2018-03-14 CE35000 Exercises PDF

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CE 35000 Fluid Mechanics – Spring 2018 Excercises 2

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2018-03-13

Floating cylindrical tank (Munson 2.144)

Thin-walled, d = 1 [m] diameter cylindrical tank is H = 3 [m] tall and has a mass of m = 90 [kg ] (Figure 1). The closed top of the tank has a pressure gauge and the open bottom is lowered into  water (ρw = 1000 kg m−3 ) and  position (h = 0.6 [m] portion sticking out from the water)  held in by a steel block (ρblock = 7840 kg m−3 ). Assume that the air in the tank is compressed at constant +patm = pgauge temperature ( pρatm ). The atmospheric pressure is patm = 101.33 [kP a]. ρcom air Determine: • the reading on the pressure gauge at the top of the tank; • the volume of the steel block NOTE: The first part of the √problem leads to a second order equation: that can be solved by the b2 −4ac . You can solve the second part (without completing the first following formula: x1,2 = −b± 2a part) by knowing that the height of the compressed air in the tank is hair = 2.581 [m].

patm ρair patm patm + pgau = → = ρcomp ρair ρcomp pgau + patm patm ρair H H= ρair V cyl = ρcomp V air → hair = pgau + patm ρcomp   patm pgau = ρw g (hair − h) = ρw g H −h pgau + patm (pgau + ρw gh) (pgau + ppatm ) − patm ρw gH = 0 2 pgau + (ρw gh + ppatm ) pgau − patm ρw g (H − h) = 0

Figure 1: Floating cylindrical tank pgau =

−ρw gh − patm ±

hair =

q

(ρw gh + pamt )2 + 4patm ρw g (H − h) 2

= 18.915 [kP a]

  patm d2 H = 2.581 [m] , V buoy = (hair − h) π = 1.541 m3 4 pgau + patm

ρw g (V buoy + V block ) = mg + ρblock gV block → ρw V buoy − m = (V block − ρw ) V block V block =

2

  ρw V buoy − m = 0.208 m3 ρblock − ρw

Variable area horizontal pipe (Munson 3.3)

  Water flows (ρ = 1.94 slugft−3 ) steadily through the variable area horizontal pipe shown in   Figure 2. The centerline velocity is given by ~U = 10 s−1 (1 [ft] + x)~i, where x is in feet. The pressure at section (1) is p1 = 50 [psi]. Viscous effects are neglected.

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CE 35000 Fluid Mechanics – Spring 2018 Excercises 2

a) Determine the pressure gradient needed to produce this flow.

∂p , ∂x

2018-03-13

(as a function of x)

b) Determine the pressure (p2 ) at (2) by integration of the pressure gradient obtained in (a) c) Application of the Bernoulli equation.

Figure 2: Variable area pipe

a)   ∂U ∂p ∂p = −ρ100 s−2 (1 [ft] + x) = −ρU → ∂x ∂x ∂x b) Z

l 0

Z l  −2  ∂p 1 [ft] + xdx ρ dx = −100 s ∂x 0   −2  1 2 ρ 1 [ft] x + x + C = −100 s 2 x = 0 → C = p1

   −2  1 ρ 1 [ft] l + l2 x = l → p2 = p1 − 100 s 2    −2      1 p2 = 50 [psi] − 100 s 1.95 slugft−3 3 + 32 ft2 2 = 39.896 [psi]

c)       U = 10 s−1 (1 [ft] + x) , U1 = 10 ft s−1 , U2 = 10 s−1 (1 [ft] + l) p1 + ρ

U2 U12 = p2 + ρ 2 2 2

U22 − U12 2       1 ft2 + 2 × 1 [ft] l + l2 − 1 ft2 2 −2 = p1 − ρ10 s 2   −2  12 ρ 1 [ft] l + l = p1 − 100 s 2

p2 = p1 − ρ

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Water streams (Munson 3.34)

Streams of water from two tanks impinge upon each other (Figure 3). If viscous effects are negligible and point A is a stagnation point, determine the height h.

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CE 35000 Fluid Mechanics – Spring 2018 Excercises 2

2018-03-13

hγw = p1 − (Hl − Hr ) γw p1 − Hl + Hr = 45.7 [ft] h= γw

Figure 3: Water streams

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Flow through pipe (Munson 3.92)

Inviscid and incompressible fluid flows steadily as shown in figure Figure 4. Determine h as a function of Q, H, d1 and d 2 . The density of the fluid ρ and gravitational acceleration g are known parameters.

p1 + ρ

U12 U2 + z1 ρg = p2 + ρ 2 + z2 ρg 2 2

d2 d12 π, A2 = 2 π 4 4 Q1 = U1 A1 , Q2 = U2 A2 , Q = Q1 = Q2 z1 = 0, z2 = H, A1 =

U1 = U2 U2 = U1

4Q A2 d2 = U2 22 = A1 d1 π d1

A1 d2 4Q = U1 12 = 2 π A2 d2 d2

p1 = p2 +

U 22 + Hρg − hρg 2 ρ

Figure 4: Syphon from tank

p p 4Q U12 8Q2 = hρg → U1 = 2gh → 2 = 2gh → h = 4 2 2 gd 1 π d1 π

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