CE 3500 Fluid Mechanics / Fall 2014 / City College of New York


 Roland Horatio Marsh
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1 1 Drag Coefficient The force ( F ) of the wind blowing against a building is given by F=C D ρu 2 A/2, where U is the wind speed, ρ is density of the air, A the crosssectional area of the building, and C D is a constant termed the drag coefficient. Determine the dimension of the drag coefficient. F [ MLT 2 ]=C D ρ [ ML 3 ] (U [ LT 1 ]) 2 A [ L 2 ] 2 C D [ unitless ] 1
2 2 Derivatives If u is a velocity, x a length, and t a time, what are the dimension (in MLT system) of a) u t, b) 2 u x t and c) u t dx? a) u [ LT 1 ] [ LT 2 ] t [T ] b) 2 u [ LT 1 ] x [ L ] t [T ] [T 2 ] c) u [ LT 1 ] t [T ] dx [ L ] [L 2 T 2 ]? 2
3 3 Dimensionality I f U is a velocity, determine the dimensions of Z, α and G, which appear in the dimensionally homogeneous equation: U =Z (α 1)+G α and 1.0 are unitless therefore Z has to be LT 1 as U and G has to have the same LT 1 unit. 3
4 4 Flat Plates (1.75) Two flat plates are oriented parallel above a fixed lower plate as shown in figure below. The top plate, located a distance b above the fixed plate, is pulled along with speed V. The other thin plate is located a distance cb, where 0<c<1.0, above the fixed plate. This plate moves with speed V 1, which is determined by the viscous shear forces imposed on it by the fluids on its top and bottom. The fluid on the top is twice as viscous as that on the bottom. Plot the ratio V 1 /V as a function of c. Δ u 1 =V 1, Δ u 2 =V V 1, τ 1 =μ V 1 cb, τ 2=2μ V V 1 b cb τ 1 = τ 2 μ V 1 cb =2μ V V 1 b cb V 1 (b cb)=2 (V V 1 ) cb bv 1 cb V 1 =2 cbv 2cbV 1 bv 1 +cb V 1 =2 cbv V 1 V = 2c 1+c 4
5 5 Water Layer (1.81) A layer of water flows down an inclined surface with the velocity profile shown in figure below. Determine the magnitude and direction of the shearing stress that the water exerts on fixed surface for U = 2 m s 1 and h = 0.1 m. τ=μ du dy τ=μ U d ( 2 y h y2 h ) 2 dy τ=μ 2 U h ( 1 y h ) U =2 [m s 1 ]; h=0.1m τ=μ 2 2 [m s 1 ] 0.1 [ m] ( 1 y 0.1 [m] ) τ=40 [s 1 ]μ ( 1 y 0.1 [m ] ) τ max =40 [s 1 y=0 5
6 6 Thin Glycerin Layer (1.82) A thin layer of glycerin ( γ = [lb/ft 3 ] and μ = [lb s/ft 2 ]) flows down an inclined, wide plate with the velocity distribution shown in figure below. For h = 0.3 in. and α = 20, determine the surface velocity, U. Note that for equilibrium, the component of weight acting parallel to the plate surface must be balanced by the shearing force developed along the plate surface. In your analysis assume a unit width. dv =h da ; F 1 =γ dv sin α ; F 2 = τ da ; F 1 =F 2 γ dv sin α=τ bottom da τ bottom =γ h sin α τ bottom =γ hsin α τ=μ du d ( 2 y dy τ=μ U h y2 h ) 2 dy ρ hsin α=μ 2U h U = ρh2 sin α U = 2μ U = [ ] ft 1 (0.025 [ ft ] ) 2 sin 20 =0.285 [ ft s 1 ] [ s ] τ=μ 2 U h ( 1 y h ) ; τ bottom=μ 2U y= [lb ft 3 ] ( 1 [ ft ] 0.3 [in ] 12 [in ] ) [lb s ft 2 ] 2 sin 20 6
7 7 Pivot Bearing (1.86) A pivot bearing ( = 0.5 [mm]) used on the shaft of an electrical instrument is shown in figure below. An oil with a viscosity of μ = 0.5 [poise] fills the [mm] gap ( Δ s ) between the rotating shaft and the stationary base. Determine the frictional torque on the shaft when it rotates at 5000 [rpm]. dt =r df ; df =τ da ; τ=μ du dy =μ r ω Δ s ; da=2 r π dr sin α T = dt = r df = T =[ 2 πμ ω 4 Δ ssin α r4]l=0 = 1 2 L r τ da= rμ r ω Δ s 2 r π πμ ω Δ s sin α 4 μ=0.5 [g cm 1 s 1 ] ; Δ s=0.025 [mm ] ; ω=5000 [rpm ] T = 1 2 π 0.5 [ g cm 1 s 1 ]2 π5000 [min 1 ] [mm ]sin 30 dr sin α =2 πμ ω r 3 dr Δ ssin α (0.5 [ mm ] ) 4 = [ N m ] 7
8 8 Circular Plate (1.89) A 30 cm diameter circular plate is placed over a fixed bottom plate with a 3 mm gap between the two plates filled with glycerin ( μ = [Pa s]) as shown in figure below. Determine the torque required to rotate the circular plate slowly at 2 rpm. Assume that the velocity distribution in the gap is linear and that the shear stress on the edge of the rotating plate is negligible. dt =r df ; df=τ da ; τ=μ du dy =μ r ω ; da=2r π dr Δ s T = r =0 dt = r df = T =[ 2 πμ ω 4 Δ s r4]l=0 = 1 2 T = 1 2 T = 1 2 L πμ ω Δ s r τ da= r μ r ω πμ ω 2 r πdr=2 r 3 dr Δ s Δ s 4 π1.412 [ Pa s ] 4 π [min 1 ] 3 [mm ] π1.412 [ N m 2 s ]4 π [min 1 ] 60 [s min 1 ] [ m ] (15 [cm ] ) 4 ( [ m ] ) 4 =0.078 [ N m ] 8
9 9 UTube Mercury Manometer (2.36) A Utube mercury manometer is connected to a closed pressurized tank as illustrated. If the air pressure is 2 psi determine the differential reading h. The specific weight of the air is negligible. P air γ H 2 O (h+h water )+ γ Hg h=p air SG Hg γ H 2 O (h )=γ H 2 O (h+h water ) 13.6 h=h+ 4 [ ft ] h= 4 [ ft ] =0.317 [ ft ]
10 10 Tank Compartments (2.38) Compartments A and B of the tank shown in figure below are closed and filled with air and a liquid with a specific gravity equal to S.G.=a. Determine the manometer reading ( h[ L] ), if the barometric pressure ( P 0 ) is and the pressure gage reads P g [ FL 2 ]. The effect of the weight of the air is negligible. P g γ w h+ γ l h+γ m z=p 0 P g P 0 + γ m z=h ( γ w γ l ) h= P g P 0 + γ m z γ w γ l h= P g P 0 + z γ w SG m γ w (1 SG l ) 10
11 11 Open Utube in Motion (2.156) The open Utube of the figure below is partially filled with a liquid. When this device is accelerated with a horizontal acceleration, a[ L T 2 ], a differential reading, h [ L], develops between the manometer legs which are spaced a distance, l [ L], apart. Determine the relationship between a, h, and l. p z = ρ(a z+ g) and p x = ρ a x dp= p x dx+ p z dz along a line of constant pressure dp=0, e.g. interfaces of liquid and air: p x dx+ p z dz=0 or ρ a x dx ρ(a z + g)dz=0 dz dx = a x (a z + g) dz then the slope of the liquid surface is a negative value, dx = h l hence, h = a x l (a z + g) and for a z=0,a x =a we have h= al g 11
12 12 otating Mercury Mirror Telescope (2.163) The largest liquid mirror telescope uses a 6ft diameter tank of mercury rotating at 7 rpm to produce its parabolicshaped mirror as shown on the figure. Determine the difference in elevation of the mercury ( Δ h ), between the edge and the center of the mirror. Δ h= ω2 r 2 2g 7 [ rpm ] 6 [ ft ] Δ h= [ ft /s 2 ] 12
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