a fluid of a constant density = 960kg/m^3 is flowing steadily through the tube shown. diameter at the sections are d1 = 100mm and d2 = 200mm. the gauge pressure at 1 is p = 200kN/m^2 and the velocity here is u1 = 5m/s
not only can you not post pictures on this web site, but you didn't even ask any questions about the figure.
To solve this problem, we can apply the continuity equation and Bernoulli's equation.
Step 1: Convert the diameter values to meters.
d1 = 100 mm = 0.1 m
d2 = 200 mm = 0.2 m
Step 2: Find the cross-sectional areas of the tube sections.
A1 = π*(d1/2)^2 = π*(0.1/2)^2 = 0.00785 m^2
A2 = π*(d2/2)^2 = π*(0.2/2)^2 = 0.0314 m^2
Step 3: Apply the continuity equation.
The continuity equation states that the mass flow rate of an incompressible fluid is constant, which means that the product of the fluid density (ρ), velocity (u), and cross-sectional area (A) is constant.
ρ1 * u1 * A1 = ρ2 * u2 * A2
Step 4: We are given the fluid density, ρ = 960 kg/m^3. We can rearrange the continuity equation to solve for u2.
ρ1 * u1 * A1 / A2 = u2
Substituting the given values, we have:
960 kg/m^3 * 5 m/s * 0.00785 m^2 / 0.0314 m^2 = u2
Simplifying, we get:
u2 = 1200 m/s
Step 5: Apply Bernoulli's equation.
Bernoulli's equation states that the sum of the pressure energy, kinetic energy, and potential energy per unit mass of a fluid is constant along a streamline.
p1 + ρ * g * h1 + 0.5 * ρ * u1^2 = p2 + ρ * g * h2 + 0.5 * ρ * u2^2
Assuming the fluid is at the same height, and neglecting the potential energy terms, we simplify Bernoulli's equation to:
p1 + 0.5 * ρ * u1^2 = p2 + 0.5 * ρ * u2^2
Substituting the given values, we have:
200 kN/m^2 + 0.5 * 960 kg/m^3 * (5 m/s)^2 = p2 + 0.5 * 960 kg/m^3 * (1200 m/s)^2
Simplifying, we get:
200,000 N/m^2 + 60,000 N/m^2 = p2 + 691,200,000 N/m^2
260,000 N/m^2 = p2 - 691,200,000 N/m^2
p2 = 260,000 N/m^2 + 691,200,000 N/m^2
p2 = 691,460,000 N/m^2
So, the gauge pressure at section 2 is p = 691,460,000 N/m^2.
To find the velocity at section 2, we can use the principle of continuity, which states that the mass flow rate of an incompressible fluid remains constant within a closed system.
The mass flow rate (m_dot) is given by the formula: m_dot = ρ * A * u
Where:
- ρ is the density of the fluid,
- A is the cross-sectional area of the tube,
- u is the velocity of the fluid.
In this case, the density (ρ) is constant at 960 kg/m^3.
Let's find the cross-sectional area at each section first:
A1 = π * (d1/2)^2 = π * (0.1/2)^2 = 0.007854 m^2
A2 = π * (d2/2)^2 = π * (0.2/2)^2 = 0.031416 m^2
Now we can use the principle of continuity to find the velocity (u2) at section 2:
m_dot = ρ * A1 * u1 = ρ * A2 * u2
Since the fluid density (ρ) is constant, we can cancel it out:
A1 * u1 = A2 * u2
Solving for u2:
u2 = (A1 * u1) / A2
u2 = (0.007854 m^2 * 5 m/s) / 0.031416 m^2
u2 ≈ 1 m/s
Therefore, the velocity at section 2 is approximately 1 m/s.