Q.Water flows steadily through a horizontal pipe that narrows from a wider section of cross-sectional area $4.0\times10^{-3}\ \text{m}^2$ to a narrower section of cross-sectional area $1.0\times10^{-3}\ \text{m}^2$. If the speed of the water in the wider section is $1.5\ \text{m/s}$, find

Given: $A_1 = 4.0\times10^{-3}\ \text{m}^2$, $A_2 = 1.0\times10^{-3}\ \text{m}^2$, $v_1 = 1.5\ \text{m/s}$, $\rho = 1000\ \text{kg/m}^3$.

1. Speed in the narrower section, from the equation of continuity, $A_1v_1 = A_2v_2$:

   $v_2 = \frac{A_1v_1}{A_2} = \frac{(4.0\times10^{-3})(1.5)}{1.0\times10^{-3}} = 6.0\ \text{m/s}$

2. Pressure difference, from Bernoulli's theorem applied at the same height (so the $\rho g h$ terms cancel), $P_1 + \tfrac12\rho v_1^2 = P_2 + \tfrac12\rho v_2^2$: $P_1 - P_2 = \frac{1}{2}\rho(v_2^2 - v_1^2) = \frac{1}{2}(1000)(6.0^2 - 1.5^2) = 500\times(36 - 2.25) = 500\times33.75$ …

Find the speed in the narrower section from the equation of continuity, $A_1v_1 = A_2v_2$, then substitute both speeds into Bernoulli's theorem (with the height terms c …

• Assuming the pressure is higher in the narrower section because the fluid is moving faster there -- Bernoulli's theorem says the opposite: pressure falls where speed rises. …