Recent content by GSXR-750

Rookie error, thanks.

I calculated input Power using the input torque of around 39.73Nm Power = Torque * AngVel = 39.73 * 10pi = 1248W That would give full speed i.e. 300rpm Then for 30sec Power = Torque * AngVel = 39.73 * 5pi = 624W

Thanks CWatters, I see this now. I have calculated very similar to the above now within a few degrees of rounding accuracy.Is Davids method for calculating input power correct?

Input torque found by multiplying the output torque by 5 to account for the gearbox and then dividing again by 0.92 to allow for the efficiency 39.84Nm Should this not be output torque divided by 5. The output shaft is running at 5 times the speed of the input shaft?

Sorry to drag this back up. I am following the question here but getting vastly different results. Bending at the 1m mark of 625mm which obviously looks wrong. Going right back to post #1, I have calculated I as being; ## I = \frac {0.03*0.06^3} {12} ## ## = 0.54*10^{-6} ##Jumping back to...

Thanks for you patience with me I got the Flow rate, mass flow rate, Reynolds and velocity head. I don't believe i have studied finding the Darcy Weisback friction factor. I think my frictional head loss is 23.29 as I was going from the water surface not the whole leg i.e. 17m height.

Issue we started with.

Initial parts That was the original

I apologise I inputted Velocity not flow. The answer should be 5441 Watts or 5.4 Kw. Hm stands for the frictional losses caused by fittings. Though I'm now looking thinking my overall process for Hf is incorrect. I now think it should be 5069 Watts

So I calculated Flow to be ##0.0019m^3s## Hm = 0.023m Hf = 1.97m Therefore using Darcys Equation is ended up with Hp to be 29.18 (very similar to my previous fluid) Which gave me a Power of 858 kW. From this I assume that Hp won't change much in this given system. Hence from my first...

Given Fluid information and heads calculated from Velocity. Using Darcy Formula

The question states using the same pump as before. So I think I need to use the power previously stated and work backwards to velocity and hence flow.

From my notes this equation looked to give a reasonable answer. It however finds the Critical Velocity, which I now see is the velocity in which the flow changes into turbulent flow. I used 2000, as I know from a previous question using the exact same system but given a flow, different Viscosity...

I am trying to calculate the flow in a system from an underground tank up through a pump into a vessel. But cannot find or derive a formula that yields a good result, I don't think I can use the same head calculations as Velocity will have changed. 1. Homework Statement ρ = 1000 kg m–3 μ =...

Homework Statement Calculate head required for the pump and then its power requirement assuming 70% efficiency. The lower storage vessel is vented to atmosphere (assume 1 bar pressure) . I have the following given information: Pipe Area = 0.00636m^3. Flow(Q)= 0.01m^3/s Average Velocity =...