A small circular hole 6.00 mm in diameter is cut in the side of a...

Question

A small circular hole 6.00 mm in diameter is cut in the side of a large water tank, 14.0 m below the water level in the tank. The top of the tank is open to the air. Find (a) the speed of efflux of the water and (b) the volume discharged per second.

Accepted Answer

Hey everyone. So in this problem we are working with volumetric flow rates. Let's see what they give us and see what they're asking for. We know that the height of the water in this concrete tank is m. A pipe is a pipe of diameter 2.54 cm is connected at the base of the tank. And we are asked to determine the speed of the water as it leads through the pipe and the volume flowing from the pipe per second. So that volume flowing per second, that's our volumetric flow rate. So we're going to right out our givens Right of 60 m diameter Of 2.54 cm. We're going to recall that. Our speed equation is given as the square root of two. G. H. And R. Volumetric flurry equation is given as Q equals V. A. So for the first part of the problem um it's just a plug and chug to find the speed. So speed equals the square root of To recall that gravity is 9.8 m/s squared. And our height was given to us in m. So that's just 60 m. We get three m/s. Looking at our possible choices, we can already eliminate these two. We found speed. We're gonna use that in our volumetric flow rate equation but we don't have area. What they did give us was the diameter of the pipe and we can recall that area is given for a circular pipe for a circle is given as pi R squared. Um So diameter is To our, so from here we can see that our is 1.27 cm. I'm just gonna put this into meters because we're gonna need that for this equation. Um so it's going to be 0. m. Alright, calculate area five times 0 one .0127 m Squared. That gives us an area of 5.07 times 10 to the negative 4th meters square. I'm gonna plug that into our equation for volumetric flow rate. Using the speed we found in part a 34.3 m/s times 5.07 times 10 to the -4 m squared. Gives us a volumetric flow rate of 1.74 Times 10 to the -2 meters cubed for a second. Looking at our answers though, the very much of flow rate is given in leaders per second. So we need to recall that the conversion between meters cubed and leaders is one liter for its 1000 l per meter cubed. So that gives us 17.4 L/s. I'm looking at our answers that gives us answer. B Alright that's a call for this problem. We'll see you in the next video

A small circular hole 6.00 mm in diameter is cut in the side of a large water tank, 14.0 m below the water level in the tank. The top of the tank is open to the air. Find (a) the speed of efflux of the water and (b) the volume discharged per second.

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