Why is 2.4-L Max Vol V1 in Diesel Engine Cylinder Example 9-60?

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The discussion centers on the designation of 2.4-L as the maximum volume (V1) rather than the displacement volume in Example 9-60. It highlights a potential error in the example where the author incorrectly states V1 as 2.4-L, while a calculation suggests it should be 2.514-L. The distinction between maximum volume and displacement volume is clarified, with displacement volume defined as V1 minus V2. Additionally, the cutoff ratio is introduced, defined as the ratio of V3 to V2. This conversation underscores the importance of accurate definitions and calculations in diesel engine examples.
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Homework Statement
Distinguishing between maximum volume and displacement volume.
Relevant Equations
displacement volume formula
1689580222034.png

1689581129463.png

Q: Why is 2.4-L referred to as the maximum volume instead of the displacement volume in Example 9-60?
note: maximum volume = V1. displacement volume=V1-V2

reference:
displacement volume
9-60
9-153
9-164
9-167
 
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tracker890 Source h said:
Homework Statement: Distinguishing between maximum volume and displacement volume.
Relevant Equations: displacement volume formula

View attachment 329356
View attachment 329357
Q: Why is 2.4-L referred to as the maximum volume instead of the displacement volume in Example 9-60?
note: maximum volume = V1. displacement volume=V1-V2

reference:
displacement volume
9-60
9-153
9-164
9-167
What is the definition of cutoff ratio?
 
tracker890 Source h said:
Q: Why is 2.4-L referred to as the maximum volume instead of the displacement volume in Example 9-60?
Post #1 has too many links to read through and it is not clear why you have included them.

However, in the model answer for Example 9.60, the author uses ##V_1= 0.0024m^3 (=2.4L)## which, as you say, is incorrect. (I get ##V_1 = 2.514L##.) Presumably that's the problem about which you are asking.

So it looks like a simple mistake by the author.
 
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Chestermiller said:
What is the definition of cutoff ratio?
$$r_c=\frac{V_3}{V_2}$$
 
Thread 'Chain falling out of a horizontal tube onto a table'
My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...

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