Pressure Drop in Regenerative Heater
A regenerative heater is packed with a bed of 6 mm cubes. The cubes are poured into the cylindrical shell of the regenerator to a depth of 3.5 m such that the bed porosity was 0.44. If air flows through this bed entering at 25oC and 7 atm abs and leaving at 200oC, calculate the pressure drop across the bed when the flow rate is 500 kg/hr per square meter of empty bed cross section. Assume average viscosity as 0.025 cP and density as 6.8 kg/m3.
Mass flow rate of Air / unit area = 500 kg/(hr.m2) = 0.139 kg/(sec.m2)
Density of Air (r) = 6.8 kg/m3
Viscosity of Air (m) = 0.025 cP = 0.025 x 10-3 kg/(m.sec)
Bed porosity (e) = 0.44
Length of bed (L) = 3.5 m
Dia of particles (Dp)= 6 mm = 0.006 m
Sphericity (F s) = 6 vp / (DpSp)
vp = volume of particle = Dp3 (for cube)
Sp = surface area of particle = 6 x Dp2 (for cube)
NRePM = DpVor/(m (1 - e) )
For laminar flow (i.e. NRePM < 10) pressure drop is given by Blake-Kozeny equation.
For turbulent flow (i.e. NRePM > 1000) pressure drop is given by Burke-Plummer equation.
For intermediate flows pressure drop is given by Ergun equation
Superficial velocity Vo = Volumetric flow rate/ cross-sectional area of bed
Superficial velocity Vo = mass flow rate per unit area / density = 0.139 / 6.8 = 0.0204 m/sec
(F s) = 6 vp / (DpSp) = 6 x 0.0063 / (0.006 x 6 x 0.0062) = 1
NRePM = 0.006 x 0.0204 x 6.8 / (0.025 x 10-3 x ( 1- 0.44 ) ) = 59.45
We shall use Ergun equation to find the pressure drop.
i.e. Dp x 0.006 x 0.443 / ( 3.5 x 6.8 x 0.02042 x ( 1 - 0.44 ) ) = 150 / 59.45 + 1.75
Dp x 0.0712 = 4.273
Dp = 4.273 /
0.0712 = 60