Sample Solution from
Fundamentals of Momentum, Heat, and Mass Transfer
6th Edition
ISBN: 9781118947463
Chapter 1
Problem 1.1P
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Interpretation Introduction

Interpretation:

The number of molecules which crosses the circular hole is to be determined.

Concept Introduction:

The formula to calculate number of molecules (N)

that crosses a unit area per unit time in only one direction is:

  N=14nv¯ ........(1)

Here, n is the number of molecules per unit volume and v¯ is the average molecular velocity.

The formula to calculate the speed of sound (c)

in a perfect gas is:

  c=kgcRT ........(2)

Here, k is the ratio of specific heats, R is the specific gas constant for a gas in kJ/(kg⋅K) , gc is the conversion factor in (kg⋅m)/(N⋅s2) , and T is temperature in K .

Expert Solution

Answer

  1.07×1018 molecules/s .

Explanation of Solution

Given information:

The gas is at standard conditions with 4×1020 molecules per in.3 . The average molecular velocity is taken approximately equal to the sound speed in a perfect gas.

The diameter of the circular hole is, d=10−3 in.

Assume that the perfect gas is air with the value of k=1.4 . For air, the value of R is taken as R=0.287 kJ/(kg⋅K) . At standard condition, the temperature is taken as, T=298 K .

Use equation (2) to calculate the speed of sound in perfect gas as:

  c=kgcRT=( 1.4)( kg ⋅m N ⋅ s 2 )( 0.287× 10 3   N -m kg ⋅ K )( 298  K )=346.028 m/s

Since,

  1 m =39.37 in.

Thus,

  c=346.028 m( 39.37 in. 1 m)/s=1.36×104 in./s

It is given that the sound speed in the perfect gas is nearly equal to the average molecular velocity.

Thus,

  v¯=c=1.36×104 in./s

Area of the circular hole is calculated as:

  A=π4d2=3.144( 10 −3 in.)2=7.85×10−7 in.2

Equation (1) gives the number of molecules that crosses a unit area per unit time in one direction. To calculate the number of molecules that crosses a particular area per unit time, multiply equation (1) by A as:

  NA=14nv¯A=14( 4× 10 20  molecules in . 3 )( 1.36× 10 4   in. s)(7.85× 10 −7  in . 2 )=1.07×1018 molecules/s

Conclusion

Thus, the number of molecules which crosses the given circular hole is, 1.07×1018 molecules/s .

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