Mecânica dos fluidos

Problem 1.2

Problem 1.3

Problem 1.4

Problem 1.6
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. Solution Given: Dimensions of a room. Find: Mass of air in lbm and kg. The data for standard air are: Rair = 53.33⋅ Then ρ = p Rair⋅ T
2

ft⋅ lbf lbm⋅ R

p = 14.7⋅ psi

T = ( 59 + 460) ⋅ R = 519⋅ R

1 lbm⋅ R 1 12⋅ in  ρ = 14.7⋅ × ⋅ × ×   2 53.33 ft⋅ lbf 519⋅ R  1⋅ ft  in lbf lbm ft
3

ρ = 0.0765

or

ρ = 1.23

kg m
3

The volume of the room is

V = 10⋅ ft × 10ft × 8ft

V = 800 ft

3

The mass of air is then

m = ρ⋅V lbm ft
3 3

m = 0.0765⋅

× 800⋅ ft

m = 61.2 lbm

m = 27.8 kg

Problem 1.7

Given: Data on nitrogen tank Find: Mass of nitrogen Solution The given or available data is: D = 6⋅ in T = ( 59 + 460) ⋅ R L = 4.25⋅ ft T = 519 R p = 204⋅ atm RN2 = 55.16⋅ ft⋅ lbf (Table A.6) lb⋅ R

The governing equation is the ideal gas equation p = ρ ⋅ RN2⋅ T where V is the tank volume and π 4 π ρ =
2

M V

V =

⋅D ⋅L
2

6 3 ×  ⋅ ft × 4.25⋅ ft = 0.834 ft V = V   4  12  Hence M = V⋅ ρ =
2

p⋅ V RN2⋅ T
3

M = 204 × 14.7⋅

lbf in
2

×

144⋅ in ft
2

× 0.834⋅ ft ×

1 lb⋅ R 1 1 lb⋅ ft ⋅ × ⋅ × 32.2⋅ 2 55.16 ft⋅ lbf 519 R s ⋅ lbf M = 0.391 slug

M = 12.6 lb

Problem 1.8

Problem 1.9

Problem 1.10

Problem 1.12

Problem 1.13

Problem 1.14

Problem 1.15

Problem 1.16
From Appendix A, the viscosity µ (N.s/m2) of water at temperature T (K) can be computed from µ = A10B/(T - C), where A = 2.414 X 10-5 N.s/m2, 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.

Solution Given: Data on water. Find: Viscosity and uncertainty in viscosity. The data provided are: A = 2.414⋅ 10