Fluid Mechanics MCQs

1

Mathematics • Fluid Mechanics

A block of concrete weighs 100 kg in air and weighs 60 kg when immersed in water. The average specific weight of the block is: A block of concrete weighs 100 kg in air and weighs 60 kg when immersed in water. The average specific weight of the block is:

A. 3.2 ton/m^3 3.2 ton/m^3 B. 2.5 ton/m^3 2.5 ton/m^3 C. 2.2 ton/m^3 2.2 ton/m^3 D. 2.1 ton/m^3 2.1 ton/m^3

सही उत्तर: B Correct Answer: B

Loss of weight in water = 100 - 60 = 40 kg, which equals the weight of displaced water. Therefore, relative density = 100/40 = 2.5 and the specific weight is 2.5 ton/m^3.

2

Mathematics • Fluid Mechanics

If the stream function is given by psi = 6x + 12y, then the speed of flow is: If the stream function is given by psi = 6x + 12y, then the speed of flow is:

A. sqrt(180) m s^-1 sqrt(180) m s^-1 B. sqrt(170) m s^-1 sqrt(170) m s^-1 C. 3sqrt(2) m s^-1 3sqrt(2) m s^-1 D. sqrt(0.5) m s^-1 sqrt(0.5) m s^-1

सही उत्तर: A Correct Answer: A

For a two-dimensional stream function, u = partial(psi)/partial(y) = 12 and v = -partial(psi)/partial(x) = -6. Thus speed = sqrt(u^2 + v^2) = sqrt(180) m s^-1.

3

Mathematics • Fluid Mechanics

The continuity equation in Cartesian coordinates is (where the symbols have their usual meanings): The continuity equation in Cartesian coordinates is (where the symbols have their usual meanings):

A. partial(rho)/partial(x) + partial(rho u)/partial(t) + partial(rho v)/partial(y) + partial(rho w)/partial(z) = 0 partial(rho)/partial(x) + partial(rho u)/partial(t) + partial(rho v)/partial(y) + partial(rho w)/partial(z) = 0 B. partial^2(rho)/partial(t)^2 + partial^2(rho u)/partial(x)^2 + partial^2(rho v)/partial(y)^2 + partial^2(rho w)/partial(z)^2 = 0 partial^2(rho)/partial(t)^2 + partial^2(rho u)/partial(x)^2 + partial^2(rho v)/partial(y)^2 + partial^2(rho w)/partial(z)^2 = 0 C. partial^2(rho)/partial(t)^2 + partial(rho u)/partial(x) + partial(rho v)/partial(y) + partial(rho w)/partial(z) = 0 partial^2(rho)/partial(t)^2 + partial(rho u)/partial(x) + partial(rho v)/partial(y) + partial(rho w)/partial(z) = 0 D. partial(rho)/partial(t) + partial(rho u)/partial(x) + partial(rho v)/partial(y) + partial(rho w)/partial(z) = 0 partial(rho)/partial(t) + partial(rho u)/partial(x) + partial(rho v)/partial(y) + partial(rho w)/partial(z) = 0

सही उत्तर: D Correct Answer: D

Conservation of mass gives partial(rho)/partial(t) + div(rho V) = 0, which expands to option D in Cartesian coordinates.

4

Mathematics • Fluid Mechanics

For the flow of a viscous Newtonian fluid between two parallel plates located at y = 0 and y = h, the upper plate is moving with velocity U. The flow field is: For the flow of a viscous Newtonian fluid between two parallel plates located at y = 0 and y = h, the upper plate is moving with velocity U. The flow field is:

A. u(y) = Uy/h u(y) = Uy/h B. u(y) = Uy/(4h) u(y) = Uy/(4h) C. u(y) = U(y)/(2h) u(y) = U(y)/(2h) D. u(y) = Uy/(3h) u(y) = Uy/(3h)

सही उत्तर: A Correct Answer: A

For simple Couette flow with no slip, u(0) = 0 and u(h) = U. The velocity profile is linear, so u(y) = Uy/h.

5

Mathematics • Fluid Mechanics

For a viscous flow of a fluid through a pipe with circular cross-section given by r = a, the pressure gradient is constant, P = -partial(p)/partial(x). If u_x, u_y and u_z are the velocity components, respectively, then the velocity field is: For a viscous flow of a fluid through a pipe with circular cross-section given by r = a, the pressure gradient is constant, P = -partial(p)/partial(x). If u_x, u_y and u_z are the velocity components, respectively, then the velocity field is:

A. (P/(4mu))(a^2 - r^2), 0, 0 (P/(4mu))(a^2 - r^2), 0, 0 B. (P/(2mu))(a^2 + r^2), 0, 0 (P/(2mu))(a^2 + r^2), 0, 0 C. (P/mu)(a^2 - r^2), 0, 0 (P/mu)(a^2 - r^2), 0, 0 D. (P/(3mu))(a^2 + r^2), 0, 0 (P/(3mu))(a^2 + r^2), 0, 0

सही उत्तर: A Correct Answer: A

For steady Hagen-Poiseuille flow in a circular pipe, the axial velocity profile is u_x = (P/(4mu))(a^2 - r^2), while the transverse velocity components are zero.

6

Mathematics • Fluid Mechanics

The metacentric height of a rectangular boat 6 x 6 x 12, a quarter portion of which is submerged in water, is: The metacentric height of a rectangular boat 6 x 6 x 12, a quarter portion of which is submerged in water, is:

A. 1.5 1.5 B. 3.5 3.5 C. 4.5 4.5 D. 2.5 2.5

सही उत्तर: B Correct Answer: B

Using the displaced volume, centre of buoyancy and water-plane moment of inertia for the rectangular boat gives a metacentric height of 3.5 units.

7

Mathematics • Fluid Mechanics

For a blood vessel of constant radius R, length L and driving force P = P1 - P2, where P1 and P2 denote the pressures at either end of length L, the average flow is equal to half the maximum velocity and the resistance (P2 - P1)/v is proportional to: For a blood vessel of constant radius R, length L and driving force P = P1 - P2, where P1 and P2 denote the pressures at either end of length L, the average flow is equal to half the maximum velocity and the resistance (P2 - P1)/v is proportional to:

A. L/R^3 L/R^3 B. L/R^4 L/R^4 C. L/R^2 L/R^2 D. L/R L/R

सही उत्तर: B Correct Answer: B

Poiseuille's law gives volumetric flow proportional to the pressure difference times R^4/L. Therefore hydraulic resistance is proportional to L/R^4.

8

Mathematics • Fluid Mechanics

In the one-dimensional equation of motion, Bernoulli's equation is given by, where v^2/2 is kinetic energy, p/rho is flow energy and gz is potential energy, all per unit mass: In the one-dimensional equation of motion, Bernoulli's equation is given by, where v^2/2 is kinetic energy, p/rho is flow energy and gz is potential energy, all per unit mass:

A. v^2/2 - p/rho - gz = constant v^2/2 - p/rho - gz = constant B. v^2/2 + p/rho - gz = constant v^2/2 + p/rho - gz = constant C. v^2/2 - p/rho + gz = constant v^2/2 - p/rho + gz = constant D. v^2/2 + p/rho + gz = constant v^2/2 + p/rho + gz = constant

सही उत्तर: D Correct Answer: D

Bernoulli's equation states that kinetic energy, pressure-flow energy and gravitational potential energy per unit mass sum to a constant along a streamline.

9

Mathematics • Fluid Mechanics

A heavy car lands at the bottom of a lake on its wheels. Its door is 1.2 m high and 1 m wide, and the top edge of the door is 8 m below the free surface of the water. The hydrostatic force on the door is: A heavy car lands at the bottom of a lake on its wheels. Its door is 1.2 m high and 1 m wide, and the top edge of the door is 8 m below the free surface of the water. The hydrostatic force on the door is:

A. 101.3 kN 101.3 kN B. 104.3 kN 104.3 kN C. 102.3 kN 102.3 kN D. 100.4 kN 100.4 kN

सही उत्तर: A Correct Answer: A

The centroid of the rectangular door is at depth 8.6 m and its area is 1.2 m^2. Thus F=rho g A h_c = 1000 x 9.81 x 1.2 x 8.6, approximately 101.3 kN.

Fluid Mechanics MCQs

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