Answer:
The weight of the tank when it is empty is 501.38 lbf
Explanation:
The velocity of outlet is equal to:
[tex]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} }[/tex]
Where
vi = 24 ft/s
di = 6 in = 0.5 ft
do = 6 in = 0.5 ft
[tex]v_{o} =\frac{0.5^{2}*24 }{2*0.5^{2} } =12ft/s[/tex]
The total weight of the water excluding the tank weight is equal to:
[tex]W=pv_{i}^{2} +pghA_{tank} -pv_{o}^{2} =p(v_{i}+gh(\frac{\pi d_{tank}^{2} }{4} -v_{o}^{2} ))[/tex]
p = 62.43 lb/ft³
dtank = 20 in = 1.67 ft
Replacing:
[tex]W=62.43(24^{2} +(32.17*2.5*(\frac{\pi 1.67^{2} }{4} )-12^{2} )=37929.5 lb=83.62lbf[/tex]
The weight of the tank is:
Wtank = 585 - 83.62 = 501.38 lbf
