Question 4.P.9: A vacuum system is required to handle 10 g/s of vapour (mole...
A vacuum system is required to handle 10 g/s of vapour (molecular weight 56 kg/kmol) so as to maintain a pressure of 1.5 kN/m² in a vessel situated 30 m from the vacuum pump. If the pump is able to maintain a pressure of 0.15 kN/m² at its suction point, what diameter of pipe is required? The temperature is 290 K, and isothermal conditions may be assumed in the pipe, whose surface can be taken as smooth. The ideal gas law is followed. Gas viscosity = 0.01 mN s/m².
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:
(1.274)^2 \ln (1.5 / 0.15)+\left(0.15^2-1.5^2\right) \times 10^6 /\left(2 \times 1.5 \times 10^3 \times 28.7\right)+(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}}.