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Solution Manual For Munson, Young and Okiishi's Fundamentals of Fluid Mechanics, 8th Edition

Solution Manual For Munson, Young and Okiishi's Fundamentals of Fluid Mechanics, 8th Edition simplifies even the toughest textbook questions with step-by-step solutions and easy explanations.

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    Chapter 1 1-1 (a) 98 Btu/(hr-ft-F) x1.7307 = 170 W/(m-K) (b) 0.24 Btu/(Ibm-F) x4186.8=1.0 kJ/kg-K (c) 0.04 Ibm/(ft-hr) 3600 sec/hr x1.488 = 16.5 2 Ns m μ (d) 1050 Btu Ibm x 4 1 9.48x10 − J Btu x 2.20462 Ibm kg = 2.44 MJ kg (e) 12,000 Btu Ibm x 1 3.412 = 3.52 kW (f) 14.7 2 Ibf in x 6894.76 = 101 kPa 1-2 (a) 120 kPa x 2 lbf / in 6.89476kPa = 17.4 lbf/in 2 (b) 100 W m K − x 0.5778 = 57.8 Btu/hr-ft-F (c) 0.8 2 W m K − x 0.1761 = 0.14 Btu/hr-ft 2 -F (d) 10 -6 N-s/m 2 x 1 1.488 = 6.7 x 10 -7 lbm ft sec − (e) 1200 kW x 3412 = 4.1 x 10 -6 Btu/hr

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    2 (f) 1000 kJ kg x 1 Btu 1.055 kJ x 1 kg 2.2046 lbm = 430 Btu lbm 1-3 Hp = 50 (ft) x 0.3048 ( m ft ) = 15.2 m ∆ P = 15.2 m 1000 Pa/kPa x 9.807 1 ( N kg ) x 1000 (kg/m 3 ) = 149 kPa 1-4 P = ∆ 4 12 (ft) x 0.3048 ( m ft ) x 9.807 1 ( N kg ) x 1000 ( 3 kg m ) ∆ P = 996 Pa 1.0 kPa ≈ 1-5 TOTAL BILL = ENERGY CHARGE + DEMAND CHARGE + METER CHARGE ( ) ( ) ( ) ( ) 96,000 kw - hrs 0.045 $ / kw hr + 624 kw 11 50 $ / kw − − + $68 = $4,320 + $7,176 + $68 = $11,564 1-6 7 AM to 6 PM 11 hrs/day, 5 days/wk hrs days (11) (22) 242 hrs / month day months = students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that

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    3 ( ) ( ) ( ) 624 kw ratio = 1.57 96,000 kw hr 242 hr = ⎛ ⎞ − ⎜ ⎟ ⎝ ⎠ 1-7 This is a trial and error solution since eq. 1-1 cannot be solved explicitly for i. Answer converges at just over 4.2% using eq. 1-1 1-8 Determine present worth of savings using eq. 1-1 ( ) ( ) ( ) 12 12 0.012 $1000 1- 1+ 12 P = 0.012 12 P $134,000 − ⎡ ⎤ ⎛ ⎞ ⎢ ⎥ ⎜ ⎟ ⎝ ⎠ ⎢ ⎥ ⎣ ⎦ ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ = 1-9 (a) Q VA =  = 2 x 3.08 x 10 -3 = 6.16 x 10 -3 m 3 /s m 6.16 x 10 Q ρ =   = -3 x 998 = 6.15 kg/s (b) A= 4 π (0.3) 2 = 7.07 x 10 -2 m 2 Q 7.07x10 =  -2 x 4 = 0.283 m 3 / s; ρ = 1.255 kq/m 3 m = 1.225 x 0.283 = 0.347 kg/s  1-10 V = 3x10x20 = 600m 3 students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that

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    4 = 600 x i Q  1 4 x 1 3600 = 4.17 x 10 -2 m 3 /s 1-11 p p 3 q = mc T c = 4.183 kJ/(kg-K) = 983.2 kg/m ρ ∆   1-11 (cont’d) ( ) ( ) ( ) ( ) 3 c 3 m kg kJ q = 1 983.2 4.183 5 20,564 s kg K m q = 20,564 kw = −   kJ s students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted 1-12 = wat q  air q −  11,200(1)(10) = = 2 5000x60x14.7x144x0.24(t 50 ) (53.35x510) − 11,200 = 5601.5 (t 2 -50); t 2 = (11,200/5601.5) + 50 = 70 F 1-13 Diagram as in 1-12 above. q  wat = - q  air 1.5 (4186)(90-t 2 ) = 2.4 (1.225)(1.0)(30-20)(1000) 6279(90-t 2 ) = 29,400

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    - l purposes only to students enrolled in courses for which the textbook has been adopted. this work beyond that permitted 5 t 2 = 90 - 29,400 6279 = 85.3 C 1-14 q hA(t =  s - ) t ∞ A= π (1/12) x 10 = 2.618 ft 2 s t = t sur 212 F ≈ = 10x2.618x(212-50) = 4241 Btu/hr q  1-15 A= π x 0.25x4 = 3.14 16 m 2 hA(t q =  s - ) t ∞ h= s q A (t -t ) ∞  = 1250 3.1416(100 10) − ; h = 4.42 W/(m 2 – C) 1-16 (t p q mc =   2 -t 1 ) ; m Q x ρ =   ρ = P/RT = 14.7x144/53.35(76+460) ρ = 0.074 lbm/ft 3 m = 5000x0.074x60 = 22,208 lbm/hr  = 0.24 Btu/lbm-F p c q = 22,208x0.24(58-76) = -95,939 Btu/hr  Negative sign indicates cooling 1-17 (t 1 p m c  3 -t 1 ) + for-profit basis for testing or instructiona Any other reproduction or translation of

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    6 (t 2 p2 m c  3 -t 2 ) = 0 = p1 c p2 c t 3 = 1 1 2 2 1 2 (m t m t ) (m m ) + +     1 2 m Q 1 ρ =   = 1000x 14.7x144 53.35(460 50) + = 73.5 lbm/min 1-17 (cont’d) 2 2 m Q 2 ρ =   = 600x 14.7x144 53.35(460 50) + = 46.7 lbm/min 3 (73.5x80) (46.7 x 50) t 68.3 F (73.5 46.7) + = = + students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that

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    Problem 1.1 The force, F , of the wind blowing against a building is given by 2 2   D F C V A , where V is the wind speed,  the density of the air, A the cross - sectional area of the building, and D C is a constant termed the drag coefficient. Determine the dimensions of the drag coefficient. Solution 1.1 2 2   D A F C V or 2 2   D F C V A , where 2  F MLT , 3   ML , 1  V LT , 2 A L Thus,        -2 0 0 0 2 -2 -1 2        D MLT C M L T ML LT L Hence, D C is dimensionless .

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    Problem 1.2 The Mach number is a dimensionless ratio of the velocity of an object in a fluid to the speed of sound in the fluid. For an airplane flying at velocity V in air at absolute temperature T , the Mach number Ma is, Ma  V kRT , where k is a dimensionless constant and R is the specific gas constant for air. Show that Ma is dimensionless. Solution 1 .2 We denote the dimension of temperature by  and use Newton’s second law to get 2  ML F T . Then     2 2 2 1                             L L T T M FL ML L M T F T or     1  M .

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    Problem 1.3 Verify the dimensions, in both the FLT and MLT systems, of the following quantities which appear in Table B.1 Physical Properties of Water (BG/EE Units). (a) V olume, (b) acceleration, (c) mass, (d) moment of inertia (area), and (e) work. Solution 1.3 a) 3 volume L b) 1 2 acceleration time rate of change of velocity    LT LT T c) mass M 2 1 2 or with mass    F MLT FL T d)      2 2 4 moment of inertia area second moment of area  L L L e) work force distance   FL 2 2 2 or with work    F MLT ML T

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    Problem 1.4 Verify the dimensions, in both the FLT and MLT systems, of the following quantities which appear in Table B.1 Physical Properties of Water (BG/EE Units). (a) A ngular velocity, (b) energy, (c) moment of inertia (area), (d) power, and (e) pressure. Solution 1.4 a) 1 angular displacement angular velocity time   T b) energy ~ capacity of body to do work single work force distance    energy FL 2 or with  F MLT      2 2 2 energy   MLT L ML T c)      2 2 4 second mom moment of ine ent of are rtia a e a r a  L L L d)       1 2 1 2 3 power rate of doing work      FL FLT MLT L T ML T T e)    2 2 2 1 2 2 force pressure area F FL MLT L ML T L      

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    Problem 1.5 Verify the dimensions, in both the FLT system and the MLT system, of the following quanti- ties which appear in Table B.1 Physical Properties of Water (BG/EE Units). (a) F requency, (b) stress, (c) strain, (d) torque, and (e) work. Solution 1.5 a) -1 cycles frequency= T time b) -2 2 -2 -2 -1 -2 2 force F stress= FL area L Since F MLT , MLT stress ML T L c)   0 change in length L strain= L dimensionless length L d)     -2 2 -2 torque=force distance FL MLT L ML T  e)     -2 2 -2 work=force distance FL MLT L ML T 

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    Problem 1.6 If u is a velocity, x a length, and t a time, what are the dimensions (in the MLT system) of (a) /   u t , (b) 2 /    u x t , and (c) ( / )    u t dx ? Solution 1.6 a) 1 2     u LT LT t T b) 2 1 2 ( )( )      u LT T x t L T c)   1 2 2 (L)       LT u x L T t T

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    Problem 1.7 Verify the dimensions, in both the FLT system and the MLT system, of the following quantities which appear in Table B.1 Physical Properties of Water (BG/EE Units). (a) A cceleration, (b) stress, (c) moment of a force, (d) volume, and (e) work. Solution 1.7 a) 2 2 velocity acceleration time   L LT T b) 2 2 2 2 1 2 2 force stress area Since stress F FL L F MLT MLT ML T L       c)   2 2 2 moment of a force force distance     FL MLT L ML T d) 3 3 volume (length)  L e)   2 2 2 work force distance     FL MLT L ML T

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    Problem 1.8 If p is a pressure, V a velocity, and  a fluid density, what are the dimensions (in the MLT system) of (a) /  p , (b)  pV , and (c) 2 /  p V ? Solution 1.8 a) 2 2 2 1 2 2 2 3 3 3            p FL MLT L ML T L T ML ML ML  b)     1 2 1 3 2 3 3        pV ML T LT ML M L T c)      1 2 0 0 0 2 2 3 1 dimensionless       p ML T M L T V ML LT

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    Problem 1.9 If P is a force and x a length, what are the dimensions (in the FLT system) of (a) / dP dx , (b) 3 3 / d P dx , and (c)  P dx ? Solution 1.9 a) 2  dP F FL dx L b) 3 3 3 3  d P F FL dx L c)  Pdx FL

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