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defazajx

Answered question

2021-03-20

A cylindrical reservoir of diameter 4 feet and height 6 feet ishalf full of water weighing 62.5 lbs/ft3. The work done,in ft-lb, in emptying the water over the top is equal to

ρ

Answer & Explanation

d2saint0

Skilled2021-03-22Added 89 answers

Given:
The weight density is:

ρ_{w} = 62.5 \frac{l b}{f t^{3}}

The differential of volume is:

d v = π r^{2} d y = π {(2)}^{2} d y = 4 π d y

.
The differential of weight is:

d F = ρ_{w} d v = 4 π ρ_{w} d y

The differential of work is the weight times the distance thedisk of water must travel to top:

d w = (6 - y) d F = 4 π ρ_{w} (6 - y) d y

.
The work is thus:

w = 4 π ρ_{w} \int_{0}^{3} (6 - y) d y = 4 π ρ_{w} {(6 y - \frac{1}{2} y^{2})}_{0}^{3} = 4 π ρ_{w} (6 (3) - \frac{1}{2} (9)) = 4 π ρ_{w} (18 - \frac{9}{2})

= 4 π ρ_{w} (\frac{36 - 9}{2})

= 4 π ρ_{w} (\frac{27}{2})

= 54 π ρ_{w}

or we have

w = 54 π (62.5) = 3.375 π f t = l b \approx 10, 602.88 f t - l b

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