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  antekL1 22.6.2011 (15:08)

  The same volume of oil which is entering the pipe from the left side has to leave from the right side in the same time t.
  Let call "S" the pipe crosssection on the left side and "s" the crosssection area on the right. (I know pipe diameters so I know both areas. Imagine a cylinder of length L of oil entering the pipe in the time t.
  Its lenght L = Vin * t, where Vin is the velocity at the entrance. The oil volume is
  V = S * L = S * Vin * t.
  The same volume has to leave the pipe with the velocity Vout = 12 m/s.
  V = s * Vout * t
  Therefore:
  S * Vin * t = s * Vout * t and Vin = Vout * s / S.
  Using diameters D = 80 mm and d = 40 mm the above equation may be expressed as:
  V_{in} = V_{out}\frac{d^2}{D^2} = 12\cdot\frac{40^2}{80^2} = 3\,\,\mbos{m/s}
  Please note I didn't need to convert D and d into meters !
  
  The mass flow rate equals oil density times volume per second. I'm taking the same cylinder as above and set t = 1s. So the volume per second is Vout * s. Let call Delta m the mass flow.
  \Delta m = \rho\cdot V_{out}\cdot s = \rho\cdot V_{out}\cdot \frac{\pi d^2}{4}
  d is the diameter which should be expressed in meters, d = 0.04 m.
  \Delta m = 850\cdot 12\cdot \frac{\pi \cdot 0.04^2}{4} \,\approx\,12.82\,\,\mbox{kg/s}
  Please note the units: they are kg / s.
  {[}\Delta m] = \frac{kg}{m^3}\cdot\frac{m}{s}\cdot m^2 = \frac{kg}{s}
  
  Best greetings - Antek

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