Solution Manual Ch 15
Problem 15.1 In Active Example 15.1, what is the velocity of the container when it has reached the position s = 2 m? A s Solution: The 180-kg container A starts from rest at s = 0. The horizontal force (in newtons) is F = 700 − 150s.
ANSWERS TO END-OF-CHAPTER QUESTIONS. Assets-in-place, also. Instructor's resource CD-ROM (in the file “Solution for FM11 Ch 15-11 Build a. Access 11th Edition Chapter 15.Q solutions now. Our solutions are written by Chegg experts so you can be assured of the highest. Home / study / solutions manuals. (11th Edition) View more editions. Solutions for Chapter 15.Q. P, CH15.Q, CH15.1, CH15.2, CH15.3, CH15.4, CH15.5, CH15.6, CH15.7, CH16. Chapter 15 Solutions, Page 1/20. 15-1 Given: Uncrowned. For the pinion, CH = 1. 15-1, CR = 1.118. Thus, from Eq.
The coefficient of kinetic friction is µk = 0.26. U12 = T2 − T1 2 (700 − 150s − 0.26180(9.81))d s = 0 700(2) − 1 (180 kg)v22 − 0 2 1 (150)(2)2 − (0.26180(9.81)2) = 90v22 2 v2 = 1.42 m/s. Problem 15.2 The mass of the Sikorsky UH-60A helicopter is 9300 kg.
It takes off vertically with its rotor exerting a constant upward thrust of 112 kN. Use the principle of work and energy to determine how far it has risen when its velocity is 6 m/s. Strategy: Be sure to draw the free-body diagram of the helicopter. Solution: U12 = T2 − T1 (112000 − 93009.81)Nh = 1 (9300 kg)(6 m/s)2 2 h = 8.06 m. C 2008 Pearson Education South Asia Pte Ltd.
Solution Manual Ch 15 Madison
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Please purchase PDF Split-Merge on www.verypdf.com to remove this watermark. 179 Problem 15.3 The 20 -N box is at rest on the horizontal surface when the constant force F = 5 N is applied. The coefficient of kinetic friction between the box and the surface is µk = 0.2. Determine how fast the box is moving when it has moved 2 m from its initial position (a) by applying Newton’s second law; (b) by applying the principle of work and energy.
15-1 Chapter 15 15.1 For Ra=d/u 2u d u R K ap −=∂ ∂= which can vary more than Kp in Eq. 15-2, because the new Kp depends on both d and u. 15.2 By definition, the ratio station sets (um – um0) = KR (dm – dm0) Thus 2 1 2 2 1 2 2 0 0 − −= d u K K dK uK dd uu K mm mm R (1) For constant gain KR, the values of u and d in Eq. 1 are taken to be at the desired steady state so that u/d=Rd, the desired ratio. Moreover, the transmitter gains are 21 mA)420( dS K −=, 22 mA)420( uS K −= Substituting for K1, K2 and u/d into (1) gives: 2 2 2 2 u d dd d u R S SRR S SK Solution Manual for Process Dynamics and Control, 2nd edition, Copyright © 2004 by Dale E.
Solution Manual Ch 15 Phoenix
Seborg, Thomas F. Edgar and Duncan A. Mellichamp Revised: 1-3-04 15-2 15.3 (a) The block diagram is the same as in Fig.
Solution Manual Ch 15 Az
15.11 where Y ≡ H2, Ym ≡ H2m, Ysp ≡ H2sp, D ≡ Q1, Dm ≡ Q1m, and U ≡ Q3. B) (A steady-state mass balance on both tanks gives 0 = q1 – q3 or Q1 = Q3 (in deviation variables) (1) From the block diagram, at steady state: Q3 = Kv Kf Kt Q1 From (1) and (2), Kf = 1 v tK K (2) c) (No, because Eq. 1 above does not involve q2. 15.4 (b) From the block diagram, exact feedforward compensation for Q1 would result when Q1 + Q2 = 0 Substituting Q2 = KV Gf Kt Q1, Gf = − 1 v tK K 15-3 (c) Same as part (b), because the feedforward loop does not have any dynamic elements.