CHAPTER 1 Basic Considerations

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1 CHAPTER Basic Considerations FE-type Exam Review Problems: Problems. to. Chapter / Basic Considerations. (C) m = F/a or kg = N/m/s = N s /m. (B) [μ] = [τ/(/dy)] = (F/L )/(L/T)/L = F. T/L. (A) Pa =.6 0 Pa =.6 npa.4 (C) The mass is the same on earth and the moon, so we calculate the mass using the weight given on earth as: m = W/g = 50 N/9.8 m/s = kg Hence, the weight on the moon is: W = mg = = N The shear stress is e to the component of the force acting tangential to the area:.5 (C) Fshear = Fsinθ = 400sin 0 = 00 N Fshear 00 N τ = = = 84 0 Pa or 84 kpa A m.6 (B) 5.6 C.7 (D).8 (A) Using Eqn. (.5.): ρ water ( T 4) (80 4) = 000 = 000 = 968 kg/m The shear stress is given by: τ = μ dr We determine /dr from the given expression for u as: d 0( 500r ) 50, 000r dr = dr = At the wall r = cm = 0.0 m. Substituting r in the above equation we get: 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or plicated, or posted to a publicly accessible website, in whole or in part.

2 dr = 50,000r = 000 /s The density of water at 0 C is 0 N s/m Now substitute in the equation for shear stress to get τ = μ = 0 N s/m 000 /s = N/m = Pa dr.9 (D) Using Eqn. (.5.6), β = 0 (for clean glass tube), and σ = N/m for water (Table B. in Appendix B) we write: 4σ cosβ N/m h = = = m or 00 cm ρgd kg/m 9.8 m/s 0 0 m where we used N = kg m/s.0 (C) Density. (C) Assume propane (C H 8 ) behaves as an ideal gas. First, determine the gas R 8.4 kj/kmol K constant for propane using u R = = = kj/kg M 44. kg/kmol pv then, m = RT 800 kn/m 4 m = = kg 60 kg kj/(kg K) (0 + 7) K Consider water and ice as the system. Hence, the change in energy for the system is zero. That is, the change in energy for water should be equal to the change in energy for the ice. So, we write Δ Eice =Δ Ewater. (B) m 0 kj/kg = m c Δ T ice water water The mass of ice is calculate using 6 = ρ V = 5 cubes 000 kg/m 40 0 m /cube = 0. kg mice ( ) ( ) Where we assumed the density of ice to be equal to that of water, namely 000 kg/m. Ice is actually slightly lighter than water, but it is not necessary for such accuracy in this problem. Similarly, the mass of water is calculated using m = ρv = 000 kg/m liters 0 m /liter = kg water ( ) ( ) Solving for the temperature change for water we get 0. kg 0 kj/kg = kg 4.8 kj/kg K ΔT Δ T = 7.66 C 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or plicated, or posted to a publicly accessible website, in whole or in part.

3 . (D) Since a dog s whistle proces sound waves at a high frequency, the speed of sound is c= RT = 87 J/kg K K = 04 m/s where we used J/kg = m /s. Dimensions, Units, and Physical Quantities.6 a) density = M L FT / L 4 = = FT / L L c) power = F velocity = F L/T = FL/T e) mass flux = MT / FT / L A LT = = FT/ L.8 b) N = [C] kg [C] = N/kg = (kg m/s )/kg = m/s.0 m m kg + c + km = f. Since all terms must have the same dimensions (units) s s 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. a) N c) Pa e) 5. 0 m.4 a) c) e) cm m hr 5 0 cm/hr = 0 = m/s hr 00 cm 600 s W 500 hp = 500 hp = 7, 85 W hp kn 00 cm 0 = = 000 kn/cm N/m cm m.6 The mass is the same on the earth and the moon, so we calculate the mass, then calculate the weight on the moon: m = 7 kg W moon = (7 kg) (.6) = 44.0 N 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or plicated, or posted to a publicly accessible website, in whole or in part.

4 Pressure and Temperature.8.0. Use the values from Table B. in the Appendix: b) At an elevation of 000 m the atmospheric pressure is kpa. Hence, the absolute pressure is = 4. kpa d) At an elevation of 0,000 m the atmospheric pressure is 6.49 kpa, and the absolute pressure is = 78.8 kpa p = p o e gz/rt = 0 e ( )/[(87) (5 + 7)] = 6.8 kpa From Table B., at 4000 m: p = 6.6 kpa. The percent error is % error = 00 =.95 % 6.6 Using Table B. and linear interpolation we write: 0,600 0,000 T =.48 + (.6 6.7) =. K,000 0,000 or (. 7.5) 5 9 = 5.8 C.4 4 The normal force e pressure is: F = (0,000 N/m ) 0. 0 m =.4 N n 4 The tangential force e to shear stress is: F = 0 N/m 0. 0 m = N The total force is F = F n Ft + =.40 N The angle with respect to the normal direction is θ = tan = t Density and Specific Weight.6 Using Eq..5. we have ρ = 000 (T 4) /80 = 000 (70 4) /80 = 976 kg/m γ = 9800 (T 4) /8 = 9800 (70 4) /80 = 9560 N/m Using Table B. the density and specific weight at 70 C are ρ = kg/m γ = = 959. N/m % error for ρ = % error for γ = = 0.0% 00 = 0.% 4 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or plicated, or posted to a publicly accessible website, in whole or in part.

5 .8 b) 6 γ V,400 N/m m m = = = 0.65 kg g 9.77 m/s Viscosity Assume carbon dioxide is an ideal gas at the given conditions, then ρ = p 00 kn/m.95 kg/m RT = = ( 0.89 kj/kg K)( K).40 W mg γ = = = ρg =.95 kg/m 9.8 m/s = 8.6 kg/m s = 8.6 N/m V V 5 From Fig. B. at 90 C, μ 0 N s/m, so that the kinematic viscosity is μ ν = = = ρ.95 kg/m 5 0 N s/m m /s The kinematic viscosity cannot be read from Fig. B. since the pressure is not at 00 kpa. The shear stress can be calculated using τ = μ / dy. From the given velocity distribution, u y y dy = = 0(0.05 ) we get 0(0.05 y).4 From Table B. at 0 C for water, μ =.08 0 N s/m So, at the lower plate where y = 0, we have dy ( ) = 0(0.05 0) = 6 s τ = = N/m At the upper plate where y = 0.05 m, dy y= 0.05 = 0( ) = 6 s τ = N/m 5 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or plicated, or posted to a publicly accessible website, in whole or in part.

6 The shear stress can be calculated using τ = μ / dr. From the given velocity u = 6( r ro ) we get 6( rro ). dr = Hence, rro dr = At the centerline, r = 0, so = 0, and hence τ = 0. dr At r = 0.5 cm, = rro = = 00 s, dr ( ) τ = 00 μ = 00 0 =. N/m At the wall, r = 0.5 cm, ( ) τ = 6400 μ = = 6.4 N/m ( ) = rro = = 6400 s, dr ( ) π R ωlμ Use Eq..5.8 to calculate the torque, T = h where h = (.6.54) = 0.0 cm = m π The angular velocity ω = 000 rpm = 09.4 rad/s 60 The viscosity of SAE-0 oil at C is μ = Ns/m (Figure B.) π (.7 0 m) 09.4 rad/s. m Ns/m T = =.0 N m (0.0 0 )m power = Tω = = 650 W = 0.65 kw.48 Assume a linear velocity in the fluid between the rotating disk and solid surface. The velocity of the fluid at the rotating disk is V = rω, and at the solid surface V = 0. So, r ω =, where h is the spacing between the disk and solid surface, dy h and ω = π = 4.9 rad/s. The torque needed to rotate the disk is T = shear force moment arm τ dr Due to the area element shown, dt = df r = τda r r where τ = shear stress in the fluid at the rotating disk and rω da = πrdr dt = μ πrdr r = μ πr dr dy h 6 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or plicated, or posted to a publicly accessible website, in whole or in part.

7 R πμω πμω T = r dr = R h h 0 4 The viscosity of water at 6 C is μ =. 0 Ns/m T 0.5 π (. 0 Ns/m )( 4.9 rad/s) πμω R h 0 m 4 = = = ( ) N m.50 If τ = μ dy = constant, and μ = AeB/T = Ae By/K = Ae Cy, then Ae Cy = constant. dy dy = DeCy, where D is a constant. Multiply by dy and integrate to get the velocity profile y Cy D Cy y Cy u = De dy= e = E e C 0 0 ( ) where A, B, C, D, E, and K are constants. Compressibility.54 The sound will travel across the lake at the speed of sound in water. The speed of B sound in water is calculated using c =, where B is the bulk molus of ρ elasticity. Assuming (T = 0 C) and using Table B. we find 7 B = 0 Pa, and ρ = kg/m c = = kg/m 7 0 N/m 45 m/s The distance across the lake is, L = cδt = = 90 m.56 b) Using Table B. and T = 7 C we find The speed of sound in water is calculated using: c B 7 c = 6 0 / 99 = 508 m/s ρ = ( ) 7 B = 6 0 N/m, ρ = 99 kg/m 7 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or plicated, or posted to a publicly accessible website, in whole or in part.

8 Surface Tension σ For a spherical droplet the pressure is given by p = R Using Table B. at 5 C the surface tension σ = N/m σ N/m p = = = kpa R 5 0 m 4σ N/m For bubbles: p = = = kpa R 5 0 m For a spherical droplet the net force e to the pressure difference Δ p between the inside and outside of the droplet is balanced by the surface tension force, which is expressed as: σ 0.05 N/m Δ p = pinside poutside = = 6 0 kpa R 5 0 m Hence, p = p + 0 kpa = 8000 kpa + 0 kpa = 800 kpa inside outside In order to achieve this high pressure in the droplet, diesel fuel is usually pumped to a pressure of about 000 bar before it is injected into the engine. See Example.4: 4σcosβ h = = ρ gd ( ) ( ) kg/m 9.8 m/s ( 0 m) N/m cos0 = m =0.45 mm Note that the minus sign indicates a capillary drop rather than a capillary rise in the tube. Draw a free-body diagram of the floating needle as shown in the figure. The weight of the needle and the surface tension force must balance: W = σ L or ρgv = σ L.64 The volume of the needle is ρ = 4 d = πd L g L 8σ πρg σ V πd = L 4 σl W σl needle 8 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or plicated, or posted to a publicly accessible website, in whole or in part.

9 .66 There is a surface tension force on the outside and on the inside of the ring. Each surface tension force = σ π D. Neglecting the weight of the ring, the free-body diagram of the ring shows that F = σπ D F D Vapor Pressure To determine the temperature, we can determine the absolute pressure and then use property tables for water. The absolute pressure is p = = kpa. From Table B., at 50 C water has a vapor pressure of. 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. is 7.8 kpa. This would be the minimum pressure that could be obtained since the water would vaporize below this pressure. The inlet pressure to a pump cannot be less than 0 kpa absolute. Assuming atmospheric pressure to be 00 kpa, we have 0,000 kpa + 00 kpa = 600x x = 6.8 km. Ideal Gas Assume air is an ideal gas and calculate the density inside and outside the house T = 5 C, p = 0. kpa, T = 5 C, and p = 85 kpa using in in out out.74 ρ 0. kn/m in = p.6 kg/m RT = 0.87 kj/kg K (5 + 7 K) = 85 ρ out = =.9 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. This is infiltration, also known as the chimney effect. 9 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or plicated, or posted to a publicly accessible website, in whole or in part.

10 The weight can be calculated using, W = γ V = ρgv.76 Assume air in the room is at 0 C and 00 kpa and is an ideal gas: p 00 ρ = = =.89 kg/m RT W = γv = ρgv =.89 kg/m 9.8 m/s (0 0 4 m ) = 9 N The pressure holding up the mass is 00 kpa. Hence, using pa = W, we have 00,000 N/m m = m 9.8 m/s This gives the mass of the air m= 0,94 kg Since air is an ideal gas we can write pv = mrt or.78 V ( ) mrt 0,94 kg 0.87 kj/kg K 5+7 K = = = 846 m p 00 kpa Assuming a spherical volume V = πd 6, which gives d m = = 5. m π 0 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or plicated, or posted to a publicly accessible website, in whole or in part.

11 First Law The fist law of the thermodynamics is applied to the mass: Q W =Δ PE+Δ KE+Δ U In this case, since there is no change in potential and internal energy, i.e., Δ PE = 0, and Δ U = 0, and there is no heat transfer to or from the system, Q = 0. The above equation simplifies to: W =ΔKE where W = Fdl a) F W = 00 N = 00 N 0 m =000 N m or J.80 Note that the work is negative in this case since it is done on the system. Substituting in the first law equation we get: ( ) 000 N 000 N m = mv V V m = (0 m/s) + = 9.5 m/s 5 kg b) F = 0s W = 0sds = 0s =000 N m 0 ( ) 000 N 000 N m = mv V V m = (0 m/s) + = 5.8 m/s 5 kg 0 = = 0 c) 00cos( π 0) 00cos( π 0) F s W s ds ( ) ( ) W = 00 0 π sin 0π 0 =4000 π N m ( ) 4000 π 4000 N m N π = mv V V m = (0 m/s) + = 6.4 m/s 5 kg.8 Applying the first law to the system (auto + water) we write: Q W =Δ PE+Δ KE+Δ U In this case, the kinetic energy of the automobile is converted into internal energy in the water. There is no heat transfer, Q = 0, no work W = 0, and no change in potential energy, Δ PE = 0. The first law equation reces to: [ ] mv =Δ U = mcδ T where, m HO HO HO = ρv = = kg/m 000 cm 0 m cm kg 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or plicated, or posted to a publicly accessible website, in whole or in part.

12 For water c = 480 J/kg K. Substituting in the above equation kg m/s = kg 480 J/kg K ΔT 600 Δ T = 69. C For a closed system the work is W = pdv For air (which is an ideal gas), pv = mrt, and mrt p = V.84 W pdv mrt dv mrt dv mrt ln V = = = = V V V V p Since T = constant, then for this process p V = p V or = V p p Substituting in the expression for work we get: W = mrt p W = kg ( 87 J kg K) 94 K ln = 6,980 J = 6.98 kj The st law states that Q W = mδ u = mc Δ T = 0 since Δ T = 0. Q= W = 6.98 kj v.86 For a closed system the work is W = pdv If p = constant, then W p( V V) = Since air is an ideal gas, then p V = mrt, and pv = mrt ( ) ( ) W = mr T T = mr T T = mrt = kg 0.87 kj/kg K 4 K = 4 kj 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or plicated, or posted to a publicly accessible website, in whole or in part.

13 Isentropic Flow Since the process is adiabatic, we assume an isentropic process to estimate the maximum final pressure:.88 k/ k.4/0.4 T 4 p p T 9 = = ( ) = 904 kpa abs or 804 kpa gage. Note: We assumed p atm = 00 kpa since it was not given. Also, a measured pressure is a gage pressure. Speed of Sound.90.9 b) c = krt = = 66.9 m/s d) c = krt = = 0 m/s Note: We must use the units on R to be J/kg. K in the above equations. b) The sound will travel at the speed of sound c= krt = = 4 m/s The distance is calculated using L= cδ= t 4 8. = 854 m 0 Cengage Learning. All Rights Reserved. May not be scanned, copied or plicated, or posted to a publicly accessible website, in whole or in part.

CHAPTER 1 Basic Considerations

CHAPTER 1 Basic Considerations FE-type Exam Review Problems: Problems 1.1 to 1.1. 1.1 (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) 8