Question
Water flows through a horizontal pipe whose cross-sectional area changes from 4 cm2 at point A to 1 cm2 at point B. The speed at A is 2 m/s and the gauge pressure at A is 30 kPa. Find the speed at B and the gauge pressure at B. Take ρwater=1000 kg/m3.
Solution — Step by Step
For an incompressible fluid, A1v1=A2v2:
vB=ABAAvA=14×2=8 m/s
Heights cancel in horizontal flow, so:
PA+21ρvA2=PB+21ρvB2
21ρvA2=21(1000)(2)2=2000 Pa
21ρvB2=21(1000)(8)2=32000 Pa
PB=PA+21ρvA2−21ρvB2
PB=30000+2000−32000=0 Pa (gauge)
Final answers: vB=8 m/s, PB=0 Pa (i.e., atmospheric).
Why This Works
Continuity is mass conservation — the same mass per second must pass every cross-section. So the fluid speeds up where the pipe narrows.
Bernoulli is energy conservation per unit volume — pressure energy + kinetic energy + potential energy is constant along a streamline. Where the fluid moves faster, pressure must drop. This is the Venturi effect, the principle behind aerofoils, atomisers and carburettors.
Alternative Method
Using the pressure difference formula directly: PA−PB=21ρ(vB2−vA2)=21(1000)(64−4)=30000 Pa. So PB=30000−30000=0 Pa gauge. Same answer.
Common Mistake
Forgetting that gauge pressure 0 means the fluid is at atmospheric pressure — not at vacuum. If the speed at B were any higher, the gauge pressure would go negative, signalling cavitation risk in real pipes.
In NEET, the Venturi-tube question appears almost every year. Memorise the form ΔP=21ρ(v22−v12) for horizontal flow — saves substitution time.