An inverted cone (base above the vertex) is 2 m high and has a base radius of 1/2 m. If the tank is full, how much work is required to pump the water to a level 1 m above the top of the tank?

avissidep 2020-10-28 Answered
An inverted cone (base above the vertex) is 2 m high and has a base radius of 12 m. If the tank is full, how much work is required to pump the water to a level 1 m above the top of the tank?
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Expert Answer

cyhuddwyr9
Answered 2020-10-29 Author has 90 answers
Step 1
Given, an inverted cone is 2 m high and has a base radius of 12 m. If the tank is full, then how much work is required to pump the water to a level 1 m above the top of the tank.
Step 2: Concept Used
If force is a function F(x) of position x then in moving from x = a to x = b work done is
W=abF(x)dx
Step 3: Calculation
Consider the water tank conical in shape. We will make a small horizontal section of the water at depth h and thickness dh and also assume radius at depth h is w.
From similarity, we will get
w12=2h2
2w=2h2
w=2h4
The weight of the slice under consideration is
Weight=volume×density×gravitationalconstant
Weight=π×w2×dh×62.4(Densityofwater=62.4lbft3)
Weight=62.4πw2dh
Now by definition the work done is
W=12dW
=12(62.4πw2dh)h
=62.4π×11612(2h)2hdh
=62.4π×11612(4+h24h)hdh
=62.4π×11612(4h+h34h2)dh
=62.4π×116{4h22+h444h33}12
=62.4π×116{162+1643234214+43}
=62.4π×116{8+4323214+43}
=1.625πlb
Hence work done is 1.635πlb.
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