Answer:
a) [tex]V_2=8m/s[/tex]
b) [tex]P_2=9.54*10^4 Pa[/tex]
Explanation
From the question we are told that:
Initial Area of pipe [tex]A_1=8.00 cm^2[/tex]
Initial Fluid flow speed [tex]r_1 =320 cm/s,\approx 320*10^{-2}[/tex]
Initial Pressure of [tex]\rho_1 =1.40*10^5 Pa[/tex]
Final area of pipe [tex]A_2 =3.70 *10^{-2} cm^2[/tex]
Density of acid [tex]\rho=1660kg/m^3[/tex]
a)
Generally the equation for continuity is mathematically given by
[tex]A_1V_1=A_2V_2\\\\V_2=\frac{A_1*V_1}{A_2}[/tex]
Since volume is directly proportional to rate of flow
[tex]V_2=\frac{8*320}{3.20} *10^{-2}[/tex]
[tex]V_2=8m/s[/tex]
b)
Generally the Bernoulli's equation is mathematically given by
[tex]p_1+\frac{1}{2}\rho v_1^2+\rho gh_1=P_2+\frac{1}{2}\rho v_2^2+\rho gh_2\\\\with\ h_1=h_2\\\\p_1+\frac{1}{2}\rho v_1^2=P_2+\frac{1}{2}\rho v_2^2[/tex]
Therefore
[tex]P_2=P_1+\frac{1}{2}\rho(v_1^2-V_2^2)\\\\P_2=(1.40*10^5)+\frac{1}{2}(1660)(v_1^2-V_2^2)[/tex]
[tex]P_2=9.54*10^4 Pa[/tex]
