NAST Questions
Question 1
A) Define Newton's law of viscosity. Explain types of fluid. [8]
B) Two horizontal plates are placed 1.25 cm apart, the space between them being filled with oil of viscosity 14 poise. Calculate the shear stress in oil if the upper plate is moved with a velocity of 2.5 m/s. [7]
Question 2
A) Define Pascal's law. Derive expression for Pascal's law. [8]
B) An open tank contains water up to a depth of 2 m and above it an oil of sp. gr. 0.9 for a depth of 1 m. Find the pressure intensity
i) at the interface of two liquids
ii) at the bottom of the tank. [3+4]
Question 3
A) Define hydrostatic force. Derive an expression for hydrostatic force on a vertical plane surface submerged in liquid. Also determine the expression for the center of pressure. [1+6]
B) A rectangular plane surface 3 m wide and 4 m deep lies in water in such a way that its plane makes an angle of 30° with the free surface of water. Determine the total pressure force and position of center of pressure, when the upper edge is 2 m below the free surface. [8]
Question 4
A) A tank containing water up to 500 mm is moving vertically upward with a constant acceleration of 2.45 m/s². Find the force exerted by the water on the side of the tank. Also calculate the force on the side of the tank when the width of the tank is 2 m. Also calculate for the tank moving downward vertically with a constant acceleration of 2.45 m/s² and also for the tank at stationary case. [3+3+3]
B) Define buoyancy, center of buoyancy, metacenter, and metacentric height. [1.5+1.5+1.5+1.5]
Answer 1
A) Newton's Law of Viscosity
Newton's law of viscosity describes the relationship between the shear stress (τ) and the velocity gradient (du/dy) in a fluid. It is expressed as:
τ = μ du/dy
where:
- τ is the shear stress
- μ is the dynamic viscosity of the fluid
- du/dy is the velocity gradient perpendicular to the direction of flow
Types of Fluids
Fluids are classified into two main types based on their behavior under stress:
1. Newtonian Fluids
- Follow Newton's law of viscosity
- Shear stress is directly proportional to the velocity gradient
- Examples: water, most gases
2. Non-Newtonian Fluids
- Do not follow Newton's law of viscosity
- Complex relationship between shear stress and velocity gradient
- Subcategories: shear-thinning, shear-thickening
- Examples: ketchup, toothpaste, blood
B) Calculation of Shear Stress
The formula for shear stress in a Newtonian fluid:
τ = μ du/dy
Given:
- Distance between plates (d) = 1.25 cm = 0.0125 m
- Viscosity (μ) = 14 poise = 1.4 Pa·s
- Velocity (u) = 2.5 m/s
Velocity gradient (du/dy):
du/dy = 2.5 / 0.0125 = 200 s-1
Shear stress (τ):
τ = 1.4 Pa·s × 200 s-1 = 280 Pa
The shear stress in the oil is 280 Pa.
Answer 2
A) Pascal's Law
Pascal's Law states that any change in pressure applied to an enclosed fluid is transmitted undiminished throughout the fluid and to the container's walls. In simpler terms, pressure changes at one point within an enclosed fluid affect the entire fluid equally in all directions.
Derivation of Pascal's Law
Imagine a small element inside the fluid. Forces acting on it are pressure forces on its surfaces. For the fluid to be in equilibrium, the net force in any direction must be zero. Applying this principle to all directions, we get:
\frac{F_1}{A_1} = \frac{F_2}{A_2} = \text{constant}
This implies that pressure at a specific depth in a fluid is the same in all directions.B) Pressure Intensity Calculation
Given:
- Water depth (h1) = 2 meters
- Oil depth (h2) = 1 meter
- Specific gravity of oil (sp. gr.) = 0.9
i) Pressure Intensity at the Interface
Pressure at the interface of two liquids is calculated using the hydrostatic pressure formula:
P_1 = P_{\text{atm}} + \rho_1 g h_1
P_2 = P_{\text{atm}} + \rho_2 g h_2
where: * P1 and P2 are pressures in water and oil, respectively * Patm is atmospheric pressure * ρ1 and ρ2 are densities of water and oil, respectively * g is acceleration due to gravity Pressure at the interface (Pinterface) is identical for both fluids:Pinterface = P1 = P2
ii) Pressure Intensity at the Bottom
Pressure at the tank's bottom is the sum of pressures in water and oil:
Pbottom = P1 + P2
Substitute the expressions for P1 and P2 into the equation. Finally, convert pressure to pressure intensity by dividing by the total height of the fluid column:Pressure Intensity = Pbottom / (h1 + h2)
Perform calculations with the given values to find the pressure intensity.Answer 3
A) Hydrostatic Force and Expression Derivation
Hydrostatic Force
Hydrostatic force is the force exerted by a fluid at rest on a submerged or partially submerged surface, acting perpendicular to the surface.
Expression for Hydrostatic Force on a Vertical Plane Surface
For a vertical plane surface submerged in a fluid, the hydrostatic force (F_h) is calculated as:
F_h = P \cdot A
where:
- P is the hydrostatic pressure at the centroid of the surface
- A is the area of the submerged surface
Hydrostatic pressure (P) at a depth (h) in a fluid is given by:
P = \rho \cdot g \cdot h
where:
- \rho is the fluid density
- g is the acceleration due to gravity
- h is the depth of the centroid
Expression for the Center of Pressure
The center of pressure (h_c) is the depth where the resultant force acts. For a vertical surface:
h_c = \frac{I_g}{A}
where:
- I_g is the moment of inertia about the base
- A is the area of the surface
B) Calculation for Given Scenario
Given:
- Width (w) = 3 m
- Depth (d) = 4 m
- Angle with free surface (\theta) = 30°
- Upper edge depth (h) = 2 m
i) Total Pressure Force
Use the formula:
F_h = P \cdot A
Calculate hydrostatic pressure at the centroid:
h = h_{\text{upper edge}} + \frac{d}{2}
P = \rho \cdot g \cdot h
Calculate the area (A) and find the total pressure force.
ii) Position of Center of Pressure
Determine the moment of inertia (I_g) and use:
h_c = \frac{I_g}{A}
Perform calculations to find the total pressure force and the position of the center of pressure.
Answer 4
A) Force Exerted by Water on the Side of the Tank
Given:
- Height of water (h) = 500 mm = 0.5 m
- Acceleration (a) = 2.45 m/s²
i) Tank Moving Upward
The force exerted by the water on the side of the tank is:
F_{\text{total}} = \rho \cdot g \cdot h + \rho \cdot a \cdot h
ii) Tank Moving Downward
The force exerted by the water on the side of the tank is:
F_{\text{total}} = \rho \cdot g \cdot h - \rho \cdot a \cdot h
iii) Tank at Stationary Case
The force exerted by the water on the side of the tank is:
F_{\text{stationary}} = \rho \cdot g \cdot h
iv) Tank Width Increased
If the width of the tank is W = 2 m, the force exerted on the side of the tank is:
F_{\text{total}} = \rho \cdot g \cdot h + \rho \cdot a \cdot h \cdot \frac{W}{2}
B) Definitions
Buoyancy
Buoyancy is the upward force exerted by a fluid on an object immersed in it, equal to the weight of the displaced fluid.
Center of Buoyancy
The center of buoyancy is the centroid of the displaced fluid volume, where the buoyant force acts vertically upward.
Metacenter
The metacenter is the point in a floating body where the buoyant force intersects the centerline of the body when tilted.
Metacentric Height
Metacentric height is the vertical distance between the center of gravity (G) and the metacenter (M) of a floating body, measuring its stability.
A) Density of Metallic Body
Given:
- The metallic body floats at the interface of mercury and water.
- 40% of its volume is submerged in mercury, and 60% is in water.
Let's denote:
- V_m as the volume submerged in mercury
- V_w as the volume submerged in water
- ρ_m as the density of mercury
- ρ_w as the density of water
- ρ as the density of the metallic body
The total volume is:
V_{\text{total}} = V_m + V_w
40% in mercury and 60% in water:
V_m = 0.4 \cdot V_{\text{total}}
V_w = 0.6 \cdot V_{\text{total}}
Buoyant forces:
F_{\text{buoyant, mercury}} = \rho_m \cdot g \cdot V_m
F_{\text{buoyant, water}} = \rho_w \cdot g \cdot V_w
Since the body floats, buoyant forces equal its weight:
F_{\text{buoyant, mercury}} = F_{\text{weight}} = \rho \cdot g \cdot V_{\text{total}}
F_{\text{buoyant, water}} = F_{\text{weight}} = \rho \cdot g \cdot V_{\text{total}}
Solving for ρ:
ρ = \frac{\rho_m \cdot V_m}{V_{\text{total}}} = 0.4 \cdot \rho_m
B) Metacentric Height of Wooden Block
Given:
- Specific gravity (SG) of wood = 0.7
- Dimensions: length (L) = 2m, width (B) = 1m, height (H) = 0.8m
Specific gravity:
\text{SG} = \frac{\rho_{\text{wood}}}{\rho_{\text{water}}}
Density of wood:
\rho_{\text{wood}} = \text{SG} \cdot \rho_{\text{water}}
Weight of the block:
W = \rho_{\text{water}} \cdot g \cdot V_{\text{submerged}}
Volume submerged:
V_{\text{submerged}} = L \cdot B \cdot H_{\text{submerged}}
Metacentric height (GM):
r), azimuthal (θ), and axial (z).
Continuity equation in Cartesian coordinates:
\frac{\partial(\rho u)}{\partial x} + \frac{\partial(\rho v)}{\partial y} + \frac{\partial(\rho w)}{\partial z} = 0
Expressing velocities in cylindrical polar coordinates:
u = u_r \cos(\theta) - u_{\theta} \sin(\theta)
v = u_r \sin(\theta) + u_{\theta} \cos(\theta)
w = u_z
Substitution and manipulation yield the continuity equation in cylindrical polar coordinates:
\frac{1}{r} \frac{\partial (r \rho u_r)}{\partial r} + \frac{1}{r} \frac{\partial (\rho u_{\theta})}{\partial \theta} + \frac{\partial (\rho u_z)}{\partial z} = 0
B) Differential Manometer
Given:
- Specific gravity of liquid in Pipe A (SG_A) = 1.5
- Specific gravity of liquid in Pipe B (SG_B) = 0.9
- Pressure at point A (P_A) = 1 kgf/cm²
- Pressure at point B (P_B) = 1.8 kgf/cm²
Pressure difference between A and B (ΔP):
ΔP = P_A - P_B
Difference in mercury level (Δh) in the manometer:
ΔP = ρ_Hg g Δh
where:
- ρ_Hg is the density of mercury
- g is the acceleration due to gravity
Calculate density of mercury using its specific gravity:
ρ_Hg = SG_Hg
i. Stability Condition on Floating Body
The stability of a floating body is crucial for understanding its equilibrium and behavior on the surface of a fluid. The stability condition involves the concept of the metacentric height (GM).
GM is the vertical distance between:
- The center of gravity (G)
- The metacenter (M), where the buoyant force intersects the centerline of the floating body
Stability Conditions
- GM and Stability:
- If GM > 0, the metacenter is above the center of gravity, indicating a stable equilibrium.
- If GM < 0, the metacenter is below the center of gravity, indicating an unstable equilibrium.
- GM and Tilting:
- A larger GM provides greater stability against tilting.
- Stability increases with an increase in GM.
- Stability in Real-world Applications:
- Ships and boats are designed to have a positive GM to ensure stability during various conditions, such as waves and loading/unloading.
ii. Continuity Equation
The continuity equation is a fundamental principle in fluid dynamics that expresses conservation of mass:
The mass flow rate into a system equals the mass flow rate out of the system for steady, incompressible flow.
Equation:
\frac{\partial (\rho A u)}{\partial x} + \frac{\partial (\rho A v)}{\partial y} + \frac{\partial (\rho A w)}{\partial z} = 0
Where:
- \rho is the density of the fluid
- A is the cross-sectional area of the flow
- u, v, w are the velocity components in the respective directions
The continuity equation ensures mass conservation, providing a basis for understanding and analyzing fluid flow.
iii. Gauge Pressure, Absolute Pressure, Vacuum Pressure
- Gauge Pressure:
- Pressure measured above atmospheric pressure.
- Equation: \text{Gauge Pressure} = \text{Absolute Pressure} - \text{Atmospheric Pressure}
- Commonly used in pressure measurements in devices like pressure gauges.
- Absolute Pressure:
- Total pressure at a point, including atmospheric pressure.
- Equation: \text{Absolute Pressure} = \text{Gauge Pressure} + \text{Atmospheric Pressure}
- Important in fluid dynamics and thermodynamics.
- Vacuum Pressure:
- Pressure below atmospheric pressure.
- Expressed as a negative gauge pressure or as an absolute pressure less than atmospheric pressure.
- Commonly used in applications involving vacuum