Holooly Plus Logo

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 thicknessY. (a) Neglecting fluid stress exerted on the circular underside, obtain an expression for the torqueT 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 is0.20 \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 is90 \mathrm{rpm} ?

The 'Blue Check Mark' means that this solution was answered by an expert.
Learn more on how do we answer questions.

(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}

Related Answered Questions

Question: 2.SP.10

Verified Answer:

Table A.1 at 10^{\circ} \mathrm{C}: \quad \...
Question: 2.SP.11

Verified Answer:

From Appendix A, Table A.3, the pressure of the st...