Chapter 2. Fluid Mechanics

𝐹𝑉 = (0.823)(9.81π‘˜π‘/π‘š3 )(1.389π‘š2 )(2.50π‘š) 𝐹𝑉 = 28.1 π‘˜π‘ π‘ƒπ‘Žπ‘Ÿπ‘Ž β„Žπ‘Žπ‘™π‘™π‘Žπ‘Ÿ 𝐹𝐻 𝐹𝐻 = 0 … … π‘π‘œπ‘Ÿ π‘žπ‘’π‘’ π‘’π‘ π‘Žπ‘  π‘“π‘’π‘’π‘Ÿπ‘§π‘Žπ‘  π‘’π‘ π‘‘π‘Žπ‘› 𝑒𝑛 π‘’π‘žπ‘’π‘–π‘™π‘–π‘π‘Ÿπ‘–

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𝐹𝑉 = (0.823)(9.81π‘˜π‘/π‘š3 )(1.389π‘š2 )(2.50π‘š) 𝐹𝑉 = 28.1 π‘˜π‘ π‘ƒπ‘Žπ‘Ÿπ‘Ž β„Žπ‘Žπ‘™π‘™π‘Žπ‘Ÿ 𝐹𝐻 𝐹𝐻 = 0 … … π‘π‘œπ‘Ÿ π‘žπ‘’π‘’ π‘’π‘ π‘Žπ‘  π‘“π‘’π‘’π‘Ÿπ‘§π‘Žπ‘  π‘’π‘ π‘‘π‘Žπ‘› 𝑒𝑛 π‘’π‘žπ‘’π‘–π‘™π‘–π‘π‘Ÿπ‘–π‘œ Ahora para hallar la FRESULTANTE 2

2 𝐹𝑅 = √𝐹𝐻 2 + 𝐹𝑉 2 = √(0)2 + (28.1π‘˜π‘)2

𝐹𝑅 = 28.1 π‘˜π‘ PROBLEMA 4.49 Consulte la figura 4.49. La superficie mide 5.00 pies de longitud.

FIGURA problema SoluciΓ³n.

4.49 4.49.

y1 = Rsin15Β° = 3.882ft s = R βˆ’ y1 = 15 βˆ’ 3.882 β†’ 𝒔 = 𝟏𝟏. πŸπŸπŸ– 𝒇𝒕 hc = β„Ž + y1 +

𝑠 = 10 + 3.882 + 5.559 β†’ 𝐑𝐜 = πŸπŸ—. πŸ’πŸ’πŸπ’‡π’• 2

FH = 𝛾hc 𝑠𝑀 = (62.4)(19.441)(11.118)(5) β†’ 𝐅𝐇 = πŸ”πŸ•. πŸ’πŸ‘πŸ•π’π’ƒ hp = hc +

𝑠2 = 19.441 + 0.53 β†’ 𝐑𝐩 = πŸπŸ—. πŸ—πŸ•πŸπ’‡π’• 12hc FV = 𝛾𝑉 = 𝛾𝐴 𝑇 𝑀

𝐴1 = (14.489𝑓𝑑)(10𝑓𝑑) β†’ π‘¨πŸ = πŸπŸ’πŸ’. πŸ–πŸ—π’‡π’•πŸ 𝐴2 =

𝑦1 π‘…π‘π‘œπ‘ 15Β° (3.882)(14.189) = β†’ π‘¨πŸ = πŸπŸ–. πŸπŸπ’‡π’•πŸ 2 2

𝐴3 = πœ‹π‘… 2

75 75 = (15)2 β†’ π‘¨πŸ‘ = πŸπŸ’πŸ•. πŸπŸ”π’‡π’•πŸ 360 360

𝐴 𝑇 = 𝐴1 + 𝐴2 + 𝐴3 β†’ 𝐴 𝑇 = 320.27𝑓𝑑 2 𝐹𝑉 = 𝛾𝐴 𝑇 𝑀 = (62.4)(320.27)(5) β†’ 𝑭𝑽 = πŸ—πŸ—πŸ—πŸπŸ“π’π’ƒ π‘₯1 =

14.489 = 7.245𝑓𝑑 2

2 π‘₯1 = (14.489) = 9.659𝑓𝑑 3 π‘₯3 = 𝑏𝑠𝑖𝑛37.5Β° π‘‘π‘œπ‘›π‘‘π‘’ 𝑏 =

38.197𝑅𝑠𝑖𝑛(37.5Β°) = 9.301𝑓𝑑 37.5Β°

π‘₯3 = 9.30 sin(37.5Β°) = 5.662𝑓𝑑 π‘₯=

𝐴1 π‘₯1 + 𝐴2 π‘₯2 + 𝐴3 π‘₯3 = 6738𝑓𝑑 𝐴𝑇

𝐹𝑅 = √(𝐹𝐻 )2 + (𝐹𝑉 )2 = √(67437)2 + (99925)2 = 120550𝑙𝑏 𝐹𝑉 99925 πœ‘ = π‘Žπ‘Ÿπ‘π‘‘π‘Žπ‘” ( ) = π‘Žπ‘Ÿπ‘π‘‘π‘Žπ‘” ( ) β†’ 𝝋 = πŸ“πŸ“. πŸ—πŸ– β‰… πŸ“πŸ”Β° 𝐹𝐻 67437

PROBLEMA 4.51 Se muestra una superficie curva que detiene un cuerpo de fluido estΓ‘tico. Calcule la magnitud de las componentes horizontal y vertical de la fuerza que el fluido ejerce sobre dicha superficie. DespuΓ©s calcule la magnitud de la fuerza resultante, asΓ­ como su direcciΓ³n. Demuestre que la fuerza resultante actΓΊa sobre la superficie curva. La superficie de interΓ©s es una porciΓ³n de un cilindro con la misma longitud que la superficie dada en el enunciado del problema.

La superficie mide 4.00m de longitud Sol. Rsen30Β°=3.00m x1 A1

h=5.20m x2 hp

A2

R=6.00m

y=Rcos30Β°=5.196m

15Β° 15Β°

x3 b

30Β° A3

s

FH

Fv Ξ± Fr DATOS: J: gravedad especΓ­fica

J=0.72

Fv: fuerza vertical FH: fuerza horizontal FR: fuerza resultante Sol: S=R-y=6.00m-5.196m=0.804m hc=h+y+s/2=5.20m+5.196m+0.402m=10.798m Hallamos la fuerza horizontal: FH=J*s*w*hc=(0.72)(9.81m/s2)(0.804m)(4.00m)(10.798m) FH=245.3kN

hp=hc+s2/12hc=10.798m+(0.804m)2/12(10.798m)=10.798m+0.0050m hp=10.83m Hallamos la fuerza vertical: Fv=JV=JAw A1=(5.20m)(3.00m)=15.60m2 A2=(3.00m)/2*(5.196m)=7.794m2 A3= πœ‹ *R2*(30/360)= πœ‹ *(6.00m)2/12=9.425m2 AT=A1+A2+A3=32.819m2 Entonces Fv=JAw Fv=(0.72)(9.81m/s2)(32.819m2)(4.00m) Fv=245.3kN X1=3.00m/2=1.5m X2=2*(3.00m)/3=2.00m X3=b*sen15Β°=(38.197*R*sen15Β°/15)*sen15Β°=1.023m X=(A1*X1+A2*X2+A3*X3)/AT=1482m Hallamos la fuerza resultante FR2=FH2+Fv2= (245.3KN)2+ (927.2KN)2 FR=959.1KN

(Fuerza resultante)

DirecciΓ³n: Ξ±=tan-1(Fv/ FH)=tan-1(927.2KN/245.3KN)=75.2Β° PROBLEMA 4.52

𝐹𝑣 = 𝑦𝐴𝑀 𝐴 = 𝐴1 + 𝐴2 = (1.20)(2.80) + 𝐹𝑣 = (9.81

πœ‹(1.20)2 = 4.491π‘š2 4

π‘˜π‘ ) (4.491π‘š2 )(1.50π‘š) = 66.1π‘˜π‘ π‘š3

π‘₯1 = 0.5(1.20) = 0.60π‘š ; π‘₯2 = 0.424(1.20) = 0.509π‘š π‘₯Μ… =

𝐴1 π‘₯1 + 𝐴2 π‘₯2 (3.36)(0.60) + (1.13)(0.509) = = 0.577π‘š 𝐴𝑇 4.491 β„Žπ‘ = β„Ž +

𝑠 1.20 = 2.80 + = 3.40π‘š 2 2

𝐹𝐻 = π‘¦π‘ π‘€β„Žπ‘ = (9.81)(1.20)(1.50)(3.40) = 60.0π‘˜π‘ 𝑠2 1.202 β„Žπ‘ = β„Žπ‘ + = 3.40 + = 3.435π‘š 12β„Žπ‘ 12(3.40) 𝐹𝑅 = βˆšπΉπ‘‰2 + 𝐹𝐻2 = √66.12 + 60.02 = 89.3π‘˜π‘ βˆ… = π‘‘π‘Žπ‘›βˆ’1

𝐹𝑣 66.1 = π‘‘π‘Žπ‘›βˆ’1 ( ) = 47.8Β° 𝐹𝐻 60.0

PROBLEMA 4.53 Calcular las componentes Horizontal y vertical de la fuerza en la superficie curva

Dato: Ancho = 2.50m; π‘ƒπ‘’π‘ π‘œ π‘’π‘ π‘π‘’π‘π‘–π‘“π‘–π‘π‘œ 𝑑𝑒𝑙 π‘Žπ‘”π‘’π‘Ž = 9.81

𝐾𝑁 π‘š3

Hallamos: 𝐴1 = (1.20π‘š)(2.80π‘š) = 3.36π‘š2

𝐴2 = 𝑅2 βˆ’ πœ‹

𝑅2 = 0.309π‘š2 4

𝐴 = 𝐴1 + 𝐴2 = 3.669π‘š2 𝐹𝑣 = π‘ƒπ‘’π‘ π‘œ = (π‘ƒπ‘’π‘ π‘œ π‘’π‘ π‘π‘’π‘π‘–π‘“π‘–π‘π‘œ 𝑑𝑒𝑙 π‘Žπ‘”π‘’π‘Ž)(π‘‰π‘œπ‘™π‘’π‘šπ‘’π‘›) = (9.81

𝐾𝑁 ) (3.669π‘š2 )(2.50π‘š) = 54𝐾𝑁 π‘š3

La ubicaciΓ³n del centroide se encuentra por medio de la tΓ©cnica del Γ‘rea compuesta De la fig: 𝑋1 =

1.20π‘š = 0.6π‘š 2

𝑋2 = 0.2234𝑅 = 0.268π‘š La ubicaciΓ³n del centroide para el Γ‘rea compuesta es 𝑋=

(𝐴1)(𝑋1) + (𝐴2)(𝑋2) (3.36π‘š2 )(0.6π‘š) + (0.309π‘š2 )(0.268π‘š) = = 0.572π‘š 𝐴 3.669π‘š2

La profundidad del centroide es: β„Žπ‘ = β„Ž +

𝑠 1.20 = 2.80 + = 3.40π‘š 2 2

La Fuerza horizontal: πΉβ„Ž = (π‘ƒπ‘’π‘ π‘œ π‘’π‘ π‘π‘’π‘π‘–π‘“π‘–π‘π‘œ 𝑑𝑒𝑙 π‘Žπ‘”π‘’π‘Ž)(π‘Žπ‘›π‘β„Žπ‘œ)(𝑠)(β„Žπ‘) = (9.81

𝐾𝑁 ) (2.50m)(1.20π‘š)(3.40π‘š) = 60𝐾𝑁 π‘š3

L a profundidad al centro de presiones: β„Žπ‘ = β„Ž +

𝑠2 1.202 = 3.40 + = 3.435π‘š 12β„Žπ‘ 12(3.40)

La fuerza resultante en la superficie es: 2

2

πΉπ‘Ÿ = βˆšπΉπ‘£ 2 + πΉβ„Ž2 = √54𝐾𝑁 2 + 60𝐾𝑁 2 = 80.7𝐾𝑁

PROBLEMA 4.54 Calcule la magnitud de las componentes horizontal y vertical de la fuerza que el fluido ejerce sobre dicha superficie. DespuΓ©s calcule la magnitud de las fuerzas resultante; asΓ­ como su direcciΓ³n Demuestre que la fuerza resultante actΓΊa sobre la superficie curva. En cada caso, la superficie de interΓ©s es una porciΓ³n de un cilindro con la misma longitud que la superficie dada.