Chapter 7, Problem 11P

a.

To determine

To find : how long will it take for the pile to reach the top of the silo

Answer to Problem 11P

It take $9.8hours$ to reach the top of the silo

Explanation of Solution

Given information : The question is

The ore resembles pink talcum powder and is poured from a conveyor at the top of the silo. The silo is a cylinder $100ft$ high with a radius $200ft$ . The conveyor carries $60,000\pi f{t}^3/h$ and the ore maintains a conical shape whose radius is $1.5$ times its height. If at a certain time $t$ , the pile is $60ft$

Calculation :

The ore is forming a conical shape, the volume of cone is

  $\begin{array}{l}V=\frac{1}{3}\pi r^{2}h\\ r=1.5h\\ V=\frac{1}{3}\pi {\left(1.5h\right)}^{2}h\\ V=\frac{1}{3}\pi \left(2.25h^2\right)h\\ V=0.75\pi h^3\end{array}$

Differentiate both sides with respect to time

  $\frac{dV}{dt}=0.75\pi \frac{d\left(h^3\right)}{dt}$

The conveyor is bringing the ore at rate of $60,000\pi f{t}^3/h$

  $\begin{array}{l}60000\pi =0.75\pi \frac{d\left(h^3\right)}{dt}\cdot \frac{dh}{dt}\\ 60000\pi =0.75\pi \cdot 3h^2\cdot \frac{dh}{dt}\\ 60000\pi =2.25\pi h^2\cdot \frac{dh}{dt}\\ 60000\pi dt=2.25\pi h^{2}dh\\ {\displaystyle \int 60000\pi dt={\displaystyle \int 2.25\pi h^{2}dh}}\\ 60000\pi t=2.25\pi \frac{h^{2+1}}{2+1}+C\\ 60000\pi t=0.75\pi h^3+C\\ t\left(h\right)=\frac{0.75\pi h^3}{60000\pi t}+\frac{C}{60000\pi t}\\ t\left(h\right)=\frac{h^3}{80000}+\frac{C}{60000\pi t}\\ \text{Reqired time}=t\left(100\right)-t\left(60\right)\\ \text{Reqired time}=\frac{{100}^3}{80000}-\frac{{60}^3}{80000}\\ \text{Reqired time}=12.5-2.7=9.8hours\end{array}$

b.

To determine

How fast is the floor area of the pile growing at the height

Answer to Problem 11P

The floor area left is $31900\pi sqft$

The floor area of pile increasing at the rate of $2000f{t}^2/hour$

Explanation of Solution

Given information : The question is

The ore resembles pink talcum powder and is poured from a conveyor at the top of the silo. The silo is a cylinder $100ft$ high with a radius $200ft$ . The conveyor carries $60,000\pi f{t}^3/h$ and the ore maintains a conical shape whose radius is

  $1.5$ times its height. When the pile is $60ft$ high

Calculation :

When the pile is $60ft$ high, the radius of pile is $1.5\times 60=90ft$

Since the radius of silo is $200ft$

The floor area left $=\pi {\left(\text{radius of silo}\right)}^2-\pi {\left(\text{radius of the pile}\right)}^2$

The floor area left $=\pi {\left(200\right)}^2-\pi {\left(90\right)}^2=3190sq.ft$

Area of the pile

  $\begin{array}{l}A=\pi r^2\\ r=1.5h\\ A=\pi {\left(1.5h\right)}^2\\ A=2.25\pi h^2\end{array}$

Differentiating both sides with respect to $t$

  $\begin{array}{l}\frac{dA}{dt}=2.25\pi \frac{d\left(h^2\right)}{dt}\\ \frac{dA}{dt}=2.25\pi \frac{d\left(h^2\right)}{dt}\cdot \frac{dh}{dt}\\ \frac{dA}{dt}=4.5\pi h\cdot \frac{dh}{dt}\end{array}$

The volume of the cone

  $\begin{array}{l}V=\frac{1}{3}\pi r^{2}h\\ r=1.5h\\ V=\frac{1}{3}\pi {\left(1.5h\right)}^{2}h\\ V=\frac{1}{3}\pi \left(2.25h^2\right)h\\ V=0.75\pi h^3\\ \frac{dV}{dt}=2.25\pi \frac{d\left(h^3\right)}{dt}\\ 60000\pi =0.75\pi \frac{d\left(h^3\right)}{dt}\cdot \frac{dh}{dt}\\ 60000\pi =0.75\pi \cdot 3h^2\cdot \frac{dh}{dt}\\ 60000\pi =2.25\pi h^2\cdot \frac{dh}{dt}\\ \frac{60000}{2.25h^2}=\frac{dh}{dt}\\ \frac{dA}{dt}=4.5\pi h\cdot \frac{60000}{2.25h^2}\\ \frac{dA}{dt}=\frac{120,000\pi }{h}\\ h=60\\ \frac{dA}{dt}=\frac{120,000\pi }{60}\\ \frac{dA}{dt}=2000\pi f{t}^2/hour\end{array}$

c.

To determine

How long will it take for pile to reach at the top of the silo under these condition

Answer to Problem 11P

Final answer

Explanation of Solution

Given information : The question is

A loader start removing the ore at the rate of $20,000\pi f{t}^3/h$ when the height of the pile reaches $90ft$

Calculation :

The volume of the conical shape

  $\begin{array}{l}V=\frac{1}{3}\pi r^{2}h\\ r=1.5h\\ V=\frac{1}{3}\pi {\left(1.5h\right)}^{2}h\\ V=\frac{1}{3}\pi \left(2.25h^2\right)h\\ V=0.75\pi h^3\\ \frac{dV}{dt}=0.75\pi \frac{d\left(h^3\right)}{dt}\\ 40000\pi =0.75\pi \frac{d\left(h^3\right)}{dt}\cdot \frac{dh}{dt}\\ 40000\pi =0.75\pi \cdot 3h^2\cdot \frac{dh}{dt}\\ 40000\pi =2.25\pi h^2\cdot \frac{dh}{dt}\\ {\displaystyle \int 40000\pi dt={\displaystyle \int 2.25\pi h^{2}dh}}\\ 40000\pi t=2.25\pi \frac{h^{2+1}}{2+1}+C\\ 40000\pi t=0.75\pi h^3+C\\ t\left(h\right)=\frac{0.75\pi h^3}{40000\pi t}+\frac{C}{40000\pi t}\\ t\left(h\right)=\frac{3h^3}{160,000}+\frac{C}{40000\pi t}\\ \text{Reqired time}=t\left(100\right)-t\left(90\right)\\ \text{Reqired time}=\frac{3{\left(100\right)}^3}{160,000}-\frac{3{\left(90\right)}^3}{40000\pi }\\ \text{Reqired time}\approx \text{5}\text{.08hours}\end{array}$