Help setting up a CHEM E problem

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SUMMARY

The discussion focuses on determining the minimum inside diameter of a pipe required to feed three fillers into a pharmaceutical emulsion without exceeding a velocity of 1.2 m/s. The total mass flow rate from the fillers is calculated to be 14,100 kg/h, which converts to 235 kg/s. The volumetric flow rate is then derived as 0.224 m³/s. The final calculation by a class partner suggests that the required pipe diameter is 2.75 cm, based on the mass flow rates and the density of the emulsion.

PREREQUISITES
  • Understanding of fluid dynamics principles, specifically mass flow rate and volumetric flow rate.
  • Knowledge of the relationship between velocity, area, and flow rate (v1A1 = v2A2).
  • Familiarity with unit conversions, particularly between kg/h, kg/s, and m³/s.
  • Basic understanding of pipe flow and shear stress in fluid mechanics.
NEXT STEPS
  • Research the application of the continuity equation in fluid dynamics.
  • Learn about the effects of shear stress on fluid flow in pipes.
  • Study the principles of pipe sizing for various flow rates and velocities.
  • Explore the use of computational fluid dynamics (CFD) tools for pipe flow analysis.
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Students in chemical engineering, process engineers, and professionals involved in the design of fluid transport systems in pharmaceuticals.

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[SOLVED] help setting up a CHEM E problem

Homework Statement



Three fillers used in packaging a pharmaceutical emulsion (ro = 1050kg/m^3) are fed from a single product line. One filler has a capacity of 2700kg/h, the second operates at
4600 kg/h and the third at 6800 kg/h. The emulsion is susceptible to damage at velocities greater that 1.2m/s, due to shear at the pipe wall. What size (ie minimum inside diameter) in cm should be used to feed the fillers through a single line?

Homework Equations


ro = mass flow rate / volumetric flow rate and do i use the v1A2=v2A2 where v = velocity and A = area to determine the size of thepipe.

b]3. The Attempt at a Solution [/b]

ok so i add the mass flow rates and get 14,100 kg/h then i thought i should make this kg/s so i got 235 kg/s then into volumetric flow 235/1050 = 0.224 m^3/s

now i draw a blank why can't i remember how to make m^3 into m do i cube root or ?? and then do i use the v1a1=v2a2 do set up my pipe??
 
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well I made a mistake reading the problem i was assuming that the three fillers were supplying the single pipe. The only thing to determine is what size of pipe is needed to supply the three fillers not to exceed 1.2 m/s.

My class partner has made this conclusion

(2700 kg/h) / (3600 s) / (1050 kg/m^3) / (1.2 m/s) / pi / sqrt *2

so we have the pipe equal to 2.75 cm when all is said and done does anyone agree with this conclusion
 

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