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In a Diesel engine, atmospheric air is quickly compressed to about 1/20 of its original volume. Estimate the temperature of the air after compression, and explain why a Diesel engine does not require spark plugs.

Short Answer

Expert verified

The temperature of air compression is 971.14 K.

Compressed temp is much above the auto ignition temp of diesel so diesel engines don't need any spark plugs.

Step by step solution

01

Given information

Atmospheric air is quickly compressed to about 1/20 of its original volume.

02

Explanation

The expression below explains the relationship between V and T as

VTfR2=Constant..............................(1)

Where V = volume and T = Temp

Consider the air molecules are diatomic, so its degree of freedom is

f=5( 3 translation + 2 rotational).

Write the equation for final and initial conditions and equate them to find final temp as below

ViTif2=VfTff2Tff2=ViVfTif2...................................(2)Tf=ViVf2fTi.....................................(3)

Substitute values and calculate to get final temp

Tf=VoVo2025293KTf=971.14K

We know that the auto ignition temperature for diesel is 210oC

So the fuel will automatically ignite when the air is compressed as temp is much above the auto ignition temp.

For this reason diesel engines don't need any spark plugs.


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Most popular questions from this chapter

A frying pan is quickly heated on the stovetop 200∘CIt has an iron handle that is 20cmlong. Estimate how much time should pass before the end of the handle is too hot to grab with your bare hand. (Hint: The cross-sectional area of the handle doesn't matter. The density of iron is about7.9g/cm3and its specific heat is 0.45J/g⋅∘C).

Measured heat capacities of solids and liquids are almost always at constant pressure, not constant volume. To see why, estimate the pressure needed to keep Vfixed as Tincreases, as follows.

(a) First imagine slightly increasing the temperature of a material at constant pressure. Write the change in volume,dV1, in terms of dTand the thermal expansion coefficient βintroduced in Problem 1.7.

(b) Now imagine slightly compressing the material, holding its temperature fixed. Write the change in volume for this process, dV2, in terms of dPand the isothermal compressibility κT, defined as

κT≡−1V∂V∂PT

(c) Finally, imagine that you compress the material just enough in part (b) to offset the expansion in part (a). Then the ratio of dPtodTis equal to (∂P/∂T)V, since there is no net change in volume. Express this partial derivative in terms of βandκT. Then express it more abstractly in terms of the partial derivatives used to define βandκT. For the second expression you should obtain

∂P∂TV=−(∂V/∂T)P(∂V/∂P)T

This result is actually a purely mathematical relation, true for any three quantities that are related in such a way that any two determine the third.

(d) Compute β,κT,and(∂P/∂T)Vfor an ideal gas, and check that the three expressions satisfy the identity you found in part (c).

(e) For water at 25∘C,β=2.57×10−4K−1andκT=4.52×10−10Pa−1. Suppose you increase the temperature of some water from 20∘Cto30∘C. How much pressure must you apply to prevent it from expanding? Repeat the calculation for mercury, for which (at25∘C)β=1.81×10−4K−1andκT=4.04×10−11Pa−1

Given the choice, would you rather measure the heat capacities of these substances at constant vor at constant p?

Calculate the heat capacity of liquid water per molecule, in terms of K . Suppose (incorrectly) that all the thermal energy of water is stored in quadratic degrees of freedom. How many degrees of freedom would each molecule have to have?

Estimate how long it should take to bring a cup of water to boiling temperature in a typical 600 -watt microwave oven, assuming that all the energy ends up in the water. (Assume any reasonable initial temperature for the water.) Explain why no heat is involved in this process.

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