Help setting up a CHEM E problem

[smatt_31]

[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??

 

[smatt_31]

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

 

[piercas]

To solve this problem, we first need to determine the total volumetric flow rate for the three fillers. As you correctly calculated, the total mass flow rate is 14,100 kg/h, which can be converted to 3.9167 kg/s. Then, using the density of the emulsion, we can calculate the total volumetric flow rate to be 0.00373 m^3/s.

To determine the minimum inside diameter of the pipe, we can use the equation v1A1 = v2A2, where A is the cross-sectional area of the pipe and v is the velocity. Since we know the maximum velocity allowed is 1.2 m/s, we can set up the equation as 1.2A1 = 0.00373, where A1 is the cross-sectional area of the pipe.

To convert from m^3 to m, we can use the formula A = πr^2, where r is the radius of the pipe. So, we can rewrite the equation as 1.2πr^2 = 0.00373. Solving for r, we get a radius of 0.0208 m or 2.08 cm.

Therefore, the minimum inside diameter of the pipe should be 4.16 cm (since diameter is equal to twice the radius). This will ensure that the velocity of the emulsion does not exceed 1.2 m/s and cause any damage.