An example of the use of energy balance to find a result easily
Suppose we have water flowing in a pipe, and suddenly turn off the tap. The water at the front is brought to an immediate halt, and this requires an impulsive pressure rise. If water were incompressible, and the pipe rigid, the whole mass of moving water would have to stop instantaneously, and we would find out what happens when an irresistible force meets an immovable pipe. Fortunately, water is compressible, and pipes are elastic, so the situation is not that desperate.
Let's assume that the pipe cannot strain, and look only at the water. Since water is compressible, pressure pulses move at the speed of sound, c = (κ/ρ)1/2, where κ is the modulus of compression, and ρ the density. When the water hits the closed tap, a pressure wave is propagated back into the flowing water, halting any motion as it arrives. When the pressure wave reaches the end of the flowing water, say a distance L away, all the water has been stopped. The wave then reflects from this end, according to the arrangements there, and returns down the pipe, reaching the tap at time t = 2L/c. It is again reflected, and the same thing happens. All the kinetic energy of the water has been transformed into the energy of a wave, which bounces back and forth. The regular arrivals are like the blows of a hammer, and the phenomenon is called water hammer.
We could find the pressure by considering the dynamics of a thin slice of water, but a much easier way is to equate the kinetic energy of the water when the tap was closed to the strain energy of the water at rest, compressed by the extra pressure. If p is the pressure, then dV = (dp/κ)V, from the definition of the modulus of compression. The work done is p dV = (V/κ)p dp by an increment of pressure from p to p + dp. The total work done in compressing to a pressure P is then W = P2V/2κ. If energy is conserved, this equals T = ρVv2/2, where v is the velocity of the water. From this equation we find that P = (ρκ)1/2v.
For water, κ = 3 x 105 psi, and ρ = 1.122 slug/ft3 (this is 62.4 pcf divided by 32.2 ft/s2), so P = 63.7v psi, where v is in ft/s. The speed of sound is 4721 ft/s from these values. There is a variation with temperature, but these figures are good enough for estimates under normal conditions.
Actual pipes are not rigid. When the tap is closed, the increase in pressure stretches the pipe longitudinally, and a wave propagates back in the pipe that travels more rapidly than the wave in the water. This makes a hammer of higher frequency. The more important effect is the diametral stretching of the pipe, which effectively reduces the modulus of compressibility to κ', where 1/κ' = 1/κ + (d/4tY)(5 - 4σ), where d is the diameter of the pipe, t its wall thickness, Y Young's modulus, and σ is Poisson's ratio (0.28 for iron and steel). If longitudinal extension is not allowed, the formula is 1/κ' = 1/κ + d/tY.
If the far end of the pipe is open, the reflected wave is of the contrary phase (the boundary condition is p = 0), and reconverts the energy to kinetic as it returns, but now the water is moving in the opposite direction. When this negative pressure wave reaches the tap at time 2L/c, it is reflected with the same phase, and returns to the open end, stopping the water as it does so. Now it is reflected as a positive pressure wave, entering the water at rest, and setting it into motion. At t = 4L/c, the water is moving as at the beginning, and a positive pressure wave is beginning to return to the open end. The pressures are relative to the static pressure in the pipe. From this Figure it is possible to work out how the pressure varies with time at any point along the pipe. At the tap, it is a square wave. Close to the beginning of the pipe, it is a series of brief pulses of alternating sign at intervals of 2L/c.
Friction, and losses on reflection, cause the waves to attenuate, and eventually disappear when the energy has been turned to heat or mechanical radiation.
There are also consequences from the variations in momentum of the water, that result in forces on the pipes. If the pipe is straight, these forces are longitudinal, but there will be sidewise forces if the pipe is bent.
Drysdale, et. al., The Mechanical Properties of Fluids (London: Blackie and Son, 1925), Chapter VI (by Prof. A. H. Gibson).
Composed by J. B. Calvert
Created 13 July 2000