CHAPTER 1 Basic Considerations

Transcription

1 CHAPTER 1 Basic Considerations FE-type Exam Review Problems: Problems 1.1 to (C) m F/a or kg N/m/s N. s /m. 1. (B) [µ] [τ//dy] (F/L )/(L/T)/L F. T/L. Chapter 1 / Basic Considerations 1. (A) npa. 1.4 (C) The mass is the same on earth and the moon: τ µ µ [4(8 r)] µ r. dr 1.5 (C) Fshear F sinθ 400 sin N. Fshear 100 N τ Pa or 84 kpa A m 1.6 (B) 1.7 (D) ρ water ( T 4) (80 4) kg/m (A) 1.9 (D) τ µ µ[ r] Pa. dr 4σ cosβ N/m 1 h m or 00 cm. ρ gd kg/m 9.81 m/s m We used kg N s /m 1.10 (C) 1.11 (C) m pv RT 800 kn/m 4 m kg kj/(kg K) (10 + 7) K 017 Cengage Learning. May not be scanned, copied or plicated, or posted to a publicly accessible website, in whole or in part. 1

2 1.1 (B) Eice Ewater. mice 0 mwater cwater T. 6 5 (40 10 ) ( 10 ) T. T 7.66 C. We assumed the density of ice to be equal to that of water, namely 1000 kg/m. Ice is actually slightly lighter than water, but it is not necessary for such accuracy in this problem. 1.1 (D) For this high-frequency wave, c RT m/s. Chapter 1 Problems: Dimensions, Units, and Physical Quantities 1.16 a) density M FT / L 4 FT / L L L c) power F velocity F L/T FL/T e) mass flux M T FT / L FT / L T 1.18 b) N [C] kg. [C] N/kg kg m/s kg m/s 1.0 kg m m s + c m s + k f. Since all terms must have the same dimensions (units) we require: [c] kg/s, [k] kg/s N s / m s N / m, [f] kg m / s N. Note: we could express the units on c as [c] kg / s N s / m s N s / m 1. a) N c) Pa e) m 0/ a) 0 cm/hr m/s 600 c) 50 Hp W e) 000 kn/cm 10 6 N/cm 100 cm /m N/m 1.6 The mass is the same on the earth and the moon: 60 m Wmoon lb. Pressure and Temperature 1.8 Use the values from Table B. in the Appendix. b) kpa. d) kpa. 017 Cengage Learning. May not be scanned, copied or plicated, or posted to a publicly accessible website, in whole or in part.

3 1.0 p po e gz/rt 101 e /87 (15 + 7) 6.8 kpa From Table B., at 4000 m: p 61.6 kpa. The percent error is % error % T 48 +,000 0,000 5,000 0,000 ( ) 59 F or ( 59 ) C F (10000) N n 4 Ft N F F n Ft θ tan N Density and Specific Weight 1.6 ρ 1000 (T 4) / (70 4) / kg/m γ 9800 (T 4) / (70 4) / N/m % error for ρ % % error for γ % b) m Viscosity kg Assume carbon dioxide is an ideal gas at the given conditions, then p 00 kn/m ρ.915 kg/m RT kj/kg K 90 7 K ( )( + ) W mg γ ρ g.915 kg/m 9.81 m/s 8.6 kg/m s 8.6 N/m V V 5 From Fig. B.1 at 90 C, µ 10 N s/m, so that the kinematic viscosity is µ ν ρ.915 kg/m 5 10 N s/m m /s The kinematic viscosity cannot be read from Fig. B.; the pressure is not 100 kpa. 017 Cengage Learning. May not be scanned, copied or plicated, or posted to a publicly accessible website, in whole or in part.

4 1.4 The shear stress can be calculated using τ µ / dy. From the given velocity distribution, dy u( y) 10(0.05 y y ) 10(0.05 y) From Table B.1 at 10 C, µ N s/m so, at the lower plate where y 0, dy y 0 ( ) 10(0.05 0) 6 s τ N/m At the upper plate where y 0.05 m, dy y ( ) 6 s τ N/m ( 1/1) τ µ µ [ r / r0 ] µ r / r0. τr 0 0, (1/1) dr τr /100. Pa, (0.5 /100) τr / Pa (0.5 /100) 1.46 Use Eq.1.5.8: T π ω µ ( ) 000 π R L h power Tω 550 π 0.5/ / hp ft-lb. rω 1.48 Assume a linear velocity so. Due to the area dy h element shown, dt df r τda r µ πr dr r. dy τ r dr T R 0 µωπ r dr h π 4 4 π.6 10 (/1) πµω R 60 h / ft-lb. 017 Cengage Learning. May not be scanned, copied or plicated, or posted to a publicly accessible website, in whole or in part. 4

5 1.50 If τ µ dy cons t and µ AeB/T Ae By/K Ae Cy, then Cy Ae dy cons t. Finally, or u(y) Compressibility dy De Cy. D y e Cy E (e Cy 1) where A, B, C, D, E, and K are constants. 0 C 1.54 Use c 1450 m/s. L c t m 1.56 b) c 7, / fps Surface Tension 1.58 p σ R Pa or 9.6 kpa. Bubbles: p 4σ/R 59. kpa 1.60 The droplet is assumed to be spherical. The pressure inside the droplet is greater than the outside pressure of 8000 kpa. The difference is given by Eq : σ 0.05 N/m p pinside poutside 6 10 kpa r 5 10 m Hence, p inside p + 10 kpa kpa outside In order to achieve this high pressure in the droplet, diesel fuel is usually pumped to a pressure of about kpa before it is injected into the engine. 1.6 See Example 1.4: h 4σ cosβ 4 0.0cos10 ρ gd / ft or in 1.64 Draw a free-body diagram: The force must balance: π W σl or d L ρ g σ L. 4 σl W σl needle d 8σ πρg 017 Cengage Learning. May not be scanned, copied or plicated, or posted to a publicly accessible website, in whole or in part. 5

6 1.66 Each surface tension force σ π D. There is a force on the outside and one on the inside of the ring. F σπd neglecting the weight of the ring. D F Vapor Pressure 1.68 The absolute pressure is p kpa. At 50 C water has a vapor pressure of 1. kpa; so T 50 C is a maximum temperature. The water would boil above this temperature At 40 C the vapor pressure from Table B.1 is 7.4 kpa. This would be the minimum pressure that could be obtained since the water would vaporize below this pressure. 1.7 The inlet pressure to a pump cannot be less than 0 kpa absolute. Assuming atmospheric pressure to be 100 kpa, we have Ideal Gas x. x 16.8 km ρ 101. in p 1.6 kg/m. RT 0.87 (15 + 7) 85 ρ out 1.19 kg/m Yes. The heavier air outside enters at the bottom and the lighter air inside exits at the top. A circulation is set up and the air moves from the outside in and the inside out: infiltration. This is the chimney effect W p RT 100 V g (10 0 4) N The pressure holding up the mass is 100 kpa. Hence, using pa W, we have m m kg. Hence, m pv RT π r / r 1.6 m or d 5. m Cengage Learning. May not be scanned, copied or plicated, or posted to a publicly accessible website, in whole or in part. 6

7 The First Law W1- KE. a) ( V f 10 ). V f m/s. c) 10 0 πs 1 00 cos ds 15( Vf 10 ). 0 0 π 1 00sin 15( Vf 10 ). Vf 16.4 m/s. π 1 E E1. mv mhoc T T. T 69. C. 600 We used c 4180 J/kg. C from Table B.5. (See Problem 1.75 for a units check.) 1.84 W pd V mrt V dv mrt dv V mrt ln V V since, for the T const process, p1 V 1 p V. Finally, 4 1 W ln 78,10 ft-lb.. The 1 st law states that 017 Cengage Learning. May not be scanned, copied or plicated, or posted to a publicly accessible website, in whole or in part. 7 mrt ln p p 1 1 Q W uɶ mc T 0. Q W 78, 10 ft-lb or 101 Btu. v T 1.86 W pd V p( V V 1 T 1). If p const, so if T T1, V 1 V then V V 1 and W p(v 1 V 1) p V 1 mrt1. b) W kj 1.88 We assume an isentropic process for the maximum pressure: k / k 1 1.4/ 0.4 T 4 p p1 ( ) 904 kpa abs or 804 kpa gage. T1 9 Note: We assumed patm 100 kpa since it was not given. Also, a measured pressure is a gage pressure. Speed of Sound 1.90 b) c krt m/s d) c krt m/s Note: We must use the units on R to be J/kg. K in the above equations.

8 1.9 b) c krt m/s. L c t m. 017 Cengage Learning. May not be scanned, copied or plicated, or posted to a publicly accessible website, in whole or in part. 8