The pipeline is schematically shown in Fig. 6.3. Let 1 and 2 designate the bottom and top end of the pipeline respectively.

Given data:

Diameter of pipe at section 1            D_1=20 \mathrm{~cm}=0.2 \mathrm{~m}

Diameter of pipe at section 2            D_2=10 \mathrm{~cm}=0.1 \mathrm{~m}

Cross-sectional area at section 1 is    A_1=\frac{\pi}{4} D_1^2=\frac{\pi}{4}(0.2)^2=0.0314 \mathrm{~m}^2

Cross-sectional area at section 2 is  A_2=\frac{\pi}{4} D_2^2=\frac{\pi}{4}(0.1)^2=0.00785 \mathrm{~m}^2

Discharge                    Q=30 \text { litres } / \mathrm{s}=30 \times 10^{-3} \mathrm{~m}^3 / \mathrm{s}=0.03 \mathrm{~m}^3 / \mathrm{s}

Difference in datum head between sections 1 andDifference in datum head between sections 1 and          z_2-z_1=2 \mathrm{~m}

Average velocity at section 1 is        V_1=\frac{Q}{A_1}=\frac{0.03 \mathrm{~m}^3 / \mathrm{s}}{0.0314 \mathrm{~m}^2}=0.955 \mathrm{~m} / \mathrm{s}

Average velocity at section 2 is      V_2=\frac{Q}{A_2}=\frac{0.03 \mathrm{~m}^3 / \mathrm{s}}{0.00785 \mathrm{~m}^2}=3.82 \mathrm{~m} / \mathrm{s}

Applying Bernoulli’s equation between sections 1 and 2 along a streamline, one can write

\frac{p_1}{\rho g}+\frac{V_1^2}{2 g}+z_1=\frac{p_2}{\rho g}+\frac{V_2^2}{2 g}+z_2

or              \frac{p_1-p_2}{\rho g}=\frac{V_2^2-V_1^2}{2 g}+z_2-z_1

or              \frac{p_1-p_2}{\rho g}=\frac{3.82^2-0.955^2}{2 \times 9.81}+2=0.697+2=2.697 \mathrm{~m} \text { of water }

or              p_1-p_2=2.697 \rho g=2.697 \times 1000 \times 9.81 \mathrm{~N} / \mathrm{m}^2

or              p_1-p_2=26457.57 \mathrm{~N} / \mathrm{m}^2=26.458 \mathrm{kN} / \mathrm{m}^2