Question

Water flows at speed of 4 m/s through a horizontal pipe of diameter 3.1 cm . The gauge
pressure P1 of the water in the pipe is 1.8 atm .
A short segment of the pipe is constricted to
a smaller diameter of 2.3 cm . What is the gauge pressure of the water
flowing through the constricted segment? Atmospheric pressure is 1.013 × 105 Pa .
The density of water is 1000 kg/m3.
The viscosity
of water is negligible.
Answer in units of atm.
Don't round answer.

Answer 1

$$A_1v_1=A_2v_2$$

$$v_2=v_1\frac{d_1^2}{d_2^2}=4\frac{3.1^2}{2.3^2}=7.27\frac{m}{s}$$

$$P_1+0.5\rho v_1^2=P_2+0.5\rho v_2^2$$

$$P_2=P_1+0.5\rho( v_1^2-v_2^2)$$

$$P_2=1.8(1.013\cdot10^5)+0.5(1000)( 4^2-7.27^2)=1.64\cdot10^5\ Pa$$

$$P_2=1.6181\ atm$$