When a plastics plant is redesigned, a tank exit is now facing the wrong directi
ID: 1858911 • Letter: W
Question
When a plastics plant is redesigned, a tank exit is now facing the wrong direction and must be connected to a pipe of diameter D = 0.2 m with a 180 degree bend in a horizontal plane. Liquid with S = 1.0 is sto flow through the bend at 0.3 m^3/s. The pressure in the liquid is 250 kPa gauge. The volume of the bend is 0.1 m^3. The steel of the bend material weighs 51 kg. Neglecting friction or other losses in the pipe, what is the tension force in the bolts holding the bend in place?
State all assumptions and show working.
Explanation / Answer
Area A = pi/4*D^2 = 3.14/4*0.2^2 = 0.0314 m^2
Velocity V = Q/A = 0.3/0.0314 = 9.554 m/s
Thrust = momentum change = m(v-(-v)) = 2mv = 2*(0.3*1000)*9.554 = 5732.5 N
Pressure force = 250*1000*0.0314 = 7850 N
Net horizontal force = 7850+5732.5 = 13582.5 N
Downward force due to weight = 51*9.81 + 0.1*1000*9.81 = 1481.31 N
Net force = sqrt(13582.5^2 + 1481.3^2) = 13663 N
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