# Question 2.SP.9: In Fig. S2.9 oil of absolute viscosityμ fills the small gap ......

In Fig. S2.9 oil of absolute viscosity$\mu$ fills the small gap of thickness$Y$. (a) Neglecting fluid stress exerted on the circular underside, obtain an expression for the torque$T$ required to rotate the truncated cone at constant speed$\omega$. (b) What is the rate of heat generation, in joules per second, if the oil’s absolute viscosity is$0.20 \mathrm{~N} \cdot \mathrm{s} / \mathrm{m}^2, \alpha=45^{\circ}, a=45 \mathrm{~mm}, b=60 \mathrm{~mm}$,$Y=0.2 \mathrm{~mm}$, and the speed of rotation is$90 \mathrm{rpm}$ ?

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(a) $U=\omega r ; \quad$  for small gap $Y, \frac{d u}{d y}=\frac{U}{Y}=\frac{\omega r}{Y}$

Eq.(2.9) : $\tau=\mu \frac{d u}{d y}=\frac{\mu \omega r}{Y} ; \quad d A=2 \pi r d s=\frac{2 \pi r d y}{\cos \alpha}$

\begin{aligned} & \text { From Eq. (2.9): } \quad d F=\tau d A=\frac{\mu \omega r}{Y}\left(\frac{2 \pi r d y}{\cos \alpha}\right) \\\\ & d T=r d F=\frac{2 \pi \mu \omega}{Y \cos \alpha} r^3 d y ; \quad r=y \tan \alpha \\\\ & d T=\frac{2 \pi \mu \omega \tan ^3 \alpha}{Y \cos \alpha} y^3 d y \\\\ & T=\frac{2 \pi \mu \omega \tan ^3 \alpha}{Y \cos \alpha} \int_a^{a+b} y^3 d y ;\left.\quad \frac{y^4}{4}\right|_a ^{a+b}=\left[\frac{(a+b)^4}{4}-\frac{a^4}{4}\right] \\\\ & T=\frac{2 \pi \mu \omega \tan ^3 \alpha}{4 Y \cos \alpha}\left[(a+b)^4-a^4\right] \\\\ & \text { (b) }\left[(a+b)^4-a^4\right]=(0.105 \mathrm{~m})^4-(0.045 \mathrm{~m})^4=0.0001175 \mathrm{~m}^4 \\\\ & \omega=\left(90 \frac{\mathrm{rev}}{\mathrm{min}}\right)\left(2 \pi \frac{\text { radians }}{\mathrm{rev}}\right)\left(\frac{1 \mathrm{~min}}{60 \mathrm{~s}}\right)=3 \pi \mathrm{rad} / \mathrm{s}=3 \pi \mathrm{s}^{-1} \\\\ & \text { Heat generation rate }=\text { power }=T \omega=\frac{2 \pi \mu \omega^2 \tan ^3 \alpha}{4 Y \cos \alpha}\left[(a+b)^4-a^4\right] \\\\ & =\frac{2 \pi\left(0.20 \mathrm{~N} \cdot \mathrm{s} / \mathrm{m}^2\right)\left(3 \pi \mathrm{s}^{-1}\right)^2(1)^3\left[0.0001175 \mathrm{~m}^4\right]}{4\left(2 \times 10^{-4} \mathrm{~m}\right) \cos 45^{\circ}} \\\\ & =23.2 \mathrm{~N} \cdot \mathrm{m} / \mathrm{s}=23.2 \mathrm{~J} / \mathrm{s} \\\\ & \end{aligned}

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