Question 4.P.9: A vacuum system is required to handle 10 g/s of vapour (mole...

Use is made of equation 4.55 to solve this problem. It is necessary to assume a value of the pipe diameter d in order to calculate values of G/A, the Reynolds number and $R / \rho u^2$.
If d = 0.10 m, $A=(\pi / 4)(0.10)^2=0.00785  m ^2$

∴                     $G / A=\left(10 \times 10^{-3} / 0.00785\right)=1.274  kg / m ^2 s$

and       $R e=d(G / A) / \mu=0.10 \times 1.274 /\left(0.01 \times 10^{-3}\right)=1.274 \times 10^4$

For a smooth pipe, $R / \rho u^2=0.0035$, from Fig. 3.7.
Specific volume at inlet, $v_1=(22.4 / 56)(290 / 273)(101.3 / 1.5)=28.7  m ^3 / kg$

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

Substituting gives:

$+(0.0035)(30 / 0.10)(1.274)^2=-16.3$

and the chosen value of d is too large.

A further assumed value of d = 0.05 m gives a value of the right hand side of equation 4.55 of 25.9 and the procedure is repeated until this value is zero.
This occurs when d = 0.08 m or $\underline{\underline{80  mm}}$.