{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Chapter 9

# Chapter 9 - Problem 9.1(In Excel Solution Governing...

This preview shows pages 1–17. Sign up to view the full content.

Problem 9.1 (In Excel) Solution Governing equations: The critical Reynolds number for transition to turbulence is Re crit = ρ VL crit / µ = 500000 The critical length is then L crit = 500000 µ /V ρ Tabulated or graphical data: µ = 3.79E-07 lbf.s/ft 2 ρ = 0.00234 slug/ft 3 (Table A.9, 68 o F) Computed results: V (mph) L crit (ft) 10 5.52 13 4.42 15 3.68 18 3.16 20 2.76 30 1.84 40 1.38 50 1.10 60 0.920 70 0.789 80 0.690 90 0.614 Length of Laminar Boundary Layer on the Roof of a Minivan 0 1 2 3 4 5 6 0 2 04 06 08 0 1 0 0 V (mph) L crit (ft)

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Problem 9.4 (In Excel) Solution Governing equations: The critical Reynolds number for transition to turbulence is Re crit = ρ UL crit / µ = 500000 The critical length is then L crit = 500000 µ /U ρ For air at sea level and 10 km, we can use tabulated data for density ρ from Table A.3. For the viscosity µ , use the Sutherland correlation (Eq. A.1) µ = bT 1/2 /(1+ S / T ) b = 1.46E-06 kg/m.s.K 1/2 S = 110.4 K Air (sea level, T = 288.2 K): Air (10 K, T = 223.3 K): Water (20 o C): ρ = 1.225 kg/m 3 ρ = 0.414 kg/m 3 ρ = 998 (Table A.3) (Table A.3) µ = 1.01E-03 µ = 1.79E-05 N.s/m 2 µ = 1.46E-05 N.s/m 2 (Table A.8) (Sutherland) (Sutherland)
Computed results: Water Air (Sea level) Air (10 km) L crit (m) L crit (m) L crit (m) 0.05 10.12 146.09 352.53 0.10 5.06 73.05 176.26 0.5 1.01 14.61 35.25 1.0 0.506 7.30 17.63 5.0 0.101 1.46 3.53 15 0.0337 0.487 1.18 20 0.0253 0.365 0.881 25 0.0202 0.292 0.705 30 0.0169 0.243 0.588 50 0.0101 0.146 0.353 100 0.00506 0.0730 0.176 200 0.00253 0.0365 0.0881 1000 0.00051 0.0073 0.0176 U (m/s) Length of Laminar Boundary Layer for Water and Air 0.0 0.0 1.0 100.0 1.E-02 1.E+00 1.E+02 1.E+04 U (m/s) L crit (m) Water Air (Sea level) Air (10 km)

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Problem 9.5 (In Excel) Solution Governing equations: The critical Reynolds number for transition to turbulence is Re crit = ρ UL crit / µ = 500000 The critical length is then L crit = 500000 µ /U ρ Let L 0 be the length at sea level (density ρ 0 and viscosity µ 0 ). Then L crit / L 0 =( µ / µ 0 )/( ρ / ρ 0 ) The viscosity of air increases with temperature so generally decreases with elevation; the density also decreases with elevation, but much more rapidly. Hence we expect that the length ratio increases with elevation For the density ρ , we use data from Table A.3. For the viscosity µ , we use the Sutherland correlation (Eq. A.1) µ = bT 1/2 /(1+ S / T ) b = 1.46E-06 kg/m.s.K 1/2 S = 110.4 K
Computed results: z (km) T (K) ρ / ρ 0 µ / µ 0 L crit / L 0 0.0 288.2 1.0000 1.000 1.000 0.5 284.9 0.9529 0.991 1.04 1.0 281.7 0.9075 0.982 1.08 1.5 278.4 0.8638 0.973 1.13 2.0 275.2 0.8217 0.965 1.17 2.5 271.9 0.7812 0.955 1.22 3.0 268.7 0.7423 0.947 1.28 3.5 265.4 0.7048 0.937 1.33 4.0 262.2 0.6689 0.928 1.39 4.5 258.9 0.6343 0.919 1.45 5.0 255.7 0.6012 0.910 1.51 6.0 249.2 0.5389 0.891 1.65 7.0 242.7 0.4817 0.872 1.81 8.0 236.2 0.4292 0.853 1.99 9.0 229.7 0.3813 0.834 2.19 10.0 223.3 0.3376 0.815 2.41 11.0 216.8 0.2978 0.795 2.67 12.0 216.7 0.2546 0.795 3.12 13.0 216.7 0.2176 0.795 3.65 14.0 216.7 0.1860 0.795 4.27 15.0 216.7 0.1590 0.795 5.00 16.0 216.7 0.1359 0.795 5.85 17.0 216.7 0.1162 0.795 6.84 18.0 216.7 0.0993 0.795 8.00 19.0 216.7 0.0849 0.795 9.36 20.0 216.7 0.0726 0.795 10.9 22.0 218.6 0.0527 0.800 15.2 24.0 220.6 0.0383 0.806 21.0 26.0 222.5 0.0280 0.812 29.0 28.0 224.5 0.0205 0.818 40.0 30.0 226.5 0.0150 0.824 54.8

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Length of Laminar Boundary Layer versus Elevation 0 10 20 30 40 50 60 0 1 02 03 0 z (m) L/L 0

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
(1) Steady flow (2) No pressure force (3) No body force in x direction (4) Uniform flow at a Assumptions: Momentum Mass Governing equations: b1 m = δ 25 mm = L3 m = U3 m s = ρ 800 kg m 3 = The given data is Solution Find: Mass flow rate across ab ; drag Given: Data on fluid and boundary layer geometry Problem 9.12

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
We are able to compute the boundary layer drag even though we do not know the viscosity because it is the viscosity that creates the boundary layer in the first place R x 30 N = R x 1 6 −ρ U 2 b ⋅δ = R x ρ U 2 b 1 2 ρ U 2 + 1 3 ρ U 2 + = R x ρ U 2 b 1 2 ρ U b U + 0 1 y ρ U 2 b
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 264

Chapter 9 - Problem 9.1(In Excel Solution Governing...

This preview shows document pages 1 - 17. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online