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3 years ago

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A river barge, whose cross section is approximately rectangular, carries a load of grain. The barge is 27 ft wide and 93 ft long

. When unloaded its draft (depth of submergence) is 5 ft, and with the load of grain the draft is 7 ft. Determine: (a) the unloaded weight of the barge, and (b) the weight of the grain.

1 answer:

Monica [59]3 years ago

8 0

Answer:

a) for equilibrium

sum of F vertical is zero

$F_{vertical} =0$

so that

$W_{b} =F_{B}=r_{h2o} *(submerged volume)\\=(62.4lb/ft^3)(5ft*27ft*93ft)\\=783432 lb$

b)

$F_{vertical} =0\\W_{b}+ W_{mg} =r_{h2o} *(submerged volume)\\W_{g} =(62.4lb/ft^3)*(7ft*27ft*93ft)-783432\\W_{g} =313372lb$

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Answer:

Explanation:

Given:

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Temperature of air, $T_{\infinity}=20^{o}C$

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The nusselt number is determined from the relation given by:

$Nu_{cyl}= 0.3 + \frac{0.62Re^{0.5}Pr^{\frac{1}{3}}}{[1+(\frac{0.4}{Pr})^{\frac{2}{3}}]^{\frac{1}{4}}}[1+(\frac{Re}{282000})^{\frac{5}{8}}]^{\frac{4}{5}}$

$Nu_{cyl}= 0.3 + \frac{0.62(576.92)^{0.5}(0.70275)^{\frac{1}{3}}}{[1+(\frac{0.4}{(0.70275)})^{\frac{2}{3}}]^{\frac{1}{4}}}[1+(\frac{576.92}{282000})^{\frac{5}{8}}]^{\frac{4}{5}}\\\\=12.11$

The covective heat transfer coefficient is given by:

$Nu_{cyl}=\frac{hD}{k}$

Rewrite and solve for $h$

$h=\frac{Nu_{cyl}\timesk}{D}\\\\=\frac{12.11\times0.03443}{3\times10^{-3}}\\\\=138.98 W/m^{2}.K$

The rate of heat transfer from the wire to the air per meter length is determined from the equation is given by:

$Q=hA_{s}(T_{s}-T{\infin})\\\\=h\times(\pi\timesDL)\times(T_{s}-T{\infinity})\\\\=138.92\times(\pi\times3\times10^{-3}\times1)\times(280-20)\\\\=340.42W/m$

The rate of heat transfer from the wire to the air per meter length is $Q=340.42W/m$

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