Verified answer

The outlet temperature [tex]T_{m}[/tex]  is [tex]35^{o}[/tex]  and the total heat transfer rate is 15982 w

Explanation:

Given:

The tube surface temperature [tex]T_{s} =150^{o} c[/tex]

Air temperature [tex]T_{m,i} =20^{o}c[/tex]

Tube Diameter D = 50 mm

Flow rate `m=0.5 kg/s

Tube length  L = 25 m

Using the given data,

Surface area of the tube = [tex]\pi DL[/tex]

⇒[tex](3.14)(0.05)(25)[/tex]

⇒[tex]3.927 m^{2}[/tex]

Let we take outlet temperature = [tex]T_{o}[/tex]

Temperature of surface [tex]T_{s} = 150^{o} c = 423 k[/tex]

Inlet temperature = [tex]20^{o} c = 293 k[/tex]

[tex]T_{o}= 293 k[/tex]

[tex]T_{o} = (T_{3} - T_{i})[/tex] ×[tex]e^{-\frac{\pi DLh}{m.c_{p} } }[/tex]

Where,

h= heat transfer coefficient

m=flow rate

cp=specific heat of oil

As we know that specific heat capacity is 2131 J/kg

[tex]m = 0.5 kg/s[/tex]

Now determine the type of flow using Reynold's number(Re)

[tex]Re = \frac{4m}{\pi Du}[/tex]

where u= Viscosity of oil

[tex]u = 0.032 kg/ ms[/tex]

[tex]Re = \frac{(4)(0.5)}{(3.14)(0.05)(0.082)}[/tex]

⇒ [tex]398.089[/tex]  this is a laminar flow

Here we also want to calculate prandtl no(Pr)

[tex]Pr = Cp.u/k[/tex]

⇒[tex]\frac{(2131)(0.032)}{0.138}[/tex]

⇒[tex]490[/tex]

Now,

[tex]x = (0.05)Pr.Re.D[/tex]

⇒[tex](0.05)(0.05)(398.089)(490)[/tex]

⇒[tex]487.65 m[/tex]

Now Nusselt no = Nu

[tex]Nu= 3.66 + \frac{(0.0668)(D/L)(Re)(Pr)}{(1+0.04(D/L)Re(Pr)^{2} }[/tex]

[tex]Nu = 3.66 + \frac{26}{(1+2.14)}[/tex]

⇒[tex]11.95[/tex]

And finally,

[tex]h = Nu (k)/D[/tex]

Where k is a thermal conductivity

[tex]h = 32.982 w/m^{3}k[/tex]

Substitute the value of [tex]T_{3} , T_{i} , h[/tex]

Then we get,

[tex]T_{o}= 35^{o} c[/tex]

Heat = q which is given

[tex]q = m . Cp(T_{o} - T_{i} )[/tex]

⇒[tex]15982 w[/tex]

Final answer:

The outlet temperature [tex]T_{m}[/tex]  is  [tex]35^{o} c[/tex] and the total heat transfer rate is 15982 w

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