Final answer:
To calculate the pressure at the entrance end of the pipe, we can apply Bernoulli's principle and the equation of continuity. First, we can find the speed of the liquid at the exit end of the pipe using the equation of continuity. Then, we can use Bernoulli's equation to calculate the pressure at the entrance end of the pipe. The pressure at the entrance end of the pipe is approximately 1.08 × 10^5 Pa.
Step-by-step explanation:
To calculate the pressure at the entrance end of the pipe, we can apply Bernoulli's principle, which states that the total energy of a fluid flowing in a pipe is constant. Bernoulli's equation is given by:
P1 + (1/2)ρv1^2 + ρgh1 = P2 + (1/2)ρv2^2 + ρgh2
In this equation, P1 and P2 are the pressures at the entrance and exit ends of the pipe, ρ is the density of the liquid, v1 and v2 are the speeds of the liquid at the entrance and exit ends, g is the acceleration due to gravity, and h1 and h2 are the heights of the liquid at the entrance and exit ends, respectively.
Using the given values, we can calculate:
- ρ = 1136 kg/m³
- v1 = 1.06 m/s
- v2 = ?
- D1 = 0.26 m
- D2 = 0.05 m
- P2 = 1.2 atm = 1.2 × 101325 Pa
- h2 - h1 = 4.25 m
First, we need to calculate v2 using the equation of continuity, which states that the mass flow rate is constant:
A1v1 = A2v2
Using the diameters of the pipe at the entrance and exit ends, we can find the areas:
- A1 = (π/4)(D1^2)
- A2 = (π/4)(D2^2)
Substituting the given values, we can solve for v2:
A1v1 = A2v2
[(π/4)(0.26^2)](1.06) = [(π/4)(0.05^2)]v2
v2 = [(π/4)(0.26^2)](1.06) / [(π/4)(0.05^2)]
v2 ≈ 22.22 m/s
Now that we have v2, we can substitute all the values into Bernoulli's equation and solve for P1:
P1 + (1/2)ρv1^2 + ρgh1 = P2 + (1/2)ρv2^2 + ρgh2
Substituting the given values:
P1 + (1/2)(1136)(1.06^2) + (1136)(9.8)(0) = (1.2 × 101325) + (1/2)(1136)(22.22^2) + (1136)(9.8)(4.25)
Solving for P1:
P1 ≈ 1.08 × 10^5 Pa
Therefore, the pressure at the entrance end of the pipe is approximately 1.08 × 10^5 Pa.