Question

water in the tank is 2.5 ft. Water enters the tank through a top port of diameter 6 in. at a velocity of 24 ft/s. The water leaves the tank through two exit ports each with a diameter of 6 in. If the scale shows a reading of 585 lbf, calculate the weight of the tank when it is empty

Answer 1

Answer:

The weight of the tank when it is empty is 501.38 lbf

Explanation:

The velocity of outlet is equal to:

$A_{i} v_{i}=2A_{o} v_{o} \\\frac{\pi d_{i}^{2} v_{i} }{4} =\frac{2\pi d_{o}^{2} v_{o} }{4}\\v_{o} =\frac{d_{i}^{2} v_{i} }{2d_{o}^{2} }$

Where

vi = 24 ft/s

di = 6 in = 0.5 ft

do = 6 in = 0.5 ft

$v_{o} =\frac{0.5^{2}*24 }{2*0.5^{2} } =12ft/s$

The total weight of the water excluding the tank weight is equal to:

$W=pv_{i}^{2} +pghA_{tank} -pv_{o}^{2} =p(v_{i}+gh(\frac{\pi d_{tank}^{2} }{4} -v_{o}^{2} ))$

p = 62.43 lb/ft³

dtank = 20 in = 1.67 ft

Replacing:

$W=62.43(24^{2} +(32.17*2.5*(\frac{\pi 1.67^{2} }{4} )-12^{2} )=37929.5 lb=83.62lbf$

The weight of the tank is:

Wtank = 585 - 83.62 = 501.38 lbf