Answer:
[tex]1.28\cdot 10^5 Pa[/tex]
Explanation:
First of all, we need to find the speed of the fluid in the second section of the tube. This can be done by using the continuity equation:
[tex]A_1 v_1 = A_2 v_2[/tex]
where
[tex]A_1 = 8.00 cm^2[/tex] is the area of the first section
[tex]v_1 = 220 cm/s = 2.20 m/s[/tex] is the speed of the fluid in the first section
[tex]A_2 = 4.00 cm^2[/tex] is the area of the second section
v2 is the speed in the second section of the pipe
Solving for v2,
[tex]v_2 = \frac{A_1 v_1}{A_2}=\frac{(8.00)(2.20)}{4.00}=4.40 m/s[/tex]
Now we can solve the problem by using Bernoulli's equation:
[tex]p_1 + \frac{1}{2}\rho v_1 ^2 =p_2 + \frac{1}{2}\rho v_2^2[/tex]
where:
[tex]p_1 = 1.40\cdot 10^5 Pa[/tex] is the pressure in the first section of the pipe
[tex]\rho = 1660 kg/m^3[/tex] is the density of the fluid
[tex]v_1 = 2.20 m/s[/tex]
[tex]p_2[/tex] is the pressure in the second section of the pipe
[tex]v_2 = 4.40 m/s[/tex]
Solving the equation, we find
[tex]p_2 = p_1 +\frac{1}{2}\rho (v_1^2-v_2^2)=1.40\cdot 10^5 +\frac{1}{2}(1660)(2.20^2-4.40^2)=1.28\cdot 10^5 Pa[/tex]
