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Modern Engineering Thermodynamics

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658 CHAPTER 16: Compressible Fluid Flow<br />

or<br />

<br />

∂h<br />

= v = 1 ∂p<br />

s<br />

ρ<br />

(16.8)<br />

Multiplying Eq. (16.6) by Eq. (16.7) and dividing by Eq. (16.8) gives a relation for the sonic velocity c in terms<br />

of the measurable properties p and ρ:<br />

<br />

<br />

<br />

∂V ∂h ∂p<br />

= ∂p <br />

= ρðc/g c Þðc/ρÞ = c 2 /g c<br />

∂ρ<br />

s<br />

∂V s ∂h<br />

s<br />

∂ρ<br />

s<br />

or<br />

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi<br />

<br />

∂p<br />

c = g c<br />

∂ρ<br />

s<br />

(16.9)<br />

Equation (16.9) is a valid equation for the isentropic sonic velocity in any compressible substance. In particular,<br />

in the case of an ideal gas, Eq. (7.39) relates pressure and density for an isentropic process by<br />

pv k = pρ –k = constant (7.39)<br />

Solving for p = constant × ρ k and taking the partial differential of p with respect to ρ gives<br />

<br />

∂p<br />

= kp/ρ = kRT<br />

∂ρ<br />

s<br />

where we use the ideal gas law, p = ρRT. Then, from Eq. (16.9), the speed of sound (sonic velocity) in an ideal<br />

gas is<br />

p<br />

= ffiffiffiffiffiffiffiffiffiffiffiffi<br />

kg c RT<br />

(16.10)<br />

and from Eq. (16.5),<br />

c ideal<br />

gas<br />

M ideal<br />

gas<br />

V<br />

= p ffiffiffiffiffiffiffiffiffiffiffiffi<br />

(16.11)<br />

kg c RT<br />

EXAMPLE 16.4<br />

Methane (CH 4 ) gas at 35.0°C flows through a pipe with a velocity of 300. m/s. Determine the Mach number of the<br />

methane.<br />

Solution<br />

The velocity of sound in methane at 35.0°C is given by Eq. (16.10). Using Table C.13b in Thermodynamic Tables to accompany<br />

<strong>Modern</strong> <strong>Engineering</strong> <strong>Thermodynamics</strong> for the values of the specific heat ratio and the gas constant for methane, we get<br />

p<br />

c methane =<br />

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi<br />

ðk methane Þðg c ÞðR methane ÞðTÞ<br />

p<br />

= ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi<br />

ð1:30Þð1Þð518 J/kgCKÞð35:0 + 273:15 KÞ<br />

= 456 m/s<br />

then, Eq. (16.5) (or (16.11)) gives the Mach number as<br />

M methane =<br />

300: m/s<br />

456 m/s = 0:658<br />

Exercises<br />

7. Determine the Mach number of the methane in Example 16.4 if its velocity is reduced from 300. m/s to 3.00 m/s and<br />

all the other variables remained unchanged. Answer: M methane = 6.58 × 10 –3 .<br />

8. If the methane in the pipe in Example 16.4 is cooled to 0.00°C and all the other variables remain unchanged, determine<br />

the new Mach number of the methane. Answer: M methane = 0.700.<br />

9. If the gas in Example 16.4 is changed from methane to argon, determine the Mach number of the argon assuming all<br />

the other variables remain unchanged. Answer: M argon = 0.917.

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