In Fig. S2.9 oil of absolute viscosityμ fills the small gap of thicknessY. (a) Neglecting fluid stress exerted on the circular underside, obtain an expression for the torqueT required to rotate the truncated cone at constant speedω. (b) What is the rate of heat generation, in joules per second, if

Question

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

Accepted Answer

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

Eq.(2.9) : [latex] au=\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}[/latex]

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

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

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