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Civil Engineering Hydraulics Mechanics of Fluids
Simple Piping Problems
Pipe Geometries In
the text, problems are worked out for pipes of noncircular cross sections. While the derivations are fine and work well, you can also use an equivalent hydraulic diameter to work on these problems.
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Simple Piping Problems
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Pipe Geometries An annulus is developed by putting one pipe inside of another so that the center of the pipes are aligned. This is a very common setup in heat exchangers. The blue area is the flow area we are concerned with.
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Simple Piping Problems
Pipe Geometries Outer pipe is a 4-nominal, schedule 40 pipe. Inner pipe is a 2-nominal, schedule 40 pipe. we can get the inner and outer diameters for the pipes from Table C.1.
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Pipe Geometries To calculate a hydraulic diameter, we need the area and the wetted perimeter.
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Pipe Geometries So the equivalent Dh is
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Pipe Geometries Calculate the average flow velocity using our normal method
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Pipe Geometries Now we can calculate the Re for the flow
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Pipe Geometries For a cast iron pipe the ε is 0.00085 from Table 5.2 so we can calculate the ε/Dh
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Simple Piping Problems
Pipe Geometries We can use the Moody Diagram to calculate the friction factor, f
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Moody Diagram
f is approximately 0.032
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Simple Piping Problems
Pipe Geometries Now we can use this friction factor to calculate the head loss in this section of pipe. In this case, all the pressure loss was due to friction.
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Pipe Geometries This is the most common way to work this type of problem. If the pipes made of different materials you could develop an average relative roughness
! = D average 13
! ! + " Dinner D outer D inner " Douter + " Dinner
" Douter
Simple Piping Problems
Pipe Geometries Example problem 5.5 gives a different type of problem where we know the pressure drop over a length of pipe and we want to find the volumetric flow rate.
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Pipe Geometries Benzene flows through a 12-nominal schedule 80 wrought iron pipe. The pressure drop measured at points 350 m apart is 34 kPa. Determine the flow rate through the pipe.
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Pipe Geometries Using something like Solver in EXCEL, this isn’t a difficult problem. However, if you need to do this problem by an iterative method, it can be solved.
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Pipe Geometries Since the pipe has a constant cross section and is horizontal, all the of the pressure loss is due to friction so we have the expression
!v 2 fL p1 ! p2 = 2 Dh 17
Simple Piping Problems
Pipe Geometries The problem we have is the f is a function of v through the Re and we don’t have a direct solution
!v 2 fL p1 ! p2 = 2 Dh 18
Simple Piping Problems
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Pipe Geometries From what is given in the problem and from what we can look up in tables we have
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Simple Piping Problems
Pipe Geometries Rearranging the Bernoulli expression !p =
!v 2 fL 2 Dh
2 Dh !p =v2 f !L
v= 20
2 Dh !p f !L
Simple Piping Problems
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Pipe Geometries Now we can start by assuming a friction factor, using the roughness ratio, and getting a Re from there.
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Simple Piping Problems
Moody Diagram Your assumed f must be such that the point on the relative roughness curve is within the diagram.
Assumed f is 0.015
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Pipe Geometries From the assumed f, calculate the v and the Re
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Simple Piping Problems
Pipe Geometries Now, using this Re, you can go back to the Moody diagram and calculate a new estimate for f
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Moody Diagram
New estimate for f is 0.0142
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Simple Piping Problems
Pipe Geometries And you can iterate the process based on the new estimate for f
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Moody Diagram Using this Moody diagram, you really can’t get much closer an estimate than we already have. Since the new Re is so close to the Re we just used, this would not be a bad termination point.
New estimate for f is 0.0142
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Simple Piping Problems
Pipe Geometries So out volumetric flow rate is
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Pipe Geometries There are expressions for approximating the friction factor that do not rely on the Moody diagram. These are all empirical equations but they do have the ability to be solved directly. The forms are given on page 246 and 247 of the 4th edition of the text and page 267 of the 3rd edition (only two are given in the 3rd edition). 29
Simple Piping Problems
Pipe Geometries One of the ones not given in the 3rd edition is the Swamee-Jain equation
f =
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0.250 ' ! ! 5.74 $ * + 0.9 & , )log # 3.7 D " Re % + (
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Simple Piping Problems
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Pipe Geometries One of the ones not given in the 3rd edition is the Swamee-Jain equation f =
0.250 ' ! ! 5.74 $ * + 0.9 & , )log # ( " 3.7 D Re % +
! # Re = # # #" e 31
2
1
$ 0.9 & 5.74 & 0.250 ! & f & 3.7 D %
Simple Piping Problems
Pipe Geometries Using the equation for f rather than the Moody diagram for iteration
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Pipe Geometries Second Iteration
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Pipe Geometries Third Iteration
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Pipe Geometries A third type of problem is where we are given the flow and the head loss and asked to choose a pipe size. Example 5.7
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Pipe Geometries
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Pipe Geometries
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Pipe Geometries p1 ! p2 =
!v 2 fL 2 Dh
Collect all the known information on the left side 2"p f =v2 !L Dh Since we are dealing with a pipe with a circular cross section, Dh is the pipe diameter Q Q v= = A !D2 4 2
This just sets up all the calculations as functions to make things a bit easier.
Simple Piping Problems
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Pipe Geometries
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This just sets up all the calculations as functions to make things a bit easier.
This just sets up all the calculations as functions to make things a bit easier.
Simple Piping Problems
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Pipe Geometries
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Simple Piping Problems
Pipe Geometries
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This just sets up all the calculations as functions to make things a bit easier.
This just sets up all the calculations as functions to make things a bit easier.
Simple Piping Problems
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Pipe Geometries So
to carry the flow what this pressure drop or lower, we would need a pipe with an inside diameter of 0.515 ft as a minimum. This is a pipe with an ID of 6.176 in. To get an internal diameter at least this large, you would specify a 8 nominal pipe