{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

HW 6 - Homework#6 ENED 1091 HW#6 Due at noon Problem 1 The...

Info icon This preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Homework #6 ENED 1091 HW#6 Due March 17, 2016 at noon Problem 1 : The figure below shows measurements of the flow rate of fluid into a tank taken every 30 seconds. The volume of fluid in the tank is simply the initial volume plus the integral of the flow rate (assuming no outflow from the tank): (a) Assume the initial volume is 0 gallons. Using the data points given in the graph, estimate the volume at 120 seconds using the trapezoidal rule. Work: t = [0 30 60 90 120 150 180]; flow_rate = [0 0.5 0.7 0.3 0.5 0.5 0.3]; for  a = 1:length(t)-3     integ(a) = (flow_rate(a+1)+flow_rate(a))*((t(a+1)-t(a))/2);     integ = sum(integ); end   fprintf( 'The volume is: %i \n' ,integ) 1
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Homework #6 Answer: 52.50 gallons (b) Repeat part (a) using Simpson’s Rule. Work: integ = 0; t = [0 30 60 90 120 150 180]; flow_rate = [0 0.5 0.7 0.3 0.5 0.5 0.3]; for  c=1:2:length(t)/2     integ = integ+((flow_rate(c)+(4*flow_rate(c+1))+flow_rate(c+2))*(t(c+2)- t(c))/6); end fprintf( 'The volume is: %i \n' ,integ) Answer: 51.00 gallons (c) Assume the initial volume is 400 gallons . Using the given data points, estimate the volume at 180 seconds using the trapezoidal rule. Work: t = [0 30 60 90 120 150 180]; flow_rate = [0 0.5 0.7 0.3 0.5 0.5 0.3]; for  a = 1:length(t)-1     integ(a) = (flow_rate(a+1)+flow_rate(a))*((t(a+1)-t(a))/2); end   integd = 400+sum(integ); fprintf( 'The volume is: %0.2f \n' ,integd) Answer: 479.50 gallons (d) Repeat part (c), again assuming an initial volume of 400 gallons , using Simpson’s Rule. Work: Estimation = 400 + changet/3 ( f(x0) + 4f(x1) + 2f(x2) + 4f(x3) + 2f(x4) + … + f(xn) ) = 400 + 30/3 ( f(0) + 4f(30) + 2f(60) + 4f(90) + 2f(120) + 4f(150) + f(180) ) = 400 + 10 (0 + 4*.5 + 2*.7 + 2*.3 + 4*.5 + 2*.5 + .3) = 400 + 10 (2 + 1.4 + .6 + 2 + 1 + .3) = 400 + 10 (7.3) = 400 + 73 = 473 Answer: 473.00 gallons Problem 2 : Consider liquid flowing into a cylindrical tank as shown in the diagram below. 2
Image of page 2
Liquid Level Height of Tank: 10 ft. Radius of Tank: 2 ft. Homework #6 (a) Calculate the volume of the tank in cubic ft. Work: 2^2*pi*10 = 125.66 Answer: 125.66 ft^3 (b) Calculate the capacity (or volume) in gallons Work: 125.66*7.47 = 940.03 Answer: 940.03 gallons (c) Assuming 400 gallons of liquid in the tank, calculate the liquid level in ft.
Image of page 3

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern