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Chapter 4

Q. 4.P.10

In a vacuum system, air is flowing isothermally at 290 K through a 150 mm diameter pipeline 30 m long. If the relative roughness of the pipewall e/d is 0.002 and the downstream pressure is 130 N/m², what will the upstream pressure be if the flowrate of air is 0.025 kg/s? Assume that the ideal gas law applies and that the viscosity of air is constant at 0.018 mN s/m².
What error would be introduced if the change in kinetic energy of the gas as a result of expansion were neglected?

Step-by-Step

Verified Solution

As the upstream and mean specific volumes v_1 and v_m are required in equations 4.55 and 4.56 respectively, use is made of equation 4.57:

(G / A)^2 \ln \left(P_1 / P_2\right)+\left(P_2^2-P_1^2\right) /(2 R T / M)+4\left(R / \rho u^2\right)(l / d)(G / A)^2=0

R = 8.314 kJ/kmol K and hence:

2 R T / M=\left(2 \times 8.314 \times 10^3 \times 290\right) / 29=1.66 \times 10^5  J / kg

The second term has units of \left( N / m ^2\right)^2 /( J / kg )= kg ^2 / s ^2  m ^4 which is consistent with the other terms.

A=(\pi / 4)(0.15)^2=0.0176 m ^2

∴                  G / A=(0.025 / 0.0176)=1.414

and              R e=d(G / A) / \mu=(0.15 \times 1.414) /\left(0.018 \times 10^{-3}\right)=1.18 \times 10^4

For smooth pipes and R e=1.18 \times 10^4, R / \rho u^2=0.0040 from Fig. 3.7. Substituting in equation 4.57 gives:

(1.414)^2 \ln \left(P_1 / 130\right)+\left(130^2-P_1^2\right) / 1.66 \times 10^5+4 \times 0.0040(30 / 0.15)(1.414)^2=0

Solving by trial and error, the upstream pressure, P_1=\underline{\underline{1.36  kN / m ^2}}
If the kinetic energy term is neglected, equation 4.57 becomes:

\left(P_2^2-P_1^2\right) /(2 R T / M)+4\left(R / \rho u^2\right)(l / d)(G / A)^2=0 and P_1=\underline{\underline{1.04  kN / m ^2}}

Thus a considerable error would be introduced by this simplifying assumption.