Question

An rectangular tank 5 m long, 2 m wide, and 8 m high is filled to a depth of 5 m of water. Set up the integral that would compute the work needed to pump the water out of a spout located at the top of the tank. Use the location of the origin and the direction of the positive axis as specified in the different parts. Note: Do not input values for the constant p9
(A) The origin is at the top of the tank with with positive y-values going down pg dy
(B) The origin is at the bottom of the tank with positive y-values going up. pg dy
(C) The origin is at the water level of the tank with positive y-values going down.dy

Answers

  1. Solution:

    Given data:

    A rectangular tank with dimensions 5 m as length, wide of 2 m and height of 8 m is filled with water tot a depth of 5 m.

    A). If we select an elemental part of the tank of thickness dy at a depth y from the top, its mass will be :

    m = p.v

    ⇒ m = p.(Ady)

    ⇒ m = p.( 5 x 2 dy)

    ⇒ m = 10 p dy

    Now work done in moving mass from a depth y to the top will,

    dw = m.g.y

    dw = (10 p dy).g.y

    or dw = 10 p.g.y.dy

    Total work done is :

    W= \int_{2}^{8} 10 p.g.y.dy

    $ W=10 pg[\frac{8^2-2^2}{2}]$

    $ W=10 pg[\frac{60}{2}]$

    W = 300 p.g

    B). Origin at the bottom and consider an elemental part of the tank at a height y from the bottom, then its mass is :

    ⇒ m = p.( 5 x 2 dy)

    ⇒ m = 10 p dy

    Now, work done in moving this mass from a height y above origin to a height 8 m above the origin is,

    dw = mg(8 – y)

    dw = (10 p dy).g.(8 – y)

    dw = 10 pg.(8 – y). dy

    The total work done is

    W= \int_{0}^{5} 10 p.g.(8-y).dy

    W=10 p.g \int_{0}^{5} (8-y).dy

    W=10 p.g \left [ 8y-\frac{y^2}{2} \right ]_0^5

    W=10 p.g \left [ 8(5)-\frac{(5)^2}{2} \right ]

    W = 10 pg (40-12.5)

    W = 275 pg

    C). Origin at the water level and consider y axis positive in the downward direction.

    The work done to move and elemental ,as from a depth y below the water level to the top of the tank is,

    dw = 10 pg (y+2) dy

    The total work done is :

    W=10 p.g \int_{0}^{5}(y+2) dy

    W=10 p.g \left[ \frac{y^2}{2}+2y \right ]_0^5

    W=10 p.g \left[ \frac{5^2}{2}+2(5) \right ]

    W = 10 p g (12.5 + 10)

    W = 225 pg

     

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