Answer:
[tex]P_2 = -1.9 \times 10^8 Pa[/tex]
Explanation:
As it is given that flow rate in the pipe is 20 cm^3/s
so we have
[tex]Q = A_1v_1 = A_2v_2[/tex]
at the upper end the area is given as
[tex]A_1 = \pi r_1^2[/tex]
[tex]A_1 = \pi(0.02)^2 = 1.26 \times 10^{-3} cm^2[/tex]
Also at the other end
[tex]A_2 = \pi r_2^2 [/tex]
[tex]A_2 = \pi(0.01)^2 = 0.314 \times 10^{-3} cm^2[/tex]
now the speed at two ends is given as
[tex]v_1 = \frac{20}{1.26 \times 10^{-3}}[/tex]
[tex]v_1 = 159.15 m/s[/tex]
[tex]v_2 = \frac{20}{0.314 \times 10^{-3}}[/tex]
[tex]v_2 = 637 m/s[/tex]
now by Bernoulli's theorem we have
[tex]P_1 + \frac{1}{2}\rho v_1^2 + \rho g h_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho g h_2[/tex]
now we have
[tex]0.112(1.013 \times 10^5) + \frac{1}{2}1000(159.15)^2 + 1000(9.81)(2.65sin40) = P_2 + \frac{1}{2}(1000)(637)^2 + 0[/tex]
Now we have
[tex]P_2 = -1.9 \times 10^8 Pa[/tex]
