## Chapter 4

## Q. 4.P.13

A 150 mm gas main is used for transferring gas (molecular weight 13 kg/kmol and kinematic viscosity 0.25 cm²/s) at 295 K from a plant to a storage station 100 m away, at a rate of 1 m³/s. Calculate the pressure drop if the pipe can be considered to be smooth.

If the maximum permissible pressure drop is 10 kN/m², is it possible to increase the flowrate by 25%?

## Step-by-Step

## Verified Solution

If the flow of 1 m³/s is at STP, the specific volume of the gas is:

(22.4 / 13)=1.723 m ^3 / kg.

The mass flowrate, G=(1.0 / 1.723)=0.58 kg / s.

Cross-sectional area, A=(\pi / 4)(0.15)^2=0.0176 m ^2

∴ G / A=32.82 kg / m ^2 s

\mu / \rho=0.25 cm ^2 / s =0.25 \times 10^{-4} m ^2 / s

and \mu=\left(0.25 \times 10^{-4}\right)(1 / 1.723)=1.45 \times 10^{-5} N s / m ^2

∴ R e=\left(0.15 \times 32.82 / 1.45 \times 10^{-5}\right)=3.4 \times 10^5

For smooth pipes, R / \rho u^2=0.0017, from Fig. 3.7.

The pressure drop due to friction is:

4\left(R / \rho u^2\right)(l / d)(G / A)^2=4(0.0017)(100 / 0.15)(32.82)^2=4883 kg ^2 / m ^4 s ^2

and: -\Delta P=(4883 / 1.723)=2834 N / m ^2 or \underline{\underline{2.83 kN / m ^2}}.

If the flow is increased by 25%, G=(1.25 \times 0.58)=0.725 kg / s

G / A=41.19 kg / m ^2 s

and: R e=(0.15 \times 41.9) /\left(1.45 \times 10^5\right)=4.3 \times 10^5

and, from Fig. 3.7, R / \rho u^2=0.00165

The pressure drop =4(0.00165)(100 / 0.15)(41.19)^2 1.723=\underline{\underline{4.33 kN / m ^2}} (which is less than 10 kN/m²)

It is therefore possible to increase the flowrate by 25%.