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FLUID MECHANICS I Solution 4 Question 1: Problem P2.139

Figure 1: Free surface for fluid in solid body acceleration. Since there is no acceleration in the y direction: pA

= pB + ρghB

pC

= pD + ρghD

where pB = pD = patm = 0[P a gage]. From Table A.3 for glycerin ρ = 1260[kg/m3 ], so pA = 3460[P a gage]. 1

For the solid body motion in the x direction: ρax = −

∂p ∂x

which can be integrated from point A to point C: Z x=L Z x=L ∂p ρax dx = − dx ∂x x=0 x=0 ρax L = −(pC − pA ) ρax L = −ρg(hD − hB ) Rearranging for ax gives: ax = g

hB − hD = 1.27[m/s2 ] L

Notice that the value of the acceleration because the density of the fluid affects both the fluid inertia and the variation of pressure with depth.

Question 2: Problem P2.156 Since pressure increases with increasing depth and pressure increases with increasing radius for fluid in solid body rotation, the minimum pressure in the U tube is at point D. Notice also that pB − pC = pA − pD since the heights of each leg of the U tube are identical. The equation of motion in the radial direction is: ρar = −ρΩ2 r = −

∂p ∂r

which can be integrated from point C to point B: Z r=R Z r=R ∂p − ρΩ2 rdr = − dr r=0 r=0 ∂r R2 −ρΩ2 = −(pB − pC ) 2 Rearranging for Ω gives: s Ω = s =

2(pB − pC ) ρR2 2(pA − pD ) ρR2

From Table A.5 for water at 50◦ C the vapour pressure is pvp = 12340[P a]. Atmospheric pressure is patm = 2116[psf a] = 101340[P a]. Using the correlation supplied in Table A.1 the density of water at 50◦ C is ρ = 988[kg/m3 ]. Final Answer: Ω = 44[rad/s]. 2

Figure 2: U tube rotating about the DC axis.

Question 3 Figure 8.14b page 546 in the textbook shows the streamlines for the flow of air at 20[◦ C] in a corner. The velocity field for this flow is approximately: u(x, y) = v(x, y) =

V0 x L V0 − y L

where V0 = 10[m/s] is a velocity scale, L = 2[m] is a length scale, x is positive to the right, and y is positive upwards. Gravity acts in the z direction. Friction plays a negligible role in establishing the pressure field of this flow. 1. Show that the pressure field for this approximately: p(x, y) = p0 −

¢ ρV02 ¡ 2 x + y2 2 2L

where p0 is the pressure at (0[m], 0[m]). 3

If this is a valid pressure field the x and y components of the equations of motion must balance. For the x equation of motion: Du Dt ∂p ∂x

∂u ∂u ∂u ∂u V2 +u +v +w = 02 x ∂t ∂x ∂y ∂z L V02 −ρ 2 x L

≡ = ?

Du z}|{ = Dt V2 Therefore ρ 02 x = L ρ

∂p x Equation ∂x V2 ρ 02 x L −

Similarly for the y equation of motion: Dv Dt ∂p ∂y

∂v ∂v ∂v ∂v V2 +u +v +w = 02 y ∂t ∂x ∂y ∂z L V02 −ρ 2 y L

≡ = ?

Dv Dt V2 Therefore ρ 02 y L

z}|{ =

ρ

=

∂p y Equation ∂y V2 ρ 02 y L −

2. If the pressure at (2[m], 0[m]) is atmospheric pressure (105 [P a]), what is p0 . p(2[m], 0[m]) = patm patm

¡ ¢ ρV02 (2[m])2 + (0[m])2 2 2(L = 2[m]) ρV02 = p0 − 2 = p0 −

Final Answer: p0 = patm +

ρV02 2

= pa tm + 60[P a].

3. Estimate the net force due to pressure acting on the wall 0[m] < x < 2[m]. The wall has a unit depth in the z direction and atmospheric pressure acts on the lower face of the wall.

δfy

=

Fy

=

−[p(x, 0) − patm ]dx · 1 Z 2[m] − [p(x, 0) − patm ]dx · 1 Z

=

0[m] 2[m]

−

[p0 − patm − 0[m]

4

¢ ρV02 ¡ 2 x + (0[m])2 ]dx 2 2L

Figure 3: Freebody diagram of pressure forces on the wall.

2[m]

=

− [p0 − patm ]x|0[m] +

=

−4

¯2[m] ρV02 3 ¯¯ x 6L2 ¯0[m]

4 ρV02 ρV02 + L2 3 L2

Final answer: The net force due to pressure on the wall is Fy = − 83 −80[N/m] which acts downwards.

5

ρV02 L2

=

Figure 4: Geometry of tank and coordinate system.

Question 4: Problem P2.141a,b To calculate the acceleration consider the equations of motion, beginning with the x equation of motion: ρax

=

⇒ pA

=

or ax

=

∂p − ρ|gx | ∂x pC + ρ(ax + |gx |)L pA − pC − |gx | ρL −

To estimate the pressure difference (pA − pC ) use the y equation of motion: 0

=

−

∂p − ρ|gy | since v = 0 ∂y 6

⇒ pA and pC ⇒ pA − pC

= = =

patm + ρ|gy |hA patm + ρ|gy |hC ρ|gy |(hA − hC )

Substituting gives: ax

= = =

pA − pC − |gx | ρL hA − hC |gy | − |gx | L hA − hC g cos θ − g sin θ L

Final answer: ax = −3.80[m/s2 ]. The tank is deaccelerating as it moves up the ramp. The tank contains water and is 50[cm] wide. What is the net force due to pressure acting on the rear wall of the tank (the wall running from point A to the free surface of the water). You may assume that water has stopped sloshing. See the above figure for the freebody diagram of the rear wall. δfx = [patm − p(y)]w · dy p(y) = patm + ρ|gy |(hA − y) Z hA Fx = [patm − p(y)]w · dy 0[m]

Z

hA

= −ρ|gy |w = −ρ|gy |w

(hA − y)dy 0[m] h2A

2

h2A

Final answer: Fx = −ρg cos θw 2 = −166N . The force due to pressure acts to push the rear wall of the tank outwards.

Question 5

Figure 5: Schematic of laminar boundary layer on a flat plate. The sketch shows the laminar boundary layer region very close to a flat plate with flow of air over the plate. The laminar velocity field is approximately: ¶ µ y y2 u(x, y) = u0 2 − 2 δ δ 7

v(x, y)

= u0

δ x

µ

y2 y3 − 3 2 2δ 3δ

¶

where δ is the thickness of the region above the plate in which the flow is influenced by the plate and u0 is the fluid speed far above the plate (also referred to as the freestream speed). For laminar flow, the thickness is a function of position along the plate: r x δ(x) = 5 ν u0 where ν is the kinematic viscosity of the air. 1. Derive an expression for the drag force that acts on the top side of the plate. The plate has a length L and a width (into the page) of b.

τw (x) ≡ ≈ = dFx

= =

D

=

⇒D

=

¯ ∂u ¯¯ µ ∂y ¯y=0 u0 µ2 δ(x) 1 µu0 3/2 2.5 ν 1/2 x1/2 τw (x)bdx Elemental force due to air acting on plate b µu0 3/2 dx 2.5 ν 1/2 x1/2 Z L b µu0 3/2 dx 1/2 x1/2 0 2.5 ν 2b µu0 3/2 1/2 L Final Answer 2.5 ν 1/2

2. Estimate the drag force when air at 20[◦ C] flows over a plate 0.5[m] long and 2.0[m] wide. The freestream speed is 3[m/s]. From Table A.2 for air at 20 [◦ C] ρ = 1.20[kg/m3 ], µ = 1.80×10−5 [N s/m2 ], and ν = 1.50 × 10−5 [N s/m2 ]. Final Answer: D = 0.031[N ]. 3. When the plate is flipped so it is perpendicular to the flow, the drag force is due to pressure. In this case, a reasonable estimate of the drag force is D = 0.6ρu20 Lb where ρ is the air density. Compare the size of the drag force in this case to that which exists when the flow is parallel to the plate. 4. Final Answer: D = 6.5[N ] which is considerably greater than the drag force on the plate when the flow is parallel to the plate.

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