Degarmo's Materials And Processes In Manufacturing
13th Edition
ISBN: 9781119492825
Author: Black, J. Temple, Kohser, Ronald A., Author.
Publisher: Wiley,
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Chapter 28, Problem 34RQ
To determine
The inclined angle in honing and purpose of it.
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During some actual expansion and compression processes in piston–cylinder devices, the gases have been observed to satisfy the relationship PVn = C, where n and C are constants. Calculate the work done when a gas expands from 350 kPa and 0.03 m3 to a final volume of 0.2 m3 for the case of n = 1.5.
The work done in this case is kJ.
Carbon dioxide contained in a piston–cylinder device is compressed from 0.3 to 0.1 m3. During the process, the pressure and volume are related by P = aV–2, where a = 6 kPa·m6. Calculate the work done on carbon dioxide during this process.
The work done on carbon dioxide during this process is kJ.
The volume of 1 kg of helium in a piston–cylinder device is initially 5 m3. Now helium is compressed to 3 m3 while its pressure is maintained constant at 130 kPa. Determine the initial and final temperatures of helium as well as the work required to compress it, in kJ. The gas constant of helium is R = 2.0769 kJ/kg·K.
The initial temperature of helium is K.
The final temperature of helium is K.
The work required to compress helium is kJ.
Chapter 28 Solutions
Degarmo's Materials And Processes In Manufacturing
Ch. 28 - What are machining processes that use abrasive...Ch. 28 - What is attrition in an abrasive grit?Ch. 28 - Why is friability an important grit property?Ch. 28 - Explain the relationship between grit size and...Ch. 28 - Why is aluminum oxide used more frequently than...Ch. 28 - Why is CBN superior to silicon carbide as an...Ch. 28 - What materials commonly are used as bonding agents...Ch. 28 - Why is the grade of a bond in a grinding wheel...Ch. 28 - How does grade differ from structure in a grinding...Ch. 28 - Prob. 10RQ
Ch. 28 - How does loading differ from glazing?Ch. 28 - What is meant by the statement that grinding is a...Ch. 28 - What is accomplished in dressing a grinding wheel?Ch. 28 - How does abrasive machining differ from ordinary...Ch. 28 - Prob. 15RQCh. 28 - How is the feed of the workpiece controlled in...Ch. 28 - Why is grain spacing important in grinding wheels?Ch. 28 - Prob. 18RQCh. 28 - How does plunge-cut grinding compare to...Ch. 28 - Prob. 20RQCh. 28 - What is the purpose of low-stress grinding?Ch. 28 - How is low-stress grinding done compared to...Ch. 28 - Prob. 23RQCh. 28 - Why are centerless grinders so popular in industry...Ch. 28 - Explain how an SEM micrograph is made. Check the...Ch. 28 - Why are vacuum chucks and magnetic chucks widely...Ch. 28 - How does creep feed grinding differ from...Ch. 28 - Why does a lap not wear, even though it is softer...Ch. 28 - How do honing stones differ from grinding wheels?Ch. 28 - What is meant by charging a lap?Ch. 28 - Why is a honing head permitted to float in a hole...Ch. 28 - How does a coated abrasive differ from an abrasive...Ch. 28 - Why are the bottoms of chips shown in Figure 28.9...Ch. 28 - Prob. 34RQCh. 28 - What are the common causes of grinding accidents?Ch. 28 - What other machine tool does a surface grinder...Ch. 28 - Figure 28.11 showed residual stress distributions...Ch. 28 - In grinding, what is infeed versus cross feed?Ch. 28 - One of the problems with waterjet cutting is that...Ch. 28 - In AWC, what keeps the abrasive jet from machining...Ch. 28 - Prob. 1PCh. 28 - Explain why it is that a small particle of a...Ch. 28 - In grinding, both the wheel and workpiece are...
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- A piston-cylinder device initially contains 0.4 kg of nitrogen gas at 160 kPa and 140°C. Nitrogen is now expanded isothermally to a pressure of 80 kPa. Determine the boundary work done during this process. The properties of nitrogen are R= 0.2968 kJ/kg-K and k= 1.4. N₂ 160 kPa 140°C The boundary work done during this process is KJ.arrow_forward! Required information An abrasive cutoff wheel has a diameter of 5 in, is 1/16 in thick, and has a 3/4-in bore. The wheel weighs 4.80 oz and runs at 11,700 rev/min. The wheel material is isotropic, with a Poisson's ratio of 0.20, and has an ultimate strength of 12 kpsi. Choose the correct equation from the following options: Multiple Choice о σmax= (314) (4r2 — r²) - о σmax = p² (3+) (4r² + r²) 16 σmax = (314) (4r² + r²) σmax = (314) (4² - r²)arrow_forwardI don't know how to solve thisarrow_forward
- I am not able to solve this question. Each part doesn't make sense to me.arrow_forwardExercises Find the solution of the following Differential Equations 1) y" + y = 3x² 3) "+2y+3y=27x 5) y"+y=6sin(x) 7) y"+4y+4y = 18 cosh(x) 9) (4)-5y"+4y = 10 cos(x) 11) y"+y=x²+x 13) y"-2y+y=e* 15) y+2y"-y'-2y=1-4x³ 2) y"+2y' + y = x² 4) "+y=-30 sin(4x) 6) y"+4y+3y=sin(x)+2 cos(x) 8) y"-2y+2y= 2e* cos(x) 10) y+y-2y=3e* 12) y"-y=e* 14) y"+y+y=x+4x³ +12x² 16) y"-2y+2y=2e* cos(x)arrow_forwardQu. 15 What are the indices for the Plane 1 drawn in the following sketch? Qu. 16 What are the Miller indices for the Plane shown in the following cubic unit cell? this is material engineering please show all workarrow_forward
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