Answer:
The absolute pressure of the water in the pipe is 1.38 x 10⁵ Pa
Explanation:
Given;
radius of the pipe, r₁ = 1.8 cm = 0.018 m
radius of the pipe, r₂ = 0.53 cm = 0.0053 m
speed of water in the pipe, v₁ = 0.75 m/s
Water absolute pressure can be determined using Bernoulli's equation;
P₁ + ¹/₂ρv₁² = P₂ + ¹/₂ρv₂²
P₁ = P₂ + ¹/₂ρv₂² - ¹/₂ρv₁²
P₁ = P₂ + ¹/₂ρ (v₂² - v₁²)
where;
ρ is density of water = 1000 kg/m³
P₂ is atmospheric pressure = 1.01 x 10⁵ Pa
From continuity equation; A₁V₁ = A₂V₂
πr₁²v₁² = πr₂²v₂²
[tex]v_2 = \frac{ r_1^2v_1}{r_2^2}[/tex]
[tex]P_1 = P_2 + \frac{1}{2} \rho[(\frac{r_1^2v_1}{r_2^2} )^2 - v_1^2]\\\\P_1 = P_2 + \frac{1}{2} \rho[\frac{r_1^4v_1^2}{r_2^4} - v_1^2]\\\\P_1 = P_2 + \frac{1}{2} \rho v_1^2[\frac{r_1^4}{r_2^4} - 1]\\\\P_1 = 1.01*10^5 + \frac{1}{2}* 1000* 0.75^2[\frac{(0.018)^4}{(0.0053)^4} - 1]\\\\P_1 = 1.01*10^5 + \frac{1}{2}* 1000* 0.75^2(132.04)\\\\P_1 = 1.01*10^5 +37136.25 \\\\P_1 = 1.38 *10^5 \ Pa[/tex]
Therefore, the absolute pressure of the water in the pipe is 1.38 x 10⁵ Pa
