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Question: The level of blood in a bottle used to give blood to a patient is \(1.3m\) high above the needle whi...

The level of blood in a bottle used to give blood to a patient is 1.3m1.3m high above the needle which is 3cm3cm in length and has an internal diameter of 0.36mm0.36mm . If 4.5cm34.5c{m^3} of blood is passing through the needle per minute, calculate the viscosity of blood. The density of blood is 1020 kg m−31020\,kg\,{m^{ - 3}}. Ignore the level of blood in the bottle.

Explanation

Solution

Viscosity is the resistance of the fluid to flow; it is caused by friction within a fluid. The flow rate is given. As the blood is at some height there will be some pressure difference. The equation relating flow rate to pressure difference is ϕ=πr4∣△P∣8ηL\phi = \dfrac{{\pi {r^4}\left| {\vartriangle P} \right|}}{{8\eta L}} .
Here, ϕ\phi is the rate of flow, rr is the radius of the needle, ∣△P∣\left| {\vartriangle P} \right| is the magnitude of pressure difference, η\eta is the coefficient of viscosity and LL is the length of the needle.

Complete step by step answer:
As there is a difference in height of bottle and the needle therefore, there will be flow of blood.
The given quantities are:
Height of the bottle above needle, h=1.3mh = 1.3m
Density of blood, ρ=1020kg m−3\rho = 1020kg\,{m^{ - 3}}
Radius of the needle rr is half of diameter dd
⇒r=d2=0.36mm2=1.8×10−4m\Rightarrow r = \dfrac{d}{2} = \dfrac{{0.36mm}}{2} = 1.8 \times {10^{ - 4}}m
Length of the needle, l=3 cm= 3×10−2cml = 3\,cm = \,3 \times {10^{ - 2}}cm
Volume of blood, V=4.5cm3=4.5×10−6m3V = 4.5c{m^3} = 4.5 \times {10^{ - 6}}{m^3}
The rate is given as per minute for SI units we will convert it to per second, then we get
ϕ=Vt=4.5×10−660m3s−1\phi = \dfrac{V}{t} = \dfrac{{4.5 \times {{10}^{ - 6}}}}{{60}}{m^3}{s^{ - 1}}

The magnitude of pressure difference ∣△P∣\left| {\vartriangle P} \right| , will be given as:
∣△P∣=hρg\left| {\vartriangle P} \right| = h\rho g
⇒∣△P∣=1.3×1020×9.8\Rightarrow \left| {\vartriangle P} \right| = 1.3 \times 1020 \times 9.8 -equation 11
Now, viscosity of blood will be given as:
η=πr4∣△P∣8ϕL\eta = \dfrac{{\pi {r^4}\left| {\vartriangle P} \right|}}{{8\phi L}}
Substituting the given values in the above equation, we get:
η=227×(1.8×10−4)4(1.3×1020×9.8)8(4.5×10−660)(3×10−2)\eta = \dfrac{{22}}{7} \times \dfrac{{{{\left( {1.8 \times {{10}^{ - 4}}} \right)}^4}\left( {1.3 \times 1020 \times 9.8} \right)}}{{8\left( {\dfrac{{4.5 \times {{10}^{ - 6}}}}{{60}}} \right)\left( {3 \times {{10}^{ - 2}}} \right)}}
⇒η=0.00238Nsm−2\Rightarrow \eta = 0.00238 {\rm N}s{m^{ - 2}}
⇒η=2.38×10−3Nsm−2\Rightarrow \eta = 2.38 \times 1{0^{ - 3}} {\rm N}s{m^{ - 2}}
This will be the viscosity of blood.

Note:
Viscosity is the resistance that a flowing fluid experiences as it flows. The bottle was at a height thus there was a pressure difference and due to this pressure difference, there will be flow of blood. It is important to convert all the values in the SI unit to avoid incorrect answers. The viscosity also depends on the length of the needle, if the length of needle increases, the fluid will experience more resistance.