Question

A water pipe is inclined 40.0° below the horizontal. The radius of the pipe at the upper end is 2.00 cm. If the gauge pressure at a point at the upper end is 0.112 atm, what is the gauge pressure at a point 2.65 m downstream, where the pipe has narrowed to a 1.00 cm radius? The flow rate is 20.0? cm^3/s

Answer 1

Answer:

$P_2 = -1.9 \times 10^8 Pa$

Explanation:

As it is given that flow rate in the pipe is 20 cm^3/s

so we have

$Q = A_1v_1 = A_2v_2$

at the upper end the area is given as

$A_1 = \pi r_1^2$

$A_1 = \pi(0.02)^2 = 1.26 \times 10^{-3} cm^2$

Also at the other end

$A_2 = \pi r_2^2$

$A_2 = \pi(0.01)^2 = 0.314 \times 10^{-3} cm^2$

now the speed at two ends is given as

$v_1 = \frac{20}{1.26 \times 10^{-3}}$

$v_1 = 159.15 m/s$

$v_2 = \frac{20}{0.314 \times 10^{-3}}$

$v_2 = 637 m/s$

now by Bernoulli's theorem we have

$P_1 + \frac{1}{2}\rho v_1^2 + \rho g h_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho g h_2$

now we have

$0.112(1.013 \times 10^5) + \frac{1}{2}1000(159.15)^2 + 1000(9.81)(2.65sin40) = P_2 + \frac{1}{2}(1000)(637)^2 + 0$

Now we have

$P_2 = -1.9 \times 10^8 Pa$