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Fluid Mechanics 8 Chapter 6
forwarded, distributed, or posted on a website, in whole or part. Solution 6.149 For air take =1.2 kg/m3 and = 0.000015 kg/ms. With no elevation change and negligible tank velocity, the energy equation would yield 2 tank 1 , 0.5 and 0.7 2 atm entrance nozzle ent noz VL p p f K K K K D − = + + + Since p is given, we can use this expression plus the Moody chart to predict V and Q = AV and compare with the flow-nozzle measurements. The flow nozzle formula is:
Fluid Mechanics 8 Chapter 6
forwarded, distributed, or posted on a website, in whole or part. Solution 6.38 Just as in Prob. 6.37, apply the log-law at the center between the wall, that is, y = h/2, u = V/2. With w known, we can evaluate u* immediately: 15 /2 1 * /2 * 0.123 , ln , 998 * wm V u h uB su = = = + /2 1 0.123(0.03/2) or: ln 5.0 23.3, . 0.123 0.41 0.001/998 VSolve for Ans = + = m V 5.72 s
Fluid Mechanics 8 Chapter 9
forwarded, distributed, or posted on a website, in whole or part. Problem 9.59 Air, at stagnation conditions of 450 K and 250 kPa, flows through a nozzle. At section 1, where the area = 15 cm2, there is a normal shock wave. If the mass flow is 0.4 kgs, estimate (a) the Mach number; and (b) the stagnation pressure just downstream of the shock.
Basics of Engineering Economy 2 Chapter 6
forwarded, distributed, or posted on a website, in whole or part. Solution 6.46 (a) Two sign changes; maximum number of i* values is two (b) Cumulative Year Net Cash Flow, $ Cash Flow, $ 0 -40,000 -40,000 1 32,000 -8000 2 18,000 +10,000 3 -2000 +8000 4 -1000 +7000 There is one sign change in the cumulative cash flow series; only one positive i* value is indicated
Basics of Engineering Economy 2 Chapter 6
forwarded, distributed, or posted on a website, in whole or part. Solution 6.41 (a) Initial cost P1, Machine 1: -60,000 - (P1) = -16,000; P1 = $-44,000 Overall ROR, Machine 2: 0 = -60,000 + 16,000(P/A,i*,10); i* = 23.41% Incremental investment, 3 to 2: -72,000 - (-60,000) = -$12,000 Incremental cash flow: 4 to 3: 24,000 – 19,000 = $5000 Incremental ROR 3 to 2: 0 -12,000 + 3000(P/A, ∆i*,10); ∆i* = 21.41% Incremental ROR 4 to 3: 0 = -26,000 + 5000(P/A, ∆i*,10); ∆i*= 14.08% (b) Machines are ranked according to initial investment: 1, 2, 3, 4; MARR = 18%
Management of Organizational Behavior 10 Chapter 7
Chapter 7 Case Study Part 4: Randy Mack, Trainer, Listo Systems Randy Mack is an experienced trainer for Listo Systems. With several years on the job, Randy has been an example and mentor for many of the newer trainers. His functional skills and evaluations have historically been first rate, and he’s been able to develop rapport with the people whom he trains, regardless of what the training content may be. Recently, however, Randy has not seemed happy with his job and has mentioned a lack of career advancement. Some of his training sessions have ended in confrontations with the trainees, and
Management of Organizational Behavior 10 Chapter 7
Chapter 7 Case Study Part 5: Lashonda McCoy, Office Manager, Listo Systems For six years, Lashonda McCoy has been an office manager with Listo Systems, and has experienced the ups and downs of the growing company. A longtime valued employee of Listo Systems, Lashonda has been assigned the task of running the new customer satisfaction survey program by the incoming management team. Although she doesn’t completely understand the new program, Lashonda is excited about the way it has been structured by management, and believes that it will improve relations with their customers. She doesn’t yet know how all the parts of