The following formula for the pressure drop through a valve was found in a design manual:                            hL ¼ 522Kq2 d4 where hl ¼ the ‘‘head loss’’ in feet of fluid flowing through the...

The following formula for the pressure drop through a valve was found in a design manual:

                           hL ¼ 522Kq2 d4

where hl ¼ the ‘‘head loss’’ in feet of fluid flowing through the valve, K ¼ dimensionless resistance coefficient for the valve, q ¼ flow rate through the valve, in ft3 /s, and d ¼ diameter of the valve, in inches.

(a) Can this equation be used without changing anything if SI units are used for the variables? Explain.

(b) What are the dimensions of ‘‘522’’ in this equation? What are its units?

(c) Determine the pressure drop through a 2-in. valve with a K of 4 for water at 208C flowing at a rate of 50 gpm (gal/min), in units of (1) feet of water, (2) psi, (3) atm, (4) Pa, (5) dyn/cm2 , and (6) inches of mercury.