Shehezaada Posted November 24, 2007 Share Posted November 24, 2007 calling all Engineers here! Have this question due on Wednesday and I'm realllly confused! A cone shaped paper drinking cup (like the kind used at water fountains) has a radius R and a height H. If the water heigh in the cup is h, the water volume is given by: V=(1/3)pi((R/H^2)(h^3). Suppose that the cup's dimensions are R= 3.8 and H=10 cm. a)If the flow rate from the fountain into the cup is 33cm^3cm/sec, how long will it take to fill to the brim. Do by hand, and by MATLAB. Use MATLAB to determine how long it would take if the flow rate was q= 2(1 - e^-2t)cm^3/s Link to comment Share on other sites More sharing options...
Shehezaada Posted November 24, 2007 Author Share Posted November 24, 2007 here's my start to the problem as per ENG methods: Flow Rate in - Flow Rate Out = Flow Stored q = A(dh/dt) where A is the area of the cone from there i'm lost......is q just equal to the volume equation given? Link to comment Share on other sites More sharing options...
Shehezaada Posted November 24, 2007 Author Share Posted November 24, 2007 i solved it using a different technique but i want to solve it using differential equations! I just found the Capacity of the of the cone and divided by the flow rate...it's 4.58 seconds Link to comment Share on other sites More sharing options...
vroomfondel Posted November 24, 2007 Share Posted November 24, 2007 maybe i'm misunderstanding the q, but you seem to have the diff. eq. already: qdt = dV = A*dh integrate from 0 to T on LHS, and 0 to H on RHS. for part a) q is just a constant, and A is dV/dh for part b) q is the formula given, which you integrate wrt time. Link to comment Share on other sites More sharing options...
Shehezaada Posted November 25, 2007 Author Share Posted November 25, 2007 maybe i'm misunderstanding the q, but you seem to have the diff. eq. already: qdt = dV = A*dh integrate from 0 to T on LHS, and 0 to H on RHS. for part a) q is just a constant, and A is dV/dh for part b) q is the formula given, which you integrate wrt time. thanks buddy! u an engineer? this has given me good steps in the right direction!! here's my work based on ur method for part A: qdt = dV 33dt = dV Integrating both sides 33t = V 33t = pi(r^2)(h)/3 (where I have taken the total volume of cone, and not the volume of flow rate given) 33t=453.65 t= 4.58 seconds that makes sense to me..does it to you? thank you so much vroom you've been a great help! Link to comment Share on other sites More sharing options...
Shehezaada Posted November 25, 2007 Author Share Posted November 25, 2007 for the second one i'm integrating 2(1 - e^-2t) that makes 2t - e^-2t = Volume of Cone from there i get e=75.xx because e^-2t is pretty negligble.. does this make sense? Link to comment Share on other sites More sharing options...
fineleg Posted November 25, 2007 Share Posted November 25, 2007 shucks! once upon a time, i used to know all this stuff :D Link to comment Share on other sites More sharing options...
vroomfondel Posted November 25, 2007 Share Posted November 25, 2007 for part b) you should have: int(2t-2t(e^-2t)) = V => (t^2) - (t(e^-2t)) + (e^(-2t))/2 = V (by integration by parts, or working backwards from the derivative) it may be negligible for large t, but may not be for small t (small volume or high rate of flow)...so I would calculate it out. Link to comment Share on other sites More sharing options...
vroomfondel Posted November 25, 2007 Share Posted November 25, 2007 for part b) you should have: int(2t-2t(e^-2t)) = V -> (t^2) - (t(e^-2t)) + (e^(-2t))/2 = V (by integration by parts, or working backwards from the derivative) it may be negligible for large t, but may not be for small t (small volume or high rate of flow)...so I would calculate it out. Link to comment Share on other sites More sharing options...
beetle Posted November 25, 2007 Share Posted November 25, 2007 Shehezaada Tupac?????:regular_smile: Link to comment Share on other sites More sharing options...
zubinpepsi Posted November 25, 2007 Share Posted November 25, 2007 Shehezaada Tupac?????:regular_smile: :haha: beetle, who are you? Link to comment Share on other sites More sharing options...
beetle Posted November 25, 2007 Share Posted November 25, 2007 :haha: beetle, who are you? beetle the big monsterous bug I don't post there...just visit on good days.....:D Link to comment Share on other sites More sharing options...
zubinpepsi Posted November 25, 2007 Share Posted November 25, 2007 beetle the big monsterous bug I don't post there...just visit on good days.....:D :haha: :hysterical::hysterical: me too... liked to see them discuss on things now... seem to have a toned down approach now.... all the beastily attitudes have gond down the drain... im enjoying it...:two_thumbs_up: Link to comment Share on other sites More sharing options...
beetle Posted November 25, 2007 Share Posted November 25, 2007 yeah...seems like a fairly normal place.... The likes of monsee and mr. F don't seem to get as much bhav...... The rabid anti India stance in cricket seems to have gone now...as has the obsession with Phast bowling and complete disrespect for Indian bowlers...(it is still there but no ahahahah comments at least...):haha: Link to comment Share on other sites More sharing options...
Recommended Posts