Question

A fluid that has a relative density of 0.013 is passed through a pipe that has an outer diameter of 98 mm and a wall thickness of 10 mm. The mass flow rate of the fluid measured to be 660 kg/min, determine the average velocity of the fluid through the pipe. O A) 177.17 m/s B) 84.95 m/s C 225.63 m/s D) 524 m/s

1) inside D = OD - 2 (thickness t)
A (Area) =pi/4 (D^2)
Volumetric flow (V)= (C m/s) (A m^2)
C=velocity
I can't figure out the density kg/m^3
rho= 1/v (m^3/kg) - specific volume

Answer 1

As shown in the calculations, correct option is (A).

Relative density = density of filed (p)/ density of water at 4 degree c 0.013 = density of field/1000 (therefore density of water r = 1000 kg./m^3) therefore density of filed, p = (0.013 times 1000) = 13 kg/m^3 inner diameter, D = (98 - 2 times 10) = (98 - 20) = 78 mm = 78 times 10^-3 m therefore area of cross-section, A = pi D^2/4 = 3.14 times (78 times 10^-3)^2/4 m^2 = 4775.94 times 10^-6 m^2 Now, mass of flow rate = 660 kg/min = 660/60 kg/s (therefore 1 min = 60 s) again, mass of flow rate = mass/time = density times volume/time = density times area times length/time therefore mass flow rate = density times area velocity (therefore velocity = length/time) therefore 660/60 = 13 times 4775.94 times 10^-6 times velocity hence, velocity = (660/60 times 13 times 4775.94 times 10^-6) m/s 0.000177 times 10^6 m/s = 177.17 m/s