) = slugs/ft 3. ) = lb ft/s. ) = ft/s

Transcription

1 1. Make use of Tables 1. in the text book (See the last page in this assignent) to express the following quantities in SI units: (a) 10. in./in, (b) 4.81 slugs, (c).0 lb, (d) 7.1 ft/s, (e) 0.04 lb s/ft. in in Sol) (a) 10. ( ) ( ) /s in 1in. 60 s (b) 4.81 slugs ( ) 70. 1slug 4.45 N (c).0 lb ( ) 1.4 N 1lb ft 0.05 (d) 7.1 ( ). /s s 1ft lb s 4.45 N (e) 0.04 ( ) ft 1ft ) ( 1.1 N s/ 1lb Make use of Tables 1.4 in the text book (See the last page in this assignent) to express the following quantities in BG units: (a) 14. k, (b) 8.14 N/, (c) 1.61 /, (d) 0.00 N /s, (e) 5.67 /hr. 81ft Sol) (a) 14. k ( ) ft 1k (b) 8.14 (c) 1.61 N 0.5 lb 1 ( ) ( ) lb/ft 1 N.8 ft slug 1 ( ) ( ) slugs/ft 1.8 ft N 0.5 lb.8 ft (d) 0.00 ( ) ( ) 0.06 lb ft/s s 1 N ft 1hr (e) 5.67 ( ) ( ) ft/s hr 1 600s

2 . The inforation on a can of pop indicates that the can contains 55 L. The ass of a full can of pop is 0.69 while an epty can weighs 0.15 N. Deterine the specific weight, density, and specific gravity of the pop and copare your results with the corresponding values of for water at 0 o. Express your results in SI units. Sol) Inforation fro the description Volue, V 55 L L (10 - ) Total ass of full can, total 0.69 Weight of an epty can W can 0.15 N Fro the inforation, we can deterine the net ass of pop and then the density. W 0.15 can pop total g 9.81 pop 0.5 Then, (1) Density: ρ pop V / c.f. ρ 998 / (Because the textbook provides only the properties o H O@0 of water at 15.6 o, you can use those values for coparison.) () Specific weight: γ pop ρ pop g N/ c.f. γ o 9790 N/ H O@0 ρ pop () Specific gravity: SG pop ρ o 1000 H O@4 c.f. SG H O@0 o

3 4. If 1 cup of crea having a density of 1005 / is turned into cups of whipped crea, deterine the specific gravity and specific weight of the whipped crea. Sol) Mass of 1 cup of crea, crea ( 1005 ) (Volue of cup, Vcup) Since there is no change in total ass of crea, i.e. crea whipped crea Density of cups of whipped crea, (1005 ) ( V ) 1005 ρ 5 Then cup whipped crea whipped crea Vcups V cups ρ 5 whipped crea (1) Specific gravity, SG 0. 5 ρ o H N () Specific weight, γ whipped crea ρwhipped crea g (5 ) (9.81 ) 90 s

4 5. A Newtonian fluid having a specific gravity of 0.9 and a kineatic viscosity of /s flows past a fixed surface. Due to the no-slip condition (The fluid sticks to a plate surface.), the velocity at the fixed surface is zero, and the velocity profile near the surface is shown in figure. Deterine the agnitude and direction of the shearing stress developed on the plate. Express your answer in ters of U and δ, with U and δ expressed in units of eters per second and eters, respectively. Sol) Because the situation is related with the viscosity of oil, let s use the following eq. du τ μ dy By using the inforation fro the description, du τ μ dy y μu δ δ (because y 1 y u U ) δ δ Please note that the proble provides a kineatic viscosity ν, not (dynaic) viscosity μ and thus we have to use, ( ) μ νρ SG ρ o 4 10 N s/ HO@4 y U Then, τ μu 0.68U δ δ δ N/ δ : because we need the shearing stress at the plate (y 0) Direction of shearing stress (τ ) exerting on the plate due to otion of the fluid Left

5 6.. The viscosity of liquids can be easured through the use of a rotating cylinder viscoeter of the type shown in Fig. In this device the outer cylinder is fixed. The inner cylinder has a diaeter ( R i ) of 40 c and the gap between the inner and outer cylinder is 0.1 c. The experient reveals that the torque of 1.0 N is required in order to develop the angular velocity of inner cylinder (ω ) of 100 rp. (a) The applied torque will create the shearing stress (τ ) on the surface of inner cylinder. Deterine the shearing stress due to the torque of 1.0 N. (b) Deterine the velocity at the surface of inner cylinder. (c) Using the results fro Questions (a) and (b), deterine the viscosity of fluid. Assue that the fluid is Newtonian, i.e. the velocity distribution (du/dy) in the gap is linear. gap 50 c Sol) (a) Shearing stress, τ F : Force due to the easured torque (Force parallel to the surface) (Surface Area) F M R i F A (1.0 N ) (0.4 ).5 N A: Area of the outer surface of the inner cylinder A π (0.4 ) (0.5 ) 1. 6 Then, F.5 τ 1.98 N/ (ANSWER) A 1.6 (b) Velocity at the surface of inner cylinder, u inner R ω (0.4 ) (100 i rev in π rad 1rev 1in ) 60sec ( 0.4) (10.47) 4.19 /s (ANSWER) (c) Velocity of the outer cylinder, u outer 0 (Stationary cylinder) Because the fluid is Newtonian, the distribution of fluid velocity is linear. du uinner uouter s -1 dy gap Then, du τ μ 1.98 μ(4190) dy μ N s/ (ANSWER) 4190

6 7. In a test to deterine the bulk odulus of a liquid it was found that as the absolute pressure was changed fro 15 to 000 psi the volue decreased fro to in. Deterine the bulk odulus for this liquid. Sol) Let s use the definition of bulk odulus E V dp E V psi dv / V As shown in Figure, surface tension forces can be strong enough to allow a double-edge steel razor blade to float on water, but a single-edge blade will sink. Assue that the surface tension forces act at an angle θ relative to the water surface. (The ass of the double-edge blade is , and the total length of its sides is 06. Deterine the value of θ required to aintain equilibriu between the blade weight and the resultant surface tension force. (b) The ass of the single-edge blade is , and the total length of its side is 154. Explain why this blade sinks. Support your answer with the necessary calculation. Sol) (a) Fro the definition of the surface tension force, (Force) (Surface tension) (Length) Fσ σ water (06 10 ) (Fro Table 1.6, σ water N/) ( ) (06 10 ) N Then, the vertical coponent of Fσ should be balanced by the weight of a double-edge blade. F σ sin θ g ( )(9.81) sinθ o θ 4.6 (b) For a single-edge blade with the ass and the total length of its side of and 154, respectively, F σ ( ) ( ) N and by the sae anner, F σ sin θ g ( )(9.81) sinθ.6 > 1 (Ipossible) The surface tension cannot aintain equilibriu with a single-edge blade.