you sir , are my hero. Thanks for preventing the previously much inevitable failure i was about to go through this semester. much love from South Africa.
Well done - no text books fully explain the moments of inertia details. I did a questions using a semi circular section and this video confirmed the moments of inertia data I used. Thank u.
May I ask about P0 as gage pressure, cause sometimes we dont get specified pressure. Lets lets inside inclosed tank, P of the air is 200kpa, then is it gage pressure or absolute pressure?
In practice, the gage pressure is usually atmospheric pressure (101 kPa). If you are have an enclosed tank with an absolute pressure of 200 kPa, the gage pressure P0 would be (200-101)kPa = 99 kPa.
You can try it yourself. Follow the same derivation procedure, except the force on the inside of the wall is not simply gamma*h, but rather gamma*h+P0 (where P0 is a gage pressure).
Hi can you help with two question, if you could go through them it would be helpful... 1) The viscous torque produced on a disc rotating in a liquid depends upon the diameter D, the speed of rotation N, the density and the viscosity. The equation taking the form:- T= ρ N 2 D 5 φ ( ρ N D 2 μ ) In order to predict the torque on a 1.8 diameter disc rotating in water of density 1000kg/m3 at 129 rev/min a model is built to a scale of 1/4. The model is rotated in oil. Calculate the speed of rotation, in rev/min, for the model to produce dynamic similarity. The oil density is 785 kg/m3 and the dynamic viscosity of the oil is 4.7 x 10-4Pa.s. The water dynamic viscosity is 1 x 10-3Pa.s. 2) Water flows through a pipe 4cm in diameter and 38m long between two tanks with square edged non projecting entrance and exit. The flow rate is controlled by a gate valve which when fully open has a loss factor of 0.32. What is the height difference between the tanks if the volume flow rate is 3.8 x10-3 m3/s. For laminar flow f=16/Re For turbulent flow f=0.08/Re0.25 Density of water, LaTeX: hoρ, is 1000kg/m3, viscosity of water, LaTeX: \muμ, is 0.001Pa.s. If you could do this before the 22.06.2020 it would help me... thanks
hydrostatic pressure = density * gravity * height, gamma is the product of density and gravity AKA specific weight therefore Pressure = gamma *height Now, since we are dealing with an inclined surface we cant just simply take the height as it is, we consider it with respect angle of the inclined surface thus having height = y*sin(theta) next, Force is equals to the pressure * area Force= Pressure * Area therefore, Force = gamma * height *area = gamma* y*sin(theta) *area we are dealing with an elemental force acting on an elemental Area, thus dFnet = gamma * y*sin(theta) * dArea. I hope it makes sense
