Answer :
Answer:
Average wind velocity 1.91m/s
Explanation:
Given
Diameter of the bottle = 10cm
height of bottle = 30cm
Temperature of cold water = [tex]3^{o}C[/tex]
Temperature of air = [tex]27^{o}C[/tex]
Temperature of water after 45min = [tex]11^{o}C[/tex]
Average temperature of water, [tex]T=\frac{3+11}{2}=7^{o}c [/tex]
Properties of water from properties of water table at [tex]7^{o}C[/tex] are
[tex]p=999.8kg/m^{3}[/tex]
[tex]c_{p} =4200j/kg^{o}c[/tex]
Average film temperature of air, [tex]T=\frac{7+27}{2}=17^{o}c[/tex]
Properties of air from properties of air table at 1atm and [tex]17^{o}C[/tex] are
[tex]k=0.02491W/m^{o}c[/tex]
[tex]v=1.489*10^{-5}x=m^{2}/s[/tex]
[tex]p_{r}=0.7317[/tex]
mass of water in bottle [tex]m=pV=p\pi\frac{D^{2} }{4}L=999.8\pi*\frac{0.1^{2} }{4}*0.3=2.356kg[/tex]
Heat added to water [tex]Q = mc_{p}(T_{1}-T_{2})=2.356*4200*(11-3)=79162J[/tex]
Heat transfer rate [tex]\dot Q=\frac{Q}{{\vartriangle}t}=\frac{79162}{45*60}=29.32W [/tex]
Surface area of cylinder [tex]A_{x}={\pi}DL={\pi}0.1*0.3=0.09425[/tex]
Heat transfer rate of conclusion [tex]\dot Q_{conv}=hA_{x}(T_{x}-T_{\infty})[/tex]
Equating the heat transfer rates
[tex]29.32W=h(0.09425)(27-7) [/tex]
[tex]h=15.55W/m^{2}.^{o}c[/tex]
Nusselt number [tex]N_{u}=\frac{hD}{k}=\frac{15.55*0.1}{0.02491}=62.42[/tex]
Reynoids number is calculated using the equation
[tex]N_{u}=0.3+\frac{0.62Re^{0.5}Pr^{\frac{1}{3}}}{[1+(0.4/Pr)^{\frac{2}{3} } ]^{\frac{1}{4}}}[1+(\frac{Re}{28200} )^{\frac{5}{8} }]^{\frac{4}{5} } [/tex]
[tex]62.42=0.3+\frac{0.62Re^{0.5}0.7317^{\frac{1}{3}}}{[1+(0.4/0.7317)^{\frac{2}{3} } ]^{\frac{1}{4}}}[1+(\frac{Re}{28200} )^{\frac{5}{8} }]^{\frac{4}{5} } [/tex]
[tex]Re=12856 [/tex]
velocity of air is calculated using the relation
[tex]Re=\frac{VD}{v}[/tex]
[tex]12856=\frac{V(0.1)}{1.489*10^{-5}}[/tex]
[tex]V=1.91m/s[/tex]
Average velocity is [tex]1.91m/s[/tex]