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# ENGR 232 Dynamic Engineering System Homework 3 ENGR 232

Drexel

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This 11 page Study Guide was uploaded by Haikal Fouzi on Sunday January 24, 2016. The Study Guide belongs to ENGR 232 at Drexel University taught by Mr. Michael Ryan in Spring 2015. Since its upload, it has received 61 views. For similar materials see Dynamic Engineering Systems in Engineering and Tech at Drexel University.

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Date Created: 01/24/16

Drexel University, College of Engineering 2014-2015 Academic Year Drexel University Office of the Dean of the College of Engineering ENGR 232 – Dynamic Engineering Systems Homework 3 Further details on how to solve these problems could be obtained from Chp. 7 of Schaum's outlines- Differential Equations by R. Bronson and G. Costa. Questions (1) An RC circuit has an emf given (in volts) by400 cos 2t, a resistance of 100 ohms, and a capacitance of -2 10 farad. Initially there is no charge on the capacitor. Find the current in the circuit at any time t. (Hint: the equation governing the amount of electrical discharge q (in coulombs) on the capacitor which is becomes This equation is linear, and its solution is (two integrations by parts are r)quired) (2) A tank initially holds 100 gal of a brine solution containing 1 Ib of salt. At t = 0 another brine solution containing 1 Ib of salt per gallon is poured into the tank at the rate of 3 gal/min, while the well-stirred mixture leaves the tank at the same rate. Find (a) the amount of salt in the tank at any time t and (b) the time at which the mixture in the tank contains 2 Ib of salt. Hint: or Drexel University, College of Engineering 2014-2015 Academic Year Here VQ = 100, a = 1, b = 1, and e =/= 3; hence the rate of change equation becomes (3) A body weighing 64 Ibis dropped from a height of 100 ft with an initial velocity of 10 ft/sec. Assume that the air resistance is proportional to the velocity of the body. If the limiting velocity is known to be 128 ft/sec, find (a) an expression for the velocity of the body at any time t and (b) an expression for the position of the body at any time t. Hint: (4) A body at an unknown temperature is placed in a room which is held at a constant temperature of 30° F. If after 10 minutes the temperature of the body is 0° F and af ter 20 minutes the temperature of the body is 15° F, find the unknown initial temperature. Hint: (5) An RL circuit has an emf of 5 volts, a resistance of 50 ohms, an inductance of 1 henry, and no initial current. Find the current in the circuit at any time t. Hint: Here E = 5, R = 50, and L = 1; hence the equation becomes (6) A 50-gal tank initially contains 10 gal of fresh water. At t = 0, a brine solution containing 1 Ib of salt per gallon is poured into the tank at the rate of 4 gal/min, while the well-stirred mixture leaves the tank at the rate of 2 gal/min. Find (a) the amount of time required for overflow to occur and (b) the amount of salt in the tank at the moment of overflow. Hint: Drexel University, College of Engineering 2014-2015 Academic Year (7) A bacteria culture is known to grow at a rate proportional to the amount present. After one hour, 1000 strands of the bacteria are observed in the culture; and after four hours, 3000 strands. Find (a) an expression for the approximate number of strands of the bacteria present in the culture at any time t and (b) the approximate number of strands of the bacteria originally in the culture. Further Problems on Separation of variables, implicit/explicit solutions (8) A simple first order ODE with an independent variable x is given as: y'= y Show by method of separation of variable that a solution y = Ae x c is obtainable with A = e where c is an arbitrary constant. (Hint: Your solution step should not be more than 4 or 5 lines) (9) Solve following problems using method of separation of variables. Is the a ppropriate solution that you selected implicit or explicit? 2 2 ' x ' 1▯ 2t ▯▯ x 2 (i) y = (ii) i = (iii)▯▯ = 2 (iv) y' = y ln (ty) y 2i 1 ▯ Drexel University, College of Engineering 2014-2015 Academic Year Drexel University Office of the Dean of the College of Engineering ENGR 232 – Dynamic Engineering Systems Homework 3 Solutions (1) We first find the charge q and then use I = dq/dt to obtain the current. Here, E = 400cos2t, R= 100, and C= 10 ;2 hence the equation governing the amount of electrical discharge q (in coulombs) on the capacitor which is becomes This equation is linear, and its solution is (two integrations by parts are required) At t = 0, q = 0; hence, Thus Using I = dq/dt, we obtain Drexel University, College of Engineering 2014-2015 Academic Year (2) or Here VQ = 100, a = 1, b = 1, and e =/= 3; hence the rate of change equation becomes The solution to this linear differential equation is At t = 0, Q = a=l. Substituting these values intoQ we find 1 = ce° + 100, or c = -99. Then Q can be rewritten a (b) We require t when 2 = 2. Substituting 2 = 2 into we obtain from which Drexel University, College of Engineering 2014-2015 Academic Year (3) Locate the coordinate system as in Fig a below. Here w = 64 Ib. Since w = mg, it follows that mg = 64, or m = 2 slugs. Given that v: = 128 ft/sec, it follows from V= mg/k that 128 = 641k, or k = j. Substituting these values into y' + p(x)y = q(x)y , we obtain the linear differential equation Fig a which has the solution At t = 0, we are given that v = 10. Substituting these values into (_/), we have 10 = ce° + 128, or c = -118. The velocity at any time t is given by (b) Since v = dx/dt, where x is displacement, can be rewritten as This last equation, in differential form, is separable; its solution is At t = 0, we have x = 0 (see Fig. 7-5). Thus, x gives Drexel University, College of Engineering 2014-2015 Academic Year The displacement at any time t is then given by (4) From Newton's law of cooling, the time rate of change of the temperature of a body is proportional to the temperature difference between the body and its surrounding medium. It could be stated as Solving, we obtain At t= 10, we are given that T=0, which implies that At t= 20, we are given that T= 15, solving for k and c, we obtain, Substituting these values into equation for T , we have for the temperature of the body at any time t Since we require T at the initial time t = 0, it follows from the obtained T equation that Drexel University, College of Engineering 2014-2015 Academic Year (5) Here E = 5, R = 50, and L = 1. The basic equation governing the amount of current I (in amperes) in a simple KL circuit consisting of a resistance R (in ohms), an inductor L (in henries), and an electromotive force (abbreviated emf) E (in volts) is Here E = 5, R = 50, and L = 1; hence the equation becomes This equation is linear; its solution is Drexel University, College of Engineering 2014-2015 Academic Year (6) (a) Here a = 0, b = 1, e = 4,f= 2, and VQ = 10. The volume of brine in the tank at any time t is given as V0 + et -ft = 10 + 2t. We require t when 10 + 2t = 50; hence, t = 20 min. (b) or For this problem, the differential equation becomes This is a linear equation; its solution is given as At t = 0, Q = a = 0. Substituting these values into (1), we findthat c = 0. We require Q at the moment of overflow, which from part (a) is t = 20. Thus, Drexel University, College of Engineering 2014-2015 Academic Year (7) Let N(t) denote the number of bacteria strands in the culture at time.tFrom y' + p(x)y = q(x) dN/dt — kN = 0, which is both linear and separable. Itssolution is At=1, N= 1000; hence At=3, N= 3000; hence solving for k and c, we find Substituting into we obtain as an expression for the amount of the bacteria present at any time t. (b) We require N at t = 0. Substituting t = 0 into (4), we obtain N(0) = 694e(0 366)(0) = 694 (8) ▯▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ e▯ c ▯▯ ▯ x ▯▯▯▯▯▯▯▯ (9) 2 2 2 3 3 x ▯▯ x 2 y x x (i) y= ▯▯▯ ▯▯ = ▯▯▯▯ ▯ ▯▯ ▯▯▯▯▯x ▯▯▯▯▯▯▯▯▯▯▯▯▯ = ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯▯▯▯ y y 2 3 3 Explicit Drexel University, College of Engineering 2014-2015 Academic Year 1▯ 2t ▯▯ 1▯ 2t 2i2 t ▯2t 2 (ii) i = ▯▯▯ = ▯▯▯▯2i di =▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ = ▯ i = ▯▯▯▯▯ 2+c ▯▯ 2i ▯▯ 2i 2 2 2 ▯▯▯▯▯ ▯▯▯▯▯+c Explicit ▯▯ x 2 (iii) = ▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯ dx ▯▯ ▯▯▯▯ ▯▯▯▯▯▯▯▯x ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ -x = c. ▯▯ 1▯ y 2 Implicit. (iv) None Separable.

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