## Chapter 1

## Q. 1.2

Water is being drawn from a well whose water level is 1500 ft below the ground surface. Determine the significance of the compressibility of water in determining the head the pumps must supply if the friction loss in the well pipe is 25 ft, and the pump must supply 80 psi of pressure at the ground surface. Assume the bulk modulus for water remains constant and equal to 320,000 psi.

## Step-by-Step

## Verified Solution

First, it is necessary to determine the relationship between density and pressure since the effects of increasing density of the fluid are to be taken into account. The definition of a fluid’s bulk modulus E_{\underline{v}} provides this relationship since E_{\underline{v}}=-\Delta p/(\Delta\underline{V}/\underline{V})=\rho dp/d\rho, in which \underline{V} is fluid volume and \Delta\underline{V} is the change in this volume due to the pressure increase Δp. Separating variables in this equation and integrating the density from ρ_{o} (the density at atmospheric pressure, which will be taken as 1.94 slugs/ft³) to ρ and integrating the pressure from 0 (atmospheric) to p gives

\frac{\rho}{\rho_{o}}=Exp \left\lgroup\frac{p}{E_{\underline{v}}}\right\rgroup(An alternative to integrating between the two limits is to just integrate and add a constant to the resulting equation. This constant can then be determined from a known condition. In this case the known condition is that the density equals \rho_{o} = 1.94 slugs/ft³ when the pressure, p = 0.) Substituting this expression into the hydrostatic Equation 1.4b gives

\frac{dp}{dh}=\gamma=\rho g (1.4b)

\frac{dp}{dh}=g\rho=g\rho_{o}Exp \left\lgroup\frac{p}{E_{\underline{v}}}\right\rgroupAgain separating variables and integrating gives

p=-E_{\underline{v}}Ln \left\lgroup1-\frac{g\rho_{o}h}{E_{\underline{v}}}\right\rgroupThe value of h to use in this equation is the sum 1500 ft, the pressure head needed at the surface, 80 × 144/(32.2 × 1.94) = 184.41 ft, and the frictional loss of 25 ft, or h = 1709.41 ft. Substituting this value for h in the above equation gives a pressure of 742.41 psi at the pump in the well. If the compressibility of the water is ignored, the pressure is obtained from p = gρh/144 = 741.55 psi. This small difference of about 1 psi points out that the compressibility of water is not very significant for most engineering applications.

An exception is whenever the speed of pressure waves are concerned, because for such applications a small amount of free air in water can dramatically change the speed of this wave as noted above.