"A liquid with specific gravity 0.8 flows at the rate of 3 l/s through a venturimeter of diameters 6 cm and 4 cm. if the manometer fluid is mercury (sp. gr=13.6) determine the value of manometer reading,h.
The specific gravity of an unknown substance is 0.80. Will it float on or sink in gasoline?
yes
To determine the value of the manometer reading, h, in this particular scenario, we need to use the principle of continuity equation and Bernoulli's theorem.
Here are the steps to solve the problem:
Step 1: Understand the given information.
- Specific gravity of the liquid: 0.8
- Flow rate of the liquid: 3 l/s
- Diameter of the wider part of the venturimeter: 6 cm
- Diameter of the narrower part of the venturimeter: 4 cm
- Specific gravity of mercury: 13.6
Step 2: Calculate the cross-sectional areas of the venturimeter.
The cross-sectional area of a circle can be calculated using the formula A = π * r^2, where r is the radius of the circle. In this case, we have two different diameters, so we need to calculate the radii and then use the formula to find the areas of the venturimeter.
- Radius of the wider part: r1 = 6 cm / 2 = 3 cm = 0.03 m
- Area of the wider part: A1 = π * (0.03 m)^2 = 0.002827 m^2
- Radius of the narrower part: r2 = 4 cm / 2 = 2 cm = 0.02 m
- Area of the narrower part: A2 = π * (0.02 m)^2 = 0.001256 m^2
Step 3: Apply the principle of continuity equation.
According to the principle of continuity equation, the product of the cross-sectional area and the velocity of a fluid remains constant during its flow through a pipe of varying diameter. Mathematically, A1 * v1 = A2 * v2, where v1 and v2 are the velocities of the fluid at sections 1 and 2, respectively.
Step 4: Calculate the velocity of the fluid at the wider and narrower sections of the venturimeter.
We are given the flow rate, Q, as 3 l/s. To find the velocities, we can use the formula Q = A * v, where A is the cross-sectional area and v is the velocity.
- Velocity at the wider part: v1 = Q / A1 = 3 l/s / 0.002827 m^2 ≈ 1061.75 m/s
- Velocity at the narrower part: v2 = Q / A2 = 3 l/s / 0.001256 m^2 ≈ 2387.30 m/s
Step 5: Apply Bernoulli's theorem.
According to Bernoulli's theorem, the sum of the pressure energy, kinetic energy, and potential energy per unit volume of a fluid remains constant along a streamline.
- In this case, we can consider the wider section (1) as the reference level, where the pressure is atmospheric pressure (P1) and the velocity is v1.
- At the narrower section (2), the pressure is P2 and the velocity is v2.
Applying Bernoulli's equation: P1 + (1/2)ρv1^2 + ρgh1 = P2 + (1/2)ρv2^2 + ρgh2
Since the venturimeter is open to the atmosphere, we can assume the potential energy terms (ρgh1 and ρgh2) to be zero.
Additionally, the density of the liquid flowing through the venturimeter is given as ρ (symbol not mentioned in the question).
Step 6: Solve for P2.
Rearranging the Bernoulli's equation, we get:
P2 = P1 + (1/2)ρv1^2 - (1/2)ρv2^2
Step 7: Substituting the given values.
- P1 is atmospheric pressure, which is approximately 101,325 Pa (Pascal).
- ρ is the density of the fluid, which is not mentioned in the question.
- v1 = 1061.75 m/s (calculated in step 4)
- v2 = 2387.30 m/s (calculated in step 4)
Since the value of ρ is not provided, we cannot calculate the exact value of P2 without it.
Therefore, without the density of the liquid (ρ), we cannot determine the value of the manometer reading, h, in this specific scenario.