Worked examples
Q1 A steel pipe is 1500 metres long. It has an inside diameter of 150 mm and a wall thickness of 6mm. It is designed to convey 0.08 of water. For steel and for water .
Calculate :-
(i) the velocity of propagation of a pressure wave through the water if the elasticity is ignored,
(ii) the velocity of the pressure wave taking the elasticity of the pipe into account,
(iii) the velocity of flow for the design condition,
(iv) the rise in pressure that would follow the sudden and complete closure of a valve at the end of the pipe,
(v) the hoop stress in the pipe that would be produced by this pressure. Ignore the longitudinal strain.
(vi) the time that would elapse before the first wave returns to the valve,
(vii) the time taken for the first cycle to be completed by ignoring friction in the pipe and taking the initial pressure in the pipe to be 2.5 bar absolute,
(viii) the minimum time in which the valve can be closed if separation is to be avoided.
Q2 Water flows at a steady velocity of 2.5 m/s through a steel pipe of 1.5 m diameter and 12 mm wall thickness. The pipe is 1,200 m long.
Calculate :-
(i) the velocity of propagation of a pressure wave in the pipe,
(ii) the value of ,
(iii) the rise in pressure at a valve at the delivery end of the pipe if the valve is closed in 2 seconds,
(iv) the stress in the pipe wall at this pressure, and,
(v) the shortest closure time that could be used if the stress in the wall is to be limited to . Suppose that the valve can be closed in such way that, during the start of closure the rate of rise = where is the time of closure, that is, the closure is “linear”.
Q3 Water flows steadily from a reservoir through a steel pipe of 300 mm diameter, of wall thickness 8 mm, and length 2,000 m. The level in the reservoir is 40 m above that of a valve at the delivery end of the pipe.
(i) Take to be equal to 0.005 and calculate the velocity of flow in the pipe.
(ii) Using the method in Q2(v) estimate the pressure rise at the valve when it is closed in 10 seconds.
(iii) Calculate the pressure rise at the valve when the valve is suddenly closed to ultimately reduce the flow to one half of the original value.
Q4 Suppose that in Q3 the valve were to be fully closed with no flow in the pipe and that the valve was then suddenly opened. We have an expression for the retardation (or acceleration) of the water in a long pipe for cases when the time for significant changes to occur is long compared with . It is where is the head causing acceleration. Use this, the Darcy expression and the steady flow equation to estimate the time taken for the velocity of flow to reach 2 m/s. Suppose to be independent of the velocity.
Q5 Plot a graph showing how the loss of energy to friction varies with time during the acceleration phase.
Q6 Water flows steadily at a velocity of 3 m/s through a steel pipe 0f 50 mm diameter and 3.5 mm wall thickness. The pipe is 70 m long and the pressure at the downstream end of the pipe is 2 bar gauge.
(i) Calculate the velocity, , of propagation of a pressure wave in the water in the pipe given that :-
where is the density of water = , is the bulk modulus of elasticity of water = , is the diameter of the pipe in metres, is the thickness of the pipe wall in metres and is the modulus of elasticity of steel = .
(ii) If the valve is now closed suddenly calculate the rise in pressure at the valve and the time for which this pressure persists.
(iii) Take the vapour pressure of water at ambient temperatures to be zero and the atmospheric pressure to be 1 bar and estimate the further time that will elapse before the pressure at the valve rises above the vapour pressure.
Q7 A horizontal pipe that is 20 m long and of 48 mm diameter is connected to a reservoir containing water at a depth of 4 metres below the free surface.
Take the surface roughness of the pipe to be 0.045 mm, the effective bulk modulus of water in the steel pipe to be , ignore loss other than pipe friction and calculate :-
(i) the maximum flow through the pipe,
(ii) the time taken for the flow to reach 95% of this maximum after sudden opening of the valve at the end of the pipe,
(iii) the maximum pressure reached in the pipe as a result of suddenly closing the valve after this maximum pressure has been achieved.
Q8 Water flows at a stead velocity of 2.5 m/s through a pipe that is 100 m long, has a diameter of 200 mm and has a wall thickness of 5 mm.
Take the bulk modulus of water to be , the value of Young’s modulus for steel to be and calculate the pressure rise when :-
(i) the flow is reduced uniformly to zero in 5 seconds and,
(ii) a valve at the end of the pipe is closed in a very short time.
Solutions
Q1 (i) , and for a rigid pipe
(ii)
(iii)
(iv)
(v) Hoop stress . For comparison the yield stress of mild steel is so this stress is not minor.
(vi) Time , that is the time for the wave to traverse the pipe from end to end and back. time .
(vii) During the second
phase the pressure will drop to zero. Then the water is retarded by a pressure
difference of 2.5 bar from an initial velocity of 4.53 m/s We need to find the
time taken to bring the water to rest and then to give it a velocity of 4.53
m/s towards the valve.
Force on the water
The mass of water in the pipe Using impulse of force = change of momentum :-
.
Total time for a cycle is
(viii) This needs some thought but one must conclude that the safe time is the one derived from slow closure and linear change in velocity, that is :-
.
Then . Of course the time would be longer because cannot be constant.
Q2 (i)
(ii)
(iii) This is a rapid closure because the closure time is less than . The pressure rise is given by
(iv) The hoop stress
(v) The first returning
wave will reach the valve after 2.5 seconds from the start of closure. The
maximum pressure will at the valve occur at this time. The maximum pressure
will be .
In this case it is limited to the pressure that produces a hoop stress of .
We know that at 24 bar the hop stress is 150 so 8 bar will produce 50 .
Therefore , that is,
Q3 (i) For steady flow :-
.
From this :-
It follows that .
Then and .
(ii) A value of is required.
We also need a value of . . From this the maximum value of
Us8ing “linear” closure as above and ignoring the rise in pressure produced by the wave advancing into water at higher pressure the rise in pressure due to valve closure = .
(iii)
Q4 Continuing from Q3, during the period of acceleration the head available is used to produce the kinetic energy at exit, to meet the friction loss and to accelerate the water. So, applying the energy equation for some instant of time seconds after opening the valve we have :-
or
Mathcad gives
Integrating longhand gives
Q5 Loss to friction
We also have
This can be solved by using Mathcad or longhand.
Q6 (i)
(ii) The pressure rise
This pressure will persist for a time .
(iii) The further time required is that to bring the water in the pipe to rest from 3 m/s and to give it a velocity of 3 m/s back towards the valve under a pressure difference of 2 bar atmospheric pressure, that is 3 bar.
Accelerating force .
Mass of water
Acceleration
Time
Q7 (i)
Applying the energy equation to 1 in the free surface in the
tank and to 2 at discharge from the pipe :-
.
Substituting
must come from the Moody diagram and we do not know the velocity of flow for the Reynolds number. We can calculate the relative roughness from and will probably lie between . If we make a first guess at =0.005 . .
This gives the same value of . So and
(ii) For accelerating flow the head causing acceleration
This can be handled in Mathcad but for examination purposes we ca proceed longhand:-
.
(iii) The maximum rise in pressure is given by where and is the equivalent bulk modulus.
Q8 For this pipe :-
So .
(i) Compared with 0.16 seconds 5 seconds is a long time and this can be treated as a slow closure. Then :-
(ii)