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Chapter 2

Chapter 2 - MAE 3161 Fluid Mechanics Notes by Paavo Sepri...

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Unformatted text preview: @ MAE 3161: Fluid Mechanics Notes by Paavo Sepri LECTURE 4 Cha ter 3 3: FLUID STATICS In this case, the macroscogic fluid motion vanishes (averaged over moiecuiar motions), although the random moiecuiar motions do not vanish. For a macroscopically motionless ﬂuid: (1) FmZE;§;(mv)a6 , since 1726 (3.1) (2) From the definition of a fluid: shear stressmust vanish everywhere. (3.2) 3.1 ISOTROPY OF NORMAL STRESS IN A MOTIONLESS FLUX!) Here, we demonstrate that pressure at every fluid point is independent of direction for the case of a static fluid. If the fluid is macroscopically motioniess, the net force on any arbitrary fiuid voiume must vanish. Consider the garticuiar volume: In the figure, there are several F’s, which represent inward forces appiied by the external fluid onto the internal fluid in directions locally pergendicular to each surface. Note the trigonometric relations among some of the iengths: (A02 = (MY +(Ay)2 Ay 2 (As) sin(8) Ax = (As) 003(6) (3.3) Consider the 3 force components (ail must vanish): (1) z—direction: (——F;) e (117;) = 0 front-i—back m 0 (both sides have the same area) (3.4) (2) x—direction: P; + (4?; sin(6)) e 0 (3.5) (3) y-direction: P; + (—FS 008(6))wmg E O (3.6) Definition: A stress component (denoted by 0'13. ) is a force component acting on an area of unit size that has a specified orientation. Example: an 2F acting on unit area whose normal is i. x A 023,3 A A I LﬁdrOC 0:2} Apply this definition to the previous equations: Eizwamﬁmam (3.7) ” (Ayxzsz) (Ayxzsz) (AsxAz) ” (3.5) (3.3) _O. E P; ﬂecoswggpwxzxyxmwz W (AxxAz) (AxxAz) (was) (3.6) : M m i— 5 ~03” ; Then the above becomes: (Ax)(Az) (ASXAZ) NOW ~03. m *0"... + pgmyﬂ 2 Take the limit: AV we 0 , which also corresponds to: Ay ~+ 0 . Conciude: a” = a” (3.8) Similarly, we consideé the same wedge, except that it is rotated by 90° about the z-axis to conclude also: a = 0' (3.9) In summary, for a static ﬂuid: (3.10) Thus, the normal stress in a motionless fluid is indegendent of direction (5.8., is isotrogic). For a fluid in motion, this is no longer true, but we may define the averaged normal stress to be the pressure at that point: pE~(0'H+O“W+O'ZZ)/3 (3.11) 3.2 BASIC EQUATION FOR PRESSURﬁ VARIATION IN A MOTIONLESS FLUID Next, we consider a force balance on a small cube in the static fluid. 2F; = 6 = (p, w pmxAyxAz)? + (p, — p,+i,)(Ax)(Az)} —pg(Ax)(Ay>(Az)} + (p, — p,+,,)(Ax>(Ay>I€ Divide by the volume of this ﬂuid element: AV = (Ax)(Ay)(Az) 6m (10:: _px+ﬂ.x)f+ (p? —py+3y)}mpgfw§w (p2 _p:r:+.t_‘.z)‘l€A Ax Ay AZ Take the limit: AV —> 0 to conclude: 5P9 5P“- A- 6P“ M ~— ~—— ~ ——~—~k=0 3.12 ax: {3er pg; 62 ( ) Or, Vp =—pg}' . Cohsider the three parts of the vector equation (3.12) separately: 1°: 21% 0:> p : p(y,z) at most. Then: )5 A 319 k: a— m 0 :> p 2 l29(32) at most. Then: Z 4 6p y J: \$=~pg mpwwmya— {(3003615) (3:13) Po This is the basic equation in a static fluid, under the influence of gravity, which reiates pressure to density as the depth is varied. ' Here: depth a wy. MAE 3161: Fluid Mechanics Notes by Paavo Sepri mam Chapter 3 Recail from last time: 25 = “P8 => My) e p(yo)- Jp(y)gdy (3-13) This integral relationship describes how pressure varies with increasing altitude (y) from a reference iocation (yo). Note: In the textbook, altitude is given the symboi, z. This is merely a matter of convention: it depends on the definition of co—ordinate frame orientation. 3.3 PRESSURE VARIATION IN A STATIC FLUID Here, we consider 3 special cases, as foliows. 3.3.1 INCOMPRESSIBLE FLUID (p2 comm; Example: liquid water) If the ﬂuid density may be assumed to be uniformly constant, then the integral relationship given in Eq. (3.13) becomes simpler by integrating the constant integrand, to obtain: p(y)_P(yo)=-~pg(y"—yo) = pgh (3-14) ,rlLﬂgﬂ L ‘4 “(W-9) “D For sea—water: 2 £93 ﬁ (lbfxsi) ﬂi (atmxinz) ~1az‘m pg _ (64 ff )(32'2 s2)(32.2(zbm)(ﬂ))(144m2) 14.7be ‘ 33 ft 3.3.2 ISOTi-IERMAL GAS (Region in the Earth's Atmosphere) Eq. (3.13) is repeated here: 95: —pg 5y Assume the fluid is a perfect gas; then: p m 1% Combine: 2P— 2: ~ﬁ 6y RT Rea-arrange (there is only one indegendent variabie): if: = mSEVZ p RT For a region of constant and uniform temperature in the atmosphere, integrate: P “go/Wye) 1 . WWW—m— “(10);. RT . we Conclude: My) m p(y9)e M (3.15) Evaluate a typical rate of exponential decrease in pressure: ~34 32.2%)(m‘s “degmxml PW “WM—Lam 1 RT " 1716ﬁ:2 441degR ‘ 23.519”: : 23.5Kfr 0.3048m 7.16Km (3.16) Therefore: In such a region, the atmospheric pressure decreases by 1/ e for every increase in aititude of 23,500 ft. 3.3.3 ATMOSPHERE WITH A LINEAR THERMAL REGION In such a region: T02) e T(y0)+a(y_yo) ; a=const. (3.17) After integration of Eq. (3.13), the result is: my) 2 ( Te) ﬁ- 3.18 17010) TCVO) ( ) Typical exponent: :5: _ j: (szxdegR) ﬂ 3 Roz— 32‘2 2X 1716f? )(—3.38x10‘3c1egR)“5'5 3.4 EXAMPLES OF PRACTICAL APPLICATION 3.4.1 FORCES AND MOMENTS ON SUBMERGﬁD PLANAR SURFACES (See ﬁg. 3-25 in Textbook) Here, we consider the special case of a vertical wall with water on one side and one atmosphere of air on the other side. Assume no ieakage. (Examples: (1) dyke, (2) dam, (3) fish tank) We seek F; and F2 required to restrain the piste. From Eq. (3.14): p 2 p0 + pgh (1') Measure 3 downward from the hinge: h : swsi (2) Calculate the force clue to the water acting on the plate (xwdirection): 52 “S: F, mLﬁpgxs—sads at i pghdh (a) s; 0 Note: p0 acts on both sides of the plate, and thus this effect cancels. Here: L =width (out of page) ; p ~p0 m pg(s ~51) SZn-Sl (\$2 *Sl)2 2 Integrate to obtain: Fp = Lpg(%—)io 2 Next, calculate the totai moment about the hinge as exerted by the water. Again: p0 acts on both sides of the plate, and thus this effect cancels. 5'2 “’5‘: MprI(p~p0)stmL jpghmmgdh (5) s, 0 Here: S =moment arm I13 1225 52“S1 (S "S )3 (5 WS )zs Then: M :L __+___1 2L 2 1 + 2 a 1 p pg(3 2 )0 pg 3 2 Combine to obtain: Mp = Lpg(52 ~sl)z(———‘:’“ +325) (6) M S S Divide 6 b 4: J22 .Z.+_i_ 7 ( ) V( ) F (3 6) ( ) P Conclude: If 31 m 0, then the moment about the hinge is related to the appiied force by: MP 2 Fp (252 /3) , which is greater than half-way along the plate. That is, the equivalent moment applied by the water is not simply applying the water force at the center of the plate. Next, we seek to find E and F2 required to keep the plate stationary. Force balance: Fp w E v- F2 2 0 (8) Moment balance: M2 —32F2 = 0 (9) Typical numbers for a large fish tank: MM? W _ 2 1 Si FEW—:2 mLpg(SZ Si) [3+———6SZJ + m 62.41bm 322;": 2 _1_ Zﬁ (lbf)(s2) : x 4 F2410]? )( ft} s2 )0”? {3+6(12ﬁ>][32.2abm)(ﬁ>]’2‘25 10 W (10) F1 mFP~F2 :8.67x103lbf (11) @ MAE 3161: Fluid Mechanics Notes by Paavo Sepri LECTURE 6 Chagter 3 3.4.2 FORCES ACTING ON SUBMERGED BODIES Consider an arbitrary solid body that is under water as shown in the diagram: M W W Mwi sum-ﬂee l I l' L R2 / The forces acting on the body are composed of three types: (1) Surface (pressure, and if the fluid is moving, shear stress) (2) Volumetric (gravity) (3) Restraining (physical contact with other bodies). «PM ,3 .4 g '"'" 13 (N) g 2 me? If there is no motion, then the force balance is given by: 13‘ : HﬁdS+ j'ﬂdeI/JrﬁR =6 (3.19) In the present chapter, we are considering a static case (no motion). Then: (1) see (2) f. I“w”“10(599435) (3) f3 m "1938? @ Here, ﬁﬁc) represents the outward unit normal vector on the surface of the body (it has unit length and is perpendicular to the local surface everywhere). Then, Eq. (3.19) becomes: ~HpﬁdS~ p3(3c~)g}dV+FR :6 S V (This is Newton's 2“d Law of Motion for the case of no motion here.) Since 9 and are constants, the above equation may be written as: —ﬂpﬁdS—M3g}+FR :6 (3x20) S Where: MB : “Jpng (This is the total mass of the body) V Eq. (3.20) is the statement that the gravitational weight of a body is balanced by the sum of all restraining forces and the net force resulting from pressure (which varies along the surface) applied to the entire surface of the body. In order to complete a specific computation, it is necessary to express the variation of 71(55) along the Surface of that body. 3.4.3 UNIT OUTWARD NORMAL TO A SURFACE The surface of a 30 solid body may often be described by a functional former? the type: F(x, )2, z) 3 const. Example (sphere): x2 + y?“ +22 2 R2 infersecf mi'f‘ﬁ pﬂawc . 2 In 3-D, take the gradient of F: VF=§£§+£E~f+§§IQ 8x 8y 62 C29 Note: VF is locally perpendicular to the surface. We scale it as follows to find the local unit normal vector: VF nm : (VWFW (3.21) Note: 17:55 51 3.4.4 BUOYANT FORCE ON A SPHERE SUBMERGED IN A FLUID We now apply the previous concepts to calculate the buoyant force on a sphere that is submerged, say under water. Then: F(R‘)=R2=x2+y2+22 ‘ VF m 2x§+23§+22i2 VF _ 2(xi" + y} + 21;) From Eq. (3.21): ﬁ(5c’)=——mmm-— (VFWFW 2(x2 + y2 + 3% Now, introduce spherical coordinates: £2. - .im- .1: R sm(¢)sm(9) , R sm(¢)cos(l9) , R cos(¢) Therefore, for a sphere, we conclude the following expression for the unit outward normal to its surface: a m sin(¢) sin(6)§ + cos(¢)} + sin(¢) cosmyé (3.22) Apply this to calculate the buoyant force on a submerged sphere. From Eq.(3.20): 212'?! HpﬁdS = j j(p0 * pgR cos(¢))(sin(¢) 3111(9)? + comm} + sin(¢)(cos(6)12)ds S Here: p0 is the fluid pressure at the depth corresponding to the center of the sphere. In spherical co—ordinates a differential element of surface area is: dS x: R?" sin(¢r)d¢d6 9 1W? 49 F R 5341696) A? (—N R Note: 2Tsin(6)d6 m 0 and 2? cos(t9)d6 m 0 Then the above simplifies to: 27m: [Loads = l Jane — pgR cos(¢))(R2 sine) cos(¢>})d6d¢ = 22:122.??(220 w pgR case» sin(¢>cos(¢)d¢ Note: ﬁjsinﬁé) cos(¢)d¢ = "35in? (925 )1: == 0 and ’]si11(¢) cos2 (¢)d¢5 = Wis—0033 (gé )|: 2%— G i} The final result is: —ﬂpﬁdszf3-’ERBpLg}mMLg} (3.23) S Conclude: The buoyant force on a submerged sphere is equal to the weight of the disglaced fluid. Recall that this is the famous principle attributed to Archimedes. This may also be shown to. be true for a submerged body of any shage. ...
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