Question

Glycerine of density 1.25 × 10³ kg m^-3 is flowing through the conical section of pipe. The area of cross-section of the pipe at its ends is 10 cm2 and 5 cm2 and pressure drop across its length is 3 Nm^-2. The rate of flow of glycerine through the pipe is x x10^-5 m3 s^-1. The value of x is____.

Answer 1

Correct Answer: 4

To solve the problem, we need to find the rate of flow of glycerin through a conical pipe given the density, cross-sectional areas, and pressure drop. Let's break this down step by step.
Step 1: Understand the given data
- Density of glycerin, ρ = 1.25 × 10³ kg/m³
- Area at one end of the pipe, A₁ = 10 cm² = 10 × 10^-4 m² = 1 × 10^-3 m²
- Area at the other end of the pipe, A₂ = 5 cm² = 5 × 10^-4 m²
- Pressure drop, ΔP = 3 N/m²
Step 2: Apply the continuity equation
Using the principle of conservation of mass, we can apply the continuity equation:
A₁v₁ = A₂v₂
From this, we can express v₂ in terms of v₁:
v₂ = (A₁/A₂)v₁ = (10 × 10^-4 / 5 × 10^-4)v₁ = 2v₁
Step 3: Apply Bernoulli's equation
According to Bernoulli's equation, we have:
P₁ + (1/2)ρv₁² = P₂ + (1/2)ρv₂²
Rearranging gives us:
P₁ - P₂ = (1/2)ρ(v₂² - v₁²)
Substituting ΔP = P₁ - P₂ = 3 N/m² and v₂ = 2v₁:
3 = (1/2)ρ((2v₁)² - v₁²)
3 = (1/2)ρ(4v₁² - v₁²) = (1/2)ρ(3v₁²)
3 = (3/2)ρv₁²
Step 4: Solve for v₁
Substituting the value of ρ:
3 = (3/2)(1.25 × 10³)v₁²
3 = (3 × 1.25 × 10³ / 2)v₁²
v₁² = (3 × 2) / (3 × 1.25 × 10³) = 2 / (1.25 × 10³)
v₁² = 2 / 1250 = 1 / 625 (m/s)²
v₁ = √(1/625) = 1/25 = 0.04 m/s
Step 5: Calculate the rate of flow
The rate of flow Q is given by:
Q = A₁v₁
Substituting A₁ = 1 × 10^-3 m² and v₁ = 0.04 m/s:
Q = (1 × 10^-3)(0.04) = 4 × 10^-5 m³/s
Step 6: Express in terms of x
The rate of flow is given as x × 10^-5 m³/s:
Q = 4 × 10^-5 ⇒ x = 4
Final Answer
The value of x is 4.