Introduction-Fluid-Mechanics-Solution-Chapter-01

# Introduction-Fluid-Mechanics-Solution-Chapter-01 - Problem...

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m 27.8kg = m 61.2lbm = m 0.0765 lbm ft 3 800 × ft 3 = m ρ V = The mass of air is then V 800ft 3 = V1 0 f t 10 × ft 8 × ft = The volume of the room is ρ 1.23 kg m 3 = or ρ 0.0765 lbm ft 3 = ρ 14.7 lbf in 2 1 53.33 × lbm R ft lbf 1 519 R × 12 in 1f t 2 × = ρ p R air T = Then T 59 460 + () R = 519 R = p 14.7 psi = R air 53.33 ft lbf lbm R = The data for standard air are: Find: Mass of air in lbm and kg. Given: Dimensions of a room. Solution Make a guess at the order of magnitude of the mass (e.g., 0.01, 0.1, 1.0, 10, 100, or 1000 lbm or kg) of standard air that is in a room 10 ft by 10 ft by 8 ft, and then compute this mass in lbm and kg to see how close your estimate was. Problem 1.6
M 0.391slug = M 12.6lb = M 204 14.7 × lbf in 2 144 in 2 ft 2 × 0.834 × ft 3 1 55.16 × lb R ft lbf 1 519 × 1 R 32.2 × lb ft s 2 lbf = MV ρ = pV R N2 T = Hence V 0.834ft 3 = V π 4 6 12 ft 2 × 4.25 × ft = V π 4 D 2 L = where V is the tank volume ρ M V = and p ρ R N2 T = The governing equation is the ideal gas equation (Table A.6) R N2 55.16 ft lbf lb R = T 519R = T 59 460 + () R = p 204 atm = L 4.25 ft = D6 i n = The given or available data is: Solution Find: Mass of nitrogen Given: Data on nitrogen tank Problem 1.7

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T µ T () d d A 10 B TC B 2 ln 10 For the uncertainty µ T ( ) 1.005 10 3 × Ns m 2 = µ T ( ) 2.414 10 5 m 2 10 247.8 K 293 K 140 K × = Evaluating µ µ T () A1 0 B = The formula for viscosity is u T 0.085% = u T 0.25 K 293 K = The uncertainty in temperature is T 293 K = C 140 K = B 247.8 K = A 2.414 10 5 m 2 = The data provided are: Find: Viscosity and uncertainty in viscosity. Given: Data on water. Solution From Appendix A, the viscosity µ (N . s/m 2 ) of water at temperature T (K) can be computed from µ = A 10 B /( T - C ) , where A = 2.414 X 10 -5 N.s/m 2 , B = 247.8 K, and C = 140 K. Determine the viscosity of water at 20°C, and estimate its uncertainty if the uncertainty in temperature measurement is +/- 0.25°C. Problem 1.16
so u µ T () T µ T T µ T ()u T d d ln 10 ()T B TC 2 u T = Using the given data u µ T ( ) ln 10 ( ) 293 K 247.8 K 293 K 140 K 2 0.085 % = u µ T ( ) 0.61% =

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u H L H L Hu L 2 θ H θ θ 2 + = For the uncertainty H 57.7ft = H L tan θ () = The height H is given by u θ 0.667% = u θ δθ θ = The uncertainty in θ is u L 0.5% = u L δ L L = The uncertainty in L is δθ 0.2 deg = θ 30 deg = δ L 0.5 ft = L 100 ft = The data provided are: Find: Given: Data on length and angle measurements. Solution The height of a building may be estimated by measuring the horizontal distance to a point on ground and the angle from this point to the top of the building. Assuming these measurements L = 100 +/- 0.5 ft and θ = 30 +/- 0.2 degrees, estimate the height H of the building and the uncertainty in the estimate. For the same building height and measurement uncertainties, use Excel ’s Solver to determine the angle (and the corresponding distance from the building) at which measurements should be made to minimize the uncertainty in estimated height. Evaluat

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## This note was uploaded on 09/21/2009 for the course CHEM 220 taught by Professor Cumming during the Spring '09 term at Nevada.

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Introduction-Fluid-Mechanics-Solution-Chapter-01 - Problem...

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