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

Average wind velocity 1.91m/s

Step-by-step 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=(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=(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(D^(2) )/(4)L=999.8\pi*(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}=(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)=(hD)/(k)=(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^{(1)/(3)}}{[1+(0.4/Pr)^{(2)/(3) } ]^{(1)/(4)}}[1+((Re)/(28200) )^{(5)/(8) }]^{(4)/(5) }`

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

`Re=12856`

velocity of air is calculated using the relation

`Re=(VD)/(v)`

`12856=(V(0.1))/(1.489*10^(-5))`

`V=1.91m/s`

Average velocity is
`1.91m/s`