Question
Water flows through a horizontal pipe whose cross-section narrows from A1=4×10−3m2 to A2=1×10−3m2. The speed of water in the wider section is 1m/s and the pressure there is 2×105Pa. Find the speed and pressure in the narrower section. Take ρwater=103kg/m3.
Solution — Step by Step
For an incompressible fluid in a pipe, A1v1=A2v2:
v2=A2A1v1=1×10−34×10−3×1=4m/s
Same height, so ρgh terms cancel. Bernoulli gives:
P1+21ρv12=P2+21ρv22
P2=P1+21ρ(v12−v22)
P2=2×105+21×103×(1−16)
P2=2×105−7500=1.925×105Pa
Final: v2=4m/s, P2=1.925×105Pa.
Why This Works
Faster flow means lower pressure — this is the Bernoulli effect, the principle behind aircraft lift and the carburetor. As the cross-section narrows, water must speed up (continuity), and energy conservation forces pressure to drop to compensate for the higher kinetic energy.
The pressure difference here is small (about 4%) because the speeds are modest. In a fire hose nozzle or a Venturi meter, the same effect can be dramatic.
Alternative Method
Using the Venturi-meter formula directly:
P1−P2=21ρv12(A22A12−1)=21×103×1×(16−1)=7500Pa
Same drop, computed in one shot.
Common Mistake
Students apply Bernoulli without checking whether the pipe is horizontal. If there is a height difference, the ρgΔh term cannot be dropped. Always sketch the setup first.