EBK MANUFACTURING ENGINEERING & TECHNOL
7th Edition
ISBN: 9780100793439
Author: KALPAKJIAN
Publisher: YUZU
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Chapter 4, Problem 4RQ
Describe the difference between a single-phase and a two-phase system.
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Chapter 4 Solutions
EBK MANUFACTURING ENGINEERING & TECHNOL
Ch. 4 - Describe the difference between a solute and a...Ch. 4 - What is a solid solution?Ch. 4 - Prob. 3RQCh. 4 - Describe the difference between a single-phase and...Ch. 4 - What is an induction heater? What kind of part...Ch. 4 - Describe the major features of a phase diagram.Ch. 4 - What do the terms equilibrium and constitutional,...Ch. 4 - Prob. 8RQCh. 4 - What is tempering? Why is it performed?Ch. 4 - Explain what is meant by severity of quenching.
Ch. 4 - What are precipitates? Why are they significant in...Ch. 4 - Prob. 12RQCh. 4 - Prob. 13RQCh. 4 - Prob. 14RQCh. 4 - Prob. 15RQCh. 4 - You may have seen some technical literature on...Ch. 4 - Prob. 17QLPCh. 4 - What is the difference between hardness and...Ch. 4 - Prob. 19QLPCh. 4 - Prob. 20QLPCh. 4 - Prob. 21QLPCh. 4 - Describe the characteristics of (a) an alloy, (b)...Ch. 4 - Explain why carbon, among all elements, is so...Ch. 4 - Prob. 24QLPCh. 4 - In Section 4.8.2, several fluids are listed in...Ch. 4 - Why is it important to know the characteristics of...Ch. 4 - Explain why, in the abscissa of Fig. 4.16c, the...Ch. 4 - Prob. 28QLPCh. 4 - Prob. 29QLPCh. 4 - Prob. 30QLPCh. 4 - Design a heat-treating cycle for carbon steel,...Ch. 4 - Using Fig. 4.4, estimate the following quantities...Ch. 4 - Prob. 33QTPCh. 4 - Prob. 34QTPCh. 4 - Prob. 35SDPCh. 4 - Figure 4.18b shows hardness distributions in...Ch. 4 - Throughout this chapter, you have seen specific...Ch. 4 - Refer to Fig. 4.24, and think of a variety of...Ch. 4 - Inspect various parts in your car or home, and...
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- Four-bar linkage mechanism, AB=40mm, BC=60mm, CD=70mm, AD=80mm, =60°, w1=10rad/s. Determine the direction and magnitude of w3 using relative motion graphical method. A B 2 3 77777 477777arrow_forwardFour-bar linkage mechanism, AB=40mm, BC=60mm, CD=70mm, AD=80mm, =60°, w1=10rad/s. Determine the direction and magnitude of w3 using relative motion graphical method. A B 2 3 77777 477777arrow_forwardThe evaporator of a vapor compression refrigeration cycle utilizing R-123 as the refrigerant isbeing used to chill water. The evaporator is a shell and tube heat exchanger with the water flowingthrough the tubes. The water enters the heat exchanger at a temperature of 54°F. The approachtemperature difference of the evaporator is 3°R. The evaporating pressure of the refrigeration cycleis 4.8 psia and the condensing pressure is 75 psia. The refrigerant is flowing through the cycle witha flow rate of 18,000 lbm/hr. The R-123 leaves the evaporator as a saturated vapor and leaves thecondenser as a saturated liquid. Determine the following:a. The outlet temperature of the chilled waterb. The volumetric flow rate of the chilled water (gpm)c. The UA product of the evaporator (Btu/h-°F)d. The heat transfer rate between the refrigerant and the water (tons)arrow_forward
- (Read image) (Answer given)arrow_forwardProblem (17): water flowing in an open channel of a rectangular cross-section with width (b) transitions from a mild slope to a steep slope (i.e., from subcritical to supercritical flow) with normal water depths of (y₁) and (y2), respectively. Given the values of y₁ [m], y₂ [m], and b [m], calculate the discharge in the channel (Q) in [Lit/s]. Givens: y1 = 4.112 m y2 = 0.387 m b = 0.942 m Answers: ( 1 ) 1880.186 lit/s ( 2 ) 4042.945 lit/s ( 3 ) 2553.11 lit/s ( 4 ) 3130.448 lit/sarrow_forwardProblem (14): A pump is being used to lift water from an underground tank through a pipe of diameter (d) at discharge (Q). The total head loss until the pump entrance can be calculated as (h₁ = K[V²/2g]), h where (V) is the flow velocity in the pipe. The elevation difference between the pump and tank surface is (h). Given the values of h [cm], d [cm], and K [-], calculate the maximum discharge Q [Lit/s] beyond which cavitation would take place at the pump entrance. Assume Turbulent flow conditions. Givens: h = 120.31 cm d = 14.455 cm K = 8.976 Q Answers: (1) 94.917 lit/s (2) 49.048 lit/s ( 3 ) 80.722 lit/s 68.588 lit/s 4arrow_forward
- Problem (13): A pump is being used to lift water from the bottom tank to the top tank in a galvanized iron pipe at a discharge (Q). The length and diameter of the pipe section from the bottom tank to the pump are (L₁) and (d₁), respectively. The length and diameter of the pipe section from the pump to the top tank are (L2) and (d2), respectively. Given the values of Q [L/s], L₁ [m], d₁ [m], L₂ [m], d₂ [m], calculate total head loss due to friction (i.e., major loss) in the pipe (hmajor-loss) in [cm]. Givens: L₁,d₁ Pump L₂,d2 오 0.533 lit/s L1 = 6920.729 m d1 = 1.065 m L2 = 70.946 m d2 0.072 m Answers: (1) 3.069 cm (2) 3.914 cm ( 3 ) 2.519 cm ( 4 ) 1.855 cm TABLE 8.1 Equivalent Roughness for New Pipes Pipe Riveted steel Concrete Wood stave Cast iron Galvanized iron Equivalent Roughness, & Feet Millimeters 0.003-0.03 0.9-9.0 0.001-0.01 0.3-3.0 0.0006-0.003 0.18-0.9 0.00085 0.26 0.0005 0.15 0.045 0.000005 0.0015 0.0 (smooth) 0.0 (smooth) Commercial steel or wrought iron 0.00015 Drawn…arrow_forwardThe flow rate is 12.275 Liters/s and the diameter is 6.266 cm.arrow_forwardAn experimental setup is being built to study the flow in a large water main (i.e., a large pipe). The water main is expected to convey a discharge (Qp). The experimental tube will be built at a length scale of 1/20 of the actual water main. After building the experimental setup, the pressure drop per unit length in the model tube (APm/Lm) is measured. Problem (20): Given the value of APm/Lm [kPa/m], and assuming pressure coefficient similitude, calculate the drop in the pressure per unit length of the water main (APP/Lp) in [Pa/m]. Givens: AP M/L m = 590.637 kPa/m meen Answers: ( 1 ) 59.369 Pa/m ( 2 ) 73.83 Pa/m (3) 95.443 Pa/m ( 4 ) 44.444 Pa/m *******arrow_forward
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