## Chapter 4

## Q. 4.P.8

Over a 30 m length of 150 mm vacuum line carrying air at 293 K, the pressure falls from 1 kN/m² to 0.1 kN/m². If the relative roughness e/d is 0.002, what is approximate flowrate?

## Step-by-Step

## Verified Solution

The specific volume of air at 293 K and 1 kN/m² is:

v_1=(22.4 / 29)(293 / 273)(101.3 / 1.0)=83.98 m ^3 / kg

It is necessary to assume a Reynolds number to determine R / \rho u^2 and then calculate a value of G/A which should correspond to the original assumed value. Assume a Reynolds number of 1 \times 10^5.

When e/d = 0.002 and R e=10^5, R / \rho u^2=0.003 from Fig. 3.7.

(G / A)^2 \ln \left(P_1 / P_2\right)+\left(P_2^2-P_1^2\right) / 2 P_1 v_1+4\left(R / \rho u^2\right)(l / d)(G / A)^2=0 (equation 4.55)

Substituting:

(G / A)^2 \ln (1.0 / 0.1)+\left(0.1^2-1^2\right) \times 10^6 /\left(2 \times 1 \times 10^3 \times 83.98\right)

and: (G / A)=1.37 kg / m ^2 s.

The viscosity of air is 0.018 mN s/m².

∴ \operatorname{Re}=(0.15 \times 1.37) /\left(0.018 \times 10^{-3}\right)=1.14 \times 10^4

Thus the chosen value of Re is too high. When \operatorname{Re}=1 \times 10^4, R / \rho u^2=0.0041 and G / A=1.26 kg / m ^2 s.

Re now equals 1.04 \times 10^4 which agrees well with the assumed value.

Thus: G=1.26 \times(\pi / 4) \times(0.15)^2=\underline{\underline{0.022 kg / s }}