Answer:
(a) Speed in the second segment=2.25 m/s
(b) Speed in the third segment=36 m/s
(c) Volume=0.02826 L/s
Explanation:
Given Data
r₁=5×10⁻⁴m
r₂=10⁻³m
r₃=2.5×10⁻⁴ m
v₁= 9.0 m/s
For (a) speed in the second segment
From continuity equation
[tex]v_{1}A_{1}=v_{2}A_{2}\\ v_{2}=\frac{v_{1}A_{1}}{A_{2}}\\ v_{2}=\frac{v_{1}4(\pi)(r_{1})^{2}}{4(\pi)(r_{2})^{2}}\\ v_{2}=\frac{v_{1}(r_{1})^{2}}{(r_{2})^{2}}\\ v_{2}=\frac{9.0*(5*10^{-4} )^{2} }{(1*10^{-3} )^{2}}\\ v_{2}=2.25 m/s[/tex]
For (b) speed in the third segment
[tex]v_{3}=\frac{v_{1}(r_{1})^{2} }{(r_{3})^{2} }\\v_{3}=\frac{9*(5*10^{-4} )^{2} }{(2.5*10^{-4} )^{2} }\\v_{3}=36m/s[/tex]
For (c) the volume flow rate through the pipe
The volume flow rate is the same in all three segments and is given as:
[tex]V=v_{1}A_{1}\\ V=v_{1}4\pi (r_{1})^{2}\\ V=9.0*4\pi (5*10^{-4} )^{2}\\ V=2.86*10^{-5}m^{3}/s[/tex]
As per requirement of question we need to convert volume cubic meter/second to Litre/second for that we need to "multiply the volume /second value by 1000"
So
Volume=(2.86×10⁻⁵ ×1000) L/s
Volume=0.02826 L/s
