### Fluid Mechanics

# 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 25^{o}C and 7 atm abs and leaving at 200^{o}C, 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/m^{3}.

Data:

Mass flow rate of Air / unit area = 500 kg/(hr.m^{2}) = 0.139 kg/(sec.m^{2})

Density of Air (r) = 6.8 kg/m^{3}

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 (D_{p})= 6 mm = 0.006 m

Formulae:

Sphericity (F _{s}) = 6 v_{p} / (D_{p}S_{p})

v_{p} = volume of particle = D_{p}^{3} (for cube)

S_{p} = surface area of particle = 6 x D_{p}^{2} (for cube)

NRe_{PM }= D_{p}V_{o}r/(m (1 - e) )

For laminar flow (i.e. NRe_{PM} < 10) pressure drop is given by Blake-Kozeny equation.

For turbulent flow (i.e. NRe_{PM} > 1000) pressure drop is given by Burke-Plummer equation.

For intermediate flows pressure drop is given by Ergun equation

Superficial velocity V_{o} = Volumetric flow rate/ cross-sectional area of bed

Calculations:

Superficial velocity V_{o} = mass flow rate per unit area / density = 0.139 / 6.8 = 0.0204 m/sec

(F _{s}) = 6 v_{p} / (D_{p}S_{p}) = 6 x 0.006^{3} / (0.006 x 6 x 0.006^{2}) = 1

NRe_{PM }= 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 0.44^{3} / ( 3.5 x 6.8 x 0.0204^{2} x ( 1 - 0.44 ) ) = 150 / 59.45 + 1.75

Dp x 0.0712 = 4.273

Dp = 4.273 /
0.0712 = **60
N/m ^{2}**