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Access full Solution hands-on only below https://www.book4me.xyz/solution-manual-fluid-mechanics-munson-young/ trouble 1.1 The force, F , the the wind blowing versus a building is given by F  CDV 2 A 2 , where V is the wind speed,  the thickness of the air, A the cross-sectional area of the building, and also CD is a constant termed the traction coefficient. Recognize the size of the traction coefficient. equipment 1.1 F  CDV 2 A 2 or CD  F V 2 A , whereby MLT 2 ,  Thus, CD 2F ML3 , V LT 1 , A L2  MLT -2   M 0 L0T 0 2  -2 -1 2   ML  LT   together   Hence, CD is dimensionless. https://www.book4me.xyz/solution-manual-fluid-mechanics-munson-young/ difficulty 1.2 The Mach number is a dimensionless proportion of the velocity of an item in a fluid to the rate of sound in the fluid. Because that an plane flying in ~ velocity V in wait at pure temperature T , the Mach number Ma is, Ma  V , kRT whereby k is a dimensionless constant and R is the details gas continuous for air. Present that Ma is dimensionless. systems 1.2 We signify the measurement of temperature by  and use Newton’s second law to obtain F  climate L   T   M    FL   ML  1    2   M   T F  or  M   1 . L   T  L2 T2 ML T2 . https://www.book4me.xyz/solution-manual-fluid-mechanics-munson-young/ problem 1.3 Verify the dimensions, in both the FLT and MLT systems, of the following quantities which show up in Table B.1 physics Properties the Water (BG/EE Units). (a) Volume, (b) acceleration, (c) mass, (d) minute of inertia (area), and also (e) work. equipment 1.3 L3 a) volume b) acceleration  time price of adjust of velocity c) fixed MLT 2 FL1T 2 d) minute of inertia  area   2nd moment that area e) occupational  force  street or withF work-related LT 2 M or withF mass LT 1 T MLT 2 ML2T 2 FL  L2  L2  L4 https://www.book4me.xyz/solution-manual-fluid-mechanics-munson-young/ problem 1.4 Verify the dimensions, in both the FLT and also MLT systems, that the following quantities which appear in Table B.1 physical Properties of Water (BG/EE Units). (a) Angular velocity, (b) energy, (c) minute of inertia (area), (d) power, and (e) pressure. equipment 1.4 angular displacement time b) power ~ capacity of body to do occupational a) angular velocity  solitary work  pressure  distance  energy or through F MLT 2  energy T 1 FL  MLT 2   together  ML2T 2  L2  L2  L4 FL FLT 1  MLT 2   l  T 1  T  MLT 2  L2  ML1T 2 c) moment of inertia  area   second moment of area d) power  rate of doing occupational e) press  force area F 2 l FL2 ML2T 3 https://www.book4me.xyz/solution-manual-fluid-mechanics-munson-young/ problem 1.5 Verify the dimensions, in both the FLT system and also the MLT system, the the following quantities which show up in Table B.1 physical Properties that Water (BG/EE Units). (a) Frequency, (b) stress, (c) strain, (d) torque, and also (e) work. Solution 1.5 cycles time pressure F b) stress= area L2 a) frequency= T-1 FL-2 because F MLT -2 , stress and anxiety MLT -2 ML-1T-2 2 L readjust in size c) strain= length together L d) torque=force  street FL e) work=force  distance FL L0  dimensionless   MLT-2   L ML2T-2  MLT-2   L ML2T-2 https://www.book4me.xyz/solution-manual-fluid-mechanics-munson-young/ trouble 1.6 If u is a velocity, x a length, and t a time, what room the size (in the MLT system) that (a) u / t , (b)  2u / xt , and also (c)  (u / t )dx ? solution 1.6 u a) t b) LT 1 T  2u xt u c)  x t LT 2 LT 1 ( L)(T ) T 2  LT  (L) 1 T L2T 2 https://www.book4me.xyz/solution-manual-fluid-mechanics-munson-young/ trouble 1.7 Verify the dimensions, in both the FLT system and also the MLT system, that the adhering to quantities which show up in Table B.1 physics Properties that Water (BG/EE Units). (a) Acceleration, (b) stress, (c) moment of a force, (d) volume, and also (e) work. Solution 1.7 velocity l time T2 pressure F FL2 b) tension  2 area l a) acceleration  since F stress LT 2 MLT 2 MLT 2 2 together ML1T 2 c) moment of a pressure  force  street d) volume  (length)3 FL  MLT 2  l L3 e) work-related  pressure  street FL  MLT 2  l ML2T 2 ML2T 2 https://www.book4me.xyz/solution-manual-fluid-mechanics-munson-young/ problem 1.8 If p is a pressure, V a velocity, and also  a fluid density, what room the dimensions (in the MLT system) the (a) p /  , (b) pV  , and also (c) p / V 2 ? equipment 1.8 a) ns FL2  3 ML b) pV  c)  ML MLT 2 L2 3 ML 1 2 T 2  ML1T 2 3 ML  LT  ML3  1 ML1T 2 ns V   ML3  LT  1 2 L2T 2 M 2 L3T 3 M 0 L0T 0  dimensionless  https://www.book4me.xyz/solution-manual-fluid-mechanics-munson-young/ difficulty 1.9 If p is a force and x a length, what room the dimensions (in the FLT system) of (a) dP / dx , (b) d 3 ns / dx3 , and (c)  p dx ? solution 1.9 a) b) dP dx FL2 d 3P F 3 3 dx c) F together  Pdx together FL FL3 https://www.book4me.xyz/solution-manual-fluid-mechanics-munson-young/ problem 1.10 If V is a velocity, a length, and  a fluid property (the kinematic viscosity) having actually dimensions of L2T 1 , i beg your pardon of the adhering to combinations space dimensionless: (a) V  , (b) V  , (c) V 2 , (d) V  ? systems 1.10 a) V  b) V  c) V 2 d) V   LT 1   together   L2T 1  L4T 2 not dimensionless   LT 1   together  L0T 0 dimensionless   L2T 1  2  LT 1   L2T 1  L4T 3 notdimensionless   LT 1  L2 notdimensionless   l   L2T 1  https://www.book4me.xyz/solution-manual-fluid-mechanics-munson-young/ trouble 1.11 The inert flux is given by the product mV , where m is mass circulation rate and V is velocity. If mass flow rate is given in systems of mass per unit time, show that the momentum flux can be to express in systems of force. solution 1.11 M  l  l  FT 2  together  M  F     T  T  T 2  l  T 2  mV    https://www.book4me.xyz/solution-manual-fluid-mechanics-munson-young/ difficulty 1.12 an equation for the frictional pressure loss p (inches H2O) in a one duct of within diameter d  in. and also length together  ft  because that air flowing v velocity V  ft/min  is 1.82  l  V  p  0.027  1.22    d   Vo  , whereby V0 is a reference velocity same to 1000ft/min. Find the devices of the “constant” 0.027. equipment 1.12 resolving for the continuous gives 0.027  pL . 1.82  l  V   1.22    D   Vo  The units give 0.027  0.027   in. H 2O   ft  ft   min  1.22   ft  in.    min 1.82      in. H 2O  in.1.22 ft https://www.book4me.xyz/solution-manual-fluid-mechanics-munson-young/ problem 1.13 The volume price of flow, Q , v a pipeline containing a slowly moving fluid is offered by the equation  R 4 p 8 where R is the pipeline radius, p the push drop along the pipe,  a liquid property called visQ   cosity FL2T , and also the size of pipe. What space the size of the continuous  / 8 ? would girlfriend classify this equation together a basic homogeneous equation? Explain. equipment 1.13 3 1  L T    L3T 1     4  2     together   FL   8   2  FL T  together        3 1   8   together T  The consistent is  is dimensionless. 8 Yes. This is a basic homogeneous equation because it is valid in any consistent units system. https://www.book4me.xyz/solution-manual-fluid-mechanics-munson-young/ trouble 1.14 show that every term in the adhering to equation has actually units that lb/ft 3. Take into consideration u a velocity, y a length, x a length, p a pressure, and  an absolute viscosity. 0 p  2u  2 . x y systems 1.14  lb   p   ft 2   x    ft  or  p   lb   x    ft 3  ,   and also  ft    2u   lb  sec   sec  or  2    2  2  y   ft  ft      2u   lb   2    3  .  y   ft  https://www.book4me.xyz/solution-manual-fluid-mechanics-munson-young/ difficulty 1.15 The press difference, p , throughout a partial blockage in an artery (called a stenosis) is approximated through the equation 2 A  p  Kv  Ku  0  1 V 2 D  A1  V   whereby V is the blood velocity,  the blood viscosity ( FL2T ) ,  the blood density ML3 , D the artery diameter, A0 the area that the unobstructed artery, and also A1 the area of the stenosis. Identify the size of the constants K and also Ku . Would certainly this equation be valid in any type of system the units? equipment 1.15 2 A  p  K v  Ku  0  1 V 2 D  A1  V FL FT l 1  Kv  2 together T l FL2  Kv   FL2  2 2  L2   FT 2 1   l 2   K u   2  1  L   together L3   T         Ku  FL2  K and also Ku room dimensionless because each hatchet in the equation must have the same dimensions,. Yes. The equation would be valid in any kind of consistent system of units. https://www.book4me.xyz/solution-manual-fluid-mechanics-munson-young/ difficulty 1.16 Assume that the speed of sound, c , in a fluid depends on an elastic modulus, E , v dimensions FL2 , and also the fluid density,  , in the form c   E  a   b . If this is to it is in a dimensionally homogeneous equation, what are the values for a and b ? Is your an outcome consistent v the typical formula for the speed of sound? (See the equation c  E  .) systems 1.16 Substituting  c   LT 1  E   FL2    LT 1    FL2        FL4T 2 into the equation noted yields:    FL4T 2    F ab L2a4bT 2b a b Dimensional homogeneity needs that the exponent that each dimension on both sides of the equal authorize be the same. F: 0 =a+b L: 1 =-2a-4b T: -1 =2b Therefore: T: -1 =2b  b= -1/2 F: a =-b  a= 1/2 L: 1 =-2a-4b = -2(1/2) -4(-1/2)= 1✓ a 1 1 ; b 2 2 Yes, this is regular with the standard formula because that the rate of sound. https://www.book4me.xyz/solution-manual-fluid-mechanics-munson-young/ trouble 1.17 A formula to estimate the volume rate of flow, Q , flowing end a dam the length, B , is provided by the equation Q  3.09 BH 3/2 wherein H is the depth that the water above the height of the dam (called the head). This formula provides 3 Q in ft /s when B and H space in feet. Is the constant, 3.09, dimensionless? would certainly this equation be precious if units other than feet and also seconds were used? systems 1.17 Q  3.09 3 BH 2  L3T 1    3.09 L L 2 3 5  L3T 1  3.09 l  2   because each ax in the equation must have actually the very same dimensions theconstant 3.09 must have actually dimensions 1 that L2 T 1  and also is thus not dimensionless.No. due to the fact that the constant has dimensions its value will adjust with a readjust in units. No. https://www.book4me.xyz/solution-manual-fluid-mechanics-munson-young/ trouble 1.18 A commercial advertisement reflects a pearl falling in a bottle of shampoo. If the diameter D of the pearl is quite little and the shampoo saturated viscous, the drag D ~ above the pearl is provided by Stokes’s law, D  3VD , where V is the speed of the pearl and  is the liquid viscosity. Show that the hatchet on the right side the Stokes’s law has actually units that force. systems 1.18 l FT 2 l  M  l  together  M  F D   3VD     l T2  LT  T  T2 https://www.book4me.xyz/solution-manual-fluid-mechanics-munson-young/ problem 1.20 express the following quantities in SI units: (a) 10.2 in. Min , (b) 4.81 slugs , (c) 3.02 lb , (d) 73.1 ft s2 , (e) 0.0234lb  s ft 2 . equipment 1.20 a) 10.2 in.  in.   mm  2 m  1min 3 m  10.2  2.540 10     4.32 10   4.32 min  min  in.  60s  s s  kg  b) 4.81slugs    4.81slugs   1.459 10   70.2kg cheese   N  c) 3.02lb   3.02lb   4.448   13.4N lb   m   2 ft  ft  d) 73.1 2   73.1 2   3.048 101 s ft s s    s2    m   22.3 2 s   N s   2  lb  s  lb  s   N s e) 0.0234 2   0.0234 2   4.788 10 m   1.12 2 lb  s  ft ft   m   2  ft   https://www.book4me.xyz/solution-manual-fluid-mechanics-munson-young/ problem 1.21 refer the following quantities in BG units: (a) 14.2 kilometres , (b) 8.14 N m3 , (c) 1.61kg m3 , (d) 0.0320 N  m s , (e) 5.67 mm hr . equipment 1.21   ft   a) 14.2km  14.2 103m  3.281   4.66 104 ft m  lb   3  N  N  lb b) 8.14 3   8.14 3   6.366 103 ft   5.18  102 3 N  m  m  ft  3  m   slugs    3  kg  kg  slugs c) 1.61 3  1.61 3  1.940 103 ft   3.12 103 3 kg  m  m  ft   m3   ft  lb   Nm  N m  1 2 ft  lb s  d) 0.0320   0.0320   7.376 10  N  m   2.36 10  s s  s   s   e) 5.67 mm  m  ft  1hr   6 ft   5.67 103  3.281    5.17 10  hr  hr  m  3600s  s