A pump delivers 0.6 hp to water at 68°F, flowing in a 6-in-diameter asphalted cast iron horizontal pipe at V = 6 ft/s. What is the proper pipe length to match these conditions?

Question

Accepted Answer

• Approach: Find [latex]h_f[/latex] from the known power and find f from [latex]Re_d[/latex] and ε/d. Then find L.

• Water properties: For water at 68°F, Table A.3, converting to BG units, ρ = 1.94 slug/[latex]ft^3[/latex] and µ = 2.09E-5 slug/(ft-s).

Table A.3 Properties of Common Liquids at 1 atm and 20°C (68°F)
Liquid	[latex] ho, kg/m^3[/latex]	µ, kg/(m·s)	Y, N/m[latex]^*[/latex]	[latex]p_{ u}, N/m^2[/latex]	Bulk modulus K, [latex]N/m^2[/latex]	Viscosity parameter [latex]C^{\dagger}[/latex]
Ammonia	608	2.20 E-4	2.13 E-2	9.10 E+5	1.82 E+9	1.05
Benzene	881	6.51 E-4	2.88 E-2	1.01 E+4	1.47 E+9	4.34
Carbon tetrachloride	1590	9.67 E-4	2.70 E-2	1.20 E+4	1.32 E+9	4.45
Ethanol	789	1.20 E-3	2.28 E-2	5.73 E+3	1.09 E+9	5.72
Ethylene glycol	1117	2.14 E-2	4.84 E-2	1.23 E+1	3.05 E+9	11.7
Freon 12	1327	2.62 E-4	–	–	7.95 E+8	1.76
Gasoline	680	2.92 E-4	2.16 E-2	5.51 E+4	1.3 E+9	3.68
Glycerin	1260	1.49	6.33 E-2	1.43 E-2	4.35 E+9	28.0
Kerosene	804	1.92 E-3	2.8 E-2	3.11 E+3	1.41 E+9	5.56
Mercury	13,550	1.56 E-3	4.84 E-1	1.13 E-3	2.85 E+10	1.07
Methanol	791	5.98 E-4	2.25 E-2	1.34 E+4	1.03 E+9	4.63
SAE 10W oil	870	1.04 E-1[latex]^{\ddagger}[/latex]	3.6 E-2	–	1.31 E+9	15.7
SAE 10W30 oil	876	1.7 E-1[latex]^{\ddagger}[/latex]	–	–	–	14.0
SAE 30W oil	891	2.9 E-1[latex]^{\ddagger}[/latex]	3.5 E-2	–	1.38 E+9	18.3
SAE 50W oil	902	8.6 E-[latex]^{\ddagger}[/latex]	–	–	–	20.2
Water	998	1.00 E-3	7.28 E-2	2.34 E+3	2.19 E+9	Table A.1
Seawater (30%)	1025	1.07 E-3	7.28 E-2	2.34 E+3	2.33 E+9	7.28

[latex]^*[/latex]In contact with air.
[latex]^{\dagger}[/latex]The viscosity–temperature variation of these liquids may be fitted to the empirical expression

[latex]\frac{\mu}{\mu_{20^{\circ}C}} \approx \exp \left[C\left(\frac{293 K}{T K} -1 ight) ight][/latex]

with accuracy of ±6 percent in the range 0 ≤ T ≤ 100°C.
[latex]^{\ddagger}[/latex]Representative values. The SAE oil classifications allow a viscosity variation of up to ±50 percent, especially at lower temperatures.

Table A.1	Viscosity and Density of Water at 1 atm
T,°C	[latex] ho, kg/m^3[/latex]	µ, N·s/[latex]m^2[/latex]	[latex] u, m^2/s[/latex]	T,°F	[latex] ho, slug/ft^3[/latex]	[latex]\mu, Ib\cdot s/ft^2[/latex]	[latex] u, ft^2/s[/latex]
0	1000	1.788 E-3	1.788 E-6	32	1.94	3.73 E-5	1.925 E-5
10	1000	1.307 E-3	1.307 E-6	50	1.94	2.73 E-5	1.407 E-5
20	998	1.003 E-3	1.005 E-6	68	1.937	2.09 E-5	1.082 E-5
30	996	0.799 E-3	0.802 E-6	86	1.932	1.67 E-5	0.864 E-5
40	992	0.657 E-3	0.662 E-6	104	1.925	1.37 E-5	0.713 E-5
50	988	0.548 E-3	0.555 E-6	122	1.917	1.14 E-5	0.597 E-5
60	983	0.467 E-3	0.475 E-6	140	1.908	0.975 E-5	0.511 E-5
70	978	0.405 E-3	0.414 E-6	158	1.897	0.846 E-5	0.446 E-5
80	972	0.355 E-3	0.365 E-6	176	1.886	0.741 E-5	0.393 E-5
90	965	0.316 E-3	0.327 E-6	194	1.873	0.660 E-5	0.352 E-5
100	958	0.283 E-3	0.295 E-6	212	1.859	0.591 E-5	0.318 E-5
Suggested curve fits for water in the range 0 ≤ T ≤ 100°C:
[latex] ho(kg/m^3)\approx 1000-0.0178 \|T^{\circ}C-4^{\circ}C\|^{1.7} \pm 0.2\%[/latex]
[latex]\ln \frac{\mu}{\mu_0}\approx 1.704-5.306_z+{7.003_z}^2[/latex]
[latex]z=\frac{273 K}{T K} \mu_0=1.788E-3 kg/(m\cdot s)[/latex]

• Pipe roughness: From Table 6.1 for asphalted cast iron, ε = 0.0004 ft.

Table 6.1: ε
Material	Condition	ft	mm	Uncertainty, %
Steel	Sheet metal, new	0.0002	0.05	±60
	Stainless, new	7E-06	0.002	±50
	Commercial, new	0.0002	0.046	±30
	Riveted	0.01	3	±70
	Rusted	0.007	2	±50
Iron	Cast, new	0.0009	0.26	±50
	Wrought, new	0.0002	0.046	±20
	Galvanized, new	0.0005	0.15	±40
	Asphalted cast	0.0004	0.12	±50
Brass	Drawn, new	7E-06	0.002	±50
Plastic	Drawn tubing	5E-06	0.0015	±60
Glass	–	Smooth	Smooth
Concrete	Smoothed	0.0001	0.04	±60
	Rough	0.007	2	±50
Rubber	Smoothed	3E-05	0.01	±60
Wood	Stave	0.0016	0.5	±40

• Solution step 1: Find the pump head from the flow rate and the pump power:

[latex]Q = AV = \frac{\pi}{4}(0.5 ft)^2 \left(6\frac{ft}{s} ight) = 1.18 \frac{ft^3}{s}[/latex]

[latex]h_{pump} = \frac{Power}{ ho gQ} = \frac{(0.6 hp) [550(ft \cdot lbf )/(s \cdot hp)]}{(1.94 slug/ft^3) (32.2 ft/s^2) (1.18 ft^3/s)} = 4.48 ft[/latex]

• Solution step 2: Compute the friction factor from the Colebrook formula, Eq. (6.48):

$[latex]\frac{1}{f^{1/2}}=-2.0 \log \left(\frac{\epsilon /d}{3.7}+\frac{2.51}{Re_d f^{1/2}} ight)[/latex]$ (6.48)

[latex]Re_d = \frac{ ho Vd}{\mu} = \frac{(1.94)(6)(0.5)}{2.09 E-5} = 278,500 \frac{\epsilon}{d} = \frac{0.0004 ft}{0.5 ft} = 0.0008[/latex]

[latex]\frac{1}{\sqrt{f}} \approx 2.0 \log_{10} \left(\frac{\epsilon /d}{3.7} + \frac{2.51}{Re_d \sqrt{f}} ight)[/latex] yields f = 0.0198

• Solution step 3: Find the pipe length from the Darcy formula (6.10):

[latex]h_p = h_f = 4.48 ft = f \frac{L}{d} \frac{V^2}{2g} = (0.0198) \left(\frac{L}{0.5 ft} ight) \frac{(6 ft/s)^2}{2(32.2 ft/s^2)}[/latex]

Solve for L ≈ 203 ft

• Comment: This is Moody’s problem (Example 6.6) turned around so that the length is unknown.

Question 6.12: A pump delivers 0.6 hp to water at 68°F, flowing in a 6-in-d...

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