A spherical tank of water has a radius of 15 ft, with the center of the tank 50 ft above the ground. How much work will it take to fill the tank by pumping water up from ground level? Assume the water weighs 62.4 lb/ft^3. to nearest ft/lb
Any help on how to set this up and do this would be appreciated.
or point me in the direction of a tutorial on how to do these problems.
Pope
An important morsel of information is missing. To what height is the water being pumped?
The weight of the water is the volume of the sphere times the weight density, which come to about 14 million pounds. Multiply the weight by the height to which the water is pumped.
The most obvious (to me) way to fill a tank would be through a gate at the top. The top of the tank is 65 feet above the ground. From there the water would fall to a mean height of 50 feet, but you asked how much work it would take, and the pump would have to lift the water all the way to the top, which is 65 feet up.
A less likely way to do it would be to pump it to an opening at the bottom of the tank. The bottom is 35 feet above ground. However, the pump would still be lifting the water that is already in the tank. If it were done this way, then the pump would lift the centroid of the water 50 feet. This would require less work from the pump, but who fills a tank from the bottom?
Another way would be to run the hose through the top, but from there let the hose hang down to the bottom of the tank interior. The siphon effect would make this equivalent to the second method above, with the centroid being lifted to the height of 50 feet.
?
Work done = force * distance
The weight W is the force.
Find that by sphere volume * density
V = (4/3)πr^3 = (4/3)π(15)^3 cubic feet
Density given as 62.4 lb/ft^3
What do you think the distance is? Easy to work out in this example.
Key idea is that weight acts at C of G which when full is at centre of sphere
We are told that this centre is 50 ft up.
Use your calculator to finish this.
david
simple work formula .... W = Fd >>> force X distance
... F = mg >> mass X acceleration of gravity
pounds are not technically mass ... pounds are force ... so a combined unit called the foot-pound is the unit of work.
This is not a simple problem because the distance varies .. some of the water is pumped less distance (to the bottom of the tank) and some must be pumped farther (to the top of the tank) === Fortunately, because the tank is a sphere, it is symmetric .. and the distance to the middle of the tank can be used as the distance for work.
==== 1st .. find the volume of the sphere
V = (4/3)pi*r^3 = (4/3)pi*15^3
V = 14137.167 cubic feet
F = 14137.167 X 62.4 = 882,159.22 lbs
W = Fd = 882,159.22 X 50
W = 44,107,960.86 ft-lbs <<< answer