Question

For turbulent flow in a tube, the heat transfer coefficient is obtained from the Dittus-Boelter correlation. If the tube diameter is halved and the flow rate is doubled, then the percentage increment in the heat transfer coefficient will be

• A. 506.25
• B. 6.06
• C. 606.25
• D. 100

Answer 1

The correct option is A 506.25

For turbulent flow

$Nu\propto {\left(\mathrm{Re}\right)}^{0.8}\times {(Pr)}^n$

Pr = Prandtl number (cosntant)

$Nu\propto {\left(\mathrm{Re}\right)}^{0.8}$

$\frac{hD}{k}\propto {\left(\frac{\rho VD}{\mu }\right)}^{0.8}$

$h\propto V^{0.8}\times D^{-0.2}$

$V_1=V,D_1=D$

and ${\dot{m}}_1=\rho AV=\rho \times \frac{\pi }{4}{D}^2\times V$

${\dot{m}}_2=2{\dot{m}}_1=2\times \rho \times \frac{\pi }{4}{\left(\frac{D}{2}\right)}^2\times 4V$

$=\rho \times \frac{\pi }{4}{\left(\frac{D}{2}\right)}^2\times 8V\Rightarrow D_2=\frac{D}{2}\&V_2=8V$

$\frac{h_2}{h_1}={\left(\frac{V_2}{V_1}\right)}^{0.8}\times {\left(\frac{D_2}{D_1}\right)}^{-0.2}={\left(\frac{8V}{V}\right)}^{0.8}\times {\left(\frac{D/2}{D}\right)}^{-0.2}$

$\frac{h_2}{h_1}=6.0628$

Percentage increment
$=\frac{h_2-h_1}{h_1}\times 100\%=(6.0625-1)\times 100\%$

= 506.25%