ANSWER
[tex]1.45atm[/tex]
EXPLANATION
Parameters given:
Speed of water at point 1, v1 = 5.1 m/s
Diameter of pipe at point 1, d1 = 3.5 cm = 0.035 m
Pressure, P1 = 1.9 atm
Diameter of pipe at point 2, d2 = 2.4 cm = 0.024 m
Atmospheric pressure = 1.013 * 10^5 Pa
Density of water = 1000 kg/m^3
First, we have to find the speed of the water at the smaller end of the pipe. To do this, apply the Bernoulli continuity equation:
[tex]A_1v_1=A_2v_2[/tex]
where A1 and A2 are the cross-sectional areas of the pipes at point 1 and point2 respectively.
Substituting the given values into the equation above:
[tex]\begin{gathered} (\pi\cdot(\frac{0.035}{2})^2)\cdot5.1=(\pi\cdot(\frac{0.024}{2})^2)\cdot v_2 \\ \Rightarrow v_2=\frac{0.0175^2\cdot5.1}{0.012^2}^{} \\ v_2=10.85m\/s \end{gathered}[/tex]
Now, we have to apply Bernoulli's equation to find the pressure at the constricted segment:
[tex]\begin{gathered} P_1+0.5\rho v^2_1=P_2+0.5\rho v^2_2 \\ \Rightarrow P_2=P_1+0.5\rho(v^2_1-v^2_2) \end{gathered}[/tex]
Substitute the known values into the equation and solve for P2, we have:
[tex]\begin{gathered} P_2=(1.9)(1.013\cdot10^5)+(0.5)(1000)(5.1^2-10.85^2) \\ P_2=192,470+(500)(26.01-117.72)=192,470+(500)(-91.71) \\ P_2=192,470-45,855 \\ P_2=146,615Pa=\frac{146615}{1.013\cdot10^5}atm \\ P_2=1.45atm \end{gathered}[/tex]
That is the answer.
