In the literature you have found an empirical equation for the pressure drop in

Question

In the literature you have found an empirical equation for the pressure drop in a column packed with a particular type of particle. The pressure drop is given by the dimensionally incorrect equation: deltap= 4.7((u^0.15*H*rho^0.85*v^1.85)/dp^1.2). Units deltap= pressure drop, lbf/ft^2; u=fluid viscosity, lbm/fts; H=height of the column, ft; rho= density, lbm/ft^3; v= superficial velocity, ft/s; dp=effective particle diameter,ft. Imagine that you are given data for u,H, rho,v,and dp in SI units and you wish to use it directly to calculate the pressure drop in lbf/ft^2. How would you change the empirical equation for deltap to obtain another empirical equation suitable for use with SI Units? Note that your objective here is to replace the coefficient 4.7 with a new coefficient. Begin by putting the equation in dimensionally correct form, i.e, find the units associated with the coefficient 4.7, and then set up the empiricism so that it can be used with SI units

Explanation / Answer

u =fluid viscosity lb/ft.s

1 lb=0.4535kg.1ft= 0.3048m

Hence u0.15= (0.4535/0.3048)1.5 kg/m,s =1.814 kg/m.sec

deltaP= Lb/ft2 to kg/m2= 0.4535/(0.3048)2= 4.88

(deltaP)1.2= (4.88)1.2 = 6.7 kg/m2

H ft= 0.3048 m

Density in lb/ft3= kg/m3 = 0.4535/(0.3048)3 = 16.01 kg/m3

Density .85= (16.01)0.85= 10.56

Velocity in ft/s =0.3048 m/s

(Velocity)1.85= 0.111

dp1.2 = (0.3048)1.2 =0.2403

delataP= (4.7* 1.814*0.3048*10.56/0.2403) u^0.15*H*rho^0.85*v^1.85)/dp^1.2

=12.7 u^0.15*H*rho^0.85*v^1.85)/dp^1.2 N/m2

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