Question

A 10-cm-diameter, 30-cm-high cylindrical bottle contains cold water at 3°C. The bottle is placed in windy air at 27°C. The water temperature is measured to be 11°C after 45 min of cooling. Disregarding radiation effects and heat transfer from the top and bottom surfaces, estimate the average wind velocity.

Answer 1

Answer:

Average wind velocity 1.91m/s

Explanation:

Given

Diameter of the bottle = 10cm

height of bottle = 30cm

Temperature of cold water = $3^{o}C$

Temperature of air = $27^{o}C$

Temperature of water after 45min = $11^{o}C$

 

Average temperature of water, $T=\frac{3+11}{2}=7^{o}c$

Properties of water from properties of water table at  $7^{o}C$ are

$p=999.8kg/m^{3}$

$c_{p} =4200j/kg^{o}c$

Average film temperature of air, $T=\frac{7+27}{2}=17^{o}c$

Properties of air from properties of air table at 1atm and $17^{o}C$ are

$k=0.02491W/m^{o}c$

$v=1.489*10^{-5}x=m^{2}/s$

$p_{r}=0.7317$

mass of water in bottle $m=pV=p\pi\frac{D^{2} }{4}L=999.8\pi*\frac{0.1^{2} }{4}*0.3=2.356kg$

Heat added to water  $Q = mc_{p}(T_{1}-T_{2})=2.356*4200*(11-3)=79162J$

Heat transfer rate $\dot Q=\frac{Q}{{\vartriangle}t}=\frac{79162}{45*60}=29.32W$

Surface area of cylinder $A_{x}={\pi}DL={\pi}0.1*0.3=0.09425$

Heat transfer rate of conclusion $\dot Q_{conv}=hA_{x}(T_{x}-T_{\infty})$

Equating the heat transfer rates

$29.32W=h(0.09425)(27-7)$

$h=15.55W/m^{2}.^{o}c$

Nusselt number $N_{u}=\frac{hD}{k}=\frac{15.55*0.1}{0.02491}=62.42$

Reynoids number is calculated using the equation

$N_{u}=0.3+\frac{0.62Re^{0.5}Pr^{\frac{1}{3}}}{[1+(0.4/Pr)^{\frac{2}{3} } ]^{\frac{1}{4}}}[1+(\frac{Re}{28200} )^{\frac{5}{8} }]^{\frac{4}{5} }$

$62.42=0.3+\frac{0.62Re^{0.5}0.7317^{\frac{1}{3}}}{[1+(0.4/0.7317)^{\frac{2}{3} } ]^{\frac{1}{4}}}[1+(\frac{Re}{28200} )^{\frac{5}{8} }]^{\frac{4}{5} }$

$Re=12856$

velocity of air is calculated using the relation

$Re=\frac{VD}{v}$

$12856=\frac{V(0.1)}{1.489*10^{-5}}$

$V=1.91m/s$

Average velocity is $1.91m/s$