3 of liquid with an initial height of 6 feet in 60 minutes.
t = 2v/q where the dimensions are demonstrated below.
For your problem you calculate initial flow as follows:
q = (2 x 10)/60 = 0.00555 m^3/sec = 0.196 ft^3/sec [Shifting to English units in order to use available equations]
If we assume the initial level in the tank is 6 feet, then
q = (0.0438)(d^2)(h/K)^0.5 [Crane TP410]
where q =ft^3/sec, d=diameter of pipe inches, h=height ft, K=Crane resistance factor = fL/d, f = Moody friction factor, L = pipe length, d = pipe diameter (same units as pipe length).
Solving for the above, we have (d^4)/K = (522)( q^2)/h = 3.33
From here on you will need to use trial and error based on the physical situation you have (length of pipe, fittings required, etc) and find a configuration that meets the ratio (d^4)/K. Problems that will help are found in TP410.
For example a lower bound on the diameter can be found quickly by determining the diameter of a pipe with just an entrance and exit loss [K= 0.5 + 1.0 = 1.5]. This calculates to be d > 1.49 inches. Then, to start, if you assume a 2”/sch40 pipe [d=2.067”], then K max = (2.067)^4)/(3.33) = 5.5. If we subtract the entrance and exit losses, Kpipe = 4.
From a Moody chart for 2” pipe at velocity of 8.4 ft/sec , f = .022 and L = dK/f, or
L = (2.067)(4)/(.022) = 375 inches = 31+ feet.
Therefore a 2”/sch40 pipe, 33 feet long, without any elbows, etc., will drain a tank containing 10 m^3 of liquid with an initial height of 6 feet in 60 minutes.
owlscreech
Mechanical
Reference my earlier posting using t = 2v/q.
This approach only works if the surface area of the liquid remains the same during draining. If the variation of the surface area can be expressed as a function of h, an equation can written in terms of the differential of h. This equation may or may not be integrable and numerical iteration may be required.
Also as the flow rate diminishes, the flow becomes less turbulent and finally laminar at the very end of the process. The net effect of this is to increase the friction factor somewhat. In most practical applications, this is not usually a problem since it is not a question of draining the tank to the last drop. If that is important, then it would require going to a iterative process that would factor in the change in the friction factor using such approximations as Altshul-Tsal shown in Idelchik's "Flow Resistance-A Design Guide for Engineers."
This approach only works if the surface area of the liquid remains the same during draining. If the variation of the surface area can be expressed as a function of h, an equation can written in terms of the differential of h. This equation may or may not be integrable and numerical iteration may be required.
Also as the flow rate diminishes, the flow becomes less turbulent and finally laminar at the very end of the process. The net effect of this is to increase the friction factor somewhat. In most practical applications, this is not usually a problem since it is not a question of draining the tank to the last drop. If that is important, then it would require going to a iterative process that would factor in the change in the friction factor using such approximations as Altshul-Tsal shown in Idelchik's "Flow Resistance-A Design Guide for Engineers."
kanna,
There is a nice usefull calculator, the following link will get you started:
Best regards.
There is a nice usefull calculator, the following link will get you started:
Best regards.
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