Source: High school physics (Chinese)
Problem Sets:
Problem
Water flows at $v_1 = 5.0$ m/s through a pipe section with a cross-sectional area of $S_1 = 4.0$ cm$^2$. The pipe descends by a height of $h=10$ m to a lower section where the cross-sectional area is $S_2 = 0.8$ cm$^2$. The pressure in the upper section is $P_1 = 1.50 \times 10^5$ Pa. Use the density of water $\rho = 1000$ kg/m$^3$ and the acceleration due to gravity $g=9.8$ m/s$^2$.
- What is the velocity of the water in the lower pipe section?
- What is the pressure in the lower pipe section?
[Q1] $v_2 = 25$ m/s [Q2] $P_2 = -5.2 \times 10^4$ Pa
Let the upper and lower sections be denoted by subscripts 1 and 2, respectively. [Q1] The continuity equation relates the velocities and cross-sectional areas:
$$S_1 v_1 = S_2 v_2$$ $$v_2 = v_1 \frac{S_1}{S_2}$$[Q2] Bernoulli's equation relates pressure, velocity, and height between two points:
$$P_1 + \frac{1}{2}\rho v_1^2 + \rho g y_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho g y_2$$Solving for the pressure in the lower section, $P_2$:
$$P_2 = P_1 + \frac{1}{2}\rho (v_1^2 - v_2^2) + \rho g (y_1 - y_2)$$Given the height drop $h = y_1 - y_2 = 10$ m.
Substituting numerical values:
$$v_2 = (5.0 \text{ m/s}) \frac{4.0 \text{ cm}^2}{0.8 \text{ cm}^2} = 25 \text{ m/s}$$ $$P_2 = 1.50 \times 10^5 \text{ Pa} + \frac{1}{2}(1000 \text{ kg/m}^3)((5.0 \text{ m/s})^2 - (25 \text{ m/s})^2) + (1000 \text{ kg/m}^3)(9.8 \text{ m/s}^2)(10 \text{ m})$$ $$P_2 = 1.50 \times 10^5 \text{ Pa} - 3.00 \times 10^5 \text{ Pa} + 0.98 \times 10^5 \text{ Pa}$$