In part of the Ghiradelli chocolate factory, liquid chocolate is poured out of a
ID: 1855590 • Letter: I
Question
In part of the Ghiradelli chocolate factory, liquid chocolate is poured out of a pipe into a cylindrical tank, as shown below. The tank is 1.5 m in diameter and has a hole in the bottom, 2.54 cm (1 in) in diameter, through which the chocolate falls. The speed with which the chocolate exits the tank is given by ue = where h is the height of the chocolate in the tank, and g = 9.81 m/s2 is the gravitational acceleration. The density of the chocolate is rho = 2,500 kg/m3. If the mass flow rate of chocolate into the tank is 25 kg/s, find the minimum height of the tank such that it will not overflow. The average residence time of the chocolate in the tank is given by the volume of the chocolate in the tank divided by the volumetric flow rate through the tank. What is the average residence time for a chocolate particle under the above conditions? You are in charge of safety at the plant and, as a precaution, you decide to add 1.5 m of height to the tank. One day a valve breaks and the mass flow rate suddenly changes from 25 kg/s to 35 kg/s. How long will it take the tank to overflow? (Note: The tank will overflow, if you do the problem correctly.)Explanation / Answer
exit of chocolate from hole
velocity of chocolate exit(v) = (2gh)^0.5
radius of hole = 2.54/2 =1.27 cm
= 0.0127 m
area of hole (a) = pi* 0.0127 * 0.0127
= 5.067 x 10^-4 m^2
density = 2500 kg/m^3
therefore mass flow rate for exit = v * a * density
for no overflow mass flow in = mass flow out
25 = 2500 * 5.067 x 10^-4 * (2* 9.81*h)^0.5
therefore
h = 19.851 m
volume of chocolate in the tank = vloume of tank
= pi * R *R * h
R is radius of tank = 0.75 m
total volume of tank = pi * 0.75 * 0.75 * 19.851
= 35.0796 m^3
mass flow rate = 25 kg/s
vloume flow rate = 25/2500
= 0.01 m^3/s
average residence time of chocolate = 35.079637.730 / 0.01
=3507.96sec
= 58.46 minutes
b) now total height = 19.851 + 1.5
= 21.351 m
mass flow rate in = 35 kg/s
mass flow rate out = density * a * v
= 2500 * 5.067 x 10^-4 * (2* 9.81* 21.351)^0.5
= 25.926 kg/s
therefore mass remaining inside = 35 - 25.926
= 9.0731 kg/s
volume remaing inside per sec= 9.0731 / density
= 9.0731 /2500
= 3.6292 x10^-3 m^3/s
total volume of tank = pi * R *R * h
R is radius of tank = 0.75 m
total volume of tank = pi * 0.75 * 0.75 * 21.351
= 37.730 m^3
total time taken to overflow the tank = total volume of tank /volume remaing inside per sec
= 37.730 / 3.6292 x10^-3
= 10396.32 s
= 2.887 hours