Chemistry, 23.08.2019 19:20 jenstets05

Calculate the pressure drop of air flowing at 30 °c and 1 atm pressure through a bed of 1.25 cm diameter spheres, at a rate of 60 kg/min. the bed has a 125 cm diameter and 250 cm height. the porosity of the bed is 0.38. the viscosity of air is 0.0182 cp and the density is 0.001156 grams/cm

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Answer from: ngozi1122357

Explanation:

The given data is as follows.

Mass flow rate of Air = $60 kg/min \times 1 min/60 sec$

= 1 kg/s

Density of Air (r) = $0.001156 g/cm^{3} \times (1000kg/m^{3}) / (g/cm^{3})$

= 1.156 $kg/m^{3}$

Viscosity of Air (m) = $0.0182 cP \times 10^{-3} kg/ms/ cP$

= $0.0182 \times 10^{-3} kg/(ms)$

Bed porosity (e) = 0.38

Diameter of bed (D)= $125 cm \times \frac{1 m}{100 cm}$ = 1.25 m

Length of bed (L) = $250 cm \times \frac{1 m}{100 cm}$ = 2.5 m

Diameter of particles (Dp) = $1.25 cm \times \frac{1 m}{100 cm}$

= 0.0125 m

Sphericity = 1

Volumetric flow rate = $\frac{\text{mass flow rate}}{density}$

= $\frac{1 kg/s}{1.156 kg/m^{3}}$

= 0.865 $m^{3}/s$

Superficial velocity, $V_{o}$ = $\frac{0.865}{\frac{3.14}{4}} \times D^{2}$

= $\frac{0.865}{\frac{3.14}{4}} times 1.25 m \times 1.25 m$

= 0.705 m/s

$NR_{ePM}$ = $\frac{D_{p}V_{o}r}{m (1 - e)}$

NRePM = \frac{0.0125 m \times 0.705 m/s \times 1.156 kg/m^{3}}{0.0182 times 10^{-3} kg/(ms) \times (1 - 0.38)[/tex]

= 903

As we known that 10 > $NR_{ePM}$ > 1000

Below 10 means laminar flow.

Higher than 1000 is turbulent flow.

As, Reynolds number is between 10 and 100, therefore it is in transition flow.

According to Ergun equation,

$\frac{\Delta p \Phi_{s} D_{p} \varepsilon^{3}}{L_{p} V^{2}_{o} (1 - \varepsilon)} = \frac{150(1 - \varepsilon)}{\Phi_{S} D_{P} V_{o} \frac{\rho}{\mu}} + 1.75$

$\Delta p \times 1 \times 0.0125 m \times 0.383 / [2.5 m \times 1.156 kg/m^{3} \times 0.7052 m^{2}/s^{2} \times (1 - 0.38)]$

= $150 \times (1 - 0.38) / [1 \times 0.0125 m \times 0.7052 m^{2}/s^{2} \times 1.156 kg/m^{3} / 0.0182 \times 10^{-3} kg/(ms)] + 1.75$

= $\Delta p \times 7.702 \times 10^{-4}$

= 1.92

$\Delta p$ = $\frac{1.92}{7.702 \times 10^{-4}}$

$\Delta p = 2492.90 N/m^{2}$

Thus, we can conclude that the pressure drop of air under given conditions is $2492.90 N/m^{2}$ .

Answer from: Quest

answer: mercury

explanation:

Which of these is not a commonly mined metallic substance?

Answer from: Quest

the balanced equation for reaction of ca metal with o2 to produce cao is as below

2 ca+ o2→ 2 cao2

explanation

the reaction is balanced since the number of atoms in reactant side is equal to number of atoms in product side. for example they are 2 atom of ca in reactant side and 2 atoms of ca in the product side.

the process is a meet the requirement of redox reaction because : calcium is oxidized from oxidation state 0 to 2+ while oxygen is reduced from oxidation state 0 to 2-

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