Sd282 Asg4

FLUID MECHANICS I Solution 3 Question 1: Example 2.5 p80 a)free-body diagram b)seawater side y

F~ w

Fp due to water

A

F~By hAB =1.829 m

Fp due to atm air

~g

F~Bx

x

B lAB = 2.438 m

width w = 1.524 m H = 15ft = 4.572 m

0=−

∂p − ρsw g ∂y

p(y) = p(y = 0) − ρsw g y p(y = 0) = patm + ρsw gH p(y) = patm + ρsw g (H − y)

1

c)local coordinate system

y

s

n B

x

d) elemental net pressure force dF pnet = (patm − p(y)) w ds dF pnet = −ρsw g (H − y) w ds elemental moment due to pressure around B dM pB = dF pnet .s dM pB = −ρsw g (H − y) w s ds net pressure force Z

s=L

F pnet =

−ρsw g (H − y) w ds s=0

Z

y = s sinθ s=L

F pnet =

−ρsw g (H − s sinθ) w ds s=0

µ Fpnet = ρsw gw

L2 sinθ − H L 2

¶

moment due to pressure around B Z s=L M pB = −ρsw g (H − y) w s ds s=0 µ 3 ¶ L2 L sinθ − H MpB = ρsw g w 3 2 e)sum forces and moments M pB + FW .hAB = 0

(1)

F px + FW + FBx = 0 F py + fBy = 0

(2) (3)

2

with F px and F py the x and y components of the net pressure force (F pnet ). Net Pressure Force and components Fp ' −170, 708 N ³ π net ´ F px = +|F pnet |cos − θ ' +102, 735 N ³π 2 ´ F py = −|F pnet |sin − theta ' −136, 333 N 2 Other Forces ¶ L2 L − 2sinθ 3 ⇒ FW = −129, 703 N (3) FBy = −F py ⇒ FBy ' +136, 333 N (2)FBx = −F px − FW ⇒ FBx ' +26, 968 N

µ (1) FW = ρsw g w H

Notice that sinθ =

hAB lAB

≈ 37◦ .

Question 2: Midterm Exam question (Winter 2005) Gate AB is 2.5 m wide into the paper, 5 m long and hinged at B. Point A is linked to a pulley which exerts a force of 45, 000 N on the gate. The gate separates two different fluids: atmospheric air on the left hand side and water on the right hand side. All fluids are at 20◦ C. The depth of the water is h. Atmospheric conditions apply above the water surface level. The weight of the gate and friction forces are negligible. a) Draw a free-body-diagram showing all forces acting on gate AB. b) (x,y) is the global coordinate system and (s,n) is the local. The origin of both systems is located at B (see sketch). Determine the elemental net pressure force acting on an elemental surface (w ds) of gate AB and the elemental moment around B due to the net pressure, w being the with of the gate. Pressure distribution on the water side dpw = −ρw g dy pw (y) = patm + ρw g (h − y) Elemental net pressure force dF pnet = (patm − pw (y))wds dFpnet = −ρw g(h − y) w ds

3

Tension of cable s

A

Pressure force due to air above water

Pressure Force due to water

WATER n

Pressure force due to air

h

By

θ = 60◦

Bx

B

Bx and By are reaction forces at B

Elemental moment due to pressure around B dM p = dF Pnet s dMp = −ρw g(h − y)w s ds c) Calculate the net pressure force on the gate and the moment due the net pressure force around B. These will be functions of h. Net pressure force on the gate Z

s=l

F pnet =

−ρw g(h − y) w ds s=0

with l = h/sinθ. Using y = s sinθ, Z

s=l

F pnet =

−ρw g(h − s sinθ) w ds s=0

1 h2 2 sinθ = −14131.2 h2 N

Fpnet = −ρw g w F pnet Moment due to pressure around B Z

s=l

−ρw g(h − s sinθ) w s ds

Mp = s=0

l3 l2 −h ) 3 2 h3 gw (sinθ)2

M p = ρw g w(sinθ 1 Mp = − ρw 6

M p = −5439.1 h3 N.m 4

d) Compute the water level h for which the gate will start to fall. sum of the moments around B =0 Moment of the reactions forces at B = 0. −5439.1 h3 + FA L = 0 µ h=

5x45000 5439.1

¶1/3

h = 3.46 m

Question 3: Final exam question (Spring 2004) see separate PDF file Q3solution.pdf

Question 4 Problem P2.83 Free-body diagram

F~

r

y F~

~g

A patm F~By

R

WATER

p(y) θ

x

F~Bx B ~ W

O (x,y) with origin at O

centre of gravity located at xw = 0.886 m Local Coordinate system Choose (r,θ) origin at O.

5

Net elemental pressure force 0 = −dp − ρw g dy p(y) = patm + ρw g(R − y) dF pR = [−patm + patm + ρw g (R − y)[R L dθ dF pR = ρw g (R − Rsinθ)R L dθ x component dF px = ρw g (1 − sinθ)R2 L cosθdθ y component dF py = ρw g (1 − sinθ)R2 L sinθ dθ Z F px =

θ= π 2

ρw g (1 − sinθ)R2 L cosθdθ

θ=0

F px = ρw g R2 L Z F py =

θ= π 2

1 2

ρw g (1 − sinθ)R2 L sinθdθ

θ=0

F py = ρw g R2 L(1 −

π ) 4

Use ρw = 998 kg/m3 , g = 9.81 m/s2 , R = 8ft = 2.4384 m and L = 10ft = 3.048 m. A page of integrals is also provided (p10). Fpx = +88, 892 N Fpy = +38, 153 N elemental moments due to pressure around B There are two components: x and y for the moments. dM py = dF py .(R − x) dM px = dF px .y Moments due to pressure around B Z M py =

π 2

ρw g R2 (1 − sinθ)Lsinθ(R − Rcosθ)dθ

0

5 π M p y = ρw g R 3 L ( − ) 6 4 Mpy = +20780 N.m

6

Z M px =

π 2

ρw g R2 (1 − sinθ)Lcosθ Rsinθ dθ

0

1 6 Mpx = 72252 N.m M p x = ρw g R 3 L

Sum moments Sum of the forces is not needed here to calculate F. F.R + W.xW − M py − M px = 0 Note that xW would be given in exam conditions. xW = 0.886 m. W(weight) = 13, 344 N. F = 33, 304 N F acts in the negative y direction according to the coordinate system.

7

Question 5 Problem P2.86 P~

B x

O

atm θ

R water

patm

p(y)

Fcy C y

r width b = 3 m negligible weight

Fcx Fcx and Fcy= reaction forces

Figure 1: Freebody diagram of the gate. pressure distribution 0 = −dp + ρ g dy p(y) = patm + ρgy p(θ) = patm + ρgRsinθ Net element pressure force dF pr = [−patm + patm + ρgRsinθ]bRdθ dF pr = +ρ gR2 bsinθdθ dF px = +ρ gR2 bsinθcosθdθ dF py = +ρ gR2 bsin2 θdθ Elemental moment around B due to pressure dM p = dF px . (R − y) + dF py . x dM p = dF px . (R − Rsinθ) + dF py . Rcosθ dM p = ρgR3 bsinθ cosθ dθ Sum moments Z 0 = +P . R −

π 2

ρgR3 bsinθ cosθ dθ

0

Final answer 1 2 |P| = 58, 742 N |P | = ρg b R2

8

(4)

Notice that we take moments about H so that the hinge forces do not need to be explicitly calculated. P is acting in the negative x direction.

Question 6

Figure 2: Freebody diagram of plate window on tank wall. The figure shows a plate window, ABC, in the tank wall. The plate is pinned at points A, B, and C. If the tank contains water at rest, estimate the forces at A, B, and C that are required to hold the plate in place. Place the x-y co-ordinate system at point B as shown in the figure. p(y) l(y)

= patm + ρg[d + lAB − y] HydrostaticLaw = lBC − y = lAB − y Geometry

δfz δMBC

= =

[patm − p(y)]dx · dy δfz y

δMAB

=

δfz x

9

X X

Z Fz

= =

lAB

=

Z

0

[p(y) − patm ]ydxdy Z

MAB

l(y)

[p(y) − patm ]dxdy 0 l(y)

0 = FA lAB − 0

0

X

Z

0 = FA + FB + FC − Z

MBC

lAB

lAB

Z

0 = FC lBC −

l(y)

[p(y) − patm ]xdxdy 0

0

After substitutions and solution the above gives: · 2 ¸ 3 dlAB lAB FA = ρg + 6 12 · 2 ¸ 3 dlAB lAB + FC = ρg 6 8 · 2 ¸ 3 dlAB lAB FB = ρg + 6 8 Final Answer: The forces required to push the plate against the water are FA = 2.25×105 [N ], and FB = FC = 2.76 ×105 [N ]. Notice that FB = FC > FA .

10

sin(2A) = cos(2A) = tan(2A) =

2 sin A cos A cos2 A − sin2 A = 1 − 2 sin2 A = 2 cos2 A − 1 2 tan A 1 − tan2 A

(5) (6) (7)

Z

cos(ax) a Z sin(ax) x cos(ax) − x sin(ax) = a2 a Z x sin(2ax) sin2 (ax)dx = − 2 4a Z 2 x x sin(2ax) cos(2ax) x sin2 (ax)dx = − − 4 4a 8a2 Z 3 cos(ax) cos (ax) + sin3 (ax)dx = − a 3a Z 3x sin(2ax) sin(4ax) sin4 (ax)dx = − + 8 4a 32a sin(ax) dx =

−

Z cos(ax)dx

=

x cos(ax)dx Z cos2 (ax)dx

=

Z

=

Z x cos2 (ax)dx = Z cos3 (ax)dx = Z cos4 (ax) = Z sin(ax) cos(ax)dx

=

sin(px) cos(qx)dx

=

sinn (ax) cos(ax)dx

=

cosn (ax) sin(ax)dx

=

sin2 (ax) cos2 (ax)dx

=

Z Z Z Z

sin(ax) a cos(ax) x sin(ax) + a2 a x sin(2ax) + 2 4a 2 x sin(2ax) cos(2ax) x + + 4 4a 8a2 sin(ax) sin3 (ax) − a 3a 3x sin(2ax) sin(4ax) + + 8 4a 32a sin2 (ax) 2a cos((p − q)x) cos((p + q)x) − − 2(p − q) 2(p + q) sinn+1 (ax) (n + 1)a cosn+1 (ax) − (n + 1)a x sin(4ax) − 8 32a 11

(8) (9) (10) (11) (12) (13)

(14) (15) (16) (17) (18) (19) (20) (21) (22) (23) (24)