Problem 1.1 [3] Some of these substances exhibit characteristics of solids and fluids under different conditions.

Transcription

1 Proble. []. A nuber of coon substances are Tar Silly Putty Modeling clay Wax Sand Jello Toothpaste Shaving crea Soe of these aterials exhibit characteristics of both solid and fluid behavior under different conditions. Explain and give exaples. Coon Substances Tar Silly Putty Modeling clay Wax Sand Jello Toothpaste Shaving crea Soe of these substances exhibit characteristics of solids and fluids under different conditions. Explain and give exaples. Tar, Wax, and Jello behave as solids at roo teperature or below at ordinary pressures. At high pressures or over long periods, they exhibit fluid characteristics. At higher teperatures, all three liquefy and becoe viscous fluids. Modeling clay and silly putty show fluid behavior when sheared slowly. However, they fracture under suddenly applied stress, which is a characteristic of solids. Toothpaste behaves as a solid when at rest in the tube. When the tube is squeezed hard, toothpaste flows out the spout, showing fluid behavior. Shaving crea behaves siilarly. Sand acts solid when in repose (a sand pile ). However, it flows fro a spout or down a steep incline.

2 Proble. []. Give a word stateent of each of the five basic conservation laws stated in Section -4, as they apply to a syste. Five basic conservation laws stated in Section -4. Write: A word stateent of each, as they apply to a syste. Assue that laws are to be written for a syste. a. Conservation of ass The ass of a syste is constant by definition. b. Newton's second law of otion The net force acting on a syste is directly proportional to the product of the syste ass ties its acceleration. c. First law of therodynaics The change in stored energy of a syste equals the net energy added to the syste as heat and work. d. Second law of therodynaics The entropy of any isolated syste cannot decrease during any process between equilibriu states. e. Principle of angular oentu The net torque acting on a syste is equal to the rate of change of angular oentu of the syste.

3 Proble. []. Discuss the physics of skipping a stone across the water surface of a lake. Copare these echaniss with a stone as it bounces after being thrown along a roadway. Open-Ended Proble Stateent: Consider the physics of skipping a stone across the water surface of a lake. Copare these echaniss with a stone as it bounces after being thrown along a roadway. Discussion: Observation and experience suggest two behaviors when a stone is thrown along a water surface:. If the angle between the path of the stone and the water surface is steep the stone ay penetrate the water surface. Soe oentu of the stone will be converted to oentu of the water in the resulting splash. After penetrating the water surface, the high drag * of the water will slow the stone quickly. Then, because the stone is heavier than water it will sink.. If the angle between the path of the stone and the water surface is shallow the stone ay not penetrate the water surface. The splash will be saller than if the stone penetrated the water surface. This will transfer less oentu to the water, causing less reduction in speed of the stone. The only drag force on the stone will be fro friction on the water surface. The drag will be oentary, causing the stone to lose only a portion of its kinetic energy. Instead of sinking, the stone ay skip off the surface and becoe airborne again. When the stone is thrown with speed and angle just right, it ay skip several ties across the water surface. With each skip the stone loses soe forward speed. After several skips the stone loses enough forward speed to penetrate the surface and sink into the water. Observation suggests that the shape of the stone significantly affects skipping. Essentially spherical stones ay be ade to skip with considerable effort and skill fro the thrower. Flatter, ore disc-shaped stones are ore likely to skip, provided they are thrown with the flat surface(s) essentially parallel to the water surface; spin ay be used to stabilize the stone in flight. By contrast, no stone can ever penetrate the paveent of a roadway. Each collision between stone and roadway will be inelastic; friction between the road surface and stone will affect the otion of the stone only slightly. Regardless of the initial angle between the path of the stone and the surface of the roadway, the stone ay bounce several ties, then finally it will roll to a stop. The shape of the stone is unlikely to affect trajectory of bouncing fro a roadway significantly.

4 Proble.4 [].4 The barrel of a bicycle tire pup becoes quite war during use. Explain the echaniss responsible for the teperature increase. Open-Ended Proble Stateent: The barrel of a bicycle tire pup becoes quite war during use. Explain the echaniss responsible for the teperature increase. Discussion: Two phenoena are responsible for the teperature increase: () friction between the pup piston and barrel and () teperature rise of the air as it is copressed in the pup barrel. Friction between the pup piston and barrel converts echanical energy (force on the piston oving through a distance) into theral energy as a result of friction. ubricating the piston helps to provide a good seal with the pup barrel and reduces friction (and therefore force) between the piston and barrel. Teperature of the trapped air rises as it is copressed. The copression is not adiabatic because it occurs during a finite tie interval. Heat is transferred fro the war copressed air in the pup barrel to the cooler surroundings. This raises the teperature of the barrel, aking its outside surface war (or even hot!) to the touch.

5 Proble.5 [] Data on oxygen tank. Mass of oxygen. Copute tank volue, and then use oxygen density (Table A.6) to find the ass. The given or available data is: D 500 c p 7 MPa T ( 5 + 7) K T 98 K J R O 59.8 (Table A.6) kg K The governing equation is the ideal gas equation p ρ R O T and ρ M V where V is the tank volue π D V V 6 π ( 5 ) V Hence M V ρ pv M N R O T 65.4 kg K M 59 kg 59.8 N 98 K

6 Proble.6 [] Basic equation: Diensions of a roo Mass of air ρ p R air T Given or available data p 4.7psi T ( )R R air 5. V 0 ft 0 ft 8 ft V 800ft ft lbf lb R Then ρ p ρ lb R air T ft ρ slug ft ρ. kg M ρ V M 6. lb M.90 slug M 7.8 kg

7 Proble.7 [] Mass of nitrogen, and design constraints on tank diensions. External diensions. Use given geoetric data and nitrogen ass, with data fro Table A.6. The given or available data is: M 0 lb p ( 00 + ) at p.95 0 psi ft lbf T ( ) K T 954 R R N 55.6 (Table A.6) lb R The governing equation is the ideal gas equation p ρ R N T and ρ M V where V is the tank volue π D V where D 4 Cobining these equations: Hence M V ρ pv R N T p R N T π D 4 p R N T π D D 4 p π D R N T Solving for D R N T M D D p π π 55.6 ft lbf in 954 K 0 lb lb R 950 lbf ft in D. ft D.5 in D 7 in These are internal diensions; the external ones are /4 in. larger: 7.5 in D.75 in

8 Proble.8 [].8 Very sall particles oving in fluids are known to experience a drag force proportional to speed. Consider a particle of net weight W dropped in a fluid. The particle experiences a drag force, F D kv, where V is the particle speed. Deterine the tie required for the particle to accelerate fro rest to 95 percent of its terinal speed, V t, in ters of k, W, and g. Sall particle accelerating fro rest in a fluid. Net weight is W, resisting force F D kv, where V is speed. Tie required to reach 95 percent of terinal speed, V t. Consider the particle to be a syste. Apply Newton's second law. Basic equation: F y a y Assuptions:. W is net weight. Resisting force acts opposite to V Then F W kv a y y dv W dt g dv dt or Separating variables, Integrating, noting that velocity is zero initially, dv dt g( k W V) dv V gdt k W zv 0 dv W V k ln( k t V) gdt gt k W 0 W O QP V 0 z

9 k or W V e W ; V W k e kgt NM kgt W O QP But V V t as t, so V t W. Therefore k V V e t kgt W kgt When V 0.95, then e W 0.05and kgt. Thus t W/gk V t W

10 Proble.9 [].9 Consider again the sall particle of Proble.8. Express the distance required to reach 95 percent of its terinal speed in ters of g, k, and W. Sall particle accelerating fro rest in a fluid. Net weight is W, resisting force is F D kv, where V is speed. Distance required to reach 95 percent of terinal speed, V t. Consider the particle to be a syste. Apply Newton's second law. Basic equation: F y a y Assuptions:. W is net weight.. Resisting force acts opposite to V. Then, dv W dv Fy W kv ay dt g V dy or V k V dv W g dy At terinal speed, a y 0 and W V V t k. Then V V dv Vg g dy Separating variables VdV gdy V Vt Integrating, noting that velocity is zero initially 0.95Vt VdV V gy VV ln 0 t Vt V Vt V t gy V V V 0.95 t t ln ( 0.95) t ln () [ ] gy V ln V t t.05 W y Vt.05 g gt 0.95Vt 0

11 Proble.0 [] Data on sphere and forula for drag. Maxiu speed, tie to reach 95% of this speed, and plot speed as a function of tie. Use given data and data in Appendices, and integrate equation of otion by separating variables. The data provided, or available in the Appendices, are: ρ air.7 kg μ Ns ρ w 999 kg SG Sty 0.06 d 0. Then the density of the sphere is ρ Sty SG Sty ρ w ρ Sty 6 kg π d The sphere ass is M ρ Sty 6 kg ( ) 6 π M kg 6 Newton's nd law for the steady state otion becoes (ignoring buoyancy effects) Mg π V d so V ax Mg π μ d kg 9.8 π s V N s ax s Newton's nd law for the general otion is (ignoring buoyancy effects) so dv π μ d g M V Integrating and using liits dt Vt () M dv dt Mg π μ d t Mg M e π μ d π μ V d

12 Using the given data V (/s) t (s) The tie to reach 95% of axiu speed is obtained fro π μ d t Mg M e π μ d 0.95 V ax so M 0.95 V ax π μ d t ln Substituting values t 0.0s π μ d Mg The plot can also be done in Excel.

13 Proble. [4] Data on sphere and forula for drag. Diaeter of gasoline droplets that take second to fall 5 c. Use given data and data in Appendices; integrate equation of otion by separating variables. The data provided, or available in the Appendices, are: μ Ns ρ w 999 kg SG gas 0.7 ρ gas SG gas ρ w ρ gas 79 kg Newton's nd law for the sphere (ass M) is (ignoring buoyancy effects) so Integrating and using liits dv dt π μ d g M V π μ d t Mg M Vt () e π μ d M dv dt Mg π μ V d Integrating again xt () π μ d t Mg M M t + e π μ d π μ d π d Replacing M with an expression involving diaeter d M ρ gas xt () 6 ρ gas d g t + 8 μ ρ gas d 8 μ e 8 μ ρ gas d t This equation ust be solved for d so that x( s ). The answer can be obtained fro anual iteration, or by using Excel's Goal Seek. (See this in the corresponding Excel workbook.) d x () Note That the particle quickly reaches terinal speed, so that a sipler approxiate solution would be to solve Mg πμvd for d, with V 0.5 /s (allowing for the fact that M is a function of d)! t (s)

14 Proble. [4] Data on sky diver: M 70 kg k 0.5 Ns Maxiu speed; speed after 00 ; plot speed as function of tie and distance. Use given data; integrate equation of otion by separating variables. Treat the sky diver as a syste; apply Newton's nd law: Newton's nd law for the sky diver (ass M) is (ignoring buoyancy effects): M dv dt Mg kv () (a) For terinal speed V t, acceleration is zero, so Mg kv Mg 0 so V t k Ns V t 75 kg 9.8 s 0.5 N s V kg t 54. s (b) For V at y 00 we need to find V(y). Fro () M dv dt M dv dy dv MV Mg kv dy dt dt Separating variables and integrating: so ln kv Mg V V 0 kv Mg y dv g dy 0 k y or V Mg e M k k y M Hence For y 00 : k y M Vy ( ) V t e Ns kg 0.5 V( 00 ) kg s N e V( 00 ) 8.8 s s

15 60 V(/s) y() (c) For V(t) we need to integrate () with respect to t: M dv dt Mg kv Separating variables and integrating: V 0 t V dv Mg V t d 0 k so t M ln kg Mg k Mg k + V V M ln kg V t + V V t V Rearranging kg t M e Vt () V k t or Vt () V kg t tanh V t t M t M e + 60 V(/s) 40 Vt () t t(s) The two graphs can also be plotted in Excel.

16 Proble. [5] Data on sky diver: M 70 kg k vert 0.5 Ns k horiz 0.05 Ns U 0 70 s Plot of trajectory. Use given data; integrate equation of otion by separating variables. Treat the sky diver as a syste; apply Newton's nd law in horizontal and vertical directions: Vertical: Newton's nd law for the sky diver (ass M) is (ignoring buoyancy effects): M dv dt Mg k vert V () For V(t) we need to integrate () with respect to t: Separating variables and integrating: so t M ln k vert g V 0 Mg k vert Mg k vert V Mg V k vert + V V t dv dt 0 Rearranging or Vt () k vert g t M Mg e so Vt () k vert k vert g t M e + Mg tanh k vert k vert g M t For y(t) we need to integrate again: dy dt V or y V dt yt () t t Vt () dt 0 0 Mg tanh k vert k vert g t M t d Mg ln cosh k vert k vert g M t yt () Mg ln cosh k vert k vert g M t

17 600 y() yt () t t(s) Horizontal: Newton's nd law for the sky diver (ass M) is: M du dt k horiz U () For U(t) we need to integrate () with respect to t: Separating variables and integrating: U t k horiz U U d dt so M 0 U 0 k horiz t + M U U 0 Rearranging or Ut () + U 0 k horiz U 0 M t For x(t) we need to integrate again: dx dt U or x U dt xt () t t Ut () dt U 0 k horiz U 0 M t dt M ln k horiz k horiz U 0 M t + xt () M ln k horiz k horiz U 0 M t +

18 x() xt () t t(s) Plotting the trajectory: 0 y(k) x(k) These plots can also be done in Excel.

19 Proble.4 [] Data on sphere and terinal speed. Drag constant k, and tie to reach 99% of terinal speed. Use given data; integrate equation of otion by separating variables. The data provided are: M 5 0 kg V t 5 c s Newton's nd law for the general otion is (ignoring buoyancy effects) M dv dt Mg kv () Newton's nd law for the steady state otion becoes (ignoring buoyancy effects) Mg kv t so k Mg V t Mg k 5 0 s kg 9.8 V t s k Ns 0.05 To find the tie to reach 99% of V t, we need V(t). Fro, separating variables Integrating and using liits M t ln k k Mg V dv dt k g M V We ust evaluate this when V 0.99 V t V 4.95 c s t 5 0 Ns kg ln Ns N s kg 5 0 kg s s kg Ns t 0.05 s

20 Proble.5 [5] Data on sphere and terinal speed fro Proble.4. Distance traveled to reach 99% of terinal speed; plot of distance versus tie. Use given data; integrate equation of otion by separating variables. The data provided are: M 5 0 kg V t 5 c s Newton's nd law for the general otion is (ignoring buoyancy effects) M dv dt Mg kv () Newton's nd law for the steady state otion becoes (ignoring buoyancy effects) Mg k 5 0 s kg 9.8 V t s k Ns 0.05 To find the distance to reach 99% of V t, we need V(y). Fro : Separating variables VdV dy k g M V M dv dt Mg kv t so k M dy dv dv MV Mg dt dy dy kv Mg V t Integrating and using liits M g y k ln k Mg V M k V We ust evaluate this when V 0.99 V t V 4.95 c s ( ) 9.8 y 5 0 kg y s kg N s 0.9 Ns ln Ns kg N s 5 0 kg Ns s kg s 9.8 Alternatively we could use the approach of Proble.4 and first find the tie to reach terinal speed, and use this tie in y(t) to find the above value of y: s kg Ns... Fro, separating variables Integrating and using liits dv dt k g M V M k t ln V () k Mg

21 We ust evaluate this when V 0.99 V t V 4.95 c s t 5 0 Ns kg ln Ns s kg kg N s kg s Ns t 0.05 s Fro, after rearranging V dy dt Mg k k t M e Integrating and using liits y Mg k k t M M t + e k y 5 0 kg y s Ns s... kg N s Ns kg e N s kg y () t (s) This plot can also be presented in Excel.

22 Proble.6 [].6 The English perfected the longbow as a weapon after the Medieval period. In the hands of a skilled archer, the longbow was reputed to be accurate at ranges to 00 eters or ore. If the axiu altitude of an arrow is less than h 0 while traveling to a target 00 away fro the archer, and neglecting air resistance, estiate the speed and angle at which the arrow ust leave the bow. Plot the required release speed and angle as a function of height h. Plot: ong bow at range, R 00. Maxiu height of arrow is h 0. Neglect air resistance. Estiate of (a) speed, and (b) angle, of arrow leaving the bow. (a) release speed, and (b) angle, as a function of h et V u i+ v j V (cos θ i + sin θ j) ΣF g, so v v 0 gt, and t f t v0 v 0 /g y dv dt Also, v dv dy v0 g, v dv gdy, 0 gh 0 Thus h v g () Fro ΣF du u0v0 x 0, so u u0 const, and R u0t f dt g (). v0 gh (). u gr gr gr u0 v gh 8h 0 0

23 gr Then V u + v + gh and V0 gh + 8h N NM b g gr 8h V M s P s 8 s 0 O QP O Q (4) gh Fro Eq. v0 gh V0 sinθθ, sin (5) V F NM HG θ sin s I K J 0 O P Q s P Plots of V 0 V 0 (h) {Eq. 4} and θ 0 θ 0(h) {Eq. 5} are presented below

24 Proble.7 [] Basic diensions F,, t and T. Diensional representation of quantities below, and typical units in SI and English systes. (a) Power Power Energy Tie Force Distance Tie F t N s lbf ft s (b) Pressure Pressure Force Area F N lbf ft (c) Modulus of elasticity Pressure Force Area F N lbf ft (d) Angular velocity AngularVelocity Radians Tie t s s (e) Energy Energy Force Distance F N lbf ft (f) Moentu Moentu Mass Velocity M t Fro Newton's nd law Force Mass Acceleration so F M t or M Ft (g) Shear stress (h) Specific heat Hence Moentu M t Force F ShearStress Area SpecificHeat Energy Mass Teperature (i) Theral expansion coefficient TheralExpansionCoefficient Ft Ft Ns t F MT engthchange ength Teperature F Ft T T t T N s K K lbf s lbf ft ft s R R (j) Angular oentu AngularMoentu Moentu Distance Ft N s lbf ft s

25 Proble.8 [] Basic diensions M,, t and T. Diensional representation of quantities below, and typical units in SI and English systes. (a) Power Power Energy Tie Force Distance Tie F t Fro Newton's nd law Force Mass Acceleration so F Hence Power F t M t t M t M t kg s slugft s (b) Pressure Pressure Force Area F M t M t kg s slug ft s (c) Modulus of elasticity Pressure Force Area F M t M t kg s slug ft s (d) Angular velocity AngularVelocity Radians Tie t s s (e) Energy Energy Force Distance F M t M t kg s slug ft s (f) Moent of a force MoentOfForce Force ength F M t M t kg s slug ft s (g) Moentu Moentu Mass Velocity M t M t kg s slug ft s (h) Shear stress (i) Strain Force F M M kg slug ShearStress Area t t s ft s engthchange Strain Diensionless ength (j) Angular oentu AngularMoentu Moentu Distance M t M t kg s slugs ft s

26 Proble.9 [] Pressure, volue and density data in certain units Convert to different units Using data fro tables (e.g. Table G.) (a) psi 6895 Pa kpa psi 6.89 kpa psi 000 Pa (b) liter quart gal liter 0.64 gal liter 4 quart (c) lbf s ft lbf s N ft ft 47.9 Ns lbf 0.054

27 Proble.0 [] Viscosity, power, and specific energy data in certain units Convert to different units Using data fro tables (e.g. Table G.) (a) s ft 0.76 ft s s (b) 00 W hp 00 W 0.4 hp 746 W (c) kj kg kj 000 J Btu kg 0.4 Btu kg kj 055 J lb lb

28 Proble. [] Quantities in English Engineering (or custoary) units. Quantities in SI units. Use Table G. and other sources (e.g., Google) (a) 00 ft 00 ft in in in in ft 60 s s (b) (c) (d) 5 gal 65 ph 5.4 acres in gal gal in 65 ile 85 hr 9. hr ile 600 s s acre acre

29 Proble. [] Quantities in SI (or other) units. Quantities in BG units. Use Table G.. (a) in ft 58 ft in (b) 50 cc 50 c in ft ft 00 c in (c) 00 kw 000 W hp 00 kw 4 hp kw 746 W (d) 5 lbf s ft is already in BG units

30 Proble. [] Acreage of land, and water needs. Water flow rate (gp) to water crops. Use Table G. and other sources (e.g., Google) as needed. The volue flow rate needed is Q.5 in week 5 acres Perforing unit conversions Q.5 in 5 acre week.5 in 5 acre week ft acre in ft week 7 day day 4 hr hr 60 in Q 0 gp

31 Proble.4 [] Geoetry of tank, and weight of propane. Volue of propane, and tank volue; explain the discrepancy. Use Table G. and other sources (e.g., Google) as needed. The author's tank is approxiately in in diaeter, and the cylindrical part is about 8 in. The weight of propane specified is 7 lb. The tank diaeter is D in The tank cylindrical height is 8 in The ass of propane is prop 7 lb The specific gravity of propane is SG prop The density of water is ρ 998 kg prop The volue of propane is given by V prop ρ prop V prop 7 lb V prop 95 in prop SG prop ρ kg kg lb in The volue of the tank is given by a cylinder diaeter D length, πd /4 and a sphere (two halves) given by πd /6 π D π D V tank π ( in) in V tank 8 in + π ( ) 4 6 V tank 80 in The ratio of propane to tank volues is V prop 5 % V tank This sees low, and can be explained by a) tanks are not filled copletely, b) the geoetry of the tank gave an overestiate of the volue (the ends are not really heispheres, and we have not allowed for tank wall thickness).

32 Proble.5 [].5 The density of ercury is given as 6. slug/ft. Calculate the specific gravity and the specific volue in /kg of the ercury. Calculate the specific weight in lbf/ft on Earth and on the oon. Acceleration of gravity on the oon is 5.47 ft/s. Density of ercury is ρ 6. slug/ft. Acceleration of gravity on oon is g 5.47 ft/s. a. Specific gravity of ercury. b. Specific volue of ercury, in /kg. c. Specific weight on Earth. d. Specific weight on oon. Apply definitions: γ ρ, v ρ, ρ ρ g SG HO Thus SG 6. slug ft.6 ft.94 slug ft slug lb 5 v (0.048) kg 6. slug ft. lb kg slug ft lbf s On Earth, γ E lbf ft ft s slug ft slug ft lbf s On the oon, γ lbf ft ft s slug ft {Note that the ass based quantities (SG and ν) are independent of gravity.}

33 Proble.6 [] Data in given units Convert to different units (a) in in in in 7 in in 60 s s (b) s gal 60 s s gp in (c) liter in liter gal 60 s 0.64 gp in liter in (d) SCFM ft in in.70 ft hr hr

34 Proble.7 [].7 The kilogra force is coonly used in Europe as a unit of force. (As in the U.S. custoary syste, where lbf is the force exerted by a ass of lb in standard gravity, kgf is the force exerted by a ass of kg in standard gravity.) Moderate pressures, such as those for auto or truck tires, are conveniently expressed in units of kgf/c. Convert psig to these units. In European usage, kgf is the force exerted on kg ass in standard gravity. Convert psi to units of kgf/c. Apply Newton's second law. Basic equation: F a N s The force exerted on kg in standard gravity is F kg N kgf s kg Setting up a conversion fro psi to kgf/c lbf lbf N in. kgf kgf, in. in. lbf ( 54. ) c 9.8 N c or Thus kgf c psi kgf c psi psi psi psi. 5 kgf c

35 Proble.8 [] Inforation on canal geoetry. Flow speed using the Manning equation, correctly and incorrectly! Use Table G. and other sources (e.g., Google) as needed. The Manning equation is R h S0 V which assues R h in eters and V in /s. n The given data is R h 7.5 S 0 n Hence V V 86.5 (Note that we don't cancel units; we just write /s 0.04 s next to the answer! Note also this is a very high speed due to the extree slope S 0.) in ft Using the equation incorrectly: R h 7.5 R in h 4.6 ft Hence V V 9 ft (Note that we again don't cancel units; we just 0.04 s write ft/s next to the answer!) This incorrect use does not provide the correct answer V 9 ft in s ft in V 58. s which is wrong! This deonstrates that for this "engineering" equation we ust be careful in its use! To generate a Manning equation valid for R h in ft and V in ft/s, we need to do the following: V ft s V s in ft in R h ( ) S0 n in ft in V ft s R h ( ft) S0 n in ft in in ft R h ( ft) S0 in n in ft in

36 In using this equation, we ignore the units and just evaluate the conversion factor.054 Hence V ft.49 R h ( ft) S 0 s n.49 Handbooks soeties provide this for of the Manning equation for direct use with BG units. In our case we are asked to instead define a new value for n: n n BG n.49 BG where V ft R h ( ft) S0 s n BG Using this equation with Rh 4.6 ft: V V 84 ft s Converting to /s V 84 ft in V 86.6 which is the correct answer! s ft in s

37 Proble.9 [] Equation for axiu flow rate. Whether it is diensionally correct. If not, find units of 0.04 ter. Write a BG version of the equation Rearrange equation to check units of 0.04 ter. Then use conversions fro Table G. or other sources (e.g., Google) "Solving" the equation for the constant 0.04: 0.04 ax T 0 A t p 0 Substituting the units of the ters on the right, the units of the constant are kg K s Pa kg K s N Ns kg K s Hence the constant is actually c 0.04 K s For BG units we could start with the equation and convert each ter (e.g., A t ), and cobine the result into a new constant, or siply convert c directly: c 0.04 K s.8 R in K in ft c R s A t p 0 so ft ax with A t in ft, p 0 in lbf/ft, and T 0 in R. T 0 This value of c assues p is in lbf/ft. For p in psi we need an additional conversion: c R s in c.6 R in s A t p 0 ft ft ft so ax.6 with A t in ft, p 0 in psi, and T 0 in R. T 0

38 Proble.0 Equation for COP and teperature data. COP Ideal, EER, and copare to a typical Energy Star copliant EER value. Use the COP equation. Then use conversions fro Table G. or other sources (e.g., Google) to find the EER. The given data is T ( ) R T 58 R T H ( ) R T H 555 R T The COP Ideal is COP Ideal T H T The EER is a siilar easure to COP except the cooling rate (nuerator) is in BTU/hr and the electrical input (denoinator) is in W: BTU 545 BTU BTU hr hr hr EER Ideal COP Ideal W 746 W W This copares to Energy Star copliant values of about 5 BTU/hr/W! We have soe way to go! We can define the isentropic efficiency as η isen EER Actual EER Ideal Hence the isentropic efficiency of a very good AC is about.5%.

39 Proble. [] Equation for drag on a body. Diensions of C D. Use the drag equation. Then "solve" for CD and use diensions. The drag equation is F D ρ V A C D F D "Solving" for C D, and using diensions C D ρ V A F C D M t But, Fro Newton's nd law Force Mass Acceleration or F M t F Hence C D M t The drag coefficient is diensionless. M t t M 0

40 Proble. [] Equation for ean free path of a olecule. Diensions of C for a diesionally consistent equation. Use the ean free path equation. Then "solve" for C and use diensions. The ean free path equation is λ C ρ d "Solving" for C, and using diensions C λρ d M The drag constant C is diensionless. C 0 M

41 Proble. [] Equation for vibrations. Diensions of c, k and f for a diensionally consistent equation. Also, suitable units in SI and BG systes. Use the vibration equation to find the diesions of each quantity The first ter of the equation is d x dt The diensions of this are M t Each of the other ters ust also have these diensions. Hence c dx dt kx M t so c t M M t and c t M M t so k t and k kg slug kg slug kg slug ft Suitable units for c, k, and f are c: k: s s s s f: s s Note that c is a daping (viscous) friction ter, k is a spring constant, and f is a forcing function. These are ore typically expressed using F ( rather than M (ass). Fro Newton's nd law: Using this in the diensions and units for c, k, and f we findc c: Ns lbf s ft F M t or M Ft k: N Ft t lbf ft Ft k Ft t f: N lbf F f F f M t M t

42 Proble.4 [] Specific speed in custoary units Units; Specific speed in SI units The units are rp gp 4 ft or 4 ft s Using data fro tables (e.g. Table G.) N Scu rp gp ft rp gp π rad N Scu 000 rev 4 ft rad s s N Scu in 60 s gal in 60 s ft

43 Proble.5 [] "Engineering" equation for a pup SI version The diensions of ".5" are ft. The diensions of "4.5 x 0-5 " are ft/gp. Using data fro tables (e.g. Table G.), the SI versions of these coefficients can be obtained.5 ft ft ft ft gp ft gp ft gal quart 60 s 4 quart in The equation is ft gp 450 s H ( ) Q s

44 Proble.6 [].6 A container weighs.5 lbf when epty. When filled with water at 90 F, the ass of the container and its contents is.5 slug. Find the weight of water in the container, and its volue in cubic feet, using data fro Appendix A. Epty container weighing.5 lbf when epty, has a ass of.5 slug when filled with water at 90 F. a. Weight of water in the container b. Container volue in ft Basic equation: F a Weight is the force of gravity on a body, W g Then W W + W W W W g W W t HO c HO t c c ft lbf s 5. slug. 5. lbf 770. lbf s slug ft HO M M g W The volue is given by ρ ρg ρg HO HO HO Fro Table A.7, ρ.9 slug/ft ft s slug ft at T 90 F lbf 4. ft 9. slug. ft lbf s

45 Proble.7 [].7 Calculate the density of standard air in a laboratory fro the ideal gas equation of state. Estiate the experiental uncertainty in the air density calculated for standard conditions (9.9 in. of ercury and 59 F) if the uncertainty in easuring the baroeter height is ±0. in. of ercury and the uncertainty in easuring teperature is ±0.5 F. (Note that 9.9 in. of ercury corresponds to 4.7 psia.) Air at standard conditions p 9.9 in Hg, T 59 F Uncertainty: in p is ± 0. in Hg, in T is ± 0.5 F Note that 9.9 in Hg corresponds to 4.7 psia a. air density using ideal gas equation of state. b. estiate of uncertainty in calculated value. p lbf lb R ρ 4. 7 RT in 5. ft lbf ρ lb ft in R ft The uncertainty in density is given by Then u ρ F NM HG p ρ ρ p u T ρ ρ T u p ρ RT RT up ρ p RT RT ± 0. ; ± 04% T ρ T p ρ T ρ RT ρ NM p I F + KJ HG p ρrt T I KJ O QP F HG I K J O ; d pi b Tg b0. 4g + b g uρ u + u ± QP 4 u ± 0. 48% ± lb ft e j u T ± 05. ± %

46 Proble.8 [].8 Repeat the calculation of uncertainty described in Proble.7 for air in a freezer. Assue the easured baroeter height is 759 ± of ercury and the teperature is 0 ± 0.5 C. [Note that 759 of ercury corresponds to 0 kpa (abs).] Air at pressure, p 759 ± Hg and teperature, T 0 ± 0.5 C. Note that 759 Hg corresponds to 0 kpa. a. Air density using ideal gas equation of state b. Estiate of uncertainty in calculated value p RT ρ 0 0 N kg K 9. kg 87 N 5 K The uncertainty in density is given by Then u ρ F NM HG p ρ ρ p u T ρ ρ T u p ρ RT up ρ p ± ; ± 0%. RT 759 T ρ T p ρ T ρ RT ρ NM p I F + KJ HG p ρrt T I KJ O QP / F HG I K J ; u T O d pi b Tg b0. g + b 098. g uρ u + u ± QP u ± 0. 8% ±. 0 kg e j ± 05. ± 098%. 7 0

47 Proble.9 [].9 The ass of the standard Aerican golf ball is.6 ± 0.0 oz and its ean diaeter is.68 ± 0.0 in. Deterine the density and specific gravity of the Aerican golf ball. Estiate the uncertainties in the calculated values. Standard Aerican golf ball: 6. ± 00. oz ( 0 to) D 68. ± 0. 0 in. ( 0 to) a. Density and specific gravity. b. Estiate uncertainties in calculated values. Density is ass per unit volue, so 6 ρ 4 πr 4π ( D ) π D kg in. ρ 6. oz 0 kg π (. 68) in. 6 oz ( ) ρ kg and SG 0. HO ρ 000 kg The uncertainty in density is given by u ρ ± F NM HG ρ ρ u I F + KJ HG D ρ ρ D u D I KJ O QP ρ u percent ρ ρ ± 00. ; 6 ± F I HG K J F 4 I HG K J ± D ρ D πd 6 4 ρ D ρ π D 6 6 ; ud percent 4 π D Thus Finally, b g b Dg u ± u + u ρ uρ ± 89. percent ± 4. kg u u ± 89. percent ± SG { b g b g } e j ρ b g ± ρ 0 ± 4. kg ( 0 to ) SG. ± ( 0 to )

48 Proble.40 [].40 The ass flow rate in a water flow syste deterined by collecting the discharge over a tied interval is 0. kg/s. The scales used can be read to the nearest 0.05 kg and the stopwatch is accurate to 0. s. Estiate the precision with which the flow rate can be calculated for tie intervals of (a) 0 s and (b) in. Mass flow rate of water deterined by collecting discharge over a tied interval is 0. kg/s. Scales can be read to nearest 0.05 kg. Stopwatch can be read to nearest 0. s. Estiate precision of flow rate calculation for tie intervals of (a) 0 s, and (b) in. Apply ethodology of uncertainty analysis, Appendix F: Coputing equations: Thus u t ± F I HG u K J F + NM HG t t u t I K J O QP t t t and t t t QP F H G I K J NM b g O The uncertainties are expected to be ± half the least counts of the easuring instruents. Tabulating results: Tie Error Uncertainty Water Uncertainty Uncertainty Interval, in in t Collected, Error in in in t(s) t(s) (percent) (kg) (kg) (percent) (percent) 0 ± 0.0 ±.0.0 ± 0.05 ±.5 ± ± 0.0 ± ± 0.05 ± 0.08 ± 0.67 A tie interval of about 5 seconds should be chosen to reduce the uncertainty in results to ± percent.

49 Proble.4 [].4 A can of pet food has the following internal diensions: 0 height and 7 diaeter (each ± at odds of 0 to ). The label lists the ass of the contents as 97 g. Evaluate the agnitude and estiated uncertainty of the density of the pet food if the ass value is accurate to ± g at the sae odds. Pet food can H 0 ± ( 0 to) D 7 ± ( 0 to) 97 ± g ( 0 to) Magnitude and estiated uncertainty of pet food density. Density is Fro uncertainty analysis 4 ρ ρ ρ πr H π D H or (, D, H) u ρ ± F NM HG ρ ρ u I F + KJ HG D ρ ρ D u I F + KJ HG H ρ ρ H u D H I KJ O QP Evaluating, Substituting ρ 4 4 u ± ; ± 0. 5% ρ ρπdh ρπdh 97 D ρ D 4 4 ud ± ( ) ( ) ; ± 7%. ρ D ρ πdh ρ πd H 7 H ρ H 4 4 uh ± ( ) ( ) ; ± % ρ H ρ πdh ρ πdh 0 u u ρ ρ o ± [( )( 05. )] + [( )( 7. )] + [( )( )] ± 9. percent t 4 π π D H ( 7) g ρ Thus ρ 90 ± 7. kg ( 0 to ) kg 90 kg 000 g

50 Proble.4 [].4 The ass of the standard British golf ball is 45.9 ± 0. g and its ean diaeter is 4. ± 0.. Deterine the density and specific gravity of the British golf ball. Estiate the uncertainties in the calculated values. Standard British golf ball: 459. ± 0. g ( 0 to) D 4. ± 0. ( 0 to) a. Density and specific gravity b. Estiate of uncertainties in calculated values. Density is ass per unit volue, so 6 ρ 4 πr 4π ( D ) π D 6 ρ kg 60 kg π (. 0 04) and The uncertainty in density is given by Thus ρ kg SG ρho 000 kg u ρ ρ u D ρ ρ D u ρ u ρ ρ ± 0. ; 45 9 ± %. D ρ D ρ D ρ π D πd ρ u ρ D ± F NM HG I F + KJ HG D I KJ F HG I K J F H G I K J 0. ± 0. 70% 4. O QP u ± [( u ) + ( u ) ] ± ( ) + [ ( 0. 70)] u u ρ ρ SG D ±. 9% ( ± 8. 9 kg ) u ±. 9% ( ± ) ρ o t Suarizing ρ 60 ± 8. 9 kg ( 0 to ) SG 6. ± ( 0 to )

51 Proble.4 [].4 The ass flow rate of water in a tube is easured using a beaker to catch water during a tied interval. The noinal ass flow rate is 00 g/s. Assue that ass is easured using a balance with a least count of g and a axiu capacity of kg, and that the tier has a least count of 0. s. Estiate the tie intervals and uncertainties in easured ass flow rate that would result fro using 00, 500, and 000 beakers. Would there be any advantage in using the largest beaker? Assue the tare ass of the epty 000 beaker is 500 g. Noinal ass flow rate of water deterined by collecting discharge (in a beaker) over a tied interval is 00 g s Scales have capacity of kg, with least count of g. Tier has least count of 0. s. Beakers with volue of 00, 500, 000 are available tare ass of 000 beaker is 500 g. Estiate (a) tie intervals, and (b) uncertainties, in easuring ass flow rate fro using each of the three beakers. To estiate tie intervals assue beaker is filled to axiu volue in case of 00 and 500 beakers and to axiu allowable ass of water (500 g) in case of 000 beaker. Then t and t ρ Tabulating results t s 5 s 5 s Apply the ethodology of uncertainty analysis, Appendix E Coputing equation: u ± F I HG u K J F + NM HG t t u t I K J O QP The uncertainties are expected to be ± half the least counts of the easuring instruents δ ± 05. g δ t 005. s t t t and t t t P F H G I K J bg NM O P bgq

52 b g b g t u ± u + u Tabulating results: Uncertainty Beaker Water Error in Uncertainty Tie Error in in t in Volue Collected (g) in Interval t(s) (percent) (percent) () (g) (percent) t(s) ± 0.50 ± ± 0.05 ± 5.0 ± ± 0.50 ± ± 0.05 ±.0 ± ± 0.50 ± ± 0.05 ±.0 ±.0 Since the scales have a capacity of kg and the tare ass of the 000 beaker is 500 g, there is no advantage in using the larger beaker. The uncertainty in could be reduced to ± 0.50 percent by using the large beaker if a scale with greater capacity the sae least count were available

53 Proble.44 [].44 The estiated diensions of a soda can are D 66.0 ± 0.5 and H 0 ± 0.5. Measure the ass of a full can and an epty can using a kitchen scale or postal scale. Estiate the volue of soda contained in the can. Fro your easureents estiate the depth to which the can is filled and the uncertainty in the estiate. Assue the value of SG.055, as supplied by the bottler. Soda can with estiated diensions D 66.0 ± 0.5, H 0 ± 0.5. Soda has SG.055 a. volue of soda in the can (based on easured ass of full and epty can). b. estiate average depth to which the can is filled and the uncertainty in the estiate. Measureents on a can of coke give ± 050. g, 7. 5± 050. g 69 ± u g u u f e f e f ± F NM HG f u f f I F + KJ HG e u e e I KJ O QP / 0.5 g 050. ± ± , u ± e 86.5 g 7. 5 R ± N M S O Q P u + T ()(. ) NM 69 Density is ass per unit volue and SG ρ/ρη Ο so O U QP V W / ( )( ) kg 69 g 50 0 ρ ρhosg 000 kg g 6 The reference value ρh O is assued to be precise. Since SG is specified to three places beyond the decial point, assue u SG ± Then

54 u u v SG SG ± [( ) u ] + [( ) u ] v SG v ± v F I HG v u K J F + NM HG o ± [( ) ( )] + [( ) ( 0. 00)] or 0. % πd or 4 πd π ( ) u ± u I K J / O QP t / / D D u o 6 D ud π 4 D ± π D D D π D 4 4 π D t 0 0 u ± [( ) ( 0. 00)] + [( ) ( )] or.5% F NM HG o I K J O P Q P + F NM HG F HG I K J D I K J O QP / / t Note:. Printing on the can states the content as 55 l. This suggests that the iplied accuracy of the SG value ay be over stated.. Results suggest that over seven percent of the can height is void of soda.

55 Proble.45 [] Data on water Viscosity; Uncertainty in viscosity The data is: A Ns B 47.8 K C 40 K T 9 K 0.5 K The uncertainty in teperature is u T u 9 K T % Also For the uncertainty B ( T C) μ( T) A 0 Evaluating μ( T).0 0 Ns d dt μ( T) AB ln( 0) B C T 0 ( C T) Hence u μ ( T) T d ln( 0) B T u T μ( T) dt μ( T) u T Evaluating u μ ( T) % C T ( )

56 Proble.46 [].46 An enthusiast agazine publishes data fro its road tests on the lateral acceleration capability of cars. The easureents are ade using a 50-ft-diaeter skid pad. Assue the vehicle path deviates fro the circle by ± ft and that the vehicle speed is read fro a fifth-wheel speed-easuring syste to ±0.5 ph. Estiate the experiental uncertainty in a reported lateral acceleration of 0.7 g. How would you iprove the experiental procedure to reduce the uncertainty? ateral acceleration, a 0.70 g, easured on 50-ft diaeter skid pad. U V W Path deviation: ± ft easureent uncertainty Vehicle speed: ± 0.5 ph a. Estiate uncertainty in lateral acceleration. b. How could experiental procedure be iproved? ateral acceleration is given by a V /R. a v R Fro Appendix F, u ± [( u ) + ( u ) ] /. ft Fro the given data, V ar; V ar ft 4. ft / s s Then u v NM / O QP δv i s ft hr ± ± ± V hr 4. ft i 600 s and so u u u R a a R ± δ ± ft ± R 75 ft ± ( ) + ( ) ± ± 445. percent / Experiental procedure could be iproved by using a larger circle, assuing the absolute errors in easureent are constant.

57 For D 400 ft, R 00 ft. ft V ar ft 67. ft / s 458. ph s QP 05. ph ft uv ± ± ; ur ± ± ph 00 ft u ± ( ) + ( ) ± or ±.4 percent a NM O / /

58 Proble.47 [4].47 Using the noinal diensions of the soda can given in Proble.44, deterine the precision with which the diaeter and height ust be easured to estiate the volue of the can within an uncertainty of ±0.5 percent. Diensions of soda can: D 66 H 0 Measureent precision needed to allow volue to be estiated with an uncertainty of ± 0.5 percent or less. Use the ethods of Appendix F: π 4 DH Coputing equations: H D u ± uh + ud H D π DH Since, then 4 π D H 4 and π DH D δ x δ x et ud ± D and uh ± H, substituting, u 4H πd δx 4D πdh δx δx δx DH 4 H πdh D H D ± + ± + π Solving, δx δx u + ( δ x) + H D H D

59 δ x u ± ± ± 0.58 ( H) + ( D) ( 0) + ( 66) Check: δ x 0.58 uh ± ± ±.44 0 H 0 δ x 0.58 ud ± ± ±.9 0 D 66 u ± [(u ) + (u ) ] ± [(0.0044) + ( ) ] ± H D If δx represents half the least count, a iniu resolution of about δx 0. is needed.

60 Proble.9 Proble.48 [4]

61 Given data: H 57.7 ft δ 0.5 ft δθ 0. deg For this building height, we are to vary θ (and therefore ) to iniize the uncertainty u H.

62 Plotting u H vs θ θ (deg) u H 5 4.0% 0.05% 5.4% 0.% 5.00% % % 40.0% 45.% 50.5% 55.44% 60.70% 65.07% 70.6% 75.5% 80 5.% % u H % 0% 8% 6% 4% % 0% Uncertainty in Height (H 57.7 ft) vs θ θ ( o ) Optiizing using Solver θ (deg) u H % To find the optiu θ as a function of building height H we need a ore coplex Solver H (ft) θ (deg) u H % % % % % % % % % % % % % % % θ (deg) Optiu Angle vs Building Height H (ft) Use Solver to vary A θ's to iniize the total u H! Total u H 's:.%

63 Proble.50 [5].50 In the design of a edical instruent it is desired to dispense cubic illieter of liquid using a pistoncylinder syringe ade fro olded plastic. The olding operation produces plastic parts with estiated diensional uncertainties of ±0.00 in. Estiate the uncertainty in dispensed volue that results fro the uncertainties in the diensions of the device. Plot on the sae graph the uncertainty in length, diaeter, and volue dispensed as a function of cylinder diaeter D fro D 0.5 to. Deterine the ratio of stroke length to bore diaeter that gives a design with iniu uncertainty in volue dispensed. Is the result influenced by the agnitude of the diensional uncertainty? Piston-cylinder device to have. Molded plastic parts with diensional uncertainties,δ ± 0.00 in. a. Estiate of uncertainty in dispensed volue that results fro the diensional uncertainties. b. Deterine the ratio of stroke length to bore diaeter that iniizes u ; plot of the results. c. Is this result influenced by the agnitude of δ? Apply uncertainty concepts fro Appendix F: πd Coputing equation: ; u ± 4 u F I HG K J F D + NM HG Fro,, and D, so u u u D ± [ + ( D ) ] in. D u D I K J O QP The diensional uncertainty is δ ± in. 5.4 ± πd π () Assue D. Then 7. u u u D U W δ ± ± ± 508. percent D Vu δ ± ± ± 400. percent 7. ±0. 9 percent ± [( 400. ) + ( ( 508. )) ]

64 To iniize u, substitute in ters of D: F ± + ± H G I K J F + H G I K J NM δ δ πd u [( u) ( ud) ] ± δ D 4 O QP F NM HG I + KJ F H G δ D I K J O QP This will be iniu when D is such that []/ D 0, or [] D F F HG I K J πδ I HG K J I 4D + ( δ ) 6 0; D HG 4 K J ; D 4 D π F I HG K J 4 6 Thus Dopt. π F 6 F HG 4 π I K J The corresponding is opt πd π (. ) The optiu stroke-to-bore ratio is D ) opt ( see table and plot on next page). Note that δ drops out of the optiization equation. This optiu /D is independent of the agnitude of δ However, the agnitude of the optiu u increases as δ increases. Uncertainty in volue of cylinder: δ in D () () /D (---) u D (%) u (%) u ( % )

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