Convert the units of power,W = 7 hp = 7 * 745.69 = 5219.83 WCalculate the power input to the pump using the efficiency of the pump equationn=Wpump/Wshaft
Substitute 0.82 for n and 5219.83 for Wshaft0.82=Wpump/5219.83 Wpump=0.82*5219.83=4280.26 WCalculate the mass flow ratein=Wpump/(gz_2 )Where g is the acceleration due to gravity, and z_2 is the elevation of water. Substitute 4280.26 for Wpump, 9.81 m/s^2 for g, and 19m for z_2in = 4280.26 / 9.81 * 19 = 22.9640 m^3/sCalculate the volume flow rate of waterV=m/ρWhere ρ is the density of water. Substitute 22.9640 m^3/s for in and 1000 m^3/kg for ρ, we get V = 22.9640 / 1000 = 0.0230 kg/sTherefore, the volume flow rate of water is 0.0230 kg/s
Answer:
the answer is 483.75 I believe
Answer:
3.18 Nm
Explanation:
Given that:
Radius (r) = 15cm = 15/100 = 0.15m
θ = 45°
Applied force (F) = 30 N
The Torque can be obtained using the relation :
T = rF * sinθ
T = 0.15 * 30 * sin(45)
T = 0.15 * 30 * 0.7071067
T = 3.18198015
T = 3.18 Nm
1 kilowatt - hour_________________________
For some mysterious reason, we can't see the graph.
On a position/time graph, constant velocity is represented by a <em>straight line</em>. Depending on the velocity, the line may be sloping up, sloping down, or horizontal. The only thing the line on the graph <u><em>CAN't</em></u> be is vertical.