mechanics of fluids si edition 5th edition potter solution

Note: we could express the units on c as = kg / s 1.21Ġ.00194 3.281 d d2 where m is in slugs, in slug/ft3 and d in feet. Since all terms must have the same dimensions (units) we require: s = kg/s, = kg/s2 = N s 2 / m s 2 N / m, = kg m / s 2 N. = m/s2 = N/kg = kg m/s2 kg = m/s2 = m3/s m2 m2/3 = m1/3/sĪ) pressure: N/m2 = kg m/s2/m2 = kg/m s2 b) energy: N m = kgĬ) power: N m/s = kg m2/s3 kg m 1 s 2 kg / m s d) viscosity: N s/m2 = 2 s m N m kg m m kg m 2 / s 3 e) heat flux: J/s = 2 s s s J N m kg m m m 2 / K s2 f) specific heat: 2 kg K kg K s kg K 1.20

= 元 / T L2 L2 / 3 L1/ 3 T Note: the slope S0 has no dimensions.Ī) m = s2. M FT 2 / L a) density = 3 L 元 b) pressure = F/L2ĭ) energy = F×L = FL M FT 2 / L e) mass flux = FT / L T T f) flow rate = AV = L2 L/T = 元/T 1.17Ī) L = T2. 1.13Ĭhapter 1 Problems: Dimensions, Units, and Physical Quantities 1.14Ĭonservation of mass - Mass - density Newton’s second law - Momentum - velocity The first law of thermodynamics - internal energy - temperatureĪ) density = mass/volume = M / 元 b) pressure = force/area = F / L2 ML / T 2 L2 M / LT 2 c) power = force velocity = F L / T ML / T 2 L / T d) energy = force distance = ML / T 2 L ML2 / T 2 e) mass flux = ρAV = M/元 × L2 × L/T = M/T f) flow rate = AV = L2 × L/T = 元/T Ice is actually slightly lighter than water, but it is not necessary for such accuracy in this problem. We assumed the density of ice to be equal to that of water, namely 1000 kg/m3. The Differential Forms of the Fundamental LawsĬHAPTER 1 Basic Considerations FE-type Exam Review Problems: Problems 1-1 to 1-14. The Integral Forms of the Fundamental Laws

MERLE C. POTTER Michigan State University DAVID C. WIGGERT Michigan State University BASSEM RAMADAN Kettering University

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Mechanics of fluids si edition 5th edition potter solution