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

`P_2 = -1.9 * 10^8 Pa`

Step-by-step 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 * 10^(-3) cm^2`

Also at the other end

`A_2 = \pi r_2^2`

`A_2 = \pi(0.01)^2 = 0.314 * 10^(-3) cm^2`

now the speed at two ends is given as

`v_1 = (20)/(1.26 * 10^(-3))`

`v_1 = 159.15 m/s`

`v_2 = (20)/(0.314 * 10^(-3))`

`v_2 = 637 m/s`

now by Bernoulli's theorem we have

`P_1 + (1)/(2)\rho v_1^2 + \rho g h_1 = P_2 + (1)/(2)\rho v_2^2 + \rho g h_2`

now we have

`0.112(1.013 * 10^5) + (1)/(2)1000(159.15)^2 + 1000(9.81)(2.65sin40) = P_2 + (1)/(2)(1000)(637)^2 + 0`

Now we have

`P_2 = -1.9 * 10^8 Pa`