Solids, Fluids and Fluid Motion

Question 1

What is a fluid?

Answer 1

A substance which cannot resist a shear force (stress) without moving.

Question 2

What is the equation for stress on a solid?

Answer 2

τ = F/A

Question 3

What is the equation for shear strain?

Answer 3

e = dx/dy = tanθ ~ θ (matrix of terms as also have dy/dz, dx/dz etc)

Question 4

What is the equation for stress in terms of shear modulus G?

Answer 4

τ = Ge, where e is the shear strain and G is the shear modulus.

Question 5

What is the shear stress equal to for fluids and why?

Answer 5

τ = μ de/dt, because e increases steadily with tie for a viscous fluid between two plates, where v increases as you move up between the plates, where μ is the dynamic viscosity.

Question 6

How can we draw the representation of a fluid with height y for the fluid between 2 plates?

Answer 6

• Bottom at v and top at v+Δv

- Starts as square and square is pushed over, where the extra length is ΔvΔt, and the angle is θ

Question 7

How can we use the representation of the fluid of height y between 2 plates to get another equation for τ?

Answer 7

• ΔvΔt = θΔy, but θ = dx/dy = e if v=0, and if v/=0, θ = Δe (change in shear between y and y+Δy)
• Hence de/dt = dv/dy, and τ = μ dv/dy

Question 8

What is the dynamic viscosity μ dependent on for Newtonian and non-Newtonian fluids?

Answer 8

• For newtonian fluids is independent of v

- For non-newtonian, is μ = μ(v)

Question 9

What are the two ways to describe fluid dynamics?

Answer 9

Fluid particle approach (Lagrangian) and the fixed control volume approach (Eulerian).

Question 10

How does the fluid particle approach work?

Answer 10

• Small fixed mass of fluid
• v = d/dt (r-r0) = dx/dt i + dy/dt j + dz/dt k
• a = dv/dt
• Often complex

Question 11

How does the fixed control volume approach work?

Answer 11

• Describes the flow at fixed points as a function of time

- Fixed volume and is fixed in space (particles leaving and entering all the time)

Question 12

What does laminar mean?

Answer 12

Smooth surfaces of flow.

Question 13

What is a good example to show fluid dynamics?

Answer 13

• Flow between two plates with pressure difference P1 on left > P2 on right
• vy,vz = 0, vx = vx i
• Experiment shows fluid in contact with walls at rest, so vx = vx(y)

Question 14

How would you find the viscous force on a fluid element?

Answer 14

• Net viscous force Fv = Fv(top)-Fv(bottom) = (Fv(t)-Fv(b)) i
• Use F = μ dvx/dy * A, where A = ΔxΔz
• |F| = μ d^2vx/dy^2 * ΔxΔyΔz

Question 15

What is the first step in finding the pressure force on the fluid element?

Answer 15

• Fp = (P(x) - P(x+Δx)) ΔyΔz = -dP/dx ΔxΔyΔz
• Steady flow so force balances Fp+Fv = 0
• dP/dx must = μ d^2vx/dy^2

Question 16

What is the second step in finding the pressure force on a fluid element?

Answer 16

• μ d^2vx/dy^2 = -Q
• vy(y) = -Q/2μ * y^2 + A*y + B
• vy(a) = 0, vy(-a) = 0, so A = 0 and B = Qa^2/2μ
• Q = -dP/dx = (P1-P2)/L, where L is the length of the plates

Question 17

How would you find the flow rate of the fluid element?

Answer 17

• In unit time fluid moves vx and volume crossing an area ΔyΔz is vx ΔyΔz, so flow is ρ vx*ΔyΔz
• Δz = 1, so total mass/time is integral from -a to a of ρ vx(y) dy = 2ρ(P1-P2)*a^3/3μL