Refer to the diagram shown below.
vā = 1 m/s, the speed in the smaller pipe
Ļ = 1000 kg/mĀ³, the density of the water (constant)
A, mĀ² =Ā the cross-sectional area of the smaller pipe
2A, mĀ² = the cross-sectionalĀ area of the larger pipe.
Let vā =Ā the velocity of the water in the larger pipe.
The mass flow rate is constant, and it is
[tex]Q = (\rho \, \frac{kg}{m^{3}} )*(v_{1} \, \frac{m}{s} )*(A \, m^{2}) = (\rho \, \frac{kg}{m^{3}} )*(v_{2} \, \frac{m}{s} )*(2A \, m^{2})[/tex]
Because vā = 1 m/s, obtain
Ļ*vā*(2A) =Ā ĻA
2vā = 1
vā = 1/2 m/s
Answer: 0.5 m/s
