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Chapter 6 •^ Viscous Flow in Ducts
6.1 In flow past a sphere, the boundary layer becomes turbulent at about Re D ≈ 2.5E5. To what air speed in mi/h does this correspond to a golf ball whose diameter is 1.6 in? Do the pressure, temperature, and humidity of the air make any difference in your calculation?
Solution: For air at 20°C, take ρ = 1.2 kg/m^3 and μ = 1.8E−5 kg/m⋅s. Convert D = 1.6 inches
to 0.0406 m. The critical Reynolds number is
D
VD 1.2V(0.0406) m Re 2.5E5 , or V 92 1.8E 5 s
Ans.
mi 206 h
Since air density and viscosity change with pressure, temperature, and humidity, the calculation does indeed depend upon the thermodynamic state of the air.
6.2 Air at approximately 1 atm flows through a horizontal 4-cm-diameter pipe. (a) Find a formula for Q max, the maximum volume flow for which the flow remains laminar, and plot Q max versus temperature in the range 0°C ≤ T ≤ 500 °C. (b) Is your plot linear? If not, explain.
Solution: (a) First convert the Reynolds number from a velocity form to a volume flow form:
2
, therefore Re 2300 for laminar flow ( /4) d
Q Vd Q V d d
π μ^ πμ
Maximum laminar volume flow is given by (^) max Ans. (a)
Q
d
With d = 0.04 m = constant, get μ and ρ for air from Table A-2 and plot Q max versus T °C:
Fig. P6.
364 Solutions Manual • Fluid Mechanics, Fifth Edition
The curve is not quite linear because ν = μ / ρ is not quite linear with T for air in this
range. Ans. (b)
6.3 For a thin wing moving parallel to its chord line, transition to a turbulent boundary layer occurs at a “local” Reynolds number Re x , where x is the distance from the leading edge of the wing. The critical Reynolds number depends upon the intensity of turbulent fluctuations in the stream and equals 2.8E6 if the stream is very quiet. A semiempirical correlation for this case [Ref. 3 of Ch. 6] is
crit
2 1/ 1/ 2
Re x 0.
where ζ is the tunnel-turbulence intensity in percent. If V = 20 m/s in air at 20°C, use this
formula to plot the transition position on the wing versus stream turbulence for ζ between
0 and 2 percent. At what value of ζ is x crit decreased 50 percent from its value at ζ = 0?
Solution: This problem is merely to illustrate the strong effect of stream turbulence on
the transition point. For air at 20°C, take ρ = 1.2 kg/m 3 and μ = 1.8E−5 kg/m⋅s. Compute
Rex,crit from the correlation and plot xtr = μRex /[ ρ(20 m/s)] versus percent turbulence:
Fig. P6.
The value of x crit decreases by half (to 1.07 meters) at ζ ≈ 0.42%. Ans.
366 Solutions Manual • Fluid Mechanics, Fifth Edition
Depending upon whether the flow is laminar (Re D < 2300) or turbulent (Re D > 2300) use the formulas:
: entrance^ 0.06Re ; : entrance 4.4Re1/ D D
L L
Laminar Turbulent D D
Fluid ν , m^2 /s Re D Type of flow L entr / D Entrance Length
(a) Hydrogen 1.08E− 4 125 Laminar 7.5 0.6 m (b) Air 1.5E− 5 900 Laminar 54.0 4.32 m (c) Gasoline 4.3E− 7 31400 Turbulent 24.7 1.98 m (d) Water 1.0E− 6 13500 Turbulent 21.5 1.72 m (e) Mercury 1.15E− 7 117000 Turbulent 30.8 2.46 m (f) Glycerin 1.18E− 3 11.4 Laminar 0.68 0.055 m
6.7 Cola, approximated as pure water at 20°C, is to fill an 8-oz container (1 U.S. gal = 128 fl oz) through a 5-mm-diameter tube. Estimate the minimum filling time if the tube flow is to remain laminar. For what cola (water) temperature would this minimum time be 1 min?
Solution: For cola “water”, take ρ = 998 kg/m 3 and μ = 0.001 kg/m⋅s. Convert 8 fluid
ounces = (8/128)(231 in^3 ) ≈ 2.37E−4 m 3. Then, if we assume transition at Re = 2300,
3 crit crit
VD 4 Q 2300 (0.001)(0.005) m Re 2300 , or: Q 9.05E 6 D 4(998) s
Then ∆t fill = υ/Q = 2.37E−4/9.05E− 6 ≈ 26 s Ans. (a)
(b) We fill in exactly one minute if Qcrit = 2.37E−4/60 = 3.94E−6 m 3 /s. Then
3 2 crit water
m 2300 D Q 3.94E 6 if 4.36E 7 m /s s 4
From Table A-1, this kinematic viscosity occurs at T ≈ 66 ° C Ans. (b)
6.8 When water at 20°C ( ρ = 998 kg/m 3 , μ = 0.001 kg/m⋅s) flows through an 8-cm-
diameter pipe, the wall shear stress is 72 Pa. What is the axial pressure gradient ( ∂ p/ ∂ x)
if the pipe is (a) horizontal; and (b) vertical with the flow up?
Chapter 6 • Viscous Flow in Ducts 367
Solution: Equation (6.9b) applies in both cases, noting that τw is negative:
(a) : (a)
Horizontal dp^ w Pa Ans. dx R m
Pa 3600 m
2 (b) : w 3600 998(9.81) (b)
dp dz Vertical, up g Ans. dx R dx
Pa 13, 400 m
6.9 A light liquid ( ρ = 950 kg/m 3 ) flows at an average velocity of 10 m/s through a
horizontal smooth tube of diameter 5 cm. The fluid pressure is measured at 1-m intervals along the pipe, as follows:
x , m: 0 1 2 3 4 5 6 p , kPa: 304 273 255 240 226 213 200
Estimate (a) the total head loss, in meters; (b) the wall shear stress in the fully developed section of the pipe; and (c) the overall friction factor.
Solution: As sketched in Fig. 6.6 of the text, the pressure drops fast in the entrance region (31 kPa in the first meter) and levels off to a linear decrease in the “fully developed” region (13 kPa/m for this data). (a) The overall head loss, for ∆ z = 0, is defined by Eq. (6.8) of the text:
(a) (950 / )(9.81 / ) f
p Pa h Ans. g (^) kg m m s
11.2 m
(b) The wall shear stress in the fully-developed region is defined by Eq. (6.9 b ):
13000 4 4 , solve for (b) 1 0.
w w fully developed w
p Pa Ans. L m d m
| 163 Pa
(c) The overall friction factor is defined by Eq. (6.10) of the text:
2 , (^2 )
(11.2 ) (c) overall f overall (^6) (10 / )
d g m m s f h m Ans. L (^) V m m s
NOTE: The fully-developed friction factor is only 0.0137.
1
Chapter 6 • Viscous Flow in Ducts 369
NOTE: IN PROBLEMS 6.12 TO 6.99, MINOR LOSSES ARE NEGLECTED.
6.12 A 5-mm-diameter capillary tube is used as a viscometer for oils. When the flow rate is 0.071 m 3 /h, the measured pressure drop per unit length is 375 kPa/m. Estimate the viscosity of the fluid. Is the flow laminar? Can you also estimate the density of the fluid?
Solution: Assume laminar flow and use the pressure drop formula (6.12):
4 4
p?^ 8Q Pa ?8(0.071/3600) , or: 375000 , solve. L R m (0.0025)
Ans
**kg
m s** ⋅
oil (^3)
kg Guessing 900 , m 4 Q 4(900)(0.071/3600) check Re. d (0.292)(0.005)
Ans
= = ≈ 16 OK, laminar
It is not possible to find density from this data, laminar pipe flow is independent of density.
6.13 A soda straw is 20 cm long and 2 mm in diameter. It delivers cold cola, approximated as water at 10°C, at a rate of 3 cm 3 /s. (a) What is the head loss through the
straw? What is the axial pressure gradient ∂ p /∂ x if the flow is (b) vertically up or
(c) horizontal? Can the human lung deliver this much flow?
Solution: For water at 10°C, take ρ = 1000 kg/m 3 and μ = 1.307E−3 kg/m⋅s. Check Re:
4 Q 4(1000)(3E 6 m /s)^3 Re 1460 (OK, laminar flow) d (1.307E 3)(0.002)
f (^4 )
128 LQ 128(1.307E 3)(0.2)(3E 6)
Then, from Eq. (6.12), h (a) gd (1000)(9.81)(0.002)
Ans.
= = ≈ 0.204 m
If the straw is horizontal , then the pressure gradient is simply due to the head loss:
f horiz
p gh 1000(9.81)(0.204 m) (c) L L 0.2 m
Ans.
| = = ≈
Pa 9980 m
If the straw is vertical , with flow up , the head loss and elevation change add together:
f vertical
p g(h z) 1000(9.81)(0.204 0.2) (b) L L 0.
Ans.
| = = ≈
Pa 19800 m
The human lung can certainly deliver case (c) and strong lungs can develop case (b) also.
370 Solutions Manual • Fluid Mechanics, Fifth Edition
6.14 Water at 20°C is to be siphoned through a tube 1 m long and 2 mm in diameter, as in Fig. P6.14. Is there any height H for which the flow might not be laminar? What is the flow rate if H = 50 cm? Neglect the tube curvature. Fig. P6.
Solution: For water at 20°C, take ρ = 998 kg/m^3 and μ = 0.001 kg/m⋅s. Write the steady
flow energy equation between points 1 and 2 above: 2 2 2 atm atm tube 1 2 f f (^2)
p 0 p V V 32 L z z h , or: H h V g 2g g 2g 2g (^) gd
2 2
V 32(0.001)(1.0)V m Enter data in Eq. (1): 0.5 , solve V 0. 2(9.81) (^) (998)(9.81)(0.002) s
Equation (1) is quadratic in V and has only one positive root. The siphon flow rate is 3 2 H=50 cm
m Q (0.002) (0.590) 1.85E 6. 4 s
Ans
m^3 0.0067 if 50 cm h
H =
Check Re = (998)(0.590)(0.002) /(0.001) ≈1180 (OK, laminar flow)
It is possible to approach Re ≈ 2000 (possible transition to turbulent flow) for H < 1 m, for the case of the siphon bent over nearly vertical. We obtain Re = 2000 at H ≈ 0.87 m.
6.15 Professor Gordon Holloway and his students at the University of New Brunswick
went to a fast-food emporium and tried to drink chocolate shakes ( ρ ≈ 1200 kg/m 3 ,
μ ≈ 6 kg/m⋅s) through fat straws 8 mm in diameter and 30 cm long. (a) Verify that their
human lungs, which can develop approximately 3000 Pa of vacuum pressure, would be unable to drink the milkshake through the vertical straw. (b) A student cut 15 cm from his straw and proceeded to drink happily. What rate of milkshake flow was produced by this strategy?
Solution: (a) Assume the straw is barely inserted into the milkshake. Then the energy equation predicts 2 2 1 1 2 2 1 2
2 3 2
(1200 / )(9.81 / )^2
f
tube f
p V p V z z h g g g g Pa V m h kg m m s g
Solve for (^) f tube Ans. (a)
V
h m m which is impossible g
2 0.255 0.3 0 2
372 Solutions Manual • Fluid Mechanics, Fifth Edition
Solution: (a) Assume no pressure drop and neglect velocity heads. The energy equation reduces to:
2 2 1 1 2 2 1 0 0 (^ )^20 0 0 ,^ : 2 2 f f f
p V p V z L l z h h h L l
ρ g g ρ g g
- = + + + = + + + = + + + or ≈ +
4
, (^) f ,
LQ
For laminar flow h and, for uniform draining Q gd t
Solve for Ans. (a)
L
t gd L l
(b) Apply to ∆ t = 6 s. For water, take ρ = 998 kg/m^3 and μ = 0.001 kg/m⋅s. Formula (a) predicts:
3 3 2 4
Solve for (b)
kg m s m E m t s kg m m s d m Ans.
d ≈ 0.0015 m
6.18 To determine the viscosity of a liquid of specific gravity 0.95, you fill, to a depth of 12 cm, a large container which drains through a 30-cm-long vertical tube attached to the bottom. The tube diameter is 2 mm, and the rate of draining is found to be 1.9 cm 3 /s. What is your estimate of the fluid viscosity? Is the tube flow laminar? Fig. P6.
Solution: The known flow rate and diameter enable us to find the velocity in the tube:
3 2
Q E m s m V A (^) π m s
Evaluate ρ liquid = 0.95(998) = 948 kg/m^3. Write the energy equation between the top surface
and the tube exit: (^2 )
2 2 2 2
a top a top f
p V p V z h g g g g
V LV or g (^) gd
Chapter 6 • Viscous Flow in Ducts 373
Note that “L” in this expression is the tube length only (L = 30 cm).
Solve for ( )
948(0.605)(0.002) 446 ( ) d 0.
laminar flow Ans.
Vd Re laminar
**kg
m s** ⋅
6.19 An oil (SG = 0.9) issues from the pipe in Fig. P6.19 at Q = 35 ft 3 /h. What is the kinematic viscosity of the oil in ft 3 /s? Is the flow laminar?
Solution: Apply steady-flow energy:
2 2 atm atm 2 1 2 f
p 0 p V z z h ,
ρg 2g ρg 2g
(^2 )
Q 35/3600 ft where V 7.
A π(0.25 /12) s
2 2 2 f 1 2
V (7.13)
Solve h z z 10 9.21 ft 2g 2(32.2)
Assuming laminar pipe flow, use Eq. (6.12) to relate head loss to viscosity:
f (^4 )
128 LQ 128(6)(35/3600)
h 9.21 ft , solve gd (32.2)(0.5/12)
Ans.
ν ν μ ν π π ρ
ft^2 3.76E 4 s
Check Re = 4Q/( πν d) = 4(35/3600)/[ π(3.76E 4)(0.5/12)]− ≈790 (OK, laminar)
6.20 In Prob. 6.19 what will the flow rate be, in m^3 /h, if the fluid is SAE 10 oil at 20°C?
Solution: For SAE 10 oil at 20°C, take ρ = 1.69 slug/ft 3 and μ =2.17E−3 slug/ft⋅s. The
steady flow energy analysis above gives, for laminar flow,
2 f (^2 )
V 32 LV 32(2.17E 3)(6.0)V
h 10 4.41V (quadratic equation) 2(32.2) (^) gd (1.69)(32.2)(0.5/12)
μ ρ
(^2 ) ft 0.5 ft Solve for V 2.25 , Q (2.25) 0. s 4 12 s
Ans. π (^) ≈ = (^) =
m^3 0. h
Chapter 6 • Viscous Flow in Ducts 375
Solution: Determine the velocity of exit from the needle and then apply the steady-flow energy equation:
(^1 )
Q 0.
306 cm/s A ( /4)(0.025)
V
π
2 2 2 2 1 1 2 1 f1 f2 1 2 2 f
p V p V Energy: z z h h , with z z , V 0, h 0
ρg 2g ρg 2g
Assume laminar flow for the head loss and compute the pressure difference on the piston:
2 2 2 1 1 f1 (^2)
p p V 32(0.002)(0.015)(3.06) (3.06) h 5.79 m ρg 2g (^) (900)(9.81)(0.00025) 2(9.81)
2 Then F pApiston (900)(9.81)(5.79) 4 (0.01) Ans.
π = ∆ = ≈ 4.0 N
6.23 SAE 10 oil at 20°C flows in a vertical pipe of diameter 2.5 cm. It is found that the pressure is constant throughout the fluid. What is the oil flow rate in m 3 /h? Is the flow up or down?
Solution: For SAE 10 oil, take ρ = 870 kg/m 3 and μ = 0.104 kg/m⋅s. Write the energy
equation between point 1 upstream and point 2 downstream:
2 2 1 1 2 2 1 2 f 1 2 1 2
p V p V z z h , with p p and V V
ρg 2g ρg 2g
Thus h f = z 1 − z 2 > 0 by definition. Therefore, flow is down. Ans.
While flowing down, the pressure drop due to friction exactly balances the pressure rise due to gravity. Assuming laminar flow and noting that ∆z = L, the pipe length, we get
f (^4)
4 3
128 LQ
h z L, gd (8.70)(9.81)(0.025) m or: Q 7.87E 4 128(0.104) s
Ans.
m^3 2. h
6.24 Two tanks of water at 20°C are connected by a capillary tube 4 mm in diameter and 3.5 m long. The surface of tank 1 is 30 cm higher than the surface of tank 2. (a) Estimate the flow rate in m 3 /h. Is the flow laminar? (b) For what tube diameter will Re d be 500?
376 Solutions Manual • Fluid Mechanics, Fifth Edition
Solution: For water, take ρ = 998 kg/m 3 and μ = 0.001 kg/m⋅s. (a) Both tank surfaces
are at atmospheric pressure and have negligible velocity. The energy equation, when neglecting minor losses, reduces to:
4 3 2 4
f
LQ kg m s m Q z m h gd kg m m s m
3 Solve for 5.3 6 (a) m Q E Ans. s
m^3 0. h Check Re d 4 /( ) 4(998)(5.3E 6)/[ (0.001)(0.004)] (a)
Q d Ans.
Re (^) d = 1675 laminar.
(b) If Red = 500 = 4 ρ Q /( πμ d ) and ∆ z = h f, we can solve for both Q and d :
Re 500 , 0. (0.001 / ) d
kg m Q Q d
π kg m s d
or
4 3 2 4
f
kg m s m Q h m or Q d
π kg m m s d
Combine these two to solve for Q = 1.05 E − 6 m^3 / s and d = 2.67 mm Ans. (b)
6.25 For the configuration shown in Fig. P6.25, the fluid is ethyl alcohol at 20 °C, and the tanks are very wide. Find the flow rate which occurs in m^3 /h. Is the flow laminar?
Solution: For ethanol, take ρ =
789 kg/m 3 and μ = 0.0012 kg/m⋅s. Write
the energy equation from upper free surface (1) to lower free surface (2): (^) Fig. P6.
2 2 1 1 2 2 1 2 f 1 2 1 2
p V p V z z h , with p p and V V 0
ρg 2g ρg 2g
f (^1 2 4 )
128 LQ 128(0.0012)(1.2 m)Q Then h z z 0.9 m gd (789)(9.81)(0.002)
Solve for Q ≈ 1.90E 6 m /s− 3 = 0.00684 m /h.^3 Ans.
Check the Reynolds number Re = 4 ρQ/( πμd) ≈ 795 − OK, laminar flow.
378 Solutions Manual • Fluid Mechanics, Fifth Edition
6.28 For straightening and smoothing an airflow in a 50-cm-diameter duct, the duct is packed with a “honeycomb” of thin straws of length 30 cm and diameter 4 mm, as in Fig. P6.28. The inlet flow is air at 110 kPa and 20°C, moving at an average velocity of 6 m/s. Estimate the pressure drop across the honeycomb.
Solution: For air at 20 °C, take μ ≈
1.8E−5 kg/m⋅s and ρ = 1.31 kg/m 3. There
would be approximately 12000 straws, but each one would see the average velocity of 6 m/s. Thus
Fig. P6.
laminar (^2 )
32 LV 32(1.8E 5)(0.3)(6.0)
p d (0.004)
Ans.
∆ = = ≈ 65 Pa
Check Re = ρVd/ μ = (1.31)(6.0)(0.004)/(1.8E−5) ≈ 1750 OK, laminar flow.
6.29 Oil, with ρ = 890 kg/m^3 and μ = 0.07 kg/m⋅s, flows through a horizontal pipe 15 m
long. The power delivered to the flow is 1 hp. (a) What is the appropriate pipe diameter if the flow is at the laminar transition point? For this condition, what are (b) Q in m 3 /h; and
(c) τw in kPa?
Solution: (a, b) Set the Reynolds number equal to 2300 and the (laminar) power equal to 1 hp:
3 Re 2300 (890^ /^ ) or 0.181 2 / d 0.07 /
kg m Vd Vd m s kg m s
2 2 laminar (^2)
LV
Power hp W Q p d V V d
π μ π = = = ∆ = (^) =
Solve for 5.32 and (a)
m V s
= d = 0.034 m Ans.
It follows that Q = ( π/4) d^2 V = ( π/4)(0.034 m)^2 (5.32 m/s) = 0.00484 m^3 /s = 17.4 m^3 /h Ans. (b)
(c) From Eq. (6.12), the wall shear stress is
8 8(0.07 / )(5.32 / ) 88 (c) w (0.034 )
V kg m s m s Pa Ans. d m
μ τ
= = = = 0.088 kPa
Chapter 6 • Viscous Flow in Ducts 379
6.30 SAE 10 oil at 20°C flows through the 4-cm-diameter vertical pipe of Fig. P6.30. For the mercury manometer reading h = 42 cm shown, (a) calculate the volume flow rate in m 3 /h, and (b) state the direction of flow.
Solution: For SAE 10 oil, take ρ =
870 kg/m 3 and μ = 0.104 kg/m⋅s. The
pressure at the lower point (1) is considerably higher than p 2 according to the manometer reading:
Fig. P6.
p 1 − p 2 = ( ρHg − ρoil)g ∆h = (13550 − 870)(9.81)(0.42) ≈52200 Pa
∆p/( ρ oilg) = 52200/[870(9.81)] ≈6.12 m
This is more than 3 m of oil, therefore it must include a friction loss: flow is up. Ans. (b) The energy equation between (1) and (2), with V 1 = V 2 , gives
1 2 2 1 f f f (^4)
p p 128 LQ z z h , or 6.12 m 3 m h , or: h 3.12 m g gd
μ ρ πρ
(6.12 3) (870)(9.81)(0.04) 4 m^3 Compute Q 0.00536 (a) 128(0.104)(3.0) s
Ans.
m^3 19. h
Check Re = 4 Q/( ρ πμd) = 4(870)(0.00536)/[ π(0.104)(0.04)] ≈1430 (OK, laminar flow).
6.31 Light oil, ρ = 880 kg/m 3 and μ = 0.015 kg/(m⋅s), flows down a vertical 6-mm-
diameter tube due to gravity only. Estimate the volume flow rate in m 3 /h if (a) L = 1 m and (b) L = 2 m. (c) Verify that the flow is laminar.
Solution: If the flow is due to gravity only, the head loss matches the elevation change:
4 f (^4)
128 LQ gd h z L , or Q gd 128
independent of pipe length
μ πρ πρ μ
For this case, 3 Q (880)(9.81)(0.006) /[128(0.015)]^4 1.83E 5 m (a, b) s
= π ≈ − = Ans.
m^3 0. h Check Re = 4 Q/( ρ πμ d) = 4(880)(1.83E 5)/[− π(0.015)(0.006)] ≈ 228 (laminar). Ans. (c)
Chapter 6 • Viscous Flow in Ducts 381
Solution: For SAE 30 oil at 20°C, ρ = 891 kg/m 3 and μ = 0.29 kg/m⋅s. With mass flow
known, we can evaluate the pipe velocity:
2
Check ( , ) d 0.
m kg s m V A s
Re OK laminar
Apply the steady flow energy equation between A and B:
(^2 2 500000 ) , : 15 2 2 891(9.81) 891(9.81)
A A B B A B f p f p
p V p V z z h h or h h
ρ g g ρ g g
2 2
where 140.5 , Solve for 118. 891(9.81)(0.03) f pump
LV
h m h m gd
The pump power is then given by
p p^3 9.81^2 (118.9^ )
kg m gQh mgh m Ans. s (^) s
Power 3500 watts
6.34 Derive the time-averaged x -momentum equation (6.21) by direct substitution of Eqs. (6.19) into the momentum equation (6.14). It is convenient to write the convective acceleration as
u (^) (u ) (^2) (uv) (uw) t x y z
d d
which is valid because of the continuity relation, Eq. (6.14).
Solution: Into the x -momentum eqn. substitute u = u + u’, v = v + v’, etc., to obtain
2 2
2 x
(u 2uu’ u’ ) (v u vu’ v’u v’u’) (wu wu’ w’u w’u’) x y z
(p p’) g [ (u u’)] x
+^ +^ +^ +^ +^ +^ +^ +^ +^ +
Now take the time-average of the entire equation to obtain Eq. (6.21) of the text:
Ans.
- (^2 ) + ( ) + ( ) = − + (^) x + ∇ (^2 )
du p u’ u’v’ u’w’ g u dt x y z x
382 Solutions Manual • Fluid Mechanics, Fifth Edition
6.35 By analogy with Eq. (6.21) write the turbulent mean-momentum differential equation for (a) the y direction and (b) the z direction. How many turbulent stress terms appear in each equation? How many unique turbulent stresses are there for the total of three directions?
Solution: You can re-derive, as in Prob. 6.34, or just permute the axes:
y
dv p v v (a) : g u’v’ v’v’ dt y x x y y v v’w’ z z
^ ^
y
z
dw p w w (b) : g u’w’ v’w’ dt z x x y y w w’w’ z z
^ ^
z
6.36 The following turbulent-flow velocity data u ( y ), for air at 75°F and 1 atm near a smooth flat wall, were taken in the University of Rhode Island wind tunnel:
y, in: 0.025 0.035 0.047 0.055 0. u, ft/s: 51.2 54.2 56.8 57.6 59.
Estimate (a) the wall shear stress and (b) the velocity u at y = 0.22 in.
Solution: For air at 75°F and 1 atm, take ρ = 0.00230 slug/ft^3 and μ = 3.80E−7 slug/ft⋅s.
We fit each data point to the logarithmic-overlap law, Eq. (6.28):
w
u 1 uy 1 0.0023uy ln B ln 5.0, u* / u* 0.41 3.80E 7
Enter each value of u and y from the data and estimate the friction velocity u*:
y, in: 0.025 0.035 0.047 0.055 0. u*, ft/s: 3.58 3.58 3.59 3.56 3.
yu*/ ν (approx): 45 63 85 99 117
Each point gives a good estimate of u*, because each point is within the logarithmic layer in Fig. 6.10 of the text. The overall average friction velocity is
avg
2 2 w,avg
- ft u 3.57 1%, u* (0.0023)(3.57) (a) s
≈ ± τ = ρ = ≈ 2 Ans.
**lbf
ft**
