Chapter 6

Q. 6.5

Water flows through a tapering pipe as shown in Fig. 6.4. The diameter at sections 1 and 2 are 10 cm and 20 cm, respectively, and the heights above a horizontal datum are 3 and 5 m, respectively. The pressure at 1 is 30 kN/m². Water flow rate is 0.03 m³/s. Estimate the pressure at section 2.



Verified Solution

Given data:

Diameter of pipe at section 1                  D_1=10 \mathrm{~cm}=0.1 \mathrm{~m}

Diameter of pipe at section 2                  D_2=20 \mathrm{~cm}=0.2 \mathrm{~m}

Pressure at section 1                                  p_1=30 \mathrm{kN} / \mathrm{m}^2=30 \times 10^3 \mathrm{~N} / \mathrm{m}^2

Height of section 1 above datum            z_1=3 \mathrm{~m}

Height of section 2 above datum          z_2=5 \mathrm{~m}

Volume flow rate of water                    Q = 0.03 m³/s

Cross-sectional area at section 1 is      A_1=\frac{\pi}{4} D_1^2=\frac{\pi}{4}(0.1)^2=0.00785 \mathrm{~m}^2

Cross-sectional area at section 2 is      A_2=\frac{\pi}{4} D_2^2=\frac{\pi}{4}(0.2)^2=0.0314 \mathrm{~m}^2

From continuity equation, we have

Q=A_1 V_1=A_2 V_2

Thus, the average velocity at section 1 is      V_1=\frac{Q}{A_1}=\frac{0.03 \mathrm{~m}^3 / \mathrm{s}}{0.00785 \mathrm{~m}^2}=3.82 \mathrm{~m} / \mathrm{s}

Average velocity at section 2 is        V_2=\frac{Q}{A_2}=\frac{0.03 \mathrm{~m}^3 / \mathrm{s}}{0.0314 \mathrm{~m}^2}=0.955 \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

\frac{30 \times 10^3}{1000 \times 9.81}+\frac{3.82^2}{2 \times 9.81}+3=\frac{p_2}{\rho g}+\frac{0.955^2}{2 \times 9.81}+5

or              3.058+0.744+3=\frac{p_2}{\rho g}+0.046+5

or            p_2=1000 \times 9.81 \times 1.756=17226.36 \mathrm{~N} / \mathrm{m}^2 \simeq 17.226 \mathrm{kN} / \mathrm{m}^2