Pumping the blood out!

Classical Mechanics Level 5

We use an electromagnetic pump to pump the blood through a tube to maintain its quality. Using mechanical pumping causes the blood quality to degrade.

Suppose that blood is filled in a long tube of length L L and square cross-section having side w w . We maintain a constant, uniform current density J ⃗ \vec{J} due to charge constituents in the blood in one of the transverse. A magnetic field B ⃗ \vec{B} is maintained in the other transverse direction while the liquid can flow along the length. If viscosity of blood is η \eta , then the flow Q Q can be written as

Q = a b J π B w c η , Q = \frac{a}{b} \frac{J \pi B w^{c}}{\eta},

where a , b a,b are coprime positive integers. Find the value of c c .

Details and assumptions

  • Flow is defined as rate of volume of fluid passing through the tube per unit time.
  • Find the flow after it has become steady.


The answer is 4.

Kartik Sharma by Kartik Sharma , 21, India

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1 solution

Tapas Mazumdar
May 6, 2018

Unfortunately, I wasn't able to come up with the exact expression and the method to exactly evaluate the desired result. But since the problem didn't ask to find the constant term a b \dfrac ab and only the index of the term w w , this is my short solution by dimensional analysis.

Dimensions and example formula to calculate dimensions:

  • Current density, J = [ A L − 2 ] J = [AL^{-2}]

J = I S J = \dfrac{I}{S} ( I = I = current, S = S = area of cross section)

  • Magnetic field induction, B = [ M A − 1 T − 2 ] B = [MA^{-1}T^{-2}]

F = q v B sin ⁡ θ F = qvB \sin \theta ( q = q= charge, F = F= magnetic force, v = v= velocity)

  • Co-efficient of viscosity, η = [ M L − 1 T − 1 ] \eta = [ML^{-1}T^{-1}]

F = − η S d v d x F = - \eta S \dfrac{dv}{dx} ( S = S = area of contact layer, F = F= viscous force, d v d x = \dfrac{dv}{dx} = velocity gradient with respect to length)

  • Width, w = [ L ] w = [L]

  • Volumetric flow rate, Q = [ L 3 T − 1 ] Q = [L^3 T^{-1}]

Q = S v Q = Sv ( S = S= area of cross section of tube/pipe, v = v= velocity of laminar flow)

Thus we obtain,

[ L 3 T − 1 ] = [ A L − 2 ] ⋅ [ M A − 1 T − 2 ] ⋅ [ L ] c [ M L − 1 T − 1 ] ⟹ [ L 3 T − 1 ] = [ L c − 1 T − 1 ] [L^3 T^{-1}] = \dfrac{ [AL^{-2}] \cdot [MA^{-1}T^{-2}] \cdot [L]^c }{ [ML^{-1}T^{-1}] } \implies [L^3 T^{-1}] = [L^{c-1} T^{-1}]

giving c − 1 = 3 ⟹ c = 4 c-1 =3 \implies \boxed{c=4} .

P.S. If anyone can provide the exact result it would be very helpful.

0 Helpful 0 Interesting 0 Brilliant 0 Confused

did the same, I think it's an easy approach to the question

A Former Brilliant Member - 3 years, 1 month ago

Yo Tapas boi B)

Sahil Silare - 3 years ago

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