midterm_09_solution

# Midterm_09_solution - 650:460 Aerodynamics Solution to Midterm Exam Prof Doyle Knight Tel 732 445 4464 Email [email protected]/* <![CDATA[ */!function(t,e,r,n,c,a,p){try{t=document.currentScript||function(){for(t=document.getElementsByTagName('script'),e=t.length;e--;)if(t[e].getAttribute('data-cfhash'))return t[e]}();if(t&&(c=t.previousSibling)){p=t.parentNode;if(a=c.getAttribute('data-cfemail')){for(e='',r='0x'+a.substr(0,2)|0,n=2;a.length-n;n+=2)e+='%'+('0'+('0x'+a.substr(n,2)^r).toString(16)).slice(-2);p.replaceChild(document.createTextNode(decodeURIComponent(e)),c)}p.removeChild(t)}}catch(u){}}()/* ]]> */ Office hours

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Unformatted text preview: 650:460 Aerodynamics Solution to Midterm Exam Prof. Doyle Knight Tel: 732 445 4464, Email: [email protected] Office hours: Tues and Thur, 4:30 pm - 6:00 pm and by appointment Fall 2009 1 Problem No. 1 Consider the superposition of uniform flow with velocity U ∞ in the x- direction, a source of strength K (where K > 0) located at ( x , y ) = ( a , 0), and a sink of strength- K located at ( x , y ) = (- a , 0). Find the location of all stagnation points, and the conditions (if any) under which they exist. Solution The velocity potential is φ = U ∞ x- K 2 π log radicalBig ( x + a ) 2 + y 2 + K 2 π log radicalBig ( x- a ) 2 + y 2 where the second term is the sink at ( x , y ) = (- a , 0) and the third term is the source at ( x , y ) = ( a , 0). 2 Problem No. 1 At the stagnation points both u = 0 and v = 0. The x- component of velocity is ∂φ ∂ x = U ∞- K 2 π braceleftbigg ( x + a ) ( x + a ) 2 + y 2- ( x- a ) ( x- a ) 2 + y 2 bracerightbigg The y- component of velocity is...
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## This note was uploaded on 12/15/2009 for the course 650 460 taught by Professor Knight during the Spring '09 term at Rutgers.

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Midterm_09_solution - 650:460 Aerodynamics Solution to Midterm Exam Prof Doyle Knight Tel 732 445 4464 Email [email protected] Office hours

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