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Q.4.33

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An Introduction to Thermal Physics
Found in: Page 143
An Introduction to Thermal Physics

An Introduction to Thermal Physics

Book edition 1st
Author(s) Daniel V. Schroeder
Pages 356 pages
ISBN 9780201380279

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Short Answer

Table 4.5 gives experimental values of the molar enthalpy of nitrogen at 1 bar and 100 bars. Use this data to answer the following questions about a nitrogen throttling process operating between these two pressures.

(a) If the initial temperature is 300 K, what is the final temperature? (Hint: You'll have to do an interpolation between the tabulated values.)

(b) If the initial temperature is 200 K, what is the final temperature?

(c) If the initial temperature is 100 K, what is the final temperature? What fraction of the nitrogen ends up as a liquid in this case?

(d) What is the highest initial temperature at which some liquefaction takes place?

(e) What would happen if the initial temperature were 600 K? Explain.

a) The final temperature is 281.4k

b) The final temperature is 153.9K

c) The final temperature after the throttling process is 77 K. Thus, the fraction of nitrogen that becomes liquefied is 0.7376¯.

d)role="math" localid="1648594282162" The highest initial temperature at which some liquefaction takes place is 164.3 K

e) role="math" localid="1648594270697" The final temperature increases when the initial temperature is 600 K.

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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Part (a) - Step 1: To find

The final temperature

Part (a) - Step 2: Explanation

Given: A nitrogen throttling process operates between pressures 1 bar and 100 bars. The initial temperature is 300 K.

Formula: The value of the initial enthalpy at temperature 300 K and pressure 100 bars is 8174J, the enthalpy at temperature 300 K and pressure 1 bar is 8717 J, and the enthalpy at temperature 200 K and pressure 1 bar is 5800 J from table 4.5

The liquid nitrogen fraction x has the following value:

x=(8174 J)-(8717 J)(5800 J)-(8717 J)= 0.186

The expression of the final temperature TF

TF=(200 K)x+(300 K)(1-x)…(1)

Calculation: Substitute x= 0.186 in equation (1)

TF=(200 K)(0.186)+(300 K)(1-0.186) = 281.4k

Hence the final temperature is 281.4k

Part (b) - Step 3: To find

The final temperature

Part (b) - Step 4: Explanation

Given:

A nitrogen throttling process operates between pressures 1 bar and 100 bars. The initial temperature is 200 K.

Formula : The value of the initial enthalpy at temperature 200 K and pressure 100 bars is 4442 J, the enthalpy at temperature 200 Kand pressure 1bar is 5800 J, and the enthalpy at temperature 100 K and pressure 1 bar is 2856 Jfrom table 4.5.

The fraction of nitrogen x that becomes liquefied has the following value:

x=(4442 J)-(5800 J)(2856 J)-(5800 J)= 0.461

The expression of the final temperature TF

TF=(100 K)x+(200 K)(1-x)….(2)

Substitute X= 0.461 in equation (2)

TF=(100 K)(0.461)+(200 K)(1-0.461)=153.9 K

Hence the final temperature is 153.9K

Part (c) - Step 5: To find

The final temperature. and the fraction of nitrogen that becomes liquefied.

Part(c) - Step 6: Explanation

Given:

A nitrogen throttling process operates between pressures 1 bar and 100 bars. The initial temperature is 100 K.

Formula: In order to conserve the initial enthalpy, the temperature has to decrease. Since the decreased temperature lies between the enthalpies for gaseous and liquid nitrogen at pressure 1 bar and temperature 77 K, therefore, the temperature will be77 Kafter the throttling process.

The expression of the initial enthalpy H at temperature 100 K and pressure 100 bars

H=Hliqx+Hgas(1-x)

Here, x is fraction of liquefaction, Hliq is the enthalpy of liquefied nitrogen and Hgas is the enthalpy of gaseous nitrogen at pressure1bar and temperature 77 K.

Rearrange the above expression for x

x=H-HgEHfiq-Hgas… (3)

Calculation:Substitute H= -1946 J Hliq =-3407 Jandrole="math" localid="1648599827773" Hgas =2161 J in equation (3)

x=(-1946 J)-(2161 J)(-3407 J)-(2161 J)=0.7376

Thus, the final temperature after the throttling process is 77 K.

Thus, the fraction of nitrogen that becomes liquefied is 0.7376¯.

Part (d) - Step 7: To find

The highest initial temperature at which some liquefaction takes place.

Part(d) - Step 8: Explanation

Given: A nitrogen throttling process operates between pressures 1 bar and 100 bars.

Formula: The temperature at which liquefaction occurs is when the enthalpy of the liquid equals the enthalpy of the gas. It's somewhere between100 Kand 200 Kin temperature.

The value of the enthalpy of gas is 2161 J, the enthalpy at temperature 100 K is -1946 J, and the enthalpy at temperature200 K is 4442 J from table 4.5.

The value of the fraction of nitrogen x that becomes liquefied is:

x=(2161 J)-(4442 J)(-1946 J)-(4442 J)= 0.357

The expression of the highest temperature TH

TH=(100 K)x+(200 K)(1-x)…(4)

Calculation: Substitute x= 0.357 in equation (4)

TF=(100 K)(0.357)+(200 K)(1-0.357)=164.3 K

Hence the highest initial temperature at which some liquefactiontakes place is 164.3 K

Part (e) - Step 9: To find

When the initial temperature is 600 K the following is the result.

Part (e) - Step 10: Explanation

The nitrogen throttling technique works at pressures ranging from 1 bar to 100 bar.

Because the enthalpy at a pressure of 100 bars is greater than the enthalpy at a pressure of 1 bar when the initial temperature is600 K the temperature after the throttling operation will rise rather than decrease, in contrast to earlier parts.

Hence the final temperature increases when the initial temperature is 600 K.

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