Question

2. An understanding of fluid resistance in micro-channels is very important for designing a microfluidic system. Since the capillary tubes can have a variety of cross-sections, researchers have always been interested in understanding the resistance to water flow in micro-channels of various cross-sections. Fluid resistance R is defined as the ratio of pressure drop AP to the volumetric flow rate of medium Q along a specific length, say L of the channel. Evaluate the resistance to water flow in micro-channels of a rectangular cross-section (wxh) with the hydraulic diameter, dh=4A/p = 30 um, where p is the perimeter and A is the cross-sectional area of fluid flow. The dynamic viscosity of water, u at room temperature is 1001.65 x 10-6 N-s/m². What will happen to the resistance to water flow per unit length of the tube in circular as well as square cross-sections if the tube diameter is increased to 300 um? Hint: For the rectangular channel, the fluid resistance is defined as R = (12uL)/(why), where L is the tube length.

Answer 1

The question involves the concept of fluid resistance and hydraulic diameter. The detailed explanation is shown in the below images.
oben rectangular I height Ch] let us assume an Channel of width (w) for rectangular Channel R = 12ML wth Since it is an open channel plyd dra. Cen) = 4A 4 h - [p = wetted feri = W teh? G +>> to a lince love have two unknown w&h We must design it for most economical case Since R & Chress use difference) & Chate flow) R should be minimum To minimise R , 8 should be maximise, for const AP Qe Au Cua velocity) ] yecilmi = I c= cheery const c Tmi 10 i slope of bed I O = AcJAi min hysh rad ] VP

S ale d- Ac JÁT a constant Area (A) for DE DE K To max. O p should be minimise P for coni- A Pince A. Wah Tus constant for a channel ] P = 48 + 28 = 4 + 2h [arwn? Since A = wh - رط - sina h to Tw = 2h / ht. This is economical condil W = 2h width: 2 (depth]

PL Now Since da = 4when sbo 4whe fut W = 2h 4 (2h) (h), 30 talle h = 15 um 8 h 30 W = 2h = 30 um NOW R = 12 4 L - w 23 - 12 ud I who Resistance per unit length I = 1 2 4 = 12 X 1001 65 x 10-6 04 | t= 2001-65x1007 (30) (15) ² x (106) 4 [alt - N -5/m² L = 1. 18714 x 1017 Pa.m sec

for circular aon section Closed Copen Channel considered a s resistance (R) = 128 ML Fluid Here hyds au lie d'omeler dh = 4 ndx I --- - --- - = d d be du= d. diameter of tube (d) = 900 um = 100 x 10 m 4= 1001. 65 x 100N-s/m² 2 128 X 100l. 65 X10 I TX 1300 X 10 ole Resistance per un't length= 5:0383 x 10th an = 5.0383 x 1012 pam" sec. square section Collosed channel la dhe 300 um a for dhaa So side fa) = 100 un
Since it was not mentioned in the question whether the channel is open or close. I chose open for first case and closed for next 2 cases . The only difference would occur in wetted perimeter(P) and everything else would be same. For example- for a closed rectangular channel P=2(w+h) and for an open channel P=w+2h. Hope the explanation would be clear to you.