Water Flow Explained in Terms of Volume

If a cylinder has a diameter of 4 cm and is initially filled to a depth of 90 cm above a hole .4 cm in diameter. Water flows through at a speed given by v = sqrt , or square root times the quantity of 2*980*y, so v=sqrt(2*980*y), where y is the depth in cm and v is the velocity in cm/sec.

This problem can be evaluated by substituting the given numbers in, and solving for the equations given.

· First we need to find how many cm^3 flow through the hole out of the tube.

This is done by calculating the area of the hole and multiplying it by the speed /sec that water flows through the hole.

·First calculate the area of the hole. Assuming the hole is a circle we can find the area by using the formula pi (approximately 3.14)* r(radius) ^2, or 3.14(r^2).The area of the hole, if done correctly, is about .12566 cm^2. Keep in mind that the given information of the hole is that its diameter was .4 cm. Thus to find the radius (which is half the diameter) the diameter must be divided by 2.

· Then calculate the speed of the exiting water. This is done by substituting 90 cm in for y into the equation above (v=sqrt (2*980*90). Therefore, v= 420.

· Multiply 420cm* .12566, which would equal about 52.7772 cm^3/sec.

· This means we are losing roughly 52.7772 cm^3/sec.

If we are losing 52.7772 cm^3/sec from the hole. We must be losing this much from the tube as well. So all we need to do is find the area of the tube and divide the amount lost by the number from the hole by the area of the tube to find the depth lost in the tube.

· The area of the tube is found by Pi * r^2.If the diameter of the tube is 4 cm, the radius would be 2 cm. The area of the tube would be about 12.566 cm.

· We then divide the amount lost (52.7772cm^3/sec) by the area of the tube (12.566 cm) to find the depth lost. This division would be equal to about 4.2 cm/sec.

We can subtract the initial depth (90 cm) by the amount lost (4.2cm) and find the new depth. We can then use the new depth to repeat the process as needed to find the depth after x number of seconds.