Solutions Manual for
Fluid Mechanics Seventh Edition in SI Units
Frank M. White Chapter 6 Viscous Flow in Ducts
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P6.1 An engineer claims that flow of SAE 30W oil, at 20°C, through a 5-cm-diameter smooth pipe at 1 million N/h, is laminar. Do you agree? A million newtons is a lot, so this sounds like an awfully high flow rate. Solution: For SAE 30W oil at 20°C (Table A.3), take ρ = 891 kg/m3 and µ = 0.29 kg/m-s. Convert the weight flow rate to volume flow rate in SI units: w (1E6 N / h)(1 / 3600 h / s) m3 π m = = 0.0318 = (0.05m)2 V , solve V = 16.2 Q = 3 2 ρg s 4 (891kg / m )(9.81m / s ) s Calculate Re D
ρVD (891kg / m 3 )(16.2m / s)(0.05m) = = ≈ 2500 (transitional) 0.29 kg / m − s µ
This is not high, but not laminar. Ans. With careful inlet design, low disturbances, and a very smooth wall, it might still be laminar, but No, this is transitional, not definitely laminar. P6.2 Water flows through a 10-cm-diameter pipe at a rate of 10–4 m3/s. Evaluate the Reynolds number when the temperature is (a) 20°C, (b) 40°C, and (c) 60°C. Estimate the temperature at which transition to turbulence occurs. Solution: Reynolds number Re =
V =
but Now
4Q
πd2
d = 0.1 m,
Vd V
∴ Re =
Q = 10
–4
4Q π dV 3
m /s
Viscosity is a function of temperature. Temp (°C)
V of water (m 2 / s)
Re
20
1.005 × 10 –6
1,267
40
0.662 × 10 –6
1,923
60
0.475 × 10 –6
2,681
Recrit ~ 2300 happens when V = 0.554 × 10 –6 m 2 /s , which is the kinematic viscosity of water at 50°C.
2
P6.3 The present pumping rate of North Slope crude oil through the Alaska Pipeline (see the chapter-opener photo) is about 600,000 barrels per day (1 barrel = 0.159 m3). What would be the maximum rate if the flow were constrained to be laminar? Assume that Alaskan crude oil fits Fig. A.1 of the Appendix at 60°C. Solution: From Fig. A.1 for crude at 60°C, ρ = 0.86(1000) = 860 kg/m3 and µ = 0.0040 kg/m-s. From Eq. (6.2), the maximum laminar Reynolds number is about 2300. Convert the pipe diameter from 48 inches to 1.22 m. Solve for velocity:
P6.4 Following up Prob. P6.3, suppose the Alaska Pipeline were carrying 1 × 105 m3/day of SAE 30W oil at 20°C? Estimate the Reynolds number. Is the flow laminar? Solution: For SAE 30W oil at 20°C, Table A.3, ρ = 891 kg/m3, and µ = 0.29 kg/m-s. Convert the flow rate into cubic meters per second and then find the Reynolds number:
m3 m (or V = 0.99 ) Q = (1 × 10 m /day)(1 day/24 × 3600 s) = 1.157 s s 4 ρQ 4(891)(1.157) Re D = = = 3710 Ans. (Transitional , not laminar) πµ D π (0.29)(1.22m) 5
3
6.5 For flow of SAE 30W oil through a 5-cm-diameter pipe, from Fig. A.1, for what flow rate in m3/h would we expect transition to turbulence at (a) 20°C and (b) 100°C? Solution: For SAE 30W oil take and take µ = 0.29 kg/m⋅s at 20°C (Table A.3) and 0.01 kg/m-s at 100°C (Fig A.1). Write the critical Reynolds number in terms of flow rate Q: 4(891 kg/m 3 )Q ρVD 4 ρQ (a) Re crit = 2300 = = = , µ πµ D π (0.29 kg/m⋅s)(0.05 m) m3 m3 solve Q = 0.0293 = 106 Ans. (a) s h
3
6.6 In flow past a body or wall, early transition to turbulence can be induced by placing a trip wire on the wall across the flow, as in Fig. P6.6. If the trip wire in Fig. P6.6 is placed where the local velocity is U, it will trigger turbulence if Ud/ν = 850, where d is the wire diameter [Ref. 3 of Ch. 6]. If the sphere diameter is 20 cm and transition is observed at ReD = 90,000, what is the diameter of the trip wire in mm?
Fig. P6.6
Solution: For the same U and ν,
6.7 For flow of a uniform stream parallel to a sharp flat plate, transition to a turbulent boundary layer on the plate may occur at Rex = ρUx/µ ≈ 1E6, where U is the approach velocity and x is distance along the plate. If U = 2.5 m/s, determine the distance x for the following fluids at 20°C and 1 atm: (a) hydrogen; (b) air; (c) gasoline; (d) water; (e) mercury; and (f) glycerin. Solution: We are to calculate x = (Rex)(µ)/(ρU) = (1E6)(µ)/[ρ (2.5m/s)]. Make a table: FLUID
ρ – kg/m3
µ - kg/m-s
x - meters
Hydrogen
0.00839
9.05E-5
43.
Air
1.205
1.80E-5
6.0
Gasoline
680
2.92E-4
0.17
Water
998
0.0010
0.40
Mercury
13,550
1.56E-3
0.046
Glycerin
1260
1.49
470.
Clearly there are vast differences between fluid properties and their effects on flows.
4
6.8 Cola, approximated as pure water at 20°C, is to fill a 0.2-L container through a 5-mmdiameter 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/m3 and µ = 0.001 kg/m⋅s. The volume = 0.2 L = 2 × 10 –4 m3. Then, if we assume transition at Re = 2300,
Then Δtfill = υ/Q = 2E–4/9.05E−6 ≈ 22 s Ans. (a) (b) We fill in exactly one minute if Qcrit = 2E–4/60 = 3.33E−6 m3/s. Then
Q crit = 3.33E−6
m 3 2300πν D = s 4
if ν water ≈ 3.69E−7 m 2 /s
From Table A-1, this kinematic viscosity occurs at T ≈ 80°C Ans. (b)
6.9 When water at 20°C (ρ = 998 kg/m3, 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? Solution: Equation (6.9b) applies in both cases, noting that τw is negative:
1
P6.10 Is Eq. (6.9b) still applicable when the fluid is non-Newtonian? Solution: Yes, the relationship between head loss and wall shear stress is still applicable when the fluid is non-Newtonian. 6.11 A light liquid (ρ = 950 kg/m3) 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: p, kPa:
0 304
1 273
2 255
3 240
4 226
5 213
6 200
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.
5
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 Δz = 0, is defined by Eq. (6.8) of the text:
(b) The wall shear stress in the fully-developed region is defined by Eq. (6.9b):
(c) The overall friction factor is defined by Eq. (6.10) of the text:
NOTE: The fully-developed friction factor is only 0.0137.
6.12 Water at 20°C (ρ = 998 kg/m3) flows through an inclined 8-cm-diameter pipe. At sections A and B, pA = 186 kPa, VA = 3.2 m/s, zA = 24.5 m, while pB = 260 kPa, VB = 3.2 m/s, and zB = 9.1 m. Which way is the flow going? What is the head loss? Solution: Guess that the flow is from A to B and write the steady flow energy equation:
pA VA2 p V2 186000 260000 + + zA = B + B + zB + h f , or: + 24.5 = + 9.1+ h f , ρ g 2g ρ g 2g 9790 9790 or: 43.50 = 35.66 + h f , solve: h f = +7.84 m Yes, flow is from A to B. Ans. (a, b)
6.13 Water at 20°C flows upward at 4 m/s in a 6-cm-diameter pipe. The pipe length between points 1 and 2 is 5 m, and point 2 is 3 m higher. A mercury manometer, connected between 1 and 2, has a reading h = 135 mm, with p1 higher. (a) What is the pressure change (p1 − p2)? (b) What is the head loss, in meters? (c) Is the manometer reading proportional to head loss? Explain. (d) What is the friction factor of the flow?
6
Solution: A sketch of this situation is shown at right. By moving through the manometer, we obtain the pressure change between points 1 and 2, which we compare with Eq. (6.9b):
The friction factor is
f = hf
2 d 2g 0.06 m 2(9.81 m/s ) = (1.7 m) = 0.025 5 m (4 m/s)2 L V2
Ans. (d)
By comparing the manometer relation to the head-loss relation above, we find that:
NOTE: IN PROBLEMS 6.15 TO 6.108, MINOR LOSSES ARE NEGLECTED.
6.14 Assuming laminar flow, determine by how much the diameter of a pipe should be increased if it is intended to double the flow rate under a fixed head difference. Solution: For laminar flow, head loss and discharge are related by
hf =
128µ LQ πρ gd 4
Under fixed hf, Q ~ d 4 or
if
Q2 = 2Q1
That is, the diameter is to increase by 19%.
d4 Q2 = 24 Q1 d1
⇒
d2 = 21/4 = 1.19 d1
7
6.15 A 5-mm-diameter capillary tube is used as a viscometer for oils. When the flow rate is 0.071 m3/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):
Guessing ρ oil ≈ 900 check Re =
kg , m3
4 ρQ 4(900)(0.071/3600) = ≈ 16 OK, laminar Ans. πµd π (0.292)(0.005)
It is not possible to find density from this data, laminar pipe flow is independent of density.
P6.16 Estimate the percentage change in the flow rate as a result of each of the following: (a) diameter increased by 10%, (b) length reduced by 30%, and (c) head difference increased by 40%. Laminar flow is assumed. Solution: For laminar flow
Q =
πρ gd 4 hf 128µL
Change
Effect on Q
d → 1.1d
Q → 1.14 Q = 1.46Q
L → 0.7 L h f → 1.4h f
Q → 0.7 –1Q = 1.43Q Q → 1.4Q
6.17 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 cm3/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/m3 and µ = 1.307E – 3 kg/m⋅s. Check Re:
If the straw is horizontal, then the pressure gradient is simply due to the head loss:
8
If the straw is vertical, with flow up, the head loss and elevation change add together:
The human lung can certainly deliver case (c) and strong lungs can develop case (b) also.
6.18 Water at 20°C is to be siphoned through a tube 1 m long and 2 mm in diameter, as in Fig. P6.18. 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. P6.18
Solution: For water at 20°C, take ρ = 998 kg/m3 and µ = 0.001 kg/m⋅s. Write the steady flow energy equation between points 1 and 2 above: (1)
Equation (1) is quadratic in V and has only one positive root. The siphon flow rate is
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. 6.19 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/m3, 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?
9
Solution: (a) Assume the straw is barely inserted into the milkshake. Then the energy equation predicts
(b) By cutting off 15 cm of vertical length and assuming laminar flow, we obtain a new energy equation
Check the Reynolds number: Red = ρVd/µ = (1200)(0.00275)(0.008)/(6) = 0.0044 (Laminar).
P6.20 Fluid flows steadily, at volume rate Q, through a large horizontal pipe and then divides into two small pipes, the larger of which has an inside diameter of 25 mm and carries three times the flow of the smaller pipe. Both small pipes have the same length and pressure drop. If all flows are laminar, estimate the diameter of the smaller pipe. Solution: For laminar flow in a horizontal pipe, the volume flow is a simple formula, Eq. (6.12): Qlaminar
π d 4 Δp = ( ) 128 µ L
Since Δp, L, and µ are the same in the two small pipes, it follows that the flows simply vary as the 4th power of their diameters. Let pipe 1 have the 25-mm diameter. Then we compute
Q1 = ( const)(d14 ) = 3Q2 = 3(const)(d24 ) Thus d2 =
d1 1/ 4
3
=
25mm = 1.316
19.0 mm
Ans.
10
6.21 A capillary viscometer measures the time required for a specified volume υ of liquid to flow through a small-bore glass tube, as in Fig. P6.21. This transit time is then correlated with fluid viscosity. For the system shown, (a) derive an approximate formula for the time required, assuming laminar flow with no entrance and exit losses. (b) If L = 12 cm, l = 2 cm, υ = 8 cm3, and the fluid is water at 20°C, what capillary diameter D will result in a transit time t of 6 seconds?
Fig. P6.21
Solution: (a) Assume no pressure drop and neglect velocity heads. The energy equation reduces to:
For laminar flow, h f =
128 µ LQ πρ gd 4
and, for uniform draining, Q =
υ Δt
(b) Apply to Δt = 6 s. For water, take ρ = 998 kg/m3 and µ = 0.001 kg/m⋅s. Formula (a) predicts:
11
6.22 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 cm3/s. What is your estimate of the fluid viscosity? Is the tube flow laminar?
Fig. P6.22
Solution: The known flow rate and diameter enable us to find the velocity in the tube:
Evaluate ρ liquid = 0.95(998) = 948 kg/m3. Write the energy equation between the top surface and the tube exit:
Note that “L” in this expression is the tube length only (L = 30 cm).
6.23 An oil (SG = 0.9) issues from the pipe in Fig. P6.23 at Q = 1 m3/h. What is the kinematic viscosity of the oil in m2/s? Is the flow laminar? Solution: Apply steady-flow energy:
12
Fig. P6.23
where V2 =
Q 1/3600 = ≈ 2.09 m/s A π (0.013 / 2)2 Solve h f = z1 − z 2 −
V22 (2.09)2 = 3− = 2.78 m 2g 2(9.81)
Assuming laminar pipe flow, use Eq. (6.12) to relate head loss to viscosity:
h f = 2.78 m =
128ν LQ 128(1.8)(1/3600)ν µ = ≈ 3.8 × 10 –5 m2 / s 4 4 , solve ν = ρ π gd π (9.81)(0.013)
Ans.
Check Re = 4Q/(πν d) = 4(1/3600)/[π (3.8 × 10 –5 )(0.013)] ≈ 710 (OK, laminar)
P6.24 The oil tanks in Tinyland are only 160 cm high, and they discharge to the Tinyland oil truck through a smooth tube 4 mm in diameter and 55 cm long. The tube exit is open to the atmosphere and 145 cm below the tank surface. The fluid is medium fuel oil, ρ = 850 kg/m3 and
µ = 0.11 kg/m-s. Estimate the oil flow rate in cm3/h. Solution: The steady flow energy equation, with 1 at the tank surface and 2 the exit, gives
z1 = z2 +
αV 2 L V2 V2 64 0.55m 850V (0.004) + f , or : Δz = 1.45m = (2.0+ ) , Re d = 2g d 2g 2g Re d 0.004m 0.11
We have taken the energy correction factor α = 2.0 for laminar pipe flow. Solve for V = 0.10 m/s, Red = 3.1 (laminar), Q = 1.26E-6 m3/s ≈ 4500 cm3/h. The exit jet energy αV2/2g is properly included but is very small (0.001 m).
Ans.
13
6.25 In Tinyland, houses are less than a foot high! The rainfall is laminar! The drainpipe in Fig. P6.25 is only 2 mm in diameter. (a) When the gutter is full, what is the rate of draining? (b) The gutter is designed for a sudden rainstorm of up to5 mm per hour. For this condition, what is the maximum roof area that can be drained successfully? (c) What is Red? Solution: If the velocity at the gutter surface is neglected, the energy equation reduces to
Fig. P6.25
For water, take ρ = 998 kg/m3 and µ = 0.001 kg/m⋅s. (a) With Δz known, this is a quadratic equation for the pipe velocity V:
0.2 m =
V2 32(0.001 kg/m⋅s)(0.2 m)V + , 2 2(9.81 m/s ) (998 kg/m 3 )(9.81 m/s 2 )(0.002 m)2
(b) The roof area needed for maximum rainfall is 0.0107 m3/h ÷ 0.005 m/h = 2.14 m2. Ans. (b) (c) The Reynolds number of the gutter is Red = (998)(0.945)(0.002)/(0.001) = 1890 laminar. Ans. (c) 6.26 A steady push on the piston in Fig. P6.26 causes a flow rate Q = 0.15 cm3/s through the needle. The fluid has ρ = 900 kg/m3 and µ = 0.002 kg/(m⋅s). What force F is required to maintain the flow?
Fig. P6.26
Solution: Determine the velocity of exit from the needle and then apply the steady-flow energy equation:
14
Assume laminar flow for the head loss and compute the pressure difference on the piston:
6.27 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 m3/h? Is the flow up or down? Solution: For SAE 10 oil, take ρ = 870 kg/m3 and µ = 0.104 kg/m⋅s. Write the energy equation between point 1 upstream and point 2 downstream:
Thus h f = z1 − 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 Δz = L, the pipe length, we get
6.28 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 m3/h. Is the flow laminar? (b) For what tube diameter will Red be 500? Solution: For water, take ρ = 998 kg/m3 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:
15
(b) If Red = 500 = 4ρQ/(πµd) and Δz = hf, we can solve for both Q and d:
Combine these two to solve for Q = 1.05E−6 m 3/s and
d = 2.67 mm Ans. (b)
6.29 For the configuration shown in Fig. P6.29, the fluid is ethyl alcohol at 20°C, and the tanks are very wide. Find the flow rate that occurs, in m3/h. Is the flow laminar? Solution: For ethanol, take ρ = 789 kg/m3 and µ = 0.0012 kg/m⋅s. Write the energy equation from upper free surface (1) to lower free surface (2):
Fig. P6.29
Check the Reynolds number Re = 4ρQ/(πµd) ≈ 795 − OK, laminar flow.
16
P6.30 Two oil tanks are connected by two 9-m-long pipes, as in Fig. P6.30. Pipe 1 is 5 cm in diameter and is 6 m higher than pipe 2. It is found that the flow rate in pipe 2 is twice as large as the flow in pipe 1. (a) What is the diameter of pipe 2? (b) Are both pipe flows laminar? (c) What is the flow rate in pipe 2 (m3/s)? Neglect minor losses.
Solution:
(a) If we know the flows are laminar, and (L, ρ, µ) are constant, then Q ∝ D4:
From Eq.(6.12),
Q2 D = 2.0 = ( 2 )4 , hence D2 = (5 cm)(2.0)1/4 = 5.95 cm Q1 D1
Ans.(a)
We will check later in part (b) to be sure the flows are laminar. [Placing pipe 1 six meters higher was meant to be a confusing trick, since both pipes have exactly the same head loss and Δz.] (c) Find the flow rate first and then backtrack to the Reynolds numbers. For SAE 30W oil at 20°C (Table A.3), take ρ = 891 kg/m3 and µ = 0.29 kg/m-s. From the energy equation, with V1 = V2 = 0, and Eq. (6.12) for the laminar head loss, Δz = 22 −15 = 7 m = h f =
128µ LQ 128(0.29kg / m − s)(9m)Q2 = 4 πρ gD2 π (891kg / m 3 )(9.81m / s 2 )(0.0595m)4
Solve for
Q2 = 0.0072 m3 / s
Ans.(c)
In a similar manner, insert D1 = 0.05m and compute Q1 = 0.0036 m3/s = (1/2)Q1. (b) Now go back and compute the Reynolds numbers:
Re1 =
4 ρQ1 4(891)(0.0036) 4 ρQ2 4(891)(0.0072) = = 281 ; Re 2 = = = 473 Ans.(b) πµ D1 π (0.29)(0.050) πµ D2 π (0.29)(0.0595)
Both flows are laminar, which verifies our flashy calculation in part (a).
17
6.31 Let us attack Prob. 6.29 in symbolic fashion, using Fig. P6.31. All parameters are constant except the upper tank depth Z(t). Find an expression for the flow rate Q(t) as a function of Z(t). Set up a differential equation, and solve for the time t0 to drain the upper tank completely. Assume quasi-steady laminar flow. Solution: The energy equation of Prob. 6.29, using symbols only, is combined with a controlvolume mass balance for the tank to give the basic differential equation for Z(t):
Fig. P6.31
Separate the variables and integrate, combining all the constants into a single “C”:
18
6.32 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. P6.32. 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/m3. There would be approximately 12000 straws, but each one would see the average velocity of 6 m/s. Thus
Fig. P6.32
Check Re = ρVd/µ = (1.31)(6.0)(0.004)/(1.8E−5) ≈ 1750 OK, laminar flow.
P6.33 SAE 30W oil at 20°C flows through a straight pipe 25 m long, with diameter 4 cm. The average velocity is 2 m/s. (a) Is the flow laminar? Calculate (b) the pressure drop; and (c) the power required. (d) If the pipe diameter is doubled, for the same average velocity, by what percent does the required power increase? Solution: For SAE 30W oil at 20°C, Table A.3, ρ = 891 kg/m3, and µ = 0.29 kg/m-s. (a) We have enough information to calculate the Reynolds number:
(b, c) The pressure drop and power follow from the laminar formulas of Eq. (6.12):
19
(d) If D doubles to 8 cm and V remains the same at 2.0 m/s, the new pressure drop will be 72,500 Pa, and the new flow rate will be Q = 0.01005 m3/s, hence the new power will be P
=
Q Δp
=
(0.01005)(72,500)
=
729 W
Zero percent change!
This is because D2 cancels in the product P = Q Δp = 8 π µ L V2. NOTE: The flow is still laminar, ReD = 492.
Ans.(d)
6.34 SAE 10 oil at 20°C flows through the 4-cm-diameter vertical pipe of Fig. P6.34. For the mercury manometer reading h = 42 cm shown, (a) calculate the volume flow rate in m3/h, and (b) state the direction of flow. Solution: For SAE 10 oil, take ρ = 870 kg/m3 and µ = 0.104 kg/m⋅s. The pressure at the lower point (1) is considerably higher than p2 according to the manometer reading:
Fig. P6.34
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 V1 = V2, gives
P6.35 A laminar flow element or LFE (Meriam Instrument Co.) measures low gas-flow rates with a bundle of capillary tubes packed inside a large outer tube. Consider oxygen at 20°C and 3
1 atm flowing at 2.5 m /min in a 0.1-m-diameter pipe. (a) Is the flow approaching the element turbulent? (b) If there are 1000 capillary tubes, L = 0.1 m, select a tube diameter to keep Red below 1500 and also to keep the tube pressure drop no greater than 3.5 kPa. (c) Do the tubes selected in part (b) fit nicely within the approach pipe?
20 2
2
Solution: For oxygen at 20°C and 1 atm (Table A.4), take R = 260 m /(s K), hence ρ = p/RT 3
= (101350Pa)/[260(293K)] = 1.33 kg/m . Also read µ = 2.0E-5 kg/m-s. Convert Q = 2.5 3
3
m /min = 0.0417 m /s. Then the entry pipe Reynolds number is
Re D =
ρVD 4 ρQ 4(1.33 kg / m 3 )(0.0417m 3 / s) = = = 35, 300 (turbulent) Ans.(a) µ πµ D π (2 × 10 –5 kg / m-s)(0.1 m)
(b) To keep Red below 1500 and keep the (laminar) pressure drop no more than 3.5 kPa,
Re d =
ρVd ≤ 1500 and µ
Δp =
32µ LV Q / 1000 ≤ 3.5 ×10 3 Pa, where V = 2 (π / 4)d 2 d
The upper limit on Reynolds number gives
Re d = 1500 if
d = 0.00235 m = 2.35 mm ; Δp = 111 Pa
Ans.(b)
This is a satisfactory answer, since the pressure drop is no problem, quite small. One thousand of these tubes would have an area about one-half of the pipe area, so would fit nicely. Ans.(c) Increasing the tube diameter would lower Red and have even smaller pressure drop. Example: d = 0.003 m, Red = 1180, Δp = 42 Pa. These 0.003-m-diameter tubes would just barely fit into the larger pipe. One disadvantage, however, is that these tubes are short: the entrance length is longer than the tube length, and thus Δp will be larger than calculated by “fully-developed” formulas.
6.36 SAE 30 oil at 20°C flows in the 3-cmdiameter pipe in Fig. P6.36, which slopes at 37°. For the pressure measure-ments shown, determine (a) whether the flow is up or down and (b) the flow ratein m3/h. Solution: For SAE 30 oil, take ρ = 891 kg/m3 and µ = 0.29 kg/m⋅s. Evaluate the hydraulic grade lines:
Fig. P6.36
21
The head loss is the difference between hydraulic grade levels:
Finally, check Re = 4ρQ/(πµd) ≈ 68 (OK, laminar flow).
P6.37 Water at 20°C is pumped from a reservoir through a vertical tube 3 m long and 1.6 mm in diameter. The pump provides a pressure rise of 75 kPa to the flow. Neglect entrance losses. (a) Calculate the exit velocity. (b) Approximately how high will the exit water jet rise? (c) Verify that the flow is laminar. Solution: For water at 20°C, Table A.3, ρ = 998 kg/m3, and µ = 0.001 kg/m-s. The energy equation, with 1 at the bottom and 2 at the top of the tube, is:
p1 V2 p V2 V2 32 µ LV2 75 × 103 + 1 + z1 = + 0 + 0 = 2 + 2 + z2 + h f = 0 + 2 + 3 + ρg 2g 998(9.81) ρg 2g 2g ρ gD 2 2 2 Vexit Vexit 32(0.001)(3)V or : 7.66 = + 3+ ; or : 4.66 = + 3.83Vexit 2(9.81) 19.62 (998)(9.81)(0.0016)2
(a, c) The velocity head is very small (<0.3 m), so the dominant term is 3.83 Vexit. One can easily iterate, or simply use EES to find the result:
Vexit = 1.2 m / s Ans.(a) ; Re D =
ρVD (998)(1.2)(0.0016) = = 1920 laminar Ans.(c) µ 0.001
(b) Assuming frictionless flow outside the tube, the jet would rise due to the velocity head:
H rise
2 Vexit (1.2 m / s)2 = = = 0.073 m 2g 2(9.81 m / s2 )
Ans.(b)
P6.38 Reconsider the analysis in Sect. 6.4 for laminar pipe flow with boundary slip. The no-slip boundary condition is replaced by a partial-slip condition
u = –λ
du dr
at r = R,
where R is the pipe radius, and λ ≥ 0 is a constant known as the slip length. (a) Show that the velocity profile is given by 2 r2 λ dp R u(r) = – 1 – + 2 dx 4 µ R R2
22
(b) Deduce a relationship between head difference and flow rate. (c) Show that the friction factor is given by
f =
64 Re d (1 + 4 λ / R)
Solution: Axial momentum equation
k d du r = – r µ dr dr ⇒ u(r) = –k
k =–
dp dx
r2 + C1 ln r + C2 4µ
B.C.
u(0) is finite ⇒ C1 = 0 u = –λu′ at r = R ⇒ ∴ u(r) =
Flow-rate
C2 =
KR 2 2λ 1 + 4µ R
2 KR 2 2λ r – 1 + 4 µ R R
Q = 2π ∫ oRurdr =
π KR 4 4λ 1 + 8 µ R Head loss But
ΔD KL 8µ LQ = = 4λ ρg ρg πρ gR 4 (1+ ) R 2 Q = π d V / 4, R = d/2
hf =
hf =
64 µ L V 2 1 ρVd d 2g (1 + 4λ /R)
L V2 To compare this with hf = f d 2g 64 get f = , Red (1 + 4λ /R)
Red =
ρVd µ
23
6.39 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
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
Now take the time-average of the entire equation to obtain Eq. (6.21) of the text:
P6.40 In the overlap layer of Fig. 6.9a, turbulent shear is large. If we neglect viscosity, we can replace Eq. (6.24) by the approximate velocity-gradient function
du = fcn( y, τ w , ρ ) dy Show that, by dimensional analysis, this leads to the logarithmic overlap relation (6.28). Solution:
There are four variables, and we may list their dimensions in the (MLT) system:
These can be formed into a single pi group that is therefore equal to a dimensionless constant:
Rearrange this into a differential equation and then integrate:
24
We recognize the square-root term as the friction velocity v* from Eq. (6.25). If the constants are rearranged so that the logarithm has a dimensionless argument, we would obtain Eq. (6.28):
6.41 The following turbulent-flow velocity data u(y), for air at 24°C and 1 atm near a smooth flat wall, were taken in the University of Rhode Island wind tunnel: y, mm:
0.635
0.889
1.194
1.397
1.651
u, m/s:
15.6
16.5
17.3
17.6
18.0
Estimate (a) the wall shear stress and (b) the velocity u at y = 5 mm. Solution: For air at 24°C and 1 atm, take ρ = 1.185 kg/m3 and µ = 1.82 × 10 –5 kg/m-s. We fit each data point to the logarithmic-overlap law, Eq. (6.28):
1 u 1 ρu*y 1.185u*y ≈ ln +B≈ ln + 5.0, u* = τ w /ρ µ 0.41 1.82 × 10 –5 u* κ Enter each value of u and y from the data and estimate the friction velocity u*: y, mm: u*, m/s: yu*/ν (approx):
0.635 1.091 45
0.889 1.091 63
1.194 1.094 85
1.397 1.085 99
1.651 1.085 117
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 u*avg ≈ 1.089 m/s ± 1%, τ w,avg = ρu*2 = (1.185)(1.089)2 ≈ 1.406 N / m2
Ans. (a)
(b) Out at y = 0.005 m, we may estimate that the log-law still holds:
ρu*y 1.185(1.089)(0.005) 1 = ≈ 355, u ≈ u* ln(355) + 5.0 –5 µ 1.82 × 10 0.41 or: u ≈ (1.089)(19.32) ≈ 21 m / s
Ans. (b)
Figure 6.10 shows that this point (y+ ≈ 355) seems also to be within the logarithmic layer.
25
6.42 Two infinite plates a distance h apart are parallel to the xz plane with the upper plate moving at speed V, as in Fig. P6.42. There is a fluid of viscosity µ and constant pressure between the plates. Neglecting gravity and assuming incompressible turbulent flow u(y) between the plates, use the logarithmic law and appropriate boundary conditions to derive a formula for dimensionless wall shear stress versus dimensionless plate velocity. Sketch a typical shape of the profile u(y).
Fig. P6.42
Solution: The shear stress between parallel plates is constant, so the centerline velocity must be exactly u = V/2 at y = h/2. Anti-symmetric log-laws form, one with increasing velocity for 0 < y < h/2, and a reverse mirror image for h/2 < y < h, as shown below:
The match-point at the center gives us a log-law estimate of the shear stress:
This is one form of “dimensionless shear stress.” The more normal form is friction coefficient versus Reynolds number. Calculations from the log-law fit a Power-law curve-fit expression in the range 2000 < Reh < 1E5:
26
6.43 Suppose in Fig. P6.42 that h = 3 cm, the fluid is water at 20°C (ρ = 998 kg/m3, µ = 0.001 kg/m⋅s), and the flow is turbulent, so that the logarithmic law is valid. If the shear stress in the fluid is 15 Pa, estimate V in m/s. Solution: Just as in Prob. 6.42, apply the log-law at the center between the wall, that is, y = h/2, u = V/2. With τw known, we can evaluate u* immediately:
6.44 By analogy with laminar shear, τ = µ du/dy. T. V. Boussinesq in 1877 postulated that turbulent shear could also be related to the mean-velocity gradient τturb = ε du/dy, where ε is called the eddy viscosity and is much larger than µ. If the logarithmic-overlap law, Eq. (6.28), is valid with τ ≈ τw, show that ε ≈ κρu*y. Solution: Differentiate the log-law, Eq. (6.28), to find du/dy, then introduce the eddy viscosity into the turbulent stress relation:
Note that ε/µ = κy+, which is much larger than unity in the overlap region.
6.45 Theodore von Kármán in 1930 theorized that turbulent shear could be represented by τ turb = ε du/dy where ε = ρκ 2y2du/dy is called the mixing-length eddy viscosity and κ ≈ 0.41 is Kármán’s dimensionless mixing-length constant [2,3]. Assuming that τ turb ≈ τw near the wall, show that this expression can be integrated to yield the logarithmic-overlap law, Eq. (6.28). Solution: This is accomplished by straight substitution:
To convert this to the exact form of Eq. (6.28) requires fitting to experimental data.
27
P6.46 Two reservoirs, which differ in surface elevation by 40 m, are connected by 350 m of new pipe of diameter 8 cm. If the desired flow rate is at least 130 N/s of water at 20°C, may the pipe material be (a) galvanized iron, (b) commercial steel, or (c) cast iron? Neglect minor losses. Solution: Applying the extended Bernoulli equation between reservoir surfaces yields Δz = 40 m
=
f
L V2 350 m V2 = f( ) D 2g 0.08 m 2(9.81m / s 2 )
where f and V are related by the friction factor relation:
1 ε/D 2.51 ≈ −2.0 log10 ( + ) 3.7 f Re D f
where
Re D =
ρVD µ
When V is found, the weight flow rate is given by w = ρgQ where Q = AV = (πD2/4)V. For water at 20°C, take ρ = 998 kg/m3 and µ = 0.001 kg/m-s. Given the desired w = 130 N/s, solve this system of equations by EES to yield the allowed wall roughness. The results are: f = 0.0257 ; V = 2.64 m/s ; ReD = 211,000 ; εmax = 0.000203 m = 0.203 mm Any less roughness is OK. From Table 6-1, the three pipe materials have (a) galvanized: ε = 0.15 mm ; (b) commercial steel: ε = 0.046 mm ; cast iron: ε = 0.26 mm Galvanized and steel are fine, but cast iron is too rough.. Ans. (a) galvanized: 135 N/s;
Actual flow rates are
(b) steel: 152 N/s; (c) cast iron: 126 N/s (not enough)
P6.47 Fluid flows steadily, at volume rate Q, through a large horizontal pipe and then divides into two small smooth pipes, the larger of which has an inside diameter of 25 mm and carries three times the flow of the smaller pipe. Both small pipes have the same length and pressure drop. If all flows are turbulent, at Red near 104, estimate the diameter of the smaller pipe. Solution: For turbulent flow, the formulas are algebraically complicated, such as Eq. (6.38). However, in the low Reynolds number region, the Blasius power-law approximation, Eq. (6.39), applies, leading to a simple approximate formula for pressure drop, Eq. (6.41):
Δp ≈ 0.241L ρ 3/ 4 µ1/ 4 d −4.75 Q1.75
28
Since Δp, L, ρ, and µ are the same for both pipes, it follows that
Q1.75 ∝ d 4.75 , or : Qlow turbulent ∝ d 4.75/1.75 = d 19/7 Thus, Q1 = const(d1 )19/7 = 3Q2 = 3(const)(d2 )19/7 , or : d2 =
d1 37/19
=
25mm = 16.7 mm 1.499
Ans.
This is slightly smaller than the laminar-flow estimate of Prob. P6.29, where d2 ≈ 19 mm.
P6.48 A reservoir supplies water through z1 = 35 m
100 m of 30-cm-diameter cast iron pipe to a
water at 20°C
turbine that extracts 60 kW from the flow. z2 = 5 m
turbine
The water then exhausts to the atmosphere. Neglect minor losses. (a) Assuming that
Fig. P6.48
f ≈ 0.019, find the flow rate (there is a cubic polynomial). Explain why there are two solutions. (b) For extra credit, solve for the flow rate using the actual friction factors. Solution:
For water at 20°C, take ρ = 998 kg/m3 and µ = 0.001 kg/m-s. The energy
equation yields a relation between elevation, friction, and turbine power: p1 V12 p V2 + + z1 = 2 + 2 + z2 + hturb + h f ρ g 2g ρ g 2g z1 − z2 = 35 − 5m = 30m = hturb
π Power L V22 + hf = + (1 + f ) , Q = D 2V2 ρ gQ 4 D 2g
(60 kW ) 100m V22 30 m = + [1 + (0.019) ] ] (9790 N / m 3 )(π / 4)(0.3m)2 V2 0.3m 2(9.81m / s 2 ) Clean this up into a cubic polynomial: 86.7 + 0.373V 2 , or : V 3 − 80.3V + 232 = 0 V Three roots : V = 3.36 m / s ; 6.79 m / s ; −10.15 m / s 30 =
29
The third (negative) root is meaningless. The other two are correct. Either Q
=
0.480 m3/s ,
hturbine = 12.8 m , hf = 17.2 m
Q
=
0.238 m3/s ,
hturbine = 25.8 m , hf = 4.2 m
Ans.(a)
Both solutions are valid. The higher flow rate wastes a lot of water and creates 17 meters of friction loss. The lower rate uses 51% less water and has proportionately much less friction. (b) The actual friction factors are very close to the problem’s “Guess”. Thus we obtain Re = 2.04E6, f = 0.0191;
Q = 0.478 m3/s ,
Re = 1.01E6, f = 0.0193 ; Q = 0.238 m3/s ,
hturbine = 12.8 m , hf = 17.2 m hturbine = 25.7 m , hf = 4.3 m Ans.(b)
The same remarks apply: The lower flow rate is better, less friction, less water used.
6.49 Mercury at 20°C flows through 4 meters of 7-mm-diameter glass tubing at an average velocity of 5 m/s. Estimate the head loss in meters and the pressure drop in kPa. Solution: For mercury at 20°C, take ρ = 13550 kg/m3 and µ = 0.00156 kg/m⋅s. Glass tubing is considered hydraulically “smooth,” ε/d = 0. Compute the Reynolds number:
P6.50 Water at 20°C is to be delivered through a 10-cm diameter smooth pipe with an average velocity of 0.1 m/s. (a) Check that the flow is turbulent. (b) Find the frictional head loss per length of pipe. (c) If the flow is mistaken to be laminar, and the formula f = 64/Red is used, how large would be the error in calculating the answer to (b)? Solution:
At 20°C V = 1 × 10 –6 m 2 /s d = 0.1 m, V = 0.1 m/s 0.1 × 0.1 Vd = = 10 4 > 2,300 ∴ Re = V 1 × 10 –6 The flow is turbulent 4
From Moody Chart, Re = 10 , smooth pipe
⇒
f = 0.031
30
∴ hf = f hf L
L V2 d 2g V2 0.12 = 0.031 × = 1.58 × 10 –4 2gd 2 × 9.81× 0.1
=f
If flow is mistaken to be laminar
f = then
hf L
64 = 64 × 10 −4 Re
= 64 ×10
−4
0.12 × = 3.26 × 10 –5 2 × 9.81× 0.1
which is only 20% of the true value.
3
6.51 Oil, SG = 0.88 and ν = 4E−5 m2/s, flows at 0.025 m /s through a 0.15-m asphalted cast-iron pipe. The pipe is 800 m long and slopes upward at 8° in the flow direction. Compute the head loss in m and the pressure change. Solution: For asphalted cast-iron, ε = 0.12 mm, hence ε/d = 0.00012/0.15 = 0.0008. Compute V, Red, and f:
V=
0.025 1.41(0.15) = 5288; calculate 2 = 1.41 m / s; Red = 4 × 10 –5 π (0.075) then h f = f
f Moody = 0.0377
2 L V2 800 (1.41) = 0.0377 = 20.4 m Ans. (a) 0.15 2(9.81) d 2g
If the pipe slopes upward at 8°, the pressure drop must balance both friction and gravity: Δp = ρ g(h f + Δz) = 0.88(9800)[20.4 + 800 sin 8°] = 1136 kPa
Ans. (b)
P6.52 Repeat Prob. P3.5, for the same data, by using the more exact turbulent flow formulas 3 to compute the volume flow rate in m /s. Recall the problem: Water at 20°C flows through a 12.5-cm-diameter smooth pipe at a centerline velocity of 7.62 m/s. Estimate the volume flow 3 rate in m /s. Solution: For water at 20°C, Table A.3, ρ = 998 kg/m3, and µ = 0.001 kg/m-s. A bit of iteration, or EES, is needed to get the proper Reynolds number and friction factor. Our estimate in Prob. P3.5 was Vav ≈ 6.4 m/s, whence the Reynolds number is
31
Re D =
ρVD (998 kg / m3 )(6.4 m / s)(0.125 m) = ≈ 798,000 µ 0.001 kg / m-s
Then estimate f smooth ≈ 0.0121 from Eq.(6.48) Eq.(6.43) : Vav = umax (1+1.3 f )−1 = (7.62 m / s)(0.875) ≈ 6.67 m / s
Iterate once to obtain fsmooth = 0.0120, Vav = 6.67 m/s, hardly any change at all. Q = Vav A pipe = (6.67 m / s)π (0.0625)2 = 0.082 m3 / s
Ans.
This is about 4.6% greater than our simple power-law estimate in Prob. P3.5.
6.53 The gutter and smooth drainpipe in Fig. P6.53 remove rainwater from the roof of a building. The smooth drainpipe is 7 cm in diameter. (a) When the gutter is full, estimate the rate of draining. (b) The gutter is designed for a sudden rainstorm of up to 10 cm per hour. For this condition, what is the maximum roof area that can be drained successfully? Solution: If the velocity at the gutter surface is neglected, the energy equation reduces to
Fig. P6.53
Δz =
V2 L V2 2gΔz 2(9.81)(4.2) + hf , hf = f , solve V 2 = = 2g d 2g 1+ fL/d 1+ f (4.2/0.07)
For water, take ρ = 998 kg/m3 and µ = 0.001 kg/m⋅s. Guess f ≈ 0.02 to obtain the velocity estimate V ≈ 6 m/s above. Then Red ≈ ρVd/µ ≈ (998)(6)(0.07)/(0.001) ≈ 428,000 (turbulent). Then, for a smooth pipe, f ≈ 0.0135, and V is changed slightly to 6.74 m/s. After convergence, we obtain
A rainfall of 10 cm/h = (0.1 m/h)/(3600 s/h) = 2.78 × 10 –5 m/s. The required roof area is Aroof = Qdrain /Vrain = (0.026 m 3/s)/ 2.78 × 10 –5 m/s ≈ 935 m2
Ans. (b)
32
6.54 Show that if Eq. (6.33) is accurate, the position in a turbulent pipe flow where local velocity u equals average velocity V occurs exactly at r = 0.777R, independent of the Reynolds number. Solution: Simply find the log-law position y+ where u+ exactly equals V/u*:
6.55 The tank-pipe system of Fig. P6.55 is to deliver at least 11 m3/h of water at 20°C to the reservoir. What is the maximum roughness height ε allowable for the pipe? Solution: For water at 20°C, take ρ = 998 kg/m3 and µ = 0.001 kg/m⋅s. Evaluate V and Re for the expected flow rate:
Fig. P6.55
The energy equation yields the value of the head loss:
With f and Re known, we can find ε/d from the Moody chart or from Eq. (6.48):
Then ε = 0.000394(0.03) ≈ 1.2E−5 m ≈ 0.012 mm (pretty smooth)
Ans.
33 3
6.56 Ethanol at 20°C flows at 8 × 10 –3 m /s through a horizontal cast-iron pipe with L = 12 m and d = 5 cm. Neglecting entrance effects, estimate (a) the pressure gradient, dp/dx; (b) the wall shear stress, τw; and (c) the percent reduction in friction factor if the pipe walls are polished to a smooth surface. Solution: For ethanol (Table A-3) take ρ = 789 kg/m3 and µ = 0.0012 kg/m⋅s. Evaluate V = Q/A = 0.008/[π (0.05)2/4] = 4.07 m/s.
Red =
ρVd 789(4.07)(0.05) ε 0.26 mm = = 133,800, = = 0.0052 Then f Moody ≈ 0.0314 µ 0.0012 d 50 mm (b) τ w = (a)
f 0.0314 ρV 2 = (789)(4.07)2 = 51 Pa Ans. (b) 8 8
dp 4τ −4(51) Pa =− w = = −4080 dx d 0.05 m
(c) Re = 133800,
Ans. (a)
fsmooth = 0.0169, hence the reduction in f is
0.0169 1 − = 46% Ans. (c) 0.0314
6.57 The viscous sublayer (Fig. 6.10) is normally less than 1 percent of the pipe diameter and therefore very difficult to probe with a finite-sized instrument. In an effort to generate a thick sublayer for probing, Pennsylvania State University in 1964 built a pipe with a flow of glycerin. Assume a smooth 30-cm-diameter pipe with V = 18 m/s and glycerin at 20°C. Compute the sublayer thickness in mm and the pumping horsepower required at 75 percent efficiency if L = 12 m. Solution: For glycerin at 20°C, take ρ = 1260 kg/m3 and µ = 1.5 kg/m-s. Then
Re =
ρVd 1260(18)(0.3) = = 4536 (barely turbulent!) Smooth: fMoody ≈ 0.0384 1.5 µ 0.0384 Then u* = V(f/8)1/2 = 18 8
1/2
≈ 1.25 m/s
The sublayer thickness is defined by y+ ≈ 5.0 = ρyu*/µ. Thus
y sublayer ≈
5µ 5(1.5) = = 0.00476 m ≈ 4.8 mm Ans. ρ u* (1260)(1.25)
With f known, the head loss and the power required can be computed: hf = f
P=
2 L V2 12 (18) = (0.0384) ≈ 25.37 m 0.3 2(9.81) d 2g
1 ρgQh f π = (1260)(9.81) (0.3)2 (18) (25.37) = 532 kW Ans. η 0.75 4
34
6.58 The pipe flow in Fig. P6.58 is driven by pressurized air in the tank. What gage pressure p1 is needed to provide a 20°C water flow rate Q = 60 m3/h? Solution: For water at 20°C, take ρ = 998 kg/m3 and µ = 0.001 kg/m⋅s. Get V, Re, f:
Fig. P6.58
Write the energy equation between points (1) (the tank) and (2) (the open jet):
[This is a gage pressure (relative to the pressure surrounding the open jet.)]
P6.59 Water at 20°C flows by gravity through a smooth pipe from one reservoir to a lower one. The elevation difference is 60 m. The pipe is 360 m long, with a diameter of 12 cm. Calculate the expected flow rate in m3/h. Neglect minor losses. Solution: For water at 20°C, Table A.3, ρ = 998 kg/m3 and µ = 0.001 kg/m-s. With no minor losses, the gravity head matches the Moody friction loss in the pipe:
The unknown is V, since f can be found as soon as the Reynolds number is known. You could iterate your way to the answer by, say, guessing f = 0.02, getting V, repeating. Or you could put the above four equations into EES, which will promptly return the correct answer:
35
6.60* A swimming pool W by Y by h deep is to be emptied by gravity through the long pipe shown in Fig. P6.60. Assuming an average pipe friction factor fav and neglecting minor losses, derive a formula for the time to empty the tank from an initial level ho.
Fig. P6.60
Solution: With no driving pressure and negligible tank surface velocity, the energy equation can be combined with a control-volume mass conservation:
We can separate the variables and integrate for time to drain:
6.61 The reservoirs in Fig. P6.61 contain water at 20°C. If the pipe is smooth with L = 4500 m and d = 4 cm, what will the flow rate in m3/h be for Δz = 100 m? Solution: For water at 20°C, take ρ = 998 kg/m3 and µ = 0.001 kg/m⋅s. The energy equation from surface 1 to surface 2 gives
Fig. P6.61
36
Iterate with an initial guess of f ≈ 0.02, calculating V and Re and improving the guess:
Alternately, one could, of course, use EES. The above process converges to
P6.62 The Alaska Pipeline in the chapter-opener photo has a design flow rate of 3 5 1.7 × 10 m /day of crude oil at 60°C (see Fig. A.1). (a) Assuming a galvanized-iron wall, estimate the total pressure drop required for the 1300 km trip. (b) If there are nine equally spaced pumps, estimate the power each pump must deliver. Solution: From Fig. A.1 for crude oil at 60°C, ρ = 860 kg/m3 and µ = 0.004 kg/m-s. The pipe diameter is 1.22 m. For galvanized iron, ε = 0.00015 m, hence ε /D = 0.00015/1.22 = 3
0.000123. Convert the data: 1.7 × 10 5 m /day = 1.97 m3/s, and D = 1.22 m. Calculate Reynolds number:
V =
Q 1.97 m3 / s m ρVD (860)(1.69)(1.22) = = 1.69 ; Re D = = = 443000 A (π / 4)(1.22m)2 s µ 0.004
Calculate the friction factor from Eq. (6.48): 1 f
≈ − 2.0 log10 (
0.000123 2.51 + ) 3.7 443000 f
Calculate f ≈ 0.0149
(a) The total 1300-km pressure drop is given by the usual Darcy-Moody expression, Eq. (6.10):
Δp = f
Lρ 2 (1300 × 103 ) 860 V = (0.0149)[ ]( )(1.69)2 = 1.95 × 107 Pa D2 1.22 2
Ans.(a)
(b) The power delivered by each of the 9 pumps is
Power = Q
Δptotal m3 1.95 × 107 Pa = (1.97 )( ) = 4268 kW 9 s 9
Ans.(b)
37
6.63 Apply the analysis of Prob. 6.60 to the following data. Let W = 5 m, Y = 8 m, ho = 2 m, L = 15 m, D = 5 cm, and ε = 0. (a) By letting h = 1.5 m and 0.5 m as representative depths, estimate the average friction factor. Then (b) estimate the time to drain the pool. Solution: For water, take ρ = 998 kg/m3 and µ = 0.001 kg/m⋅s. The velocity in Prob. 6.60 is calculated from the energy equation:
(a) With a bit of iteration for the Moody chart, we obtain ReD = 108,000 and f ≈ 0.0177 at h = 1.5 m, and ReD = 59,000 and f ≈ .0202 at h = 0.5 m; thus the average value fav ≈ 0.019. Ans. (a) The drain formula from Prob. 6.60 then predicts:
P6.64 For the system in Prob. P6.59, a pump is used at night to drive water back to the upper reservoir. If the pump delivers 15,000 W to the water, estimate the flow rate. Solution: For water at 20°C, Table A.3, ρ = 998 kg/m3 and µ = 0.001 kg/m-s. Since the pressures and velocities cancel, the energy equation becomes
The unknown is V, since f can be found as soon as the Reynolds number is known. You could iterate your way to the answer by, say, guessing f = 0.02, getting V (from a cubic equation!), and repeating. Or put the above five equations into EES, which returns the only positive answer:
6.65 The following data were obtained for flow of 20°C water at 20 m3/hr through a badly corroded 5-cm-diameter pipe which slopes downward at an angle of 8°: p1 = 420 kPa, z1 = 12 m, p2 = 250 kPa, z2 = 3 m. Estimate (a) the roughness ratio of the pipe; and (b) the percent change in head loss if the pipe were smooth and the flow rate the same.
38
Solution: The pipe length is given indirectly as L = Δz/sinθ = (9 m)/sin8° = 64.7 m. The steady flow energy equation then gives the head loss:
Now relate the head loss to the Moody friction factor:
The estimated (and uncertain) pipe roughness is thus ε = 0.0211d ≈ 1.06 mm Ans. (a) (b) At the same Red = 141000, fsmooth = 0.0168, or 66% less head loss. Ans. (b)
P6.66 In the spirit of Haaland’s explicit pipe friction factor approximation, Eq. (6.49), Jeppson [20] proposed the following explicit formula:
1 f
≈
−2.0 log10 (
ε /d 5.74 + ) 3.7 Re 0.9 d
(a) Is this identical to Haaland’s formula and just a simple rearrangement? Explain. (b) Compare Jeppson to Haaland for a few representative values of (turbulent) Red and ε/d and their deviations compared to the Colebrook formula (6.48). Solution: (a) No, it looks like a rearrangement of Haaland’s formula, but it is not. Haaland started with Colebrook’s smooth-wall formula and added just enough ε/d effect for accuracy. Jeppson started with the rough-wall formula and added just enough Red effect for accuracy. Both are excellent approximations over the full (turbulent) range of Red and ε/d. Their predicted values of f are nearly the same and very close to the implicit Colebrook formula. Here is a table of their standard deviations of their values when subtracted from Colebrook: 1E4 < Red < 1e8
ε/d = 0.03
0.01
0.001
0.0001
0.00001
Jeppson rms error
0.000398
0.000328
0.000195
0.000067
0.000088
Haaland rms error
0.000034
0.000043
0.000129
0.000113
0.000083
As expected, Jeppson is slightly better for smooth walls, Haaland for rough walls. Both are within ±2% of the Colebrook formula over the entire range of Red and ε/d.
39
6.67 What level h must be maintained in Fig. P6.67 to deliver a flow rate of 4 × 10 –4 m3/s through the 1.27-cm commercial-steel pipe?
Fig. P6.67
Solution: For water at 20°C, take ρ = 1000 kg/m3 and µ = 0.001 kg/m-s. For commercial steel, take ε ≈ 0.000046 m, or ε/d = 0.000046/0.0127 ≈ 0.0036. Compute V=
Re =
Q 4 × 10 –4 = = 3.16 m/s; A (π /4)(0.0127)2
ρVd 1000(3.16)(0.0127) = ≈ 40100 ε /d = 0.0036, fMoody ≈ 0.0302 µ 0.001
The energy equation, with p1 = p2 and V1 ≈ 0, yields an expression for surface elevation:
h = hf +
V2 V2 L (3.16)2 24 = 1 + f 1 + 0.0302 ≈ 29.5 m Ans. = 0.0127 2g 2g d 2(9.81)
6.68 Water at 20°C is to be pumped through 600 m of pipe from reservoir 1 to 2 at a rate of 0.08 m3/s, as shown in Fig. P6.68. If the pipe is cast iron of diameter 15 cm and the pump is 75 percent efficient, what power pump is needed?
Fig. P6.68
Solution: For water at 20°C, take ρ = 1000 kg/m3 and µ = 0.001 kg/m⋅s. For cast iron, take ε ≈ 0.26 mm, or ε/d = 0.00026/0.15 ≈ 0.0017. Compute V, Re, and f:
V=
Re =
Q 0.08 = = 4.53 m/s; A (π /4)(0.15)2
ρVd 1000(4.53)(0.15) = ≈ 680000 ε /d = 0.0017, fMoody ≈ 0.0227 µ 0.001
40
The energy equation, with p1 = p2 and V1 ≈ V2 ≈ 0, yields an expression for pump head:
h pump
2 L V2 600 (4.53) = Δz + f = 36 m + 0.0227 = 131 m 0.15 2(9.81) d 2g
Power: P =
ρgQh p 1000(9.81)(0.08)(131) = = 137 kW Ans. η 0.75
6.69 A tank contains 1 m3 of water at 20°C and has a drawn-capillary outlet tube at the bottom, as in Fig. P6.69. Find the outlet volume flux Q in m3/h at this instant. Solution: For water at 20°C, take ρ = 998 kg/m3 and µ = 0.001 kg/m⋅s. For drawn tubing, take ε ≈ 0.0015 mm, or ε/d = 0.0015/40 ≈ 0.0000375. The steady-flow energy equation, with p1 = p2 and V1 ≈ 0, gives
Fig. P6.69
P6.70 Find the size of a galvanized iron pipe that is designed to carry water at 10°C over a 3 distance of 200 m at 0.1 m /s under a head difference of 10 m. Solution: At 10°C, V = 1.31 × 10 –6 m2/s, L = 200 m, Q = 0.1 m3/s, hf = 10m, Galvanized iron pipe ⇒ t = 0.15 mm
Darcy eqn ⇒
hf = f
L V 2 8LQ 2 = f, d 2g π 2 gd 5
d 5 = C1 f ,
C1 =
8LQ 2 hf π 2g
41
Reynolds number Now
Re =
C2 , d
C2 =
4Q πv
8 × 200 × 0.12 = 0.0165 10 × π 2 × 9.81 4 × 0.1 = 9.72 ×10 4 C2 = –6 π ×1.31×10
C1 =
5 d = 0.0165 f 9.72 × 10 4 Re = ⇒ d –3 1 = –2.0 log 0.15 × 10 / d + 2.51 f 1/2 3.7 Re f 1/2
Solving these equations by EES. d = 0.1994 m,
f = 0.01909,
Hence, the required pipe diameter
Re = 487501 d = 0.2 m
6.71 Repeat Prob. 6.69 to find the flow rate if the fluid is SAE 10 oil. Is the flow laminar or turbulent? Solution: For SAE 10 oil at 20°C, take ρ = 870 kg/m3 and µ = 0.104 kg/m⋅s. For drawn tubing, take ε ≈ 0.0015 mm, or ε/d = 0.0015/40 ≈ 0.0000375. Guess laminar flow:
So it is laminar flow, and Q = (π/4)(0.04)2(4.33) = 0.00544 m3/s = 19.6 m3/h. Ans.
6.72 In Prob. 6.69 the initial flow is turbulent. As the water drains out of the tank, will the flow revert to laminar motion as the tank becomes nearly empty? If so, at what tank depth? Estimate the time, in h, to drain the tank completely. Solution: Recall that ρ = 998 kg/m3, µ = 0.001 kg/m⋅s, and ε/d ≈ 0.0000375. Let Z be the depth of water in the tank (Z = 1 m in Fig. P6.69). When Z = 0, find the flow rate:
42
So even when the tank is empty, the flow is still turbulent. Ans. The time to drain the tank is
So all we need is the average value of (1/Q) during the draining period. We know Q at Z = 0 and Z = 1 m, let’s check it also at Z = 0.5 m: Calculate Qmidway ≈ 19.8 m3/h. Then estimate, by Simpson’s Rule,
6.73 Ethyl alcohol at 20°C flows through a 10-cm horizontal drawn tube 100 m long. The fully developed wall shear stress is 14 Pa. Estimate (a) the pressure drop, (b) the volume flow rate, and (c) the velocity u at r = 1 cm. Solution: For ethyl alcohol at 20°C, ρ = 789 kg/m3, µ = 0.0012 kg/m⋅s. For drawn tubing, take ε ≈ 0.0015 mm, or ε/d = 0.0015/100 ≈ 0.000015. From Eq. (6.12),
The wall shear is directly related to f, and we may iterate to find V and Q:
Then V ≈ 3.00 m/s, and Q = (π/4)(0.1)2(3.00) = 0.0236 m3/s = 85 m3/h. Ans. (b) Finally, the log-law Eq. (6.28) can estimate the velocity at r = 1 cm, “y” = R − r = 4 cm:
43
6.74 A straight 10-cm commercial-steel pipe is 1 km long and is laid on a constant slope of 5°. Water at 20°C flows downward, due to gravity only. Estimate the flow rate in m3/h. What happens if the pipe length is 2 km? Solution: For water at 20°C, take ρ = 998 kg/m3 and µ = 0.001 kg/m⋅s. If the flow is due to gravity only, then the head loss exactly balances the elevation change:
Thus the flow rate is independent of the pipe length L if laid on a constant slope. Ans. For commercial steel, take ε ≈ 0.046 mm, or ε/d ≈ 0.00046. Begin by guessing fully-rough flow for the friction factor, and iterate V and Re and f:
*6.75 The Moody chart, Fig. 6.13, is best for finding head loss (or Δp) when Q, V, d, and L are known. It is awkward for the “2nd” type of problem, finding Q when hf or Δp are known (see Ex. 6.9). Prepare a modified Moody chart whose abscissa is independent of Q and V, using ε/d as a parameter, from which one can immediately read the ordinate to find (dimensionless) Q or V. Use your chart to solve Example 6.9. Solution: This problem was mentioned analytically in the text as Eq. (6.51). The proper parameter which contains head loss only, and not flow rate, is ζ: Eq. (6.51) We simply plot Reynolds number versus ζ for various ε/d, as shown below:
44
To solve Example 6.9, a 100-m-long, 30-cm-diameter pipe with a head loss of 8 m and ε/d = 0.0002, we use that data to compute ζ = 5.3E7. The oil properties are ρ = 950 kg/m3 and ν = 2E−5 m2/s. Enter the chart above: let’s face it, the scale is very hard to read, but we estimate, at ζ = 5.3E7, that 6E4 < Red < 9E4, which translates to a flow rate of 0.28 < Q < 0.42 m3/s. Ans. (Example 6.9 gave Q = 0.342 m3/s.) 6.76 For Prob. 6.68 suppose the only pump available can deliver only 60 kW to the fluid. What is the proper pipe size in cm to maintain the 0.08 m3/s flow rate? Solution: For water at 20°C, take ρ = 1000 kg/m3 and µ = 0.001 kg/m-s. For cast iron, take ε ≈ 0.26 mm. We can’t specify ε/d because we don’t know d. The energy analysis above is correct and should be modified to replace V by Q:
h p = 36 + f But also h p =
L (4Q/π d 2 )2 600 [4(0.08)/π d 2 ]2 f = 36 + f = 36 + 0.317 5 d 2g d 2(9.81) d
Power 60 × 10 3 0.317 f = = 76.45 = 36 + , or: d 5 ≈ 0.00784f 5 ρgQ 9810(0.08) d
Guess f ≈ 0.02, calculate d, ε/d and Re and get a better f and iterate:
f ≈ 0.020, d ≈ [0.00784(0.02)]1/5 ≈ 0.173 m, Re = or Re ≈ 589000,
4 ρQ 4(1000)(0.08) = , πµd π (0.001)(0.173)
ε 0.00026 = ≈ 0.0015, Moody chart: fbetter ≈ 0.022 (repeat) d 0.173
We are nearly converged. The final solution is f ≈ 0.022, d ≈ 0.177 m ≈ 18 cm Ans.
P6.77 Ethylene glycol at 20°C flows through 80 meters of cast iron pipe of diameter 6 cm. The measured pressure drop is 250 kPa. Neglect minor losses. Using a non-iterative formulation, estimate the flow rate in m3/h. Solution: For ethylene glycol at 20°C, Table A.3, ρ = 1117 kg/m3 and µ = 0.0214 kg/m-s. The head loss is given: Δp/ρg = 250,000/[1117(9.81)] = 22.8 m. For cast iron, ε = 0.26 mm, hence the roughness ratio is ε/d = 0.26/60 = 0.00433. We can use the direct approach of Eq. (6.51):
With the Reynolds number known, we can find the velocity and flow rate:
45
6.78 It is desired to solve Prob. 6.68 for the most economical pump and cast-iron pipe system. If the pump costs $200 per kW delivered to the fluid and the pipe costs $3000 per cm of diameter, what are the minimum cost and the pipe and pump size to maintain the 0.08 m3/s flow rate? Make some simplifying assumptions. Solution: For water at 20°C, take ρ = 1000 kg/m3 and µ = 0.001 kg/m-s. For cast iron, take ε ≈ 0.00026 m. Write the energy equation (from Prob. 6.68) in terms of Q and d: Pin kW =
2 2 9810(0.08) ρgQ 0.25f 600 [4(0.08)/π d ] (Δz + h f ) = 36 + f = 28.25 + 5 2(9.81) 1000 d 1000 d
Cost = $200PkW + $3000d cm = 200(28.25 + 0.25f/d 5 ) + 3000(100d), with d in m. Clean up: Cost ≈ $5650 + 50f/d 5 + 300000d Regardless of the (unknown) value of f, this Cost relation does show a minimum. If we assume for simplicity that f is constant, we may use the differential calculus:
d(Cost) |f ≈const = −5(50)f + 3 ×10 5 , or d best ≈ (0.307 f)1/6 6 d(d) d 4 ρQ ε ≈ 636600, ≈ 0.00163 πµd d ≈ 0.0225, d better ≈ 0.163 m (converged)
Guess f ≈ 0.02, d ≈ 0.16 m, Re = Then fbetter
Result: dbest ≈ 0.163 m ≈ 16.3 cm, Costmin ≈ $15400pump + $48900pipe ≈ $64,300. Ans.
6.79 Modify Prob. P6.63 by letting the diameter be unknown. Find the proper pipe diameter for which the pool will drain in about 2 hours flat. Solution: Recall the data: Let W = 5 m, Y = 8 m, ho = 2 m, L = 15 m, and ε = 0, with water, ρ = 998 kg/m3 and µ = 0.001 kg/m⋅s. We apply the same theory as Prob. 6.57: V=
2gh 4WY , t drain ≈ 1+ fL/D π D2
2ho (1+ fav L/D) , g
fav = fcn(Re D ) for a smooth pipe.
For the present problem, tdrain = 2 hours and D is the unknown. Use an average value h = 1 m to find fav. Enter these equations on EES (or you can iterate by hand) and the final results are
*6.80 The Moody chart, Fig. 6.13, is best for finding head loss (or Δp) when Q, V, d, and L are known. It is awkward for the “3rd” type of problem, finding d when hf (or Δp) and Q are known (see Ex. 6.10). Prepare a modified Moody chart whose abscissa is independent of d, using as a parameter ε non-dimensionalized without d, from which one can immediately read the (dimensionless) ordinate to find d. Use your chart to solve Ex. 6.10.
46
Solution: An appropriate Pi group which does not contain d is β = (ghfQ3)/(Lν5). Similarly, an appropriate roughness parameter without d is σ = (εν/Q). After a lot of algebra, the Colebrook friction factor formula (6.48) becomes
A plot of this messy relation is given below.
To solve Example 6.10, a 100-m-long, unknown-diameter pipe with a head loss of 8 m, flow rate of 0.342 m3/s, and ε = 0.06 mm, we use that data to compute β = 9.8E21 and σ = 3.5E−6. The oil properties are ρ = 950 kg/m3 and ν = 2E−5 m2/s. Enter the chart above: let’s face it, the scale is very hard to read, but we estimate, at β = 9.8E21 and σ = 3.5E−6, that 6E4 < Red < 8E4, which translates to a diameter of 0.27 < d < 0.36 m. Ans. (Example 6.10 gave d = 0.3 m.) P6.81 Two reservoirs, which differ in surface elevation by 40 m, are connected by a new commercial steel pipe of diameter 8 cm. If the desired weight flow rate is 200 N/s of water at 20°C, what is the proper length of the pipe? Neglect minor losses. Solution: For water at 20°C, take ρ = 998 kg/m3 and µ = 0.001 kg/m-s. For commercial steel, ε = 0.046 mm, thus ε/d = 0.046mm/80mm = 0.000575. Find the velocity and the friction factor:
V=
w / (ρ g) 200 / [998(9.81)] m ρVD 998(4.06)(0.08) = = 4.06 , Re D = = = 324, 000 2 2 s 0.001 µ (π / 4)(0.08) (π / 4)D 1 2.51 ε/D ≈ −2.0 log10 ( + ) yields f = 0.0185 3.7 f Re D f
Then we find the pipe length from the energy equation, which is simple in this case: L V2 L (4.06)2 Δz = 40 m = f = (0.0185) , Solve L ≈ 205 m D 2g (0.08m) 2(9.81)
Ans.
47
6.82 You wish to water your garden with 30 m of 1.6 cm-diameter hose whose roughness is 0.03 cm. What will be the delivery, in m3/s, if the gage pressure at the faucet is 400 kPa? If there is no nozzle (just an open hose exit), what is the maximum horizontal distance the exit jet will carry?
Fig. P6.82
Solution: For water, take ρ = 1000 kg/m3 and µ = 0.001 kg/m-s. We are given ε/d = 0.03/1.6 ≈ 0.0188. For constant area hose, V1 = V2 and energy yields
p faucet 400 × 10 3 Pa L V2 30 V2 = h f , or: = 40.77 m = f =f , ρg 1000(9.81) d 2g 0.016 2(9.81)
or fV2 ≈ 0.427. Guess f ≈ ffully rough = 0.048, V ≈ 3 m / s, Re ≈ 48000 then fbetter ≈ 0.0483, Vfinal ≈ 2.97m/s (converged) The hose delivery then is Q = (π/4)(0.016)2(2.97) = 0.0006 m3/s. Ans. (a) From elementary particle-trajectory theory, the maximum horizontal distance X traveled by the jet occurs at θ = 45° (see figure) and is X = V2/g = (2.97)2/(9.81) ≈ 0.9 m Ans. (b), which is pitiful. You need a nozzle on the hose to increase the exit velocity.
6.83 The small turbine in Fig. P6.83 extracts 400 W of power from the water flow. Both pipes are wrought iron. Compute the flow rate Q m3/h. Why are there two solutions? Which is better?
Solution: For water, take ρ = 998 kg/m3 and µ = 0.001 kg/m⋅s. For wrought iron, take ε ≈ 0.046 mm, hence ε/d1 = 0.046/60 ≈ 0.000767 and ε/d2 = 0.046/40 ≈ 0.00115. The energy equation, with V1 ≈ 0 and p1 = p2, gives
48
If we rewrite the energy equation in terms of Q and multiply by Q, it is essentially a cubic polynomial, because for these rough walls the friction factors are almost constant:
Solve by EES or by iteration. There are three solutions, two of which are positive and the third is a meaningless negative value. The two valid (positive) solutions are: (a) Q = 0.00437 m3 / s = 15.7m3 / hr ; Re1 =92,500, f1 =0.0215 ; Re1 =138,800, f1 =0.0221 (b) Q = 0.00250 m3 / s = 9.0 m3 / hr ; Re1 =52,900, f1 =0.0232 ; Re1 =79,400, f1 =0.0232 Ans.
[The negative (meaningless) solution is Q = - 0.0069 m3/hr.] Both solutions (a) and (b) are valid mathematically. Solution (b) is preferred – the same power for 43% less water flow, and the turbine captures 16.3 m of the available 20 m head. Solution (a) is also unrealistic, because a real turbine’s power increases with water flow rate. Turbine (a) would generate more than 400 W. 6.84 Modify Prob. 6.83 into an economic analysis, as follows. Let the 40 m of wrought-iron pipe have a uniform diameter d. Let the steady water flow available be Q = 30 m3/h. The cost of the turbine is $4 per watt developed, and the cost of the piping is $75 per centimeter of diameter. The power generated may be sold for $0.08 per kilowatt hour. Find the proper pipe diameter for minimum payback time, i.e., minimum time for which the power sales will equal the initial cost of the system. Solution: With flow rate known, we need only guess a diameter and compute power from the energy equation similar to Prob. 6.83:
The Moody friction factor is computed from Re = 4ρ Q/(πµd) and ε/d = 0.066/d(mm). The payback time, in years, is then the cost divided by the annual income. For example, If d = 0.1 m, Re ≈ 106000, f ≈ 0.0200, ht ≈ 19.48 m, P = 1589.3 W Cost ≈ $7107 Income = $1,114/year Payback ≈ 6.38 years Since the piping cost is very small (<$1000), both cost and income are nearly proportional to power, hence the payback will be nearly the same (6.38 years) regardless of diameter. There is an almost invisible minimum at d ≈ 7 cm, Re ≈ 151000, f ≈ 0.0201, ht ≈ 17.0 m, Cost ≈ $6078, Income ≈ $973, Payback ≈ 6.25 years. However, as diameter d decreases, we generate less power and gain little in payback time.
49
6.85 In Fig. P6.85 the connecting pipe is commercial steel 6 cm in diameter. Estimate the flow rate, in m3/h, if the fluid is water at 20°C. Which way is the flow? Solution: For water, take ρ = 998 kg/m3 and µ = 0.001 kg/m⋅s. For commercial steel, take ε ≈ 0.046 mm, hence ε/d = 0.046/60 ≈ 0.000767. With p1, V1, and V2 all ≈ 0, the energy equation between surfaces (1) and (2) yields
Fig. P6.85
The iteration converges to f ≈ 0.0205, V ≈ 2.49 m/s, Q = (π/4)(0.06)2(2.49) = 0.00705 m3/s = 25 m3/h ← Ans.
6.86 A garden hose is used as the return line in a waterfall display at the mall. In order to select the proper pump, you need to know the hose wall roughness, which is not supplied by the manufacturer. You devise a simple experiment: attach the hose to the drain of an aboveground pool whose surface is 3 m above the hose outlet. You estimate the minor loss coefficient in the entrance region as 0.5, and the drain valve has a minor-loss equivalent length of 200 diameters when fully open. Using a bucket and stopwatch, you open the valve and measure a flow rate of 2 × 10 –4 m3/s for a hose of inside diameter 1.5 cm and length 10 m. Estimate the roughness height of the hose inside surface. Solution: First evaluate the average velocity in the hose and its Reynolds number:
Write the energy equation from surface (point 1) to outlet (point 2), assuming an energy correction factor α = 1.05:
50
The unknown is the friction factor:
For f = 0.0514 and Red = 16940, the Moody chart (Eq. 6.48) predicts ε/d ≈ 0.0206. Therefore the estimated hose-wall roughness is ε = 0.0206(1.5 cm) = 0.031 cm Ans. 6.87 The head-versus-flow-rate characteristics of a centrifugal pump are shown in Fig. P6.87. If this pump drives water at 20°C through 120 m of 30-cm-diameter cast-iron pipe, what will be the resulting flow rate, in m3/s? Solution: For water, take ρ = 998 kg/m3 and µ = 0.001 kg/m⋅s. For cast iron, take ε ≈ 0.26 mm, hence ε/d = 0.26/300 ≈ 0.000867. The head loss must match the pump head:
Fig. P6.87
6.88 The pump in Fig. P6.87 is used to deliver gasoline at 20°C through 350 m of 30-cm-diameter galvanized iron pipe. Estimate the resulting flow rate, in m3/s. (Note that the pump head is now in meters of gasoline.) Solution: For gasoline, take ρ = 680 kg/m3 and µ = 2.92E−4 kg/m⋅s. For galvanized iron, take ε ≈ 0.15 mm, hence ε/d = 0.15/300 ≈ 0.0005. Head loss matches pump head:
This converges to f ≈ 0.0168, Re ≈ 5.96E6, Q ≈ 0.603 m3/s. Ans.
51
P6.89 Fluid at 20°C flows through a horizontal galvanized-iron pipe 20 m long and 8 cm in diameter. The wall shear stress is 90 Pa. Calculate the flow rate in m3/h if the fluid is (a) glycerin; and (b) water. Solution: (a) For glycerin, take ρ = 1260 kg/m3 and µ = 1.49 kg/m⋅s. For galvanized iron, take ε ≈ 0.15 mm, hence ε/D = 0.15/80 ≈ 0.001875. But we are guessing this flow is laminar:
(b) For water, take ρ = 998 kg/m3 and µ = 0.001 kg/m⋅s. For galvanized iron, take ε ≈ 0.15 mm, hence ε/D = 0.15/80 ≈ 0.001875. Now we are guessing this flow is turbulent. At that roughness, the minimum friction factor is 0.023, which we can use for a first estimate:
The Reynolds number estimate is certainly high enough, and into the fully-rough region of the Moody chart. Iterate briefly to the final result, only slightly different:
6.90 For the system of Fig. P6.61, let Δz = 80 m and L = 185 m of cast-iron pipe. What is the pipe diameter for which the flow rate will be 7 m3/h? Solution: For water, take ρ = 998 kg/m3 and µ = 0.001 kg/m⋅s. For cast iron, take ε ≈ 0.26 mm, but d is unknown. The energy equation is simply Fig. P6.61
Iterate: fbetter ≈ 0.0372, dbetter ≈ 0.0306 m, Rebetter ≈ 80700, ε/d|better ≈ 0.00850, etc. The process converges to f ≈ 0.0367, d ≈ 0.0305 m. Ans.
52
6.91 It is desired to deliver 60 m3/h of water (ρ = 998 kg/m3, µ = 0.001 kg/m⋅s) at 20°C through a horizontal asphalted cast-iron pipe. Estimate the pipe diameter which will cause the pressure drop to be exactly 40 kPa per 100 meters of pipe length. Solution: Write out the relation between Δp and friction factor, taking “L” = 100 m:
Knowing ε = 0.12 mm, then ε/d = 0.00012/d and Red = 4ρQ/(πµd) = 21178/d. Use EES, or guess f ≈ 0.02 and iterate until the proper diameter and friction factor are found. Final convergence: f ≈ 0.0216; Red ≈ 204,000; d = 0.104 m. Ans.
P6.92 For the system in Prob. P6.59, a pump, which delivers 15,000 W to the water, is used at night to refill the upper reservoir. The pipe diameter is increased from 12 cm to provide more flow. If the resultant flow rate is 90 m3/h, estimate the new pipe size. Solution: For water at 20°C, Table A.3, ρ = 998 kg/m3 and µ = 0.001 kg/m-s. Recall that Δz = 60 m and L = 360 m. Since the pressures and velocities cancel, the energy equation becomes zlower = zupper + h f − hpump , or : hp = Δz + h f = 60m + f where Q =V
1 f
≈ 2.0 log10 (
360m V 2 D 2(9.81)
Re D f 4ρQ 4(998)(90 / 3600) ρVD ) , Re D = = , 2.51 µ πµ D π (0.001)D
m3 m3 π 2 D = 90 = 0.025 , Power = ρ g Q hp = (998)(9.81)Q hp = 15,000 W 4 h s
You could solve by iteration, guessing values of D greater than 12 cm, until Q = 90 m3/h. Or you could put the above equations into EES, which will report the answer:
Thus a 57% increase in diameter only produces a 13% increase in flow rate. Even with an indefinitely large diameter, because of the 60-meter elevation head to fight against, Q can never be greater than 92 m3/h if P = 15 kW.
53 3
P6.93 A 0.2-m-diameter pipe carried a discharge of 0.055 m /s when it was new. After one year, the discharge under the same head difference was found to have decreased by 5% because of increase in wall roughness due to deposits and corrosion. The pipe has effective roughness element ε0 = 0.02 mm when new. The annual growth of these elements is assumed to be linear and given by ε = ε0 (1 + kt) where k is a constant and t is the number of years of use of the pipe. (a) Determine the head gradient along the pipe. (b) Find the value of k. (c) Estimate the percentage decrease in the discharge after 10 years of use of the pipe. You may assume that the head gradient remains unchanged with time. Solution:
Darcy equation
hf = f
L V2 8LQ 2 =f 2 5 d 2g π gd
8 fQ 2 Hence head gradiant = 2 5 L π gd G0 0.02 When new = = 1×10 –4 , d 200 ⇒ Moody Chart fo = 0.0151 hf
∴
hf / L =
4Q0 4 × 0.055 = = 3.5 ×10 5 –6 π dV π × 0.2 ×1×10
8 × 0.0151× 0.055 2 = 0.0118 π 2 × 9.81× 0.2 5
One year later ⇒ Also,
Re 0 =
f1Q12 = f0Q02
Q1 = 0.95Q0
f1 = 0.0151 × 0.95 –2 = 0.0167 Re1 Q1–1 = Re 0 Q0–1
⇒ Re1 = 3.5 ×10 5 × 0.95 = 3.325 ×10 5 From Moody Chart G1 / d = 0.00029 ⇒ G = 0.058 mm Therefore k = ε1 / ε 0 – 1 = 1.9
Ten years later ∴
G10 = 0.02(1+1.9 ×10) = 0.4 mm
G10 / d = 0.4 / 200 = 0.002 2 f10Q10 = f0Q02
⇒
–1 Re10Q10 = Re 0 Q0–1
and
Re10 = Re 0 ( f0 / f10 )1/2 = 3.5 ×10 5 (0.0151 / f10 )1/2
Solving by iteration or by EES (with Eq. (6.48)) Re10 = 2.77 ×10 5
get
f10 = 0.0241
∴
0.0151 Q10 = 0.055 0.0241
1/2
= 0.044 m 3 /s
which is 20% lower than when the pipe was new.
54
6.94 SAE 10 oil at 20°C flows at an average velocity of 2 m/s between two smooth parallel horizontal plates 3 cm apart. Estimate (a) the centerline velocity, (b) the head loss per meter, and (c) the pressure drop per meter. Solution: For SAE 10 oil, take ρ = 870 kg/m3 and µ = 0.104 kg/m⋅s. The half-distance between plates is called “h” (see Fig. 6.37). Check Dh and Re:
The head loss and pressure drop per meter follow from laminar theory, Eq. (6.63):
6.95 A commercial-steel annulus 12 m long, with a = 2.5 cm and b = 1 cm, connects two reservoirs which differ in surface height by 6 m. Compute the flow rate in m3/s through the annulus if the fluid is water at 20°C. Solution: For water, take ρ = 1000 kg/m3 and µ = 0.001 kg/m-s. For commercial steel, take ε ≈ 0.046 mm. Compute the hydraulic diameter of the annulus: 4A = 2(2.5 −1) = 3 cm; P 12 V2 L V2 = f , or: fV2 ≈ 0.2943 hf = 6 m = f 0.03 2(9.81) Dh 2g Dh =
We can make a reasonable estimate by simply relating the Moody chart to Dh, rather than the more complicated “effective diameter” method of Eq. (6.77). Thus
0.000046 ε = ≈ 0.0015, Guess frough ≈ 0.023, V = (0.2943/0.023)1/2 ≈ 3.6 m/s Dh 0.03 Re =
ρVD h 1000(3.6)(0.03) = ≈ 108000, fbetter ≈ 0.0236, Vbetter ≈ 3.53 m/s µ 0.001
This converges to f ≈ 0.0236, V ≈ 3.53 m/s, Q = π(a2 − b2)V = 0.0058 m3/s. Ans.
55
6.96 An oil cooler consists of multiple parallel-plate passages, as shown in Fig. P6.96. The available pressure drop is 6 kPa, and the fluid is SAE 10W oil at 20°C. If the desired total flow rate is 900 m3/h, estimate the appropriate number of passages. The plate walls are hydraulically smooth.
Fig. P6.96
Solution: For SAE 10W oil, ρ = 870 kg/m3 and µ = 0.104 kg/m⋅s. The pressure drop remains 6 kPa no matter how many passages there are (ducts in parallel). Guess laminar flow, Eq. (6.63),
where h is the half-thickness between plates. If there are N passages, then b = 50 cm for all and h = 0.5 m/(2N). We find h and N such that NQ = 900 m3/h for the full set of passages. The problem is ideal for EES, but one can iterate with a calculator also. We find that 18 passages are one too many—Q only equals 835 m3/h. The better solution is:
N = 17 passages, QN = 935 m 3/h, h = 1.47 cm, Re Dh = 512 (laminar flow) P6.97 Reconsider laminar flow through a parallel-plate channel as shown in Fig. 6.14. Show that, when the walls have a slip length λ, the friction factor is given by
f =
96 Re Dh (1 + 3λ / h)
Solution: For laminar flow through a parallel-plate channel of height 2h with wall slip, the velocity profile is
u(y) = k
2 h2 y 2λ 1 − + 2µ h h
Flow-rate Q = k
where k = −
h 3 2 2λ + µ 3 h
Hydraulic diameter
D h = 4h
Reynolds number
Re Dh =
ρVDh µ
dp dx
56
Head loss
But
hf =
Q = V (2h),
Δp KL µ LQ = = 2 ρg ρg ρ gh 3 + 2λ / h) 3 h = Dh /4
96µ L V 2 1 Hence h f = ρVDh Dh 2g (1+ 3λ / h) Comparing this with h f = f get
f =
L V2 D 2g
96 Re Dh (1+ 3λ / h)
6.98 An annulus of narrow clearance causes a very large pressure drop and is useful as an accurate measurement of viscosity. If a smooth annulus 1 m long with a = 50 mm and b = 49 mm carries an oil flow at 0.001 m3/s, what is the oil viscosity if the pressure drop is 250 kPa? Solution: Assuming laminar flow, use Eq. (6.73) for the pressure drop and flow rate:
P6.99 A rectangular sheet-metal duct is 60 m long and has a fixed height H = 15 cm. The width B, however, may vary from 15 to 90 cm. A blower provides a pressure drop of 80 Pa of air at 20°C and 1 atm. What is the optimum width B that will provide the most airflow in m3/s? Solution: For air at 20°C and 1 atm, take ρ = 1.20 kg/m3 and µ = 1.8E-5 kg/m-s. The pressure drop is related to the hydraulic diameter of the duct. L = 60 m. For sheet metal, from Table 6.1, the roughness ε = 0.05 mm.
Δp = f
L ρ 2 4A 2BH ε V , where Dh = = , f related to Re Dh and Dh 2 P B+ H Dh
Solve for V =
2 Δp Dh = f ρL
2(80 Pa) Dh Dh m = 1.491 which gives V in f (1.20)(60) f s
57
The duct area A = 2BH increases with B for a fixed H, and so does the hydraulic diameter. The Reynolds number (ρVDh/µ) also increases, hence the friction factor f decreases. All of these factors make the flow rate Q increase with H. Therefore, without even making calculations, we conclude that the widest B (90 cm) produces the most flow rate. Ans. We can calculate the actual flow rate for B = 0.9 m: H = 0.15 m, Dh = 0.257 m , Re Dh = 93,500 ,
ε = 0.000195 , f = 0.0192, Dh
m3 m Giving V = 5.45 , Q = V B H = 0.736 s s
Ans.
6.100 Heat exchangers often consist of many triangular passages. Typical is Fig. P6.100, with L = 60 cm and an isosceles-triangle cross section of side length a = 2 cm and included angle β = 80°. If the average velocity is V = 2 m/s and the fluid is SAE 10 oil at 20°C, estimate the pressure drop.
Fig. P6.100
Solution: For SAE 10 oil, take ρ = 870 kg/m3 and µ = 0.104 kg/m⋅s. The Reynolds number based on side length a is Re = ρVa/µ ≈ 335, so the flow is laminar. The bottom side of the triangle is 2(2 cm)sin40° ≈ 2.57 cm. Calculate hydraulic diameter:
58
6.101 A large room uses a fan to draw in atmospheric air at 20°C through a 30 cm by 30 cm commercial-steel duct 12 m long, as in Fig. P6.101. Estimate (a) the air flow rate in m3/hr if the room pressure is 10 Pa vacuum; and (b) the room pressure if the flow rate is 1200 m3/hr. Neglect minor losses.
Fig. P6.101
Solution: For air, take ρ = 1.2 kg/m3 and µ = 1.8E−5 kg/m⋅s. For commercial steel, ε = 0.046 mm. For a square duct, Dh = side-length = 30 cm, hence ε/d = 0.046/300 = 0.000153. The (b) part is easier, with flow rate known we can evaluate velocity, Reynolds number, and friction factor:
Then the pressure drop follows immediately:
(a) If Δp = 10 Pa (vacuum) is known, we must iterate to find friction factor:
After iteration, the results converge to: V = 4.69 m/s; Red = 93800; f = 0.0190; Q = 0.422 m3/s = 1520 m3/h Ans. (a)
P6.102 In Moody’s Example 6.6, the 15-cm diameter, 60-m-long asphalted cast iron pipe has a pressure drop of about 13 kPa when the average water velocity is 1.8 m/s. Compare this to an annular cast iron pipe with an inner diameter of 15 cm and the same annular average velocity of 1.8 m/s. (a) What outer diameter would cause the flow to have the same pressure drop of 13 kPa? (b) How do the cross-section areas compare, and why? Use the hydraulic diameter approximation. Solution: Recall the Ex. 6.6 data, ε = 0.00012 m. For water at 20°C, take ρ = 1000 kg/m3 and µ = 0.001 kg/m-s. The hydraulic diameter of an annulus is Dh = 2(Ro – Ri), where Ri = 0.075 m. We know the pressure drop, hence the head loss is
L V2 60 Δp 13 ×10 3 Pa (1.8 m / s)2 hf = f = f = = = 1.325 m Dh 2g 2(Ro − 0.075) 2 × 9.81 m / s 2 ρ g 1000 × 9.81N / m 3
59
We do not know f or Ro. The additional relation is the Moody friction factor correlation:
1 ε / Dh 2.51 ≈ − 2.0 log10 ( + ) 3.7 f Re Dh f
where
Re Dh =
ρVDh (1000)(1.8)[2(Ro − 0.075)] = µ 0.001
(a) For ε = 0.00012 m, solve these two simultaneously, using EES or Excel, to obtain
f = 0.01985 ; Re Dho = 267, 000 ; Ro = 0.149 m
Ans.(a)
(b) The annular gap is 0.149 – 0.075 = 0.074 m, just about equal to the inner radius. However, the annular area is three times the area of Moody’s pipe! Ans .(b) The annular pipe has much more wall area than a hollow pipe, more friction, so more cross-section area is needed to match the pressure drop.
P6.103 Air at 20°C flows through a smooth duct of diameter 20 cm at an average velocity of 5 m/s. It then flows into a smooth square duct of side length a. Find the square duct size a for which the pressure drop per meter will be exactly the same as the circular duct? Solution: For air at 20°C and 1 atm, take ρ = 1.20 kg/m3 and µ = 1.8E-5 kg/m-s. Compute the pressure drop in the circular duct:
The square duct will have slightly different size, Reynolds number, and velocity:
Thus everything can be written in terms of the square duct size a:
Guess f equal to, say, 0.02, find the improved Reynolds number and f, finally find a: Vs = 4.70 m/s ; ReDh = 57,350 ; f = 0.0203 ; a = 0.183 m Ans. _______________________________________________________________________
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P6.104 Although analytical solutions are available for laminar flow in many duct shapes [34], what do we do about ducts of arbitrary shape? Bahrami et al. [57] propose that a better approach to the pipe result, f Re = 64, is achieved by replacing the hydraulic diameter Dh by √A, where A is the area of the cross section. Test this idea for the isosceles triangles of Table 6.4. If time is short, at least try 10°, 50°, and 80°. What do you conclude about this idea? Solution: We can see for the triangles in Table 6.4 that the values of f ReDh are all less than 64, by as much as 25%. If we denote Bahrami’s idea as DB = √A, the new Reynolds number is based on
So we multiply the values of f ReDh by the above factor and see how we close it is to 64: θ, degrees f ReB
10 73
20 63
30 61
40 62
50 65
60 72
70 85
80 116
We see that intermediate angles, 20° to 50°, work well, sharper angles not so good. Bahrami et al. [57] make some suggestions to modify the idea for too-small or too-large angles. P6.105 A fuel cell [Ref. 59] consists of air (or oxygen) and hydrogen micro ducts, separated by a membrane that promotes proton exchange for an electric current, as in Fig. P6.105. Suppose that the air side, at 20°C and approximately 1 atm, has five 1 mm by 1 mm ducts, each 1 m long. The total flow rate is 1.5E-4 kg/s. (a) Determine if the flow is laminar or turbulent. (b) Estimate the pressure drop.
Fig. P6.105. Simplified diagram of an air-hydrogen fuel cell. [Problem courtesy of Dr. Pezhman Shirvanian]
61
Solution: For air at 20°C and 1 atm, take ρ = 1.20 kg/m3 and µ = 1.8E-5 kg/m-s. The hydraulic diameter of a square duct is easy, the side length a = 1 mm. The mass flow through one duct is
(b) We could go with the simply circular-duct approximation, f = 64/Re, but we have a more exact laminar-flow result in Table 6.4 for a square duct:
6.106 A heat exchanger consists of multiple parallel-plate passages, as shown in Fig. P6.106. The available pressure drop is 2 kPa, and the fluid is water at 20°C. If the desired total flow rate is 900 m3/h, estimate the appropriate number of passages. The plate walls are hydraulically smooth.
Fig. P6.106
Solution: For water, ρ = 998 kg/m3 and µ = 0.001 kg/m⋅s. Unlike Prob. 6.96, here we expect turbulent flow. If there are N passages, then b = 50 cm for all N and the passage thickness is H = 0.5 m/N. The hydraulic diameter is Dh = 2H. The velocity in each passage is related to the pressure drop by Eq. (6.58):
Select N, find H and V and Qtotal = AV = b2V and compare to the desired flow of 900 m3/h. For example, guess N = 20, calculate f = 0.0173 and Qtotal = 2165 m3/h. The converged result is
62
6.107 A rectangular heat exchanger is to be divided into smaller sections using sheets of commercial steel 0.4 mm thick, as sketched in Fig. P6.107. The flow rate is 20 kg/s of water at 20°C. Basic dimensions are L = 1 m, W = 20 cm, and H = 10 cm. What is the proper number of square sections if the overall pressure drop is to be no more than 1600 Pa?
Fig. P6.107
Solution: For water at 20°C, take ρ = 998 kg/m3 and µ = 0.001 kg/m⋅s. For commercial steel, ε ≈ 0.046 mm. Let the short side (10 cm) be divided into “J” squares. Then the long (20 cm) side divides into “2J” squares and altogether there are N = 2J2 squares. Denote the side length of the square as “a,” which equals (10 cm)/J minus the wall thickness. The hydraulic diameter of a square exactly equals its side length, Dh = a. The total cross-section area is A = N a2. Then the pressure drop relation becomes
As a first estimate, neglect the 0.4-mm wall thickness, so a ≈ 0.1/J. Then the relation for Δp above reduces to fJ ≈ 0.32. Since f ≈ 0.036 for this turbulent Reynolds number (Re ≈ 1E4) we estimate that J ≈ 9 and in fact this is not bad even including wall thickness:
So the wall thickness increases V and decreases a so Δp is too large. Try J = 8:
[I suppose a practical person would specify J = 7, N = 98, to keep Δp < 1600 Pa.]
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P6.108 In Sec. 6.11 it was mentioned that Roman aqueduct customers obtained extra water by attaching a diffuser to their pipe exits. Fig. P6.108 shows a simulation a smooth inlet pipe, with and without a 15 diffuser expanding to a 5-cm-diameter exit. The pipe entrance is sharpedged. Calculate the flow rate (a) without, and (b) with the diffuser.
Solution: For water at 20°C, take ρ = 998 kg/m3 and µ = 0.001 kg/m-s. The energy equation between the aqueduct surface and the pipe exit yields
zsurf = z2 +
V22 V2 V2 L + h f + Σhm = z2 + 2 + 1 ( f + K entrance + K diffuser ) 2g 2g 2g D1
(a) Without the diffuser, Kdiff = 0, and V1 = V2. For a sharp edge, take Kent = 0.5. We obtain V12 2m 2m = (1 + f + 0.5) , with f = fcn(Re = ρV1 D1 / µ ) 2g 0.03m Solve : Re =115, 000 ; f = 0.0175 ; V1 = 4.48 m / s ; Qwithout = 0.00271m 3 / s Ans.(a)
(b) With the diffuser, from Fig. 6.23, for D1/D2 = 3/5 = 0.6 and 2θ = 15°, read Kdiffuser ≈ 0.2. From one-dimensional continuity, V2 = V1(3/5)2 = 0.36V1. The energy equation becomes 2m =
(0.36V1 )2 V12 2m + (1 + f + 0.5 + 0.2) 2g 2g 0.03m
Solve : Re = 134, 000 ; f = 0.0169 ; V1 = 3.84 m / s ; Qwith = 0.00316 m 3 / s
Ans.(b)
Adding the diffuser increases the flow rate by 17%. [NOTE: Don’t know if the Romans did this, but a well-rounded entrance, Kent = 0.05, would increase the flow rate by another 15%.]
64 3
P6.109 Water at 20°C flows at 23 m /h through a 5-cm-diameter commercial steel pipe that is 45 m long. According to Table 6.5, how many 90° regular screwed elbows can be added to the system before the overall pressure drop exceeds 170 kPa? 3
Solution: For water at 20°C, ρ = 1000 kg/m and µ = 0.001 kg/m-s. For commercial steel, ε = 0.046 mm, hence ε/D = 0.000046/0.05 = 0.0009. Compute the velocity, Reynolds number, 3
and friction factor for a steady flow rate of 23 m /h.
Q = 23 m 3 /h = 0.00639 m 3 /s ; V = Re D =
Q ( π / 4)D 2
=
0.00639 m3 / s ( π / 4)(0.05)2
= 3.25 m / s
ρVD (1000)(3.25)(0.05) ε = = 162500 ; = 0.0009 ; f Moody = 0.0209 µ (0.001) D
Compute the pressure drop of the pipe alone:
Δp = f
Lρ 2 45 1000 V = (0.0209)( )( )(3.25)2 = 99.34 kPa D2 0.05 2
Thus we can add elbows until their total minor loss = 170 – 99.34 = 70.66 kPa. From Table 6.5, each elbow has a loss K = 0.95. The pressure loss of each elbow is
ρ 1000 Δp elbow = K elbow V 2 = (0.95)( )(3.25)2 = 5.02 kPa 2 2 total elbow loss 70.66 Thus the no. of elbows = = = 14 elbows max Ans. loss per elbow 5.02
P6.110 A pipe of sectional area 0.1 m2 is abruptly connected to a pipe of sectional area 0.25 m2. The flow rate is 0.4 m3/s, and the pressure immediately before the connection is 10 kPa. Find (a) head loss due to the sudden enlargement and (b) the pressure immediately after the connection. Solution: a1 = 0.1 m 2 ,
a2 = 0.25 m 2
Q = 0.4 m 3 /s,
P1 = 10 kPa
v1 =
Q 0.4 = = 4 m/s a1 0.1
v2 =
Q 0.4 = = 1.6 m/s a2 0.25
Minor loss hm (SE) = (v1 –v2 )2 /2g = (4 – 1.6)2 /2 × 9.81 = 0.29 m
65
Energy equation
⇒
v2 P v2 P1 + 1 = 2 + 2 + hm ρ g 2g ρ g 2g 1 P2 = P1 + ρ (v12 – v22 ) – ρ ghm 2 1 = 10 + (4 2 – 1.6 2 ) – 9.81× 0.29 2 = 13.84 kPa
NOTE: IN PROBLEMS 6.109−6.121, MINOR LOSSES ARE INCLUDED.
6.111 In Fig. P6.111 a thick filter is being tested for losses. The flow rate in the pipe is 7 m3/min, and the upstream pressure is 120 kPa. The fluid is air at 20°C. Using the watermanometer reading, estimate the loss coefficient K of the filter.
Fig. P6.111
Solution: The upstream density is ρair = p/(RT) = 120000/[287(293)] = 1.43 kg/m3. The average velocity V (which is used to correlate loss coefficient) follows from the flow rate:
The manometer measures the pressure drop across the filter:
This pressure is correlated as a loss coefficient using Eq. (6.78):
66
6.112 A 70 percent efficient pump delivers water at 20°C from one reservoir to another 6 m higher, as in Fig. P6.112. The piping system consists of 18 m of galvanized-iron 5-cm pipe, a reentrant entrance, two screwed 90° long-radius elbows, a screwed-open gate valve, and a sharp exit. What is the input power required with and without a 6° well-designed conical expansion added to the exit? The flow rate is 40 m3/hr.
Fig. P6.112
Solution: For water at 20°C, take ρ = 1000 kg/m3 and µ = 0.001 kg/m-s. For galvanized iron, ε ≈ 0.15 mm, whence ε/d = 0.00015/0.05 ≈ 0.003. Without the 6° cone, the minor losses are:
K reentrant ≈ 1.0; K elbows ≈ 2(0.41); K gate valve ≈ 0.16; K sharp exit ≈ 1.0
Evaluate V =
Q 40/3600 ρVd 1000(5.66)(0.05) = = 5.66 m/s; Re = = ≈ 283000 2 A π (0.05) /4 µ 0.001
At this Re and roughness ratio, we find from the Moody chart that f ≈ 0.0266. Then
(a) h pump
V2 L (5.66)2 18 = Δz + 0.0266 + 1.0 + 0.82 + 0.16 + 1.0 f + ∑ K = 6 + 0.05 2g d 2(9.81)
ρgQh p (9810)(40/3600)(26.5) = η 0.70 = 4.13 kW Ans. (a)
or h pump ≈ 26.5m, Power =
(b) If we replace the sharp exit by a 6° conical diffuser, from Fig. 6.23, Kexit ≈ 0.3. Then
(5.66)2 18 hp = 6 + 0.0266 + 1.0 + .82 + .16 + 0.3 = 25.4 m 0.05 2(9.81)
then Power = (9810)(40 / 3600)(25.4)/0.7 = 3.96 kW (4% less) Ans. (b)
67
6.113 The reservoirs in Fig. P6.113 are connected by cast-iron pipes joined abruptly, with sharp-edged entrance and exit. Including minor losses, estimate the flow of water at 20°C if the surface of reservoir 1 is 14 m higher than that of reservoir 2.
Fig. P6.113
Solution: For water at 20°C, take ρ = 1000 kg/m3 and µ = 0.001 kg/m-s. Let “a” be the small pipe and “b” the larger. For wrought iron, ε ≈ 0.046 mm, whence ε/da = 0.00184 and ε/db = 0.00092. From the continuity relation,
For pipe “a” there are two minor losses: a sharp entrance, K1 = 0.5, and a sudden expansion, Fig. 6.22, Eq. (6.101), K2 = [1 − (1/2)2]2 ≈ 0.56. For pipe “b” there is one minor loss, the submerged exit, K3 ≈ 1.0. The energy equation, with equal pressures at (1) and (2) and near zero velocities at (1) and (2), yields
Va2 120 1.0 or, since Vb = Va /4, Δz = 14 = 240fa + 1.06 + fb + 2(9.81) 16 16 where fa and fb are separately related to different values of Re and ε/d. Guess to start:
fa ≈ fb ≈ 0.02: then Va = 6.73 m/s, Re a ≈ 168000, ε /d a = 0.00184, fa-2 ≈ 0.024 Vb = 1.68 m/s, Re b ≈ 84000, ε /d b = 0.00092, fb-2 ≈ 0.0223 Converges to: fa = 0.0241, fb = 0.0225, Va ≈ 6.23 m/s, Q = Va A a ≈ 0.00306 m3 /s. Ans.
P6.114 Water at 20°C flows through a smooth horizontal pipe at 12 m3/h. The pipe diameter is 5 cm, and its length is 10 m. (a) Compute the pressure drop. (b) If a gate valve is added, what gate opening h/D will reduce the flow rate by 50% for the same pressure drop? Solution: For water at 20°C, Table A.3, ρ = 998 kg/m3 and µ = 0.001 kg/m-s. (a) First compute the Reynolds number from the known flow rate and then get the (smooth) friction factor:
68
Then the pressure drop is given by
(b) The partly closed gate valve cuts the flow rate in half to 6 m3/s, or V = 0.85 m/s. This cuts the Reynolds number in half, to 42,350, which slightly increases the (smooth) friction factor. From Eq. (6.38), the new friction factor is f = 0.0217. Add the K factor of the valve to Δp:
Figure 6.18b shows that, for the gate valve, a minor loss coefficient K ≈ 10.5 is reached for a gate opening h/D ≈ 0.35. Ans.(b)
6.115 The system in Fig. P6.115 consists of 1200 m of 5 cm cast-iron pipe, two 45° and four 90° flanged long-radius elbows, a fully open flanged globe valve, and a sharp exit into a reservoir. If the elevation at point 1 is 400 m, what gage pressure is required at point 1 to deliver 0.005 m3/s of water at 20°C into the reservoir? Solution: For water at 20°C, take ρ = 998 kg/m3 and µ = 0.001 kg/m⋅s. For cast iron, take ε ≈ 0.26 mm, hence ε/d = 0.0052. With the flow rate known, we can compute V, Re:
Fig. P6.115
69
The minor losses may be listed as follows:
Then the energy equation between (1) and (2—the reservoir surface) yields
P6.116 In the analysis of the head loss due to a sudden contraction from a diameter of D to d, it is assumed that the contraction loss amounts to the loss caused by the expansion from the vena contract dmin to d (Fig. 6.22). Hence, based on Eq. (6.80), derive a formula for the 2 sudden contraction loss in terms of the coefficient of contraction Cc = (dmin/d) . Further use Eqs. (6.80) and (6.81) to express Cc as a function of d/D. Solution: On assuming that the contraction loss = loss due to expansion from dmin to d, then by Eq (6.80) 2
2 d 2 Vmin hm (SC) = 1 – min d 2 2g 2
d2 V2 = 2 – 1 dmin 2g
2 (by Vmin dmin = Vd 2 )
2
1 V2 – 1 = Cc 2g
Cc =
2 dmin d2
2
1 h K sc = 2 m = – 1 V / 2g Cc From Fig. 6.22
d d2 ≤ 0.76 0.42 1 – 2 for D D K sc = 2 d2 d > 0.76 1 – 2 for D D ∴
–1 1 + 0.65(1 – d 2 /D 2 )1/2 for d/D ≤ 0.76 Cc = (2 – d 2 /D 2 )–1 for d/D > 0.76
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6.117 The water pipe in Fig. 6.117 slopes upward at 30°. The pipe is 2.5-cm diameter and smooth. The flanged globe valve is fully open. If the mercury manometer shows a 10 3 cm deflection, what is the flow rate in m /s? Solution: For water at 20°C, take ρ = 1000 kg/m3 and µ = 0.001 kg/m-s. The pipe length and elevation change are Fig. P6.117
3m = 3.46 m; z 2 − z1 = 3tan 30° = 1.73 m, Open 1′′ globe valve: K ≈ 13 cos30° The manometer indicates the total pressure change between (1) and (2): L=
p1 − p 2 = (ρMerc − ρw )gh + ρw gΔz = (13.6 − 1)(9810) ( 0.1) + 9810(1.73) ≈ 29332 Pa The energy equation yields V2 3.46 p1 − p 2 29332 Pa = Δz + h f + h m = 1.73 + f + 13 ≈ ρg 2(9.81) 0.025 9810 N/m 3 24.72 or: V2 ≈ . Guess f ≈ 0.02, V ≈ 1.25 m/s, Re ≈ 31300, fnew ≈ 0.0233 (138.4f + 13) Rapid convergence to f ≈ 0.0233, V ≈ 1.23 m/s, Q = V(π /4)(0.025)2 ≈ 0.0006 m3/s. Ans. [NOTE that the manometer reading of 10 cm exactly balances the friction losses, and the hydrostatic pressure change ρgΔz cancels out of the energy equation.] 6.118 In Fig. P6.118 the pipe is galvanized iron. Estimate the percentage increase in the flow rate (a) if the pipe entrance is cut off flush with the wall and (b) if the butterfly valve is opened wide.
Fig. P6.118
Solution: For water at 20°C, take ρ = 998 kg/m3 and µ = 0.001 kg/m⋅s. For galvanized iron, take ε ≈ 0.15 mm, hence ε/d = 0.003. First establish minor losses as shown: Protruding entrance (Fig. 6.21a), Butterfly @ 30° (Fig 6.19) K ≈ 80 ± 20
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The energy equation, with p1 = p2, yields:
Thus the “base” flow, for our comparison, is Vo ≈ 1.086 m/s, Qo ≈ 0.00213 m3/s. If we cut off the entrance flush, we reduce Kent from 1.0 to 0.5; hardly a significant reduction in view of the huge butterfly valve loss Kvalve ≈ 80. The energy equation is
If we open the butterfly wide, Kvalve decreases from 80 to only 0.3, a huge reduction:
Obviously opening the valve has a dominant effect for this system.
6.119 The water pump in Fig. P6.119 maintains a pressure of 45 kPa at point 1. There is a filter, a half-open disk valve, and two regular screwed elbows. There are 24 m of 10-cm diameter commercial steel pipe. (a) If the flow rate is 40 m3/h, what is the loss coefficient of the filter? (b) If the disk valve is wide open and Kfilter = 7, what is the resulting flow rate?
Fig. P6.119
Solution: For water, take ρ = 1000 kg/m3 and µ = 0.001 kg/m-s. The energy equation is written from point 1 to the surface of the tank:
72
(a) From the flow rate, V1 = Q/A = (40/3600)/[(π /4)(0.1)2] = 1.41 m/s. Look up minor losses and enter into the energy equation:
45 × 10 3 Pa (1.41 m/s)2 + +0 9810 N/m 3 2(9.81 m/s2 ) (1.41)2 24 = 0 + 0 + 2.7 m + f + 2.8 + K filter + 2(0.64) + 1 2(9.81) (0.1) We can solve for Kfilter if we evaluate f. Compute ReD = (1000)(1.41)(0.1)/(0.001) = 141,000. For commercial steel, ε/D = 0.000046/0.1 = 0.00046 and V12/2g = 0.101 m. From the Moody chart, f ≈ 0.0193, and fL/D = 4.63. The energy equation above becomes:
4.59 + 0.101 = 2.7 + 0.101(4.63 + 2.8 + K filter + 1.28 + 1), Solve
Kfilter ≈ 10.0 Ans. (a)
(b) If Kfilter = 7.0 and V is unknown, we must iterate for the velocity and flow rate. The energy equation becomes, with the disk valve wide open (KValve ≈ 0):
4.59 +
V2 V 2 24 = 2.7 + + 0 + 7.0 +1.28 +1 f 2(9.81) 2(9.81) 0.1
Iterate to find
f ≈ 0.0189, Re D = 170,000, V = 1.7 m/s,
Q = AV = 0.0133 m3 /s
Ans. (b)
6.120 In Fig. P6.120 there are 40 m of 5-cm pipe, 24 m of 15-cm pipe, and 50 m of 7.5-cm pipe, all cast iron. There are three90° elbows and an open globe valve, all flanged. If the exit elevation is zero, what power is extracted by the turbine when the flow rate is 16 m3/h of water at 20°C?
Fig. P6.120
Solution: For water at 20°C, take ρ = 1000 kg/m3 and µ = 0.001 kg/m-s. For cast iron, ε ≈ 0.26 mm. The 5-cm, 15-cm, and 7.5-cm pipes have, respectively, (a) L/d = 800, ε/d = 0.0052; (b) L/d = 160, ε/d = 0.00173; (c) L/d = 667, ε/d = 0.00347 The flow rate is known, so each velocity, Reynolds number, and f can be calculated:
Va =
16/3600 1000(2.26)(0.05) = 2.26 m/s; Re a = = 113000, fa ≈ 0.0315 2 0.001 π (0.05) /4
73
Also, Vb = 0.25 m/s, Re b = 37500, fc ≈ 0.0267; Vc = 1 m/s, Re c = 75000, fc ≈ 0.0288 Finally, the minor loss coefficients may be tabulated: sharp 5-cm entrance: K = 0.5; three 5-cm 90° elbows: K = 3(0.95) 5-cm sudden expansion: K ≈ 0.79; 7.5-cm open globe valve: K ≈ 6.3 The turbine head equals the elevation difference minus losses and the exit velocity head:
h t = Δz − ∑ h f − ∑ h m − Vc2 /(2g) = 30 − −
(2.26)2 [0.0315(800) + 0.5 + 3(0.95) + 0.79] 2(9.81)
(0.25)2 (1)2 (0.0267)(160) − [0.0288(667) + 6.3 + 1] ≈ 21 m 2(9.81) 2(9.81)
The resulting turbine power = ρgQht = (9810)(16/3600)(21) ≈ 920 W. Ans. 6.121 In Fig. P6.121 the pipe entrance is sharp-edged. If the flow rate is 0.004 m3/s, what power, in W, is extracted by the turbine? Solution: For water at 20°C, take ρ =998 kg/m3 and µ = 0.001 kg/m⋅s. For cast
Fig. P6.121
iron, ε ≈ 0.26 mm, hence ε/d = 0.26/50 ≈ 0.0052. The minor loss coefficients are Entrance: K ≈ 0.5; 5-cm(≈2″) open globe valve: K ≈ 6.9. The flow rate is known, hence we can compute V, Re, and f:
The turbine head equals the elevation difference minus losses and exit velocity head:
74
P6.122 For completely turbulent flow, the frictional head loss hf in a given length of pipe is 2 known to be hf = rQ in which Q is the flow rate and r is a parameter depending only on the pipe characteristics. (a) Derive from the Darcy friction equation an expression for r in terms of the pipe characteristics. (b) Show that an equivalent pipe, which is to replace a pipe system consisting of a number of pipes connected either in series or in parallel with one another, has –1/2 –2 an effective parameter re given by (i) re = ∑ri if the pipes are in series or (ii) re = (∑ri ) if the pipes are in parallel. Solution: Darcy equation hf = f
L V2 8fL 8 fL = 2 5 Q 2 = rQ 2 r = 2 5 d 2g π gd π gd
For pipes in series, Total h f = ∑h fi = ∑riQi2 = (∑ri ) Q 2 But
h f = reQ 2
Q1 = Q2 = … = Q ⇒
re = ∑ri
For pipes in parallel, Total Q = ∑Qi = ∑(h fi / ri )1/2 = (∑ri–1/2 ) h1/2 f ⇒
h f = (∑ri–1/2 )–2 Q 2
But
h f = reQ 2
⇒
hf1 = hf 2 = … = hf re = (∑ri–1/2 )–2
6.123 For the parallel-pipe system of Fig. P6.123, each pipe is cast iron, and the pressure drop p1 − p2 = 20 kPa. Compute the total flow rate between 1 and 2 if the fluid is SAE 10 oil at 20°C.
Fig. P6.123
Solution: For SAE 10 oil at 20°C, take ρ = 870 kg/m3 and µ = 0.104 kg/m-s. For cast iron, ε ≈ 0.26 mm. Guess laminar flow in each: ? 128µ L a Q a
Δp a =
π d a4
= 20 × 10 3 =
128(0.104)(80)Q a , π (0.075)4
Q a ≈ 0.00187 m 3 /s. Check Re ≈ 265 (OK) ? 128µ L b Q b
Δp b =
π d b4
= 20 × 10 3 =
128(0.104)(60)Q b , π (0.05)4
Q b ≈ 0.00049 m 3 /s. Check Re ≈ 105 (OK) The total flow rate is Q = Q a + Q b = 0.00187 + 0.00049 ≈ 0.0024 m3 /s. Ans.
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P6.124 A 0.15-m-diameter pipe, which is 4 km long and of wall roughness 0.06 mm, delivers water between two reservoirs. (a) If the water levels of the two reservoirs differ by 50 m, estimate the discharge between the reservoirs. (b) It is now required to increase the discharge by 50%, and the proposed method is to install an additional pipe of the same diameter and material in parallel to the original pipe, from a certain point in the pipeline all the way to the lower reservoir. Determine the length of the additional pipe. You may assume rough pipes, and ignore the minor losses. Solution: L = 4000 m, Assume friction factor Darcy eqn
hf =
f = 0.016
8 fL 2 Q π 2 gd 5
Moody Chart gives
50 =
Check
50 =
⇒
8 × 0.016 × 4000 2 Q π 2 × 9.81× 0.15 5
50 =
f = 0.018
8 × 0.018 × 4000 2 Q ⇒ Q = 0.025 m 3 /s 2 5 π × 9.81× 0.15 4 × 0.025 = 2.1×10 5 Re = –6 π ×1×10 × 0.15
Moody Chart gives ∴
ε = 4 × 10 –4 d
ε = 0.06 mm,
⇒ Q = 0.027 m 3 /s 4 × 0.027 4Q = = 2.3 ×10 5 Re = –6 π vd π ×1×10 × 0.15
Check
∴
d = 0.15 m,
f = 0.0182
8 × 0.0182 × 4000 2 Q π 2 × 9.81 × 0.15 5
⇒
Q = 0.025 m 3 /s
(OK) 3
To increase discharge by 50%, new discharge Q’ = 0.0375 m /s
L1 + L2 = 4000 m Re1 = 1.5 × 2.1×10 5 = 3.15 ×10 5 ,
Moody diagram gives ∴
f1 = 0.0175,
Re 2 = 0.5 Re1 = 1.575 ×10 5
f1 = 0.019
8 f1 L1Q′2 8 f2 L2 (Q′ / 2)2 + π 2 gd 5 π 2 gd 5 8 0.0175(4000 – L )0.0375 2 + 0.019 × L × (0.0375/2)2 = 2 2 2 5 π × 9.81× 0.15
50 =
Solving, get
L2 = 2927 m
76
6.125 If the two pipes in Fig. P6.123 are instead laid in series with the same total pressure drop of 20 kPa, what will the flow rate be? The fluid is SAE 10 oil at 20°C. Solution: For SAE 10 oil at 20°C, take ρ = 870 kg/m3 and µ = 0.104 kg/m-s. Again guess laminar flow. Now, instead of Δp being the same, Qa = Qb = Q: Δp a + Δp b = 20 × 10 3 =
128µL a Q 128µL bQ 128(0.104) 80 60 + = Q 4 4 4 + π π da π db (0.05)4 (0.075)
Solve for Q ≈ 0.00039 m3 /s
Ans. Check Re a ≈ 55 (OK) and Re b ≈ 83 (OK)
In series, the flow rate is six times less than when the pipes are in parallel.
6.126 The parallel galvanized-iron pipe system of Fig. P6.126 delivers water at 20°C with a total flow rate of 0.036 m3/s. If the pump is wide open and not running, with a loss coefficient K = 1.5, determine (a) the flow rate in each pipe and (b) the overall pressure drop. Fig. P6.126
Solution: For water at 20°C, take ρ = 998 kg/m3 and µ = 0.001 kg/m⋅s. For galvanized iron, ε = 0.15 mm. Assume turbulent flow, with Δp the same for each leg:
When the friction factors are correctly found from the Moody chart, these two equations may be solved for the two velocities (or flow rates). Begin by guessing f ≈ 0.020:
The 2nd iteration converges: f1 ≈ 0.0264, V1 = 11.69 m/s, f2 ≈ 0.0282, V2 = 10.37 m/s, Q1 = A1V1 = 0.023 m3/s, Q2 = A2V2 = 0.013 m3/s. Ans. (a) The pressure drop is the same in either leg:
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*P6.127 A blower supplies standard air to a plenum that feeds two horizontal square sheetmetal ducts with sharp-edged entrances. One duct is 30 m long, with a cross-section 15 cm by 15 cm. The second duct is 60 m long. Each duct exhausts to the atmosphere. When the plenum pressure is 250 Pa gage, the volume flow in the longer duct is three times the flow in the shorter duct. Estimate both volume flows and the cross-section size of the longer duct. Solution: For standard air, take ρ = 1.2 kg/m3 and µ = 1.8 × 10 –5 kg/m-s. For sheet metal, take ε = 0.05 mm. The energy equation for this case is
p1 ρg
V12 p2 V2 + z1 = + 2 + z2 + h f + hentrance , or : 2g ρg 2g L 1 + K ent ) where K sharp−edged ≈ 0.5 Δp = ρ V 2 (1 + f Dh 2 +
We have abbreviated the duct velocity to V, without a subscript. For a square duct, the hydraulic diameter is the side length of the square. First compute the flow rate in the short duct: 250 =
ε 1.2V 2 30 (1 + f + 0.5) , f = fcn(Re D , ) h D 2 0.15 h 4
The Reynolds number for the short duct is Re = (1.2)V(0.15)/( 1.8 × 10 –5 ) = 10 V, and ε/Dh = 0.00033. The solution is
L = 30 m : Re D = 87, 000 ; f = 0.0200 ; V = 8.7 m / s ; Qshort = 0.196 m3 / s h
For the longer duct, Re = (1.2)VDh /( 1.8 × 10 –5 ), and ε/Dh = 0.00005/Dh. We don’t know Dh and must solve to make Qlong = 3Qshort. The solution is L = 60 m : Re D = 150, 000 ; f = 0.0178 ; V = 8.63 m / s ; Qlong = 0.588 m 3 / s h
Solve for
Dh,long = 0.261 m
Ans.
NOTE: It is an interesting numerical quirk that, for these duct parameters, the velocities in each duct are almost identical, regardless of the magnitude of the pressure drop.
78
6.128 In Fig. P6.128 all pipes are 8-cm-diameter cast iron. Determine the flow rate from reservoir (1) if valve C is (a) closed; and (b) open, with Kvalve = 0.5.
Fig. P6.128
Solution: For water at 20°C, take ρ = 998 kg/m3 and µ = 0.001 kg/m⋅s. For cast iron, ε ≈ 0.26 mm, hence ε/d = 0.26/80 ≈ 0.00325 for all three pipes. Note p1 = p2, V1 = V2 ≈ 0. These are long pipes, but we might wish to account for minor losses anyway: sharp entrance at A: K1 ≈ 0.5; line junction from A to B: K2 ≈ 0.9 (Table 6.5) branch junction from A to C: K3 ≈ 1.3; two submerged exits: KB = KC ≈ 1.0 If valve C is closed, we have a straight series path through A and B, with the same flow rate Q, velocity V, and friction factor f in each. The energy equation yields
Guess f ≈ f fully rough ≈ 0.027, then V ≈ 3.04 m/s, Re ≈ 998(3.04)(0.08)/(0.001) ≈ 243000, ε/d = 0.00325, then f ≈ 0.0273 (converged). Then the velocity through A and B is V = 3.03 m/s, and Q = (π /4)(0.08)2(3.03) ≈ 0.0152 m3/s. Ans. (a). If valve C is open, we have parallel flow through B and C, with QA = QB + QC and, with d constant, VA = VB + VC. The total head loss is the same for paths A-B and A-C:
79
plus the additional relation VA = VB + VC. Guess f ≈ ffully rough ≈ 0.027 for all three pipes and begin. The initial numbers work out to
Repeat once for convergence: VA ≈ 3.46 m/s, VB ≈ 1.90 m/s, VC ≈ 1.56 m/s. The flow rate from reservoir (1) is QA = (π/4)(0.08)2(3.46) ≈ 0.0174 m3/s. (14% more) Ans. (b)
6.129 For the series-parallel system of Fig. P6.129, all pipes are 8-cm-diameter asphalted cast iron. If the total pressure drop p1 − p2 = 750 kPa, find the resulting flow rate Q m3/h for water at 20°C. Neglect minor losses. Solution: For water at 20°C, take ρ = 998 kg/m3 and µ = 0.001 kg/m⋅s. For asphalted cast iron, ε ≈ 0.12 mm, hence ε/d = 0.12/80 ≈ 0.0015 for all three pipes. The head loss is the same through AC and BC:
Fig. P6.129
Since d is the same, VA + VB = VC and fA, fB, fC are found from the Moody chart. Cancel g and introduce the given data:
Now compute ReA ≈ 167000, fA ≈ 0.0230, ReB ≈ 264000, fB ≈ 0.0226, ReC ≈ 431000, and fC ≈ 0.0222. Repeat the head loss iteration and we converge: VA ≈ 2.06 m/s, VB ≈ 3.29 m/s, VC ≈ 5.35 m/s, Q = (π / 4)(0.08)2(5.35) ≈ 0.0269 m3/s. Ans.
80
6.130 A blower delivers air at 3000 m3/h to the duct circuit in Fig. P6.130. Each duct is commercial steel and of square cross-section, with side lengths a1 = a3 = 20 cm and a2 = a4 = 12 cm. Assuming sea-level air conditions, estimate the power required if the blower has an efficiency of 75%. Neglect minor losses. Solution: For air take ρ = 1.2 kg/m3 and µ = 1.8E−5 kg/m⋅s. Establish conditions in each duct:
Fig. P6.130
For commercial steel (Table 6.1) ε = 0.046 mm. Then we can find the two friction factors:
The total power required, at 75% efficiency, is thus:
81
P6.131 A single uniform pipe joins two reservoirs. Calculate the percentage increase in flow rate obtainable if, from the mid-point of this pipe, a pipe of the same diameter and material is added in parallel to it. Ignore minor losses and assume complete turbulent flow. Solution: Initial Condition
2
hf = r0Q0 1/2 Q0 = (hf /r0) Modified Condition
r1 = r0 / 2 An equivalent pipe to replace the 2 pipes in parallel
rCB = (2r1–1/2 )–2 =
r1 r = 0 4 8
An equivalent pipe to replace the whole system
rAB = r1 + rCB = ∴ ⇒
5 5r r1 = 0 4 8
5 r0Q12 = r0Q02 8 8 1/2 Q1 /Q0 = 5
hf =
8 1/2 % increase = – 1 = 26.5% 5
*6.132 For the piping system of Fig. P6.132, all pipes are concrete with a roughness of 1 mm. Neglecting minor losses, compute the overall pressure drop p1 − p2 in kPa. The flow rate is 0.5 m3/s of water at 20°C.
82
Solution: For water at 20°C, take ρ = 1000 kg/m3 and µ = 0.001 kg/m-s. Since the pipes are all different make a little table of their respective L/d and ε/d:
Fig. P6.132
(a) (b) (c) (d)
L = 300 m, 450 m 250 m 350 m
d = 0.3 m, L/d = 1000, 0.2 m 2250 0.3 m 833 0.4 m 875
ε/d = 0.00333 0.00500 0.00333 0.0025
With the flow rate known, we can find everything in pipe (a):
Va =
Qa 0.5 1000(7.07)(0.3) = = 2.12 × 10 6 , fa ≈ 0.0270 2 = 7.07 m/s, Re a = 0.001 A a (π /4)(0.3)
Then pipes (b,c,d) are in parallel, each having the same head loss and with flow rates which must add up to the total of 0.5 m3/s: h fb =
8fb L bQ 2b 8fc L cQ 2c 8fd L dQ 2d = h = = h = , and Q b + Q c + Q d = 0.5 m 3 /s fc fd π 2g d 5b π 2g d 5c π 2g d 5d
Introduce Lb, db, etc. to find that Qc = 3.70Qb(fb/fc)1/2 and Qd = 6.41Qb(fb/fd)1/2 Then the flow rates are iterated from the relation ∑Q = 0.5 = Q b [1 + 3.70(fb /fc )1/2 + 6.41(fb /fd )1/2 ] First guess: fb = fc = fd : Q b ≈ 0.045 m 3/s; Q c ≈ 0.167 m 3/s; Q d ≈ 0.288 m 3/s
Improve by computing Reb ≈ 286500, fb ≈ 0.0307, Rec ≈ 708800, fc ≈ 0.0271, Red ≈ 916700, fd ≈ 0.025. Repeat to find Qb ≈ 0.0415 m3/s, Qc ≈ 0.164 m3/s, Qd ≈ 0.295 m3/s, from which Vb ≈ 1.32 m/s, Vc ≈ 2.32 m/s, Vd ≈ 2.35 m/s. The pressure drop is L a ρVa2 L ρVb2 + fb b da 2 db 2 = 674.8 + 60.2 ≈ 735 kPa Ans.
p1 − p 2 = Δp a + Δp b = fa
83
P6.133
For the system of Prob. P6.123, again let the fluid be SAE 10W oil at 20°C, 3
with both pipes cast iron. If the flow rate in the 5-cm pipe (b) is 5 m /h, estimate the flow rate in the 7.5 cm pipe (a), in m3/h.
Solution: For SAE 10W oil at 20°C, ρ = 870 kg/m3 and µ = 0.104 kg/m-s. The oil is pretty viscous and the flow rate small, so we think ReDb will be laminar: ρVb Db 4 ρ Qb 4(870)(5 / 3600) Re Db = = = = 296 < 2300 Yes, Laminar µ π µ Db π (0.104)(0.05)
If we need it, Vb =
4Qb
π Db2
=
4(5 / 3600)
π (0.05)2
= 0.71 m / s
For laminar flow, compute the pressure drop Δpb from Eq. (6.12) and set it equal to Δpa: Δpb =
128 µ Lb Qb
π Db4
=
128(0.104)(60)(5 / 3600)
π (0.05)4
Parallel Pipes : Δpa = Δpb = 56498 = Solve for
= 56498 Pa
128 µ La Qa
π Da4
=
128(0.104)(80)Qa
π (0.075)4
Qa = 0.0053 m3 / s = 19 m 3 / h
Check Rea = 4ρQ/(πµDa) = 4(870)(0.0053)/[π(0.104)(0.075)] = 753
Ans.
Yes, laminar also.
6.134 Three cast-iron pipes are laid in parallel with these dimensions: Pipe 1: Pipe 2: Pipe 3:
L1 = 800 m L2 = 600 m L3 = 900 m
d1 = 12 cm d2 = 8 cm d3 = 10 cm
The total flow rate is 200 m3/h of water at 20°C. Determine (a) the flow rate in each pipe; and (b) the pressure drop across the system. Solution: For water at 20°C, take ρ = 998 kg/m3 and µ = 0.001 kg/m⋅s. For cast iron, ε = 0.26 mm. Then, ε/d1 = 0.00217, ε/d2 = 0.00325, and ε/d3 = 0.0026. The head losses are the same for each pipe, and the flow rates add:
84
We could either go directly to EES or begin by guessing f1 = f2 = f3, which gives Q1 = 0.0275 m3/s, Q2 = 0.0115 m3/s, and Q3 = 0.0165 m3/s. This is very close! Further iteration gives
6.135 Consider the three-reservoir system of Fig. P6.135 with the following data: L1 = 95 m
L2 = 125 m L3 = 160 m
z1 = 25 m
z2 = 115 m
z3 = 85 m
All pipes are 28-cm-diameter unfinished concrete (ε = 1 mm). Compute the steady flow rate in all pipes for water at 20°C.
Fig. P6.135
Solution: For water at 20°C, take ρ = 998 kg/m3 and µ = 0.001 kg/m⋅s. All pipes have ε/d = 1/280 = 0.00357. Let the intersection be “a.” The head loss at “a” is desired:
plus the requirement that Q1 + Q2 + Q3 = 0 or, for same d, V1 + V2 + V3 = 0 We guess ha then iterate each friction factor to find V and Q and then check if ∑Q = 0.
Repeating for ha = 80 m gives V1 = −10.75, V2 = +7.47, V3 = +2.49 m/s, ∑V = −0.79. Interpolate to ha ≈ 78 m, gives V1 = −10.55 m/s, V2 = +7.68 m/s, V3 = +2.95 m/s, or: Q1 = −0.65 m3/s, Q2 = +0.47 m3/s, Q3 = +0.18 m3/s. Ans.
85
6.136 Modify Prob. 6.135 by reducing the diameter to 15 cm, with ε = 1 mm. Compute the flow rate in each pipe. They all reduce, compared to Prob. 6.135, by a factor of about 5.2. Can you explain this? Solution: The roughness ratio increases to ε/d = 1/150 = 0.00667, and all L/d’s increase. Guess ha = 75 m: converges to f1 = 0.0333, f2 = 0.0333, f4 = 0.0334 and V1 ≈ −6.82 m/s, V2 ≈ +5.32 m/s, V3 ≈ +2.34 m/s, ∑V ≈ +0.85 We finally obtain ha ≈ 78.2 m, giving V1 = −7.04 m/s, V2 = +5.10 m/s, V3 = +1.94 m/s, or: Q1 = −0.124 m3/s, Q2 = +0.090 m3/s, Q3 = +0.034 m3/s. Ans.
*6.137 Modify Prob. 6.135 as follows. Let z3 be unknown and find its value such that the flow rate in pipe 3 is 0.2 m3/s toward the junction. (This problem is best suited for computer iteration or EES.) Solution: For water at 20°C, take ρ = 998 kg/m3 and µ = 0.001 kg/m⋅s. All pipes have ε/d = 1/280 = 0.00357. Let the intersection be “a.” The head loss at “a” is desired for each in order to check the flow rate in pipe 3. In Prob. 6.135, with z3 = 85 m, we found Q3 to be 0.18 m3/s toward the junction, pretty close. We repeat the procedure with a few new values of z3, closing to ∑Q = 0 each time:
6.138 The three-reservoir system in Fig. P6.138 delivers water at 20°C. The system data are as follows: D1 = 0.2 m
D2 = 0.15 m
D3 = 0.22 m
L1 = 550 m
L2 = 350 m
L3 = 500 m
All pipes are galvanized iron. Compute the flow rate in all pipes.
Fig. P6.138
86
Solution: For water at 20°C, take ρ = 1000 kg/m3 and µ = 0.001 kg/m-s. For galvanized iron, take ε = 0.15 mm. Then the roughness ratios are
ε /d1 = 0.00075 ε /d2 = 0.0010 ε /d3 = 0.000682 Let the intersection be “a.” The head at “a” is desired: z1 − h a =
f1L1 V12 ; d1 2g
z2 − ha =
f2 L 2 V22 ; d 2 2g
z3 − ha =
f3L 3 V32 ; plus Q1 + Q 2 + Q 3 = 0 d 3 2g
We guess ha then iterate each friction factor to find V and Q and then check if ∑Q = 0.
f1 (550)V12 , (0.2)2(9.81) solve f1 = 0.0193, V1 = −1.82 m/s
Guess h a = 15 m: 15 − 6 = 9 =
Similarly, f2 = 0.0204, V2 ≈ +2.49 m/s and of course V3 = 0. Get ∑Q = −0.013 m 3/s
Try again with a slightly lower ha to reduce Q1 and increase Q2 and Q3:
h a = 14.6 m: converges to Q1 = −0.0559 m 3 /s, Q 2 = +0.0445 m 3 /s, Q 3 = +0.0153 m 3 /s, ∑Q = +0.0039 m 3 /s Interpolate to h a = 14.7 m: Q1 = −0.056 m3 /s, Q2 = +0.044 m3 /s, Q3 = +0.013 m3 /s
Ans.
6.139 Suppose that the three cast-iron pipes in Prob. 6.134 are instead connected to meet smoothly at a point B, as shown in Fig. P6.139. The inlet pressures in each pipe are: p1 = 200 kPa; p2 = 160 kPa; p3 = 100 kPa. The fluid is water at 20°C. Neglect minor losses. Estimate the flow rate in each pipe and whether it is toward or away from point B. Solution: For water take ρ = 998 kg/m3 and µ = 0.001 kg/m⋅s. The pressure at point B Fig. P6.139 must be a known (constant) value which makes the net flow rate equal to zero at junction B. The flow clearly goes from (1) to B, and from B to (3), but we are not sure about pipe (2). For cast iron (Table 6.1), ε = 0.26 mm. Each pipe has a flow rate based upon its pressure drop:
87
where the f ’s are determined from the Moody chart for each pipe’s ε/D and ReD. The correct value of pB makes the flow rates Qi = (π/4)Di2Vi balance at junction B. EES is excellent for this type of iteration, and the final results balance for pB = 166.7 kPa:
*6.140 Modify Prob. 6.138 as follows. Let all data be the same except that pipe 1 is fitted with a butterfly valve (Fig. 6.19b). Estimate the proper valve opening angle (in degrees) for the flow rate through pipe 1 to be reduced to 0.04 m3/s toward reservoir 1. (This problem requires iteration and is best suited to a digital computer or EES.) Solution: For water at 20°C, take ρ = 1000 kg/m3 and µ = 0.001 kg/m-s. For galvanized iron, take ε = 0.15 mm. Then the roughness ratios are
ε /d1 = 0.00075 ε /d2 = 0.0010 ε /d3 = 0.000682 For a butterfly valve loss coefficient “K” (to be found). Let the junction be “J.” The head loss at “J” is desired and then to be iterated to give the proper flow rate in pipe (1):
We know z1 = 6 m, z2 = 30 m, and z3 = 15 m. From Prob. 6.138, where K = 0, the flow rate was 0.056 m3/s toward reservoir 1. After some iteration: K = 55: gives h J = 15.04 m, Q1 = −0.04 m3 /s Q 2 = 0.044 m 3 /s Q 3 = −0.004 m 3 /s Ans.
From Fig. 6.19b, a butterfly valve coefficient K ≈ 55 occurs at θopening ≈ 35°. Ans.
*6.141 In the five-pipe horizontal network of Fig. P6.141, assume that all pipes have a friction factor f = 0.025. For the given inlet and exit flow rate of 0.06 m3/s of water at 20°C, determine the flow rate and direction in all pipes. If pA = 800 kPa gage, determine the pressures at points B, C, and D. Solution: For water at 20°C, take ρ = 1000 kg/m3 and µ = 0.001 kg/m-s. Each pipe has a head loss which is known except for the square of the flow rate:
88
Fig. P6.141
Pipe AC: h f =
8fLQ 2 π 2gd 5
2
2 AC |AC = 8(0.025)(900)Q 2 5 = K ACQ AC ,
π (9.81)(0.15)
where K AC ≈ 24482
Similarly, K AB = 7746, K BC = 6012, K CD = 7746, K BD = 567354.
(Q in m /s) 3
There are two triangular closed loops, and the total head loss must be zero for each. Using the flow directions assumed on the figure P6.141 above, we have Loop A-B-C: 7746Q 2AB + 6012Q 2BC − 24482Q 2AC = 0 2 Loop B-C-D: 6012Q 2BC + 7746Q CD − 567354Q 2BD = 0
And there are three independent junctions which have zero net flow rate:
Junction A: Q AB + Q AC = 0.06; B: Q AB = Q BC + Q BD ; C: Q AC + Q BC = Q CD These are five algebraic equations to be solved for the five flow rates. The answers are: Q AB = 0.0355, Q AC = 0.0245, Q BC = 0.02863, Q CD = 0.05313, Q BD = 0.00687 m3 / s Ans. (a)
The pressures follow by starting at A (800 kPa) and subtracting off the friction losses: p B = p A − ρgK ABQ 2AB = 800 × 10 3 − 9810(7746)(0.0355)2 = 704 kPa
Similarly, p C ≈ 656 kPa and p D ≈ 440 kPa Ans. (b)
*6.142 Modify Prob. 6.141 above as follows: Let the inlet flow at A and the exit flow at D be unknown. Let pA − pB = 700 kPa. Compute the flow rate in all five pipes. Solution: Our head loss coefficients “K” from above are all the same. Head loss AB is known, plus we have two “loop” equations and two “junction” equations: p A − p B 700 × 10 3 = = 71.36 m = K ABQ 2AB = 7746Q 2AB , or Q AB = 0.096 m3 /s ρg 9810
89
Two loops: 71.36 + 6012Q 2BC − 24482Q 2AC = 0 2 6012Q 2BC + 7746Q CD − 567354Q 2BD = 0
Two junctions: QAB = 0.096 = QBC + QBD; QAC + QBC = QCD The solutions are in exactly the same ratio as the lower flow rates in Prob. 6.141:
Q AB = 0.096 m3 / s, Q BC = 0.0774 m3 / s, Q BD = 0.0186 m3 / s, Q CD = 0.1437 m3 / s, Q AC = 0.0663 m3 / s
Ans.
6.143 In Fig. P6.143 all four horizontal cast-iron pipes are 45 m long and 8 cm in diameter and meet at junction a, delivering water at 20°C. The pressures are known at four points as shown: p1 = 950 kPa p3 = 675 kPa
p2 = 350 kPa p4 = 100 kPa
Neglecting minor losses, determine the flow rate in each pipe.
Fig. P6.143
Solution: For water at 20°C, take ρ = 998 kg/m3 and µ = 0.001 kg/m⋅s. All pipes are cast iron, with ε/d = 0.26/80 = 0.00325. All pipes have L/d = 45/0.08 = 562.5. One solution method is to guess the junction pressure pa, iterate to calculate the friction factors and flow rates, and check to see if the net junction flow is zero:
Trying pa = 530 kPa gives ∑Q = −0.00296, hence iterate to pa ≈ 517 kPa:
90
*6.144 In Fig. P6.144 lengths AB and BD are 600 m and 450 m, respectively. The friction factor is 0.022 everywhere, and pA = 600 kPa gage. All pipes have a diameter of 0.15 m. For water at 20°C, determine the flow rate in all pipes and the pressures at points B, C, and D.
Fig. P6.144
Solution: For water at 20°C, take ρ = 1000 kg/m3 and µ = 0.001 kg/m-s. Each pipe has a head loss which is known except for the square of the flow rate: Pipe AC: h f =
8fLQ 2 8(0.022)(450)Q 2AC = = K ACQ 2AC , where K AC ≈ 10772 π 2gd 5 π 2 (9.81)(0.15)5
Similarly, KAB = KCD = 14363, KBD = 10772, and KBC = 17953. The solution is similar to Prob. 6.141, except that (1) the K’s are different; and (2) junctions B and C have additional flow leaving the network. The basic flow relations are: Loop ABC: 14363Q 2AB + 17953Q 2BC − 10772Q 2AC = 0 2 Loop BCD: 17953Q 2BC + 14363Q CD − 10772Q 2BD = 0
Junctions A,B,C: Q AB + Q AC = 0.04; Q AB = Q BC + Q BD + 0.02; Q AC + Q BC = Q CD + 0.01 In this era of PC “equation solvers” such as EES, it is probably not necessary to dwell upon any solution methods. For handwork, one might guess QAB, then the other four are obtained in sequence from the above relations, plus a check on the original guess for QAB. The assumed arrows are shown above. It turns out that we have guessed the direction incorrectly on QBC above, but the others are OK. The final results are: Q AB = 0.01898 m3 /s (toward B); Q AC = 0.02102 m3 /s (toward C)
Q BC = 0.004774 m3 / s (toward B); Q CD = 0.00625 m3 / s (toward D); Q BD = 0.00375 m3 / s (to D) Ans. (a) The pressures start at A, from which we subtract the friction losses in each pipe: p B = p A − ρgK ABQ 2AB = 600 × 10 3 − 9810(14363)(0.01898)2 = 549 kPa
Similarly, we obtain p C = 553 kPa; p D = 548 kPa Ans. (b)
91
6.145 A water-tunnel test section has a 1-m diameter and flow properties V = 20 m/s, p = 100 kPa, and T = 20°C. The boundary-layer blockage at the end of the section is 9 percent. If a conical diffuser is to be added at the end of the section to achieve maximum pressure recovery, what should its angle, length, exit diameter, and exit pressure be? Solution: For water at 20°C, take ρ = 998 kg/m3 and µ = 0.001 kg/m⋅s. The Reynolds number is very high, Re = ρVd/µ = (998)(20)(1)/(0.001) ≈ 2.0Ε7; much higher than the diffuser data in Fig. 6.28b (Re ≈ 1.2E5). But what can we do (?) Let’s use it anyway:
6.146 For Prob. 6.145, suppose we are limited 1.8 × 10 –5 by space to a total diffuser length of 10 meters. What should be the diffuser angle, exit diameter, and exit pressure for maximum recovery? Solution: We are limited to L/D = 10.0. From Fig. 6.28b, read Cp,max ≈ 0.62 at AR ≈ 4 and 2θ ≈ 6°. Ans. The exit diameter and pressure are
6.147 A wind-tunnel test section is 1 m square with flow properties V = 50 m/s, p = 101 kPa absolute, and T = 20°C. Boundary-layer blockage at the end of the test section is 8 percent. Find the angle, length, exit height, and exit pressure of a flat-walled diffuser added onto the section to achieve maximum pressure recovery. Solution: For air at 20°C and 101 kPa, take ρ = 1.2 kg/m3 and µ = 1.8 × 10 –5 kg/m-s. The Reynolds number is rather high, Re = ρVd/µ = (1.2)(50)(1)/ 1.8 × 10 –5 ≈ 3.3 × 10 6 ; much higher than the diffuser data in Fig. 6.28a (Re ≈ 2.8E5). But what can we do (?) Let’s use it anyway:
Then θ best ≈ 4.75°, L ≈ 17W1 ≈ 17 m, W2 ≈ (AR)W1 = 3.75(1) ≈ 3.75 m Ans. Cp ≈ 0.70 =
pe − pt p e − 101 × 10 3 = , or: p exit ≈ 102 kPa (1/2)ρV12 (1/2)(1.2)(50)2
Ans.
92
6.148 For Prob. 6.147 above, suppose we are limited by space to a total diffuser length of 10 m. What should the diffuser angle, exit height, and exit pressure be for maximum recovery? Solution: We are limited to L/W1 = 10.0. From Fig. 6.28a, read Cp,max ≈ 0.645 at AR ≈ 2.8 and 2θ ≈ 10°. Ans. The exit height and pressure are
Wl,e = (AR)W1 = (2.8)(1) ≈ 2.8 m Ans.
Cp, max ≈ 0.645 = [p e − (101 × 10 3 )]/[(1/2)(1.2)(50)2 ], or p e = 102 kPa Ans.
6.149 An airplane uses a pitot-static tube as a velocimeter. The measurements, with their uncertainties, are a static temperature of (−11 ± 3)°C, a static pressure of 60 ± 2 kPa, and a pressure difference (po − ps) = 3200 ± 60 Pa. (a) Estimate the airplane’s velocity and its uncertainty. (b) Is a compressibility correction needed? Solution: The air density is ρ = p/(RT) = (60000 Pa)/[(287 m2/s2⋅K)(262 K)] = 0.798 kg/m3. (a) Estimate the velocity from the incompressible Pitot formula, Eq. (6.97):
The overall uncertainty involves pressure difference, absolute pressure, and absolute temperature:
The uncertainty in velocity is 2%, therefore our final estimate is V ≈ 90 ± 2 m/s Ans. (a) Check the Mach number. The speed of sound is a = (kRT)1/2 = [1.4(287)(262)]1/2 = 324 m/s. Therefore Ma = V/a = 90/324 = 0.28 < 0.3. No compressibility correction is needed. Ans. (b)
6.150 For the pitot-static pressure arrange-ment of Fig. P6.150, the manometer fluid is (colored) water at 20°C. Estimate (a) the centerline velocity, (b) the pipe volume flow, and (c) the (smooth) wall shear stress.
Fig. P6.150
93
Solution: For air at 20°C and 1 atm, take ρ = 1.2 kg/m3 and µ = 1.8E−5 kg/m⋅s. For water at 20°C, take ρ = 998 kg/m3 and µ = 0.001 kg/m⋅s. The manometer reads
p o − p = (ρ water − ρ air )gh = (998 −1.2)(9.81)(0.040) ≈ 391 Pa Therefore VCL = [2Δp/ρ ]1/2 = [2(391)/1.2]1/2 ≈ 25.5 m/s
Ans. (a)
We can estimate the friction factor and then compute average velocity from Eq. (6.43):
6.151 For the 20°C water flow of Fig. P6.151, use the pitot-static arrangement to estimate (a) the centerline velocity and (b) the volume flow in the 10-cm-diameter smooth pipe. (c) What error in flow rate is caused by neglecting the 0.3-m elevation difference? Solution: For water at 20°C, take ρ = 1000 kg/m3 and µ = 0.001 kg/m-s. For the manometer reading h = 5 cm,
Fig. P6.151
p oB − p A = (SG merc − 1)(ρg)water h + ρwater g(0.3 m) but from the energy equation, p A − p B = ρwater gh f-AB − ρwater g(0.3 m) Therefore p oB − p B = (SG − 1)ρgh mano + ρgh f-AB where friction loss h f-AB ≈ f(ΔL/d)(V2 /2g)
94
Thus the pitot tube reading equals the manometer reading (of about 6 kPa) plus the friction loss between A and B (which is only about 0.2 Pa), so there is only a small error: 1/2
1/2
2Δp 2(6161) (SG − 1)ρgh = (13.56 − 1)(9810)(0.05) ≈ 6161 Pa, VCL ≈ = 1000 ρ 1000(2.98)(0.1) or VCL ≈ 3.51 m/s, so Vavg ≈ 0.85VCL ≈ 2.98 m/s, Re = ≈ 298000, 0.001 so fsmooth ≈ 0.01448, or Δp friction = f(L/d)ρV2 /2 ≈ 193 Pa If we now correct the pitot tube reading to Δp pitot ≈ 6161 + 193 = 6354 Pa, we may iterate and converge rapidly to the final estimate:
f ≈ 0.01443, VCL ≈ 3.57 m / s; Q ≈ 0.0238 m3 / s; Vavg ≈ 3.04 m / s
Ans. (a, b)
The error compared to our earlier estimate V ≈ 2.98 m/s is about 2% Ans. (c)
6.152 An engineer who took college fluid mechanics on a pass-fail basis has placed the static pressure hole far upstream of the stagnation probe, as in Fig. P6.152, thus contaminating the pitot measurement ridiculously with pipe friction losses. If the pipe flow is air at 20°C and 1 atm and the manometer fluid is Meriam red oil (SG = 0.827), estimate the air centerline velocity for the given manometer reading of 16 cm. Assume a smooth-walled tube.
Fig. P6.152
Solution: For air at 20°C and 1 atm, take ρ = 1.2 kg/m3 and µ = 1.8E−5 kg/m⋅s. Because of the high friction loss over 10 meters of length, the manometer actually shows poB less than pA, which is a bit weird but correct:
95
P6.153 Fig. P6.153 shows a jet of liquid being discharged to the atmosphere from a nozzle attached to a pipe end. The diameters of the pipe and the nozzle mouth are D and d, respectively, while the densities of the liquid and the manometer fluid are ρ and ρm, respectively. A piezometer installed at (1) registers a reading of H, while a Pitot static tube placed across (1) and (3) registers a differential manometer reading of h. Energy loss may be ignored. (a) Find the discharge rate in terms of H, D, and d. (b) Find the relation between the Pitot static tube reading h and the piezometer reading H.
Fig. P6.153 Solution: By piezometer
p1 = ρ gH
By Continuity
v1 D 2 = v2 d 2
⇒
v1 = v2 (d / D)2
By Bernoulli eqn (ignoring losses)
p1 v12 v22 + = ρ g 2g 2g ⇒
p1 = 12 ρ(v22 – v12 ) = 12 ρv22 [1 – (d / D)4 ] 1/2
1/2
Hence
2 p1 / ρ v2 = 4 1 – (d / D)
2gH = 4 1 – (d / D)
Discharge
1/2 πd2 π d 2 2gH Q= v2 = 4 1 – (d / D)4 4
Piezometer
p1 = ρgH
ρ Pitot static tube p3 – p1 = ρ gh m − 1 ρ p3 is stagnation pressure
p3 = 12 ρ v22 = p1 / [1 – (d / D)4 ]
from above.
Substituting and rearranging,
h=
H [(D / d) – 1][ ρm / ρ – 1] 4
96
6.154 Professor Walter Tunnel must measure velocity in a water tunnel. Due to budgetary restrictions, he cannot afford a pitot-static tube, so he inserts a total-head probe and a static-head probe, as shown, both in the mainstream away from the wall boundary layers. The two probes are connected to a manometer. (a) Write an expression for tunnel velocity V in terms of the parameters in the figure. (b) Is it critical that h1 be measured accurately? (c) How does part (a) differ from a pitot-static tube formula?
Fig. P6.154
Solution: Write Bernoulli from total-head inlet (1) to static-head inlet (2):
Combine this with hydrostatics through the manometer:
Introduce this into the expression for V above, for the final result:
This is exactly the same as a pitot-static tube—h1 is not important. Ans. (b, c) 6.155 Kerosene at 20°C flows at 18 m3/h in a 5-cm-diameter pipe. If a 2-cm-diameter thinplate orifice with corner taps is installed, what will the measured pressure drop be, in Pa? Solution: For kerosene at 20°C, take ρ = 804 kg/m3 and µ = 1.92E−3 kg/m⋅s. The orifice beta ratio is β = 2/5 = 0.4. The pipe velocity and Reynolds number are:
From Eqs. (6.112) and (6.113a) [corner taps], estimate Cd ≈ 0.6030. Then the orifice pressure-drop formula predicts
97
*6.156 Gasoline at 20°C flows at 105 m3/h in a 10-cm-diameter pipe. We wish to meter the flow with a thin-plate orifice and a differential pressure transducer that reads best at about 55 kPa. What is the proper β ratio for the orifice? Solution: For gasoline at 20°C, take ρ = 680 kg/m3 and µ = 2.92E−4 kg/m⋅s. This problem is similar to Example 6.21 in the text, but we don’t have to be so precise because we don’t taps, etc. The pipe velocity is know the exact geometry: corner taps,
From Fig. 6.41, which is reasonable for all orifice geometries, read Cd ≈ 0.61. Then
Checking back with Fig. 6.41, we see that this is about right, so no further iteration is needed for this level of accuracy.
6.157 The shower head in Fig. P6.157 delivers water at 50°C. An orifice-type flow reducer is to be installed. The up-stream pressure is constant at 400 kPa. What flow rate, in m3/h, results without the reducer? What reducer orifice diameter would decrease the flow by 40 percent? Solution: For water at 50°C, take ρ = 988 kg/m3 and µ = 0.548E−3 kg/m⋅s. Further assume that the shower head is a poor diffuser, so the pressure in the head is also about 400 kPa.
Fig. P6.157
Assume the outside pressure is sea-level standard, 101 kPa. From Fig. 6.41 for a ‘typical’ orifice, estimate Cd ≈ 0.61. Then, with β ≈ 0 for the small holes, each hole delivers a flow rate of
98
or Q1 hole ≈ 2.65E−5 m 3/s and Q total = 45Q1 hole ≈ 0.00119
m3 ≈ 4.28m 3 /h s
(
)
This is a large flow rate—a lot of expensive hot water. Checking back, the inlet pipe for this flow rate has ReD ≈ 183000, so Cd ≈ 0.60 would be slightly better and a repeat of the calculation would give Qno reducer ≈ 0.00117 m3/s ≈ 4.21 m3/h. Ans. A 40% reduction would give Q = 0.6(0.00117) = 7.04E−4 m3/s ÷ 45 = 1.57E−5 m3/s for each hole, which corresponds to a pressure drop
6.158 A 10-cm-diameter smooth pipe contains an orifice plate with D: taps and β = 0.5. The measured orifice pressure drop is 75 kPa for water flow at 20°C. Estimate the flow rate, in m3/h. What is the nonrecoverable head loss? Solution: For water at 20°C, take ρ = 998 kg/m3 and µ = 0.001 kg/m⋅s. We know everything in the orifice relation, Eq. (6.104), except Cd, which we can estimate (as 0.61):
This is converged: Q = 0.0249(0.605) = 0.0150 m3/s ≈ 54 m3/h. Ans. (a) (b) From Fig. 6.44, the non-recoverable head loss coefficient is K ≈ 1.8, based on Vt:
99
6.159 Water at 20°C flows through the orifice in the figure, which is monitored by a mercury manometer. If d = 3 cm, (a) what is h when the flow is 20 m3/h; and (b) what is Q when h = 58 cm? Solution: (a) Evaluate V = Q/A = 2.83 m/s and ReD = ρVD/µ = 141,000, β = 0.6, thus Cd ≈ 0.613.
Fig. P6.159
where we have introduced the manometer formula Δp = (ρmercury − ρwater)gh.
Solve this problem when h = 58 cm is known and Q is the unknown. Well, we can see that the numbers are the same as part (a), and the solution is
6.160 The 1-m-diameter tank in Fig. P6.160 is initially filled with gasoline at 20°C. There is a 2-cm-diameter orifice in the bottom. If the orifice is suddenly opened, estimate the time for the fluid level h(t) to drop from 2.0 to 1.6 meters. Solution: For gasoline at 20°C, take ρ = 680 kg/m 3 and µ = 2.92E−4 kg/m⋅s. The
Fig. P6.160
100
orifice simulates “corner taps” with β ≈ 0, so, from Eq. (6.112), Cd ≈ 0.596. From the energy equation, the pressure drop across the orifice is Δp = ρgh(t), or
Set the Q’s equal, separate the variables, and integrate to find the draining time:
6.161 A pipe connecting two reservoirs, as in Fig. P6.161, contains a thin-plate orifice. For water flow at 20°C, estimate (a) the volume flow through the pipe and (b) the pressure drop across the orifice plate. Solution: For water at 20°C, take ρ = 998 kg/m3 and µ = 0.001 kg/m⋅s. The energy equation should include the orifice head loss and the entrance and exit losses:
Fig. P6.161
The final Re = ρVD/µ ≈ 166000, and Q = (π/4)(0.05)2(3.33) ≈ 0.00653 m3/s Ans. (a) (b) The pressure drop across the orifice is given by the orifice formula: ReD = 166000, β = 0.6, Cd ≈ 0.609 (Fig. 6.41):
101
*6.162 Air flows through a 6-cm-diameter smooth pipe which has a 2-m-long per-forated section containing 500 holes (diameter 1 mm), as in Fig. P6.162. Pressure outside the pipe is sea-level standard air. If p1 = 105 kPa and Q1 = 110 m3/h, estimate p2 and Q2, assuming that the holes are approximated by thin-plate orifices. Hint: A momentum control volume may be very useful.
Fig. P6.162
Solution: For air at 20°C and 105 kPa, take ρ = 1.25 kg/m3 and µ = 1.8E−5 kg/m⋅s. Use the entrance flow rate to estimate the wall shear stress from the Moody chart:
Further assume that the pressure does not change too much, so Δporifice ≈ 105000 − 101350 ≈ 3650 Pa. Then the flow rate from the orifices is, approximately,
Then V2 = Q2/A2 = 0.01225/[(π/4)(0.06)2] ≈ 4.33 m/s. A control volume enclosing the pipe walls and sections (1) and (2) yields the x-momentum equation:
Thus p2 = 105000 + 71 ≈ 105 kPa also and above is correct: Q2 = 0.0123 m3/s. Ans.
6.163 A smooth pipe containing ethanol at 20°C flows at 7 m3/h through a Bernoulli obstruction, as in Fig. P6.163. Three piezometer tubes are installed, as shown. If the obstruction is a thin-plate orifice, estimate the piezometer levels (a) h2 and (b) h3.
102
Fig. P6.163
Solution: For ethanol at 20°C, take ρ = 789 kg/m3 and µ = 0.0012 kg/m⋅s. With the flow rate known, we can compute Reynolds number and friction factor, etc.:
From Fig. 6.44, at β = 0.6, K ≈ 1.5. Then the head loss across the orifice is
Then the piezometer change between (2) and (3) is due to Moody friction loss:
6.164 In a laboratory experiment, air at 20°C flows from a large tank through a 2-cmdiameter smooth pipe into a sea-level atmosphere, as in Fig. P6.164. The flow is metered by a long-radius nozzle of 1-cm diameter, using a manometer with Meriam red oil (SG = 0.827). The pipe is 8 m long. The measurements of tank pressure and manometer height are as follows:
Fig. P6.164
ptank, Pa (gage): 60
320
1200
2050
2470
3500
4900
hmano, mm:
38
160
295
380
575
820
6
103
Use this data to calculate the flow rates Q and Reynolds numbers ReD and make a plot of measured flow rate versus tank pressure. Is the flow laminar or turbulent? Compare the data with theoretical results obtained from the Moody chart, including minor losses. Discuss. Solution: For air take ρ =1.2 kg/m3 and µ = 0.000015 kg/m⋅s. With no elevation change and negligible tank velocity, the energy equation would yield
Since Δp is given, we can use this expression plus the Moody chart to predict V and Q = AV and compare with the flow-nozzle measurements. The flow nozzle formula is:
The friction factor is given by the smooth-pipe Moody formula, Eq. (6.48) for ε = 0. The results may be tabulated as follows, and the plot on the next page shows excellent (too good?) agreement with theory. ptank, Pa:
60
320
1200
2050
2470
3500
4900
V, m/s (nozzle data):
2.32
5.82
11.9
16.1
18.2
22.3
26.4
Q, m3/h (nozzle data):
2.39
6.22
12.9
17.6
19.9
24.5
29.1
Q, m3/h (theory):
2.31
6.25
13.3
18.0
20.0
24.2
28.9
fMoody:
0.0444
0.0331 0.0271 0.0252 0.0245 0.0234 0.0225
104
6.165 Gasoline at 20°C flows at 0.06 m3/s through a 15-cm pipe and is metered by a 9-cm-diameter long-radius flow nozzle (Fig. 6.40a). What is the expected pressure drop across the nozzle? Solution: For gasoline at 20°C, take ρ = 680 kg/m and µ = 2.92E−4 kg/m⋅s. Calculate the pipe velocity and Reynolds number:
The ISO correlation for discharge (Eq. 6.114) is used to estimate the pressure drop:
P6.166 An engineer needs to monitor a flow of 20°C gasoline at about 50±5 m3/h through a 10-cm-diameter smooth pipe. She can use an orifice plate, a long-radius flow nozzle, or a venturi nozzle, all with 5-cm-diameter throats. The only differential pressure gage available is accurate in the range 30 to 60 kPa. Disregarding flow losses, which device is best? Solution: For gasoline at 20°C, take ρ = 680 kg/m3 and µ = 2.92E-4 kg/m-s. We are given
β = 2/4 = 0.5. The flow rate is in the range 0.0125 < Q < 0.0153 m3/s. The pipe Reynolds number is in the range ReD = 412,000 ± 10%. The throat diameter is 0.05 m, and its area is (π/4)(0.05 m)2 = 0.00196 m2. Our basic “obstruction” formula is Eq. (6.104):
Q = Cd At
2 Δp 2 Δp m3 2 = C (0.00196m ) = 0.0139 ± 0.0014 d ρ(1 − β 4 ) (680kg / m 3 ){1 − (0.5)4 } s
It remains only to determine Cd for the three devices and then calculate Δp. The results are: Orifice plate, D:1/2D taps: Cd ≈ 0.605 ,
Δp = 35.4 to 53.1 kPa
Long-radius flow nozzle:
Cd ≈ 0.99 ,
Δp = 13.2 to 19.8 kPa
Venturi nozzle:
Cd ≈ 0.977 ,
Δp = 13.6 to 20.3 kPa
Ans.
Only the orifice plate, with its high losses, is compatible with the available pressure gage.
105
6.167 Kerosene at 20°C flows at 20 m3/h in an 8-cm-diameter pipe. The flow is to be metered by an ISA 1932 flow nozzle so that the pressure drop is 7 kPa. What is the proper nozzle diameter? Solution: For kerosene at 20°C, take ρ = 804 kg/m3 and µ =1.92E−3 kg/m⋅s. We cannot calculate the discharge coefficient exactly because we don’t know β, so just estimate Cd:
Now compute a better Cd from the ISA nozzle correlation, Eq. (6.115):
Iterate once to obtain a better β ≈ 0.515, d = 0.515(8 cm) ≈ 4.12 cm Ans. 6.168 Two water tanks, each with base area of 0.1 m2, are connected by a 1-cm-diameter long-radius nozzle as in Fig. P6.168. If h = 0.3 m as shown for t = 0, estimate the time for h(t) to drop to 0.075 m. Solution: For water at 20°C, take ρ = 1000 kg/m3 and µ = 0.001 kg/m-s. For a longradius nozzle with β ≈ 0, guess Cd ≈ 0.98 and Kloss ≈ 0.9 from Fig. 6.44. The elevation difference h must balance the head losses in the nozzle and submerged exit: Δz = ∑h loss
Fig. P6.168
Vt2 Vt2 = ∑K = (0.9 nozzle + 1.0 exit ) = h, solve Vt = 3.21 h 2g 2(9.81)
1 dh dh π 2 hence Q = Vt ( 0.01) ≈ 0.000252 h = − A tank = −0.05 4 2 dt dt The boldface factor 1/2 accounts for the fact that, as the left tank falls by dh, the right tank rises by the same amount, hence dh/dt changes twice as fast as for one tank alone. We can separate and integrate and find the time for h to drop from 0.3 m to 0.075 m: 0.3
∫
0.075
dh = 0.00504 h
t final
∫ 0
dt, or: t final =
2
(
0.3 − 0.075 0.00504
) ≈ 108.7 s
Ans.
P6.169 Find the minimum diameter of the throat of a venture meter that is to be installed in a horizontal pipe of internal diameter 0.15 m carrying as much as 0.05 m3/s of water. It is given that the pressure head difference between the inlet and the throat of the meter must not exceed 20 m of water.
106
Solution: Assuming ideal flow p v2 p1 v12 + = 2 + 2 ρ g 2g ρ g 2g where
v2 = v1(D/d)2 ∴
v12 [(D / d)4 −1] = ( p1 − p2 ) / ρ g = h 2g
⇒
2gh v1 = (D / d)4 −1
Ideal
1/2 π D2 π D 2 2gh Q= v1 = 4 4 (D / d)4 −1
1/2
Actual discharge 1/2 π D 2 2gh 4 (D / d)4 −1 Assume Cd = 0.98
Qact = Cd Q = Cd
Given Qact = 0.05 m 3 /s, D = 0.15 m, h ≤ 20 m 1/2 π × 0.15 2 2 × 9.81× 20 0.05 = 0.98 4 (0.15 / d)4 −1 Solving, get d = 0.057 m 0.057 = 0.38 ∴ β=d/D= 0.15 From Fig. 6.43 Cd = 0.982 repeating the above, get d = 0.057 m (OK) Hence d ≥ 0.057 m for h ≤ 20 m
P6.170 Gasoline at 20°C flows through a 6-cm-diameter pipe. It is metered by a modern venturi nozzle with d = 4 cm. The measured pressure drop is 8.5 kPa. Estimate the flow rate in m3/s. Solution: For gasoline at 20°C, from Table A.3, ρ = 680 kg/m3 and µ = 2.92E-4 kg/m-s. We are not sure of the Reynolds number, so assume for the present that Fig. 6.43 is valid, in which case, from Eq. (6.116),
Then, from Eq. (6.104), the flow rate is
Q = Cd At
m3 π 2(8500 Pa) = 0.954( )(0.04m)2 = 0.00669 Ans. 4 s ρ(1 − β 4 ) 680[1 − (4 / 6)4 ] 2 Δp
Check Re D =
4ρQ 4(680)(0.00669) = = 330,000 OK, within range. π µ D π (2.92E − 4)(0.06)
________________________________________________________________________
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P6.171 It is desired to meter methanol at 20°C flowing through 10-cm-diameter pipe. The expected flow rate is about 0.02 m3/s. Two flowmeters are available: a venturi nozzle and a thin-plate orifice, each with d = 4 cm. The differential pressure gage on hand is most accurate at about 200–300 kPa. Which meter is better for this job? Solution: For methanol at 20°C, from Table A.3, ρ = 791 kg/m3 and µ = 5.98E-4 kg/m-s. Compute the average velocity in the pipe and find the discharge coefficients for each meter, for β = (4 cm)/(10 cm) = 0.4:
π (0.1)2 Vavg ,solve for V = 2.55 m/s or Vt = 15.92 m/s 4 ρVD (791)(2.55)(0.1) Orifice : Re D = = = 337,300, β = 0.4, Fig.6.41: Cd ≈ 0.601 µ 5.98 × 10−4 Venturi nozzle : 1.5E5 < Re D < 2E6, OK , β = 0.4, Fig.6.43 : Cd ≈ 0.983 Q = 0.02 m 3 /s =
Find the expected pressure drops from Eq. (6.104):
2 Δp
Vt = 15.92 m/s = Cd
4
ρ (1− β )
Venturi nozzle :
= Cd
2 Δp (791)[1− (0.4)4 ]
= 0.0509Cd Δp
Cd = 0.983 , solve
Δp ≈ 101 kPa
Thin - plate orifice : Cd = 0.601 , solve
Δp ≈ 270 kPa
The orifice plate is the better choice, for accuracy, although the head loss is much larger.
6.172 Ethanol at 20°C flows down through a modern venturi nozzle as in Fig. P6.172. If the mercury manometer reading is 10 cm, as shown, estimate the flow rate, in m3/h. Solution: For ethanol at 20°C, take ρ = 789 kg/m3 and µ = 1.2 × 10–3 kg/m⋅s. Given β = 0.5, the discharge coefficient is Cd = 0.9858 − 0.196(0.5)4.5 ≈ 0.9771
Fig. P6.172
The 20-cm displacement of manometer taps does not affect the pressure drop reading, because both legs are filled with ethanol. Therefore we proceed directly to Δp and Q:
Δp nozzle = (ρ merc − ρ eth )gh = (13550 − 789)(9.81)(0.1) ≈ 12519 Pa 2 Δp 1/2 π 2 2(12519) Hence Q = Cd A t = 0.9771 ≈ 0.0251 m 3 /s 0.075 ( ) 4 4 4 789(1− 0.5 ) ρ (1− β ) = 90.4 m 3 /h
Ans.
108
6.173 Modify Prob. P6.172 if the fluid is air at 20°C, entering the venturi at a pressure of 120 kPa. Should a compressibility correction be used? Solution: For air at 20°C and 120 kPa, take ρ = 1.47 kg/m3 and µ = 1.8 × 10–5 kg/m⋅s. With β still equal to 0.5, Cd still equals 0.9771 as previous page. The manometer reading is
Δp nozzle = (13550 −1.47)(9.81)(0.1) ≈ 13291 Pa, π 2(13291) 2 whence Q = 0.9771 ( 0.075 ) ≈ 0.6 m3 / s 4 4 1.47(1 − 0.5 )
Ans.
From this result, the throat velocity Vt = Q/At ≈ 136 m/s, quite high, the Mach number in the throat is approximately Ma = 0.4, a (small) compressibility correction might be expected. [Making a one-dimensional subsonic-flow correction, using the methods of Chap. 9, results in a throat volume flow estimate of Q ≈ 0.623 m3/s, about 4% higher.]
6.174 Water at 20°C flows in a long horizontal commercial-steel 6-cm-diameter pipe that contains a classical Herschel venturi with a 4-cm throat. The venturi is connected to a mercury manometer whose reading is h = 40 cm. Estimate (a) the flow rate, in m3/h, and (b) the total pressure difference between points 50 cm upstream and 50 cm downstream of the venturi. Solution: For water at 20°C, take ρ = 998 kg/m3 and µ = 0.001kg/m⋅s. For commercial steel, ε ≈ 0.046 mm, hence ε/d = 0.046/60 = 0.000767. First estimate the flow rate:
At this Reynolds number, we see from Fig. 6.42 that Cd does indeed ≈ 0.985 for the Herschel venturi. Therefore, indeed, Q = 0.0137 m3/s ≈ 49 m3/h. Ans. (a) (b) 50 cm upstream and 50 cm downstream are far enough that the pressure recovers from its throat value, and the total Δp is the sum of Moody pipe loss and venturi head loss. First work out the pipe velocity, V = Q/A = (0.0137)/[(π/4)(0.06)2] ≈ 4.85 m/s. Then
109
6.175 A modern venturi nozzle is tested in a laboratory flow with water at 20°C. The pipe diameter is 5.5 cm, and the venturi throat diameter is 3.5 cm. The flow rate is measured by a weigh tank and the pressure drop by a water-mercury manometer. The mass flow rate and manometer readings are as follows: kg/s: h, mm:
0.95
1.98
2.99
5.06
8.15
3.7
15.9
36.2
102.4
264.4
Use these data to plot a calibration curve of venturi discharge coefficient versus Reynolds number. Compare with the accepted correlation, Eq. (6.116). Solution: For water at 20°C, take ρ = 998 kg/m3 and µ = 0.001 kg/m⋅s. The given data of mass flow and manometer height can readily be converted to discharge coefficient and Reynolds number:
The data can then be converted and tabulated as follows: h, m:
0.037
0.0159
0.0362
0.1024
0.2644
Cd:
0.947
0.953
0.953
0.959
0.962
ReD:
22000
46000
69000
117000
189000
These data are plotted in the graph below, similar to Fig. 6.42 of the text:
They closely resemble the “classical Herschel venturi,” but this data is actually for a modern venturi, for which we only know the value of Cd for 1.5E5 < ReD ≤ 2E5:
The two data points near this Reynolds number range are quite close to 0.960 ± 0.002.
110
P6.176 An instrument popular in the beverage industry is the target flowmeter in Fig. P6.176. A small flat disk is mounted in the center of the pipe, supported by a strong but thin rod. (a) Explain how the flowmeter works. (b) If the bending moment M of the rod is measured at the wall, derive a formula for the estimated velocity of the flow. (c) List a few advantages and disadvantages of such an instrument.
Solution: (a) The flow creates a drag force F on the disk, approximately proportional to V2. (b) The bending moment is the drag force times the pipe radius. Thus the formula
We are neglecting the drag of the thin rod, especially if it is streamlined. In Chapter 7, Fig. 7.16, we learn that the drag coefficient of a disk is about 1.2 over a wide range of Reynolds number. (c) Advantages: low cost; disk easy to keep clean; works for a wide variety of fluids; can measure flow in either direction; useful in moderately unsteady flows. Disadvantages: needs calibration because the formula is too simplified; very poor accuracy at low velocities, due to the square-root relationship; heavy flows can break the rod; gummy flows can coat the disk and change the calibration; the drag force causes a pipe head loss.
P6.177 An instrument popular in the water supply industry, sketched in Fig. P6.177 is the single jet water meter. (a) How does it work? (b) What do you think a typical calibration curve would look like? (c) Can you cite further details, for example, reliability, head loss, cost [58]?
111
Solution: (a) The single-jet meter is similar in principle to the standard turbine meter, Fig. 6.32 of the text, except that, instead of flowing axially through the turbine, it creates an inplane jet that strikes the turbine blades and turns them. There are also multi-jet meters. (b) Just as in a standard turbine meter, the single-jet meter turns the turbine at a rate nearly proportional to the flow rate. The manufacturer states a “K factor” of an equation for turning rate Ω versus volume flow rate Q, in the form Ω = K Q. At very low flow rates, the Reynolds number is very low, and the manufacturer includes an error curve (a few per cent deviation) for low rates. (c) The single-jet meter is quite reliable and prized for its accuracy at low flow rates. It is not used for large pipe sizes because, due to its offset design, it would have to be huge. ________________________________________________________________________ 6.178 Air flows at high speed through a Herschel venturi monitored by a mercury manometer, as shown in Fig. P6.178. The upstream conditions are 150 kPa and 80°C. If h = 37 cm, estimate the mass flow in kg/s. [HINT: The flow is compressible.] Solution: The upstream density is ρ1 = p1/(RT) = (150000)/[287(273 + 80)] = 1.48 kg/m3. The clue “high speed” means that we had better use the compressible venturi formula, Eq. (6.117):
Fig. P6.178
112
The pressure difference is measured by the mercury manometer:
The pressure ratio is thus (150 − 49.2)/150 = 0.67 and, for β = 2/3, we read Y ≈ 0.76 from Fig. 6.45. From Fig. 6.43 estimate Cd ≈ 0.985. The (compressible) venturi formula thus predicts:
*6.179 Modify Prob. 6.178 as follows. Find the manometer reading h for which the mass flow through the venturi is approximately 0.4 kg/s. [HINT: The flow is compressible.] Solution: This is, in fact, the answer to Prob. 6.178, but who knew? The present problem is intended as an iteration exercise, preferably with EES. We know the upstream pressure and density and the discharge coefficient, but we must iterate for Y and p2 in the basic formula:
The answer should be h = 0.37 m, Y ≈ 0.76, and Cd ≈ 0.985, as in Prob. 6.178, but the problem is extremely sensitive to the value of h. A 10% change in h causes only a 2% change in mass flow. The actual answer to Prob. 6.177 was a mass flow of 0.402 kg/s. EES reports that, for mass flow exactly equal to 0.400 kg/s, the required manometer height is h = 0.361 m. Ans.
FUNDAMENTALS OF ENGINEERING EXAM PROBLEMS: Answers FE 6.1 In flow through a straight, smooth pipe, the diameter Reynolds number for transition to turbulence is generally taken to be (a) 1500 (b) 2300 (c) 4000 (d) 250,000 (e) 500,000 FE 6.2 For flow of water at 20°C through a straight, smooth pipe at 0.06 m3/h, the pipe diameter for which transition to turbulence occurs is approximately (a) 1.0 cm (b) 1.5 cm (c) 2.0 cm (d) 2.5 cm (e) 3.0 cm FE 6.3 For flow of oil (µ = 0.1 kg/(m⋅s), SG = 0.9) through a long, straight, smooth 5-cmdiameter pipe at 14 m3/h, the pressure drop per meter is approximately (a) 2200 Pa (b) 2500 Pa (c) 10,000 Pa (d) 160 Pa (e) 2800 Pa FE 6.4 For flow of water at a Reynolds number of 1.03E6 through a 5-cm-diameter pipe of roughness height 0.5 mm, the approximate Moody friction factor is (a) 0.012 (b) 0.018 (c) 0.038 (d) 0.049 (e) 0.102
113
FE 6.5 Minor losses through valves, fittings, bends, contractions etc. are commonly modeled as proportional to (a) total head (b) static head (c) velocity head (d) pressure drop (e) velocity FE 6.6 A smooth 8-cm-diameter pipe, 200 m long, connects two reservoirs, containing water at 20°C, one of which has a surface elevation of 700 m and the other with its surface elevation at 560 m. If minor losses are neglected, the expected flow rate through the pipe is (a) 0.048 m3/h (b) 2.87 m3/h (c) 134 m3/h (d) 172 m3/h (e) 385 m3/h FE 6.7 If, in Prob. FE 6.6 the pipe is rough and the actual flow rate is 90 m3/hr, then the expected average roughness height of the pipe is approximately (a) 1.0 mm (b) 1.25 mm (c) 1.5 mm (d) 1.75 mm (e) 2.0 mm FE 6.8 Suppose in Prob. FE 6.6 the two reservoirs are connected, not by a pipe, but by a sharp-edged orifice of diameter 8 cm. Then the expected flow rate is approximately (a) 90 m3/h (b) 579 m3/h (c) 748 m3/h (d) 949 m3/h (e) 1048 m3/h FE 6.9 Oil (µ = 0.1 kg/(m⋅s), SG = 0.9) flows through a 50-m-long smooth 8-cm-diameter pipe. The maximum pressure drop for which laminar flow is expected is approximately (a) 30 kPa (b) 40 kPa (c) 50 kPa (d) 60 kPa (e) 70 kPa FE 6.10 Air at 20°C and approximately 1 atm flows through a smooth 30-cm-square duct at 0.71 m3/s. The expected pressure drop per meter of duct length is (a) 1.0 Pa (b) 2.0 Pa (c) 3.0 Pa (d) 4.0 Pa (e) 5.0 Pa FE 6.11 Water at 20°C flows at 3 cubic meters per hour through a sharp-edged 3-cmdiameter orifice in a 6-cm-diameter pipe. Estimate the expected pressure drop across the orifice. (a) 440 Pa (b) 680 Pa (c) 875 Pa (d) 1750 Pa (e) 1870 Pa FE 6.12 Water flows through a straight 10-cm-diameter pipe at a diameter Reynolds number of 250,000. If the pipe roughness is 0.06 mm, what is the approximate Moody friction factor? (a) 0.015 (b) 0.017 (c) 0.019 (d) 0.026 (e) 0.032 FE 6.13 What is the hydraulic diameter of a rectangular air-ventilation duct whose crosssection is 1 meter by 25 cm? (a) 25 cm (b) 40 cm (c) 50 cm (d) 75 cm (e) 100 cm FE 6.14 Water at 20°C flows through a pipe at 0.02 m3/s with a friction head loss of 12.95 m. What is the power required to drive this flow? (a) 0.16 kW (b) 1.88 kW (c) 2.54 kW (d) 3.41 kW (e) 4.24 kW FE 6.15 Water at 20°C flows at 0.0126 m3/s through a pipe 150 m long and 8 cm in diameter. If the friction head loss is 12 m, what is the Moody friction factor? (a) 0.010 (b) 0.015 (c) 0.020 (d) 0.025 (e) 0.030
114
COMPREHENSIVE PROBLEMS C6.1 A pitot-static probe will be used to measure the velocity distribution in a water tunnel at 20°C. The two pressure lines from the probe will be connected to a U-tube manometer which uses a liquid of specific gravity 1.7. The maximum velocity expected in the water tunnel is 2.3 m/s. Your job is to select an appropriate U-tube from a manufacturer which supplies manometers of heights 8, 12, 16, 24 and 36 inches. The cost increases significantly with manometer height. Which of these should you purchase? Solution: The pitot-static tube formula relates velocity to the difference between stagnation pressure po and static pressure ps in the water flow:
Meanwhile, the manometer reading h relates this pressure difference to the two fluids:
It would therefore be most economical to buy the 16-inch manometer. But be careful when you use it: a bit of overpressure will pop the manometer fluid out of the tube!
C6.2 A pump delivers a steady flow of water (ρ,µ) from a large tank to two other higherelevation tanks, as shown. The same pipe of diameter d and roughness ε is used throughout. All minor losses except through the valve are neglected, and the partially-closed valve has a loss coefficient Kvalve. Turbulent flow may be assumed with all kinetic energy flux correction coefficients equal to 1.06. The pump net head H is a known function of QA and hence also of VA = QA/Apipe, for example, where a and b are constants. Subscript J refers to the junction point at the tee where branch A splits into B and C. Pipe length LC is much longer than LB. It is desired to predict the pressure at J, the three pipe velocities and friction factors, and the pump head. Thus there are 8 variables: H, VA, VB, VC, fA, fB, fC, pJ. Write down the eight equations needed to resolve this problem, but do not solve, since an elaborate iteration procedure, or an equation solver such as EES, would be required. Solution: First, equation (1) is clearly the pump performance: (1)
3 Moody factors:
ε f A = fcn VA , d ε f B = fcn VB , d ε fC = fcn VC , d
Conservation of mass (constant area) at the junction J: VA = VB + VC
(2) (3) (4) (5)
115
Finally, there are three independent steady-flow energy equations: (6)
(7) (8)
Fig. PC6.2
C6.3 The water slide in the figure is to be installed in a swimming pool. The manufacturer recommends a continuous water flow of 1.39E−3 m3/s (about 22 gal/min) down the slide to ensure that customers do not burn their bottoms. An 80%-efficient pump under the slide, submerged 1 m below the water surface, feeds a 5-m-long, 4-cm-diameter hose, of roughness 0.008 cm, to the slide. The hose discharges the water at the top of the slide, 4 m above the water surface, as a free jet. Ignore minor losses and assume α = 1.06. Find the brake horsepower needed to drive the pump.
FigC6.3
116
Solution: For water take r = 998 kg/m3 and m = 0.001 kg/m⋅s. Write the steady-flow energy equation from the water surface (1) to the outlet (2) at the top of the slide:
pa α1V12 p α V2 1.39E−3 m + + z1 = a + 2 2 + z2 + h f − h pump , where V2 = = 1.106 2 ρ g 2g ρg 2g s π (0.02)
Solve for h pump = (z2 − z1 ) +
V22 L α 2 + f 2g d
Work out Red = ρVd/µ = (998)(1.106)(0.04)/0.001 = 44200, ε/d = 0.008/4 = 0.002, whence fMoody = 0.0268. Use these numbers to evaluate the pump head above:
C6.4 Suppose you build a house out in the ‘boonies,’ where you need to run a pipe to the nearest water supply, which fortunately is about 1 km above the elevation of your house. The gage pressure at the water supply is 1 MPa. You require a minimum of 3 gal/min when your end of the pipe is open to the atmosphere. To minimize cost, you want to buy the smallest possible diameter pipe with an extremely smooth surface. (a) Find the total head loss from pipe inlet to exit, neglecting minor losses. (b) Which is more important to this problem, the head loss due to elevation difference, or the head loss due to pressure drop in the pipe? (c) Find the minimum required pipe diameter. Solution: Convert 3.0 gal/min to 1.89E−4 m3/s. Let 1 be the inlet and 2 be the outlet and write the steady-flow energy equation:
Fig. C6.4
or:
117
(b) Thus, elevation drop of 1000 m is more important to head loss than Δp/ρg = 102 m. (c) To find the minimum diameter, iterate between flow rate and the Moody chart:
We are given hf = 1102 m and νwater = 1.005E−6 m2/s. We can iterate, if necessary, or use EES, which can swiftly arrive at the final result:
C6.5 Water at 20°C flows, at the same flow rate Q = 9.4E−4 m3/s, through two ducts, one a round pipe, and one an annulus, as shown. The cross-section area A of each duct is identical, and each has walls of commercial steel. Both are the same length. In the cross-sections shown, R = 15 mm and a = 25 mm. (a) Calculate the correct radius b for the annulus. (b) Compare head loss per unit length for the two ducts, first using the hydraulic diameter and second using the ‘effective diameter’ concept. (c) If the losses are different, why? Which duct is more ‘efficient’? Why?
Fig. C6.5
Solution: (a) Set the areas equal:
(b) Find the round-pipe head loss, assuming ν = 1.005E−6 m2/s:
Thus hf/L = (f/d)(V2/2g) = (0.0261/0.03)(1.332)/2/9.81 = 0.0785 (round) Ans. (b)
118
Annulus: Dh = 4A/P = 2(a-b) = 20 mm, same V = 1.33 m/s:
Effective-diameter concept: b/a = 0.8, Table 6.3: Deff = 0.667Dh = 13.3 mm. Then
NOTE: Everything here uses Deff except hf, which by definition uses Dh. We see that the annulus has about 85% more head loss than the round pipe, for the same area and flow rate! This is because the annulus has more wall area, thus more friction. Ans. (c)
C6.6 John Laufer (NACA Tech. Rep. 1174, 1954) gave velocity data for 20°C airflow in a smooth 24.7-cm-diameter pipe at Re ≈ 5 E5: u/uCL:
1.0
0.997 0.988 0.959 0.908 0.847 0.818 0.771 0.690
r/R:
0.0
0.102 0.206 0.412 0.617 0.784 0.846 0.907 0.963
The centerline velocity uCL was 30.5 m/s. Determine (a) the average velocity by numerical integration and (b) the wall shear stress from the log-law approximation. Compare with the Moody chart and with Eq. (6.43). Solution: For air at 20°C, take ρ = 1.2 kg/m3 and µ = 0.00018 kg/m⋅s. The average velocity is defined by the (dimensionless) integral
Prepare a spreadsheet with the data and carry out the integration by the trapezoidal rule:
The integral is evaluated on the spreadsheet below. The result is V/uCL ≈ 0.8356, or V ≈ (0.8356)(30.5) ≈ 25.5 m/s. Ans. (a) The wall shear stress is estimated by fitting the log-law (6.28) to each data point:
119
We know ν for air and are given u and y from the data, hence we can solve for u*. The spreadsheet gives u* ≈ 1.1 m/s ± 1%, or τw = ρu*2 = (1.2)(1.1)2 ≈ 1.45 Pa. Ans. (b) y/R
r/R
u/uCL
∫u/uCL 2π r /R dr/R
u*
1.000
0.000
1.000
.0000
—
0.898
0.102
0.997
.0104
1.126
0.794
0.206
0.988
.0421
1.128
0.588
0.412
0.959
.1654
1.126
0.383
0.617
0.908
.3613
1.112
0.216
0.784
0.847
.5657
1.099
0.154
0.846
0.818
.6498
1.101
0.093
0.907
0.771
.7347
1.098
0.037
0.963
0.690
.8111
1.097
0.000
1.000
0.000
.8356
—
We make similar estimates from the Moody chart by evaluating Re and f and iterating:
This converges to V ≈ 26.5 m/s Ans. and τ w = (f/8)ρV2 ≈ 1.42 Pa. Ans.
C6.7 Consider energy exchange in fully-developed laminar flow between parallel plates, as in Eq. (6.63). Let the pressure drop over a length L be Δp. Calculate the rate of work done by this pressure drop on the fluid in the region (0 < x < L, −h < y < +h) and compare with the integrated energy dissipated due to the viscous function Φ from Eq. (4.50) over this same region. The two should be equal. Explain why this is so. Can you relate the viscous drag force and the wall shear stress to this energy result? Solution: From Eq. (6.63), the velocity profile between the plates is parabolic:
Let the width of the flow be denoted by b. The work done by pressure drop Δp is:
Meanwhile, from Eq. (4.50), the viscous dissipation function for this fully-developed flow is:
120
Integrate this to get the total dissipated energy over the entire flow region of dimensions L by b by 2h:
The two energy terms are equal. There is no work done by the wall shear stresses (where u = 0), so the pressure work is entirely absorbed by viscous dissipation within the flow field. Ans.
C6.8 This text has presented the traditional correlations for turbulent smooth-wall friction factor, Eq. (6.38), and the law-of-the-wall, Eq. (6.28). Recently, groups at Princeton and Oregon [56] have made new friction measurements and suggest the following smooth-wall friction law:
1 f
= 1.930 log10 ( Re D
f ) − 0.537
In earlier work, they also report that better values for the constants κ and B in the log-law, Eq. (6.28), are κ ≈ 0.421 ± 0.002 and B ≈ 5.62 ± 0.08. (a) Calculate a few values of f in the range 1E4 ≤ ReD ≤ 1E8 and see how the two formulas differ. (b) Read Ref. 56 and briefly check the five papers in its bibliography. Report to the class on the general results of this work. Solution: The two formulas are practically identical except as the Reynolds number is very high or very low. The new formula was fit to new, and extensive, friction data in Ref. 56 and can thus be said to be slightly more accurate. Here is a table of calculations. ReD 3000 10000 30000 100000 300000 1000000 3.0E+06 1.0E+07 3.0E+07 1.0E+08
fPrandtl 0.04353 0.03089 0.02349 0.01799 0.01447 0.01165 0.009722 0.008104 0.006949 0.005941
f Ref.56
Difference
0.04251 0.0305 0.02344 0.01811 0.01464 0.01186 0.009938 0.008316 0.007153 0.006134
-2.41% -1.10% -0.18% 0.62% 1.22% 1.76% 2.17% 2.56% 2.86% 3.15%
They differ by no more than 3%.
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C6.9 A pipeline has been proposed to carry natural gas 2760 km from Alaska’s North Slope to Calgary, Alberta, Canada. The (smooth) pipe diameter will be 1.3 m. The gas will be at high pressure, averaging 17 MPa. (a) Why? The proposed flow rate is 100 million cubic meter per day at sea-level conditions. (b) What volume flow rate, at 20°C, would carry the same mass at the high pressure? (c) If natural gas is assumed to be methane (CH4), what is the total pressure drop? (d) If each pumping station can deliver 9,000 kW to the flow, how many stations are needed? Solution: From Table A.4, for CH4, R = 518 m2/(s2-K) and µ = 1.03E-5 kg/m-s. Sea-level density is ρo = p/RT = 101350/[518(288)] = 0.679 kg/m3. The proposed mass flow rate is
υol = 1×108 m3; Q = υol / time =(1×108 m3 ) / (24 × 3600s) = 1157m3 / s m = ρ oQ = (0.679 kg / m3 )(1157 m3 / s) = 786 kg / s (a) The high pressure means that the gas is compacted about 170 times and thus much less volume needs to be pumped. Ans.(a) (b) At 17 × 106 Pa and 20°C, the average gas density would be
p 17 ×106 Pa kg ρ = = = 112 2 2 RT m3 (518 m / s K )(293K ) To match the mass flows, we would have m standard = 890
kg kg m3 = m pipeline = (112 )Q pipeline , hence Q pipeline = 7.95 Ans.(b) s s m3
(c) For the pressure drop, first find the Moody (smooth-wall) friction factor: V pipeline =
Q 7.95m3 / s m ρVD (112)(5.99)(1.3) = = 5.99 ; Re pipeline = = µ A ( π / 4)(1.3m)2 s 1.03×10 –5 = 8.47 ×107
1 f
≈ 2.0 log(
Re D f ) 2.51
yields
f smooth ≈ 0.006066
Then the Darcy formula gives the (horizontal) pressure drop: Δptotal = f
Lρ 2 2760000m 112 V = (0.006066)( )( )(5.99)2 = 2.59 × 107 Pa Ans.(c) D2 1.3m 2
(d) Total power = Q Δp = (7.95)(2.59 × 107) = 2.06 × 108 W = 2.06 × 105 kW total. If we divide this by 9,000 kW per pump, we get 22.9 or 23 pump stations minimum. Ans. (d) Each station must increase the gas pressure by 2.59 × 107/23 = 1.126 MPa, only 7% of the average pressure in the pipeline.
