Question

Uniform Laminar Flow ($9.1) Consider the channel designed for electronic cooling in Prob. 9.28. Because of the heating, the viscosity will change through the channel. Assume the viscosity varies as 0.1 (fro-)da where uo is the viscosity at s 0 and L is the length of the channel. Find the percentage change of the pressure drop due to viscosity variation. Prob. 9.28 - An engineer is designing a very thin, horizontal channel for cooling electronic circuitry. The channel is 2 cm wide and 5 cm long. The distance between the plates is 0.2 mm. The average velocity is 5 cm/s. The fluid used has a viscosity of 1.2 cp and a density of 800 kg/m3. Assuming no change in viscosity or density, find the pressure drop in the channel and the power required to move the flow through the channel.

Answer 1

Ans 9.28) We know,

Pressure drop in two dimensional straight channel($\Delta$P) = 2 $\mu$ V L / d²

where, $\mu$ = viscosity of fluid = 1.2 cP or 0.0012 Ns/m²

V = velocity = 5 cm/s or 0.05 m /s

L = Length of plate = 5 cm or 0.05 m

d = half the distance between plates = 1.5 mm or 0.0015 m

Putting values,

     $\Delta$P = 2 x 0.0012 x 0.05 x 0.05 / 0.0015²

= 2.667 N/m²

Now, calculate Reynold number, Re = VL / $\mu$

= 0.05 x 0.05 / 0.0012

= 2.08 < 2000

Hence, since Reynold number is less than 2000 ,flow is laminar

=> h_f = 12 $\mu$ V L/ $\rho$gd²

= 12 (0.0012)(0.05)(0.05) / 800 x 9.81 x 0.003²

⁼ 5.09 x 10^-4 m

Power required (P) = $\rho$ g Q h_f

where, Q = discharge = A x V = (0.05 x 0.02 x 0.05)= 5 x 10^-5 m³/s

=> P = 800 x 9.81 x 5 x 10^-5 x 5.09 x 10^-4

= 1.997 x 10^-4 W

Ans) Pressure drop in two dimensional straight channel($\Delta$P) = 2 $\mu$ V L / d²

Also, $\mu$ = $\mu$_o exp (-0.1 s/ L)

When s = L ,

  $\mu$ = 0.0012 exp ( - 0.1)

= 0.00108 Ns/m²

=> $\Delta$ P = 2 x 0.00108 x 0.05 x 0.05 / 0.0015²

= 2.4 N/m²

Percent change in pressure = [(2.667 - 2.4 ) / 2.667] x 100 = 10 %