Recent content by ron taylor

  1. R

    Rotary to linear motion using a crank

    this is a simple torque question. if you apply 1 pound of force to a crank that is located 12 inches away from the center of the cranks shaft then you have 1 foot pound of torque or force applied to the center of the cranks shaft. by placing a gear on the other side of the cranks shaft that is...
  2. R

    Pressure differential through orifice plate

    if your test rig is only going to be used to measure flow rate through an orifice then you should remove the expansion chamber from your test and also remove all of the turns as these will affect the downstream pressure leading to the orifice and the upstream pressure leading away from the...
  3. R

    Some questions on pressure drops and head loss in pipe flow

    why pressure drop is constant through pipe? How can we know that dP/dx=constant? ....... if a pressure gauge located at one end of a pipe reads 100psi then that does not say that a pressure gauge located anywhere else along the pipe will read 100psi. any fluid traveling through a pipe will meet...
  4. R

    Help Calculating Pressure of Seawater against a flat surface

    I was pointing out that he wouldn't want to simply multiply the weight of 1 cu ft of sea water by the depth of the sea water @ 280 ft to find the average pressure of the sea water that acts against the wall. and that he would need to divide the result of your suggested math by 144 to find the...
  5. R

    Help Calculating Pressure of Seawater against a flat surface

    actually there's a really big difference. that would deliver an answer of 17999.82 lbs vertical pressure per horizontal square ft. there is no way that he could use that number unless he divided it by the 144 sq inches in the horizontal square foot. which results in 124.998 psi he was wanting...
  6. R

    Help Calculating Pressure of Seawater against a flat surface

    since you have only given the pressure at the top 0psi and at the bottom 250psi then the psi on the wall at any given point along the wall at a depth of 280 ft would be 1/2 of 250psi 250psi/2=125psi the psi of a standing column of water can be found by multiplying the weight of 1 cu in of the...
Back
Top