Solve the problem.
-A vertical right circular cylindrical tank measures $22 \mathrm { ft }$ high and $14 \mathrm { ft }$ in diameter. It is full of oil weighing $60 \mathrm { lb } / \mathrm { ft } ^ { 3 }$. How long will it take a (1/2)-horsepower (hp) motor (work output $275 \mathrm { ft } \cdot \mathrm { lb } / \mathrm { sec }$ ) to pump the oil to the level of the top of the tank? Give your answer to the nearest minute.
A) 43 min
B) 135 min
C) 271 min
D) 8128 min

Question

Quizplus · Accepted Answer

First, we need to find the volume of the tank:

Radius, $r = \frac{14}{2} = 7\mathrm{ft}$

Volume, $V = \pi r^2 h = \pi (7^2)(22) = 3388 \mathrm{ft}^3$

Next, we need to find the weight of the oil in the tank:

Weight of oil, $W = V \cdot \rho = (3388)(60) = 203,280 \mathrm{lb}$

To pump all the oil to the top of the tank, we need to do work against gravity:

Work, $W = Fd = mgh$

where F is the force needed to lift the oil, d is the distance it needs to be lifted, m is the mass of the oil, g is the acceleration due to gravity, and h is the height of the tank.

Solving for F:

$F = mg = 203,280 \cdot 32.2 = 6,549,096 \mathrm{lb}$

Now we can use the work output of the motor to find the time it takes to pump the oil to the top:

Work output, $P = \frac{dW}{dt} = 275$

Power is defined as the rate at which work is done, so we can rearrange the equation to solve for time:

$t = \frac{dW}{P} = \frac{Fd}{P} = \frac{203,280\cdot 32.2 \cdot 22}{2\cdot 275} \approx 135 \mathrm{min}$

Therefore, the answer is B.

Solve the Problem 22ft22 \mathrm { ft }22ft High And 14ft14 \mathrm { ft }14ft

Solve the Problem $22 \mathrm { ft }$ High And $14 \mathrm { ft }$