Question

Oil Flow Oil flows through a cylindrical pipe of radius 3 inches, but friction from the pipe slows the flow toward the outer edge. The speed at which the oil flows at a distance \(r\) inches from the center is 8\(\left(10-r^{2}\right)\) inches per second. (a) In a plane cross section of the pipe, a thin ring with thickness \(\Delta r\) at a distance \(r\) inches from the center approximates a rectangular strip when you straighten it out. What is the area of the strip (and hence the approximate area of the ring)? (b) Explain why we know that oil passes through this ring at approximately 8\(\left(10-r^{2}\right)(2 \pi r) \Delta r\) cubic inches per second. (c) Set up and evaluate a definite integral that will give the rate (in cubic inches per second) at which oil is flowing through the pipe.

Short answer

The rate at which oil is flowing through the pipe is \(1230π\) cubic inches per second.

Step-by-step solution

Compute the area of the ring

The area of the rectangular strip, which consists of the circumference of the ring (2πr) multiplied by the thickness Δr, gives the approximate area of this ring. Using the formula for the area of a rectangle (length × width), we can find the area \(A = 2πr Δr\).

Compute the oil flow through the ring

The speed at which oil flows through this ring (or through the approximated rectangular strip) is given as \(8(10-r^{2})\). Hence, the volume flow rate (volume/time) would be the speed times the cross-section area of flow (A), which we calculated in Step 1. So, the volume flow rate \(V_{rate} = 8(10-r^{2}) 2πr Δr\) cubic inches per second.

Setup the integral for the whole pipe

The total volume flow rate through the entire pipe would be the sum of the flow rate through all such rings. Lets model radius 'r' from 0 to 3 (the actual radius) and sum the volumes. We do this using the integration which is a sophisticated form of summing. Hence, the total volume flow rate \(V_{total}= \int_{0}^{3} 8(10-r^{2}) 2πr dr \)

Evaluate the integral

Solving this integral would give us the total volume flow rate. \(V_{total}= \int_{0}^{3} 16πr(10 - r^{2}) dr = 16π[(10/2)r^{2} - (1/4)r^{4}]_{0}^{3} = 16π [(5*9) - (81/4)] = 16π (45 - 20.25) = 16π (24.75) = 1230π cubic inches per second.