(Refer to Example Problem 11.11.) Air at 68^◦ F flows in a prototype 1300-ft-long circular pipe with a diameter of 3 ft and an absolute pipe roughness of 0.03 ft, as illustrated in Figure EP 11.12. A smaller model of the larger prototype is designed in order to study the flow characteristics of

Question

(Refer to Example Problem 11.11.) Air at [latex] 68^{\circ } F [/latex] flows in a prototype 1300-ft-long circular pipe with a diameter of 3 ft and an absolute pipe roughness of 0.03 ft, as illustrated in Figure EP 11.12. A smaller model of the larger prototype is designed in order to study the flow characteristics of transitional pipe flow. The model fluid is water at [latex] 70^{\circ } F [/latex], the velocity of water in the smaller model pipe is 0.07 ft/sec, and the model scale, λ is 0.2. The flow resistance is modeled by the friction factor, f = 8 [latex] C_{D} [/latex]. (a) Determine the pressure drop (and head loss) in the flow of the water in the model. (b) Determine the velocity flow of the air in the prototype pipe flow in order to achieve dynamic similarity between the model and the prototype. (c) Determine the pressure drop (and head loss) in the flow of the air in the prototype in order to achieve dynamic similarity between the model and the prototype.

Accepted Answer

(a) In order to determine the pressure drop (and head loss) in flow of the water in the model, the major head loss equation, Equation 11.122 [latex] h_{f} = \frac{ au _{w}L}{\gamma R_{h}} = \frac{\Delta p}{\gamma } = S_{f}L = C_{D} ho v^{2} \frac{L}{\gamma R_{h}} = \frac{v^{2}L}{C^{2}R_{h}} = f \frac{L}{D} \frac{v^{2}}{2g} = \left(\frac{vn}{R_{h}^{2/3}} ight)^{2}L [/latex], is applied as follows:

[latex] h_{f} = \frac{\Delta p}{\gamma } = f \frac{L}{D} \frac{v^{2}}{2g} [/latex]

where the friction factor, f is used to model the flow resistance. Because the Reynolds number, 2,000 < R < 4,000 (transitional pipe flow) is assumed, the friction factor, f is a function of both ɛ/D and R, as illustrated by the Moody diagram in Figure 8.1. However the Colebrook equation, Equation 8.33 [latex]C = \sqrt{\frac{8g}{f} } [/latex], presents a mathematical representation of the Moody diagram in Figure 8.1. Furthermore, in order to determine the length, diameter, and the absolute pipe roughness of the model pipe, the model scale, λ (inverse of the length ratio) is applied. The fluid properties for water are given in Table A.2 in Appendix A.

[latex] D_{p}: = 3 ft [/latex] [latex] L_{p}: = 1300 ft [/latex] [latex] \varepsilon _{p} : = 0.03 ft [/latex] [latex] \lambda : = 0.2 [/latex]

Guess value: [latex] D_{m}: = 0.1 ft [/latex] [latex] L_{m}: = 1 ft [/latex] [latex] \varepsilon _{m}: = 0.01 ft [/latex]

Given

[latex] \lambda = \frac{D_{m}}{D_{p}} [/latex] [latex] \lambda = \frac{L_{m}}{L_{p}} [/latex] [latex] \lambda = \frac{\varepsilon _{m}}{\varepsilon _{p}} [/latex]
[latex] \left ( \begin{matrix} D_{m} \ L_{m} \ \varepsilon _{m} \end{matrix} ight ) : = Find (D_{m}, L_{m}, \varepsilon _{m}) = \left ( \begin{matrix} 0.6 \ 260 \ 6 imes 10^{-3} \end{matrix} ight ) ft [/latex]
[latex] slug: = 1 lb \frac{sec^{2}}{ft} [/latex] [latex] ho _{m} : = 1.936 \frac{slug}{ft^{3}} [/latex] [latex] \mu _{m} : = 20.5 imes 10^{-6} lb \frac{sec}{ft^{2}} [/latex]

[latex] g: = 32.174 \frac{ft}{sec^{2}} [/latex] [latex] \gamma _{m}: = ho _{m}. g = 62.289 \frac{lb}{ft^{3}} [/latex] [latex] V_{m}: = 0.07 \frac{ft}{sec} [/latex]

[latex] R_{m}: = \frac{ ho _{m} .V_{m} .D_{m}}{\mu _{m}} = 3.966 imes 10^{3} [/latex]

Guess value: [latex] h_{fm}: = 1 ft [/latex] [latex] \Delta P_{m}: = 1 \frac{lb}{ft^{2}} [/latex] [latex] f_{m}: = 0.01 [/latex]

Given

[latex] h_{fm} = f_{m} \frac{L_{m}}{D_{m}} \frac{V^{2}_{m}}{2g} [/latex] [latex] \frac{1}{\sqrt{f_{m}} } = - 2 log \left(\frac{\frac{\varepsilon _{m}}{D_{m}} }{3.7} + \frac{2.51 }{R_{m}. \sqrt{f_{m}} } ight) [/latex]
[latex] \Delta P_{m} = h_{fm} . \gamma _{m} [/latex]
[latex] \left ( \begin{matrix} h_{fm} \ \Delta P_{m} \ f_{m} \end{matrix} ight ) : = Find (h_{fm, \Delta P_{m} , f_{m}}) [/latex]

[latex] h_{fm}= 1.622 imes 10^{-3} ft [/latex] [latex] \Delta P_{m} = 0.101 \frac{lb}{ft^{2}} [/latex] [latex] f_{m}= 0.049 [/latex]

(b) To determine the velocity flow of the air in the prototype pipe flow in order to achieve dynamic similarity between the model and the prototype for transitional pipe flow, the ɛ/D must remain a constant between the model and prototype as follows:

[latex] \left(\frac{\varepsilon}{D} ight)_{p} = \left(\frac{\varepsilon }{D } ight)_{m} [/latex]
[latex]\frac{\varepsilon _{p}}{D_{p}} = 0.01 [/latex] [latex]\frac{\varepsilon _{m}}{D_{m}} = 0.01 [/latex]

Furthermore, the R must remain a constant between the model and prototype as follows:

[latex] \underbrace{\left[\left(\frac{ ho vL}{\mu } ight)_{p} ight] }_{R_{p}} = \underbrace{\left[\left(\frac{ ho vL}{ \mu } ight)_{m} ight] }_{R_{m}} [/latex]

The fluid properties for air are given in Table A.5 in Appendix A.

[latex] ho _{p} : = 0.00231 \frac{slug}{ft^{3}} [/latex] [latex] \mu _{p} : = 0.376 imes 10^{-6} lb \frac{sec}{ft^{2}} [/latex]

Guess value: [latex] V_{p}: = 1 \frac{ft}{sec} [/latex] [latex]R_{p}: = 3000 [/latex]

Given
[latex] R_{p} : = \frac{ ho _{p} . V_{p}. D_{p}}{\mu _{p}}[/latex]
[latex] R_{p} = R_{m} [/latex]
[latex] \left ( \begin{matrix} V_{p} \ R_{p} \end{matrix} ight ) : = Find ( V_{p}, R_{p}) [/latex]

[latex] V_{p} = 0.215 \frac{ft}{s} [/latex] [latex]R_{p} = 3.966 imes 10^{3} [/latex]

(c) To determine the pressure drop (and head loss) in the flow of the air in the prototype in order to achieve dynamic similarity between the model and the prototype for transitional pipe flow, the friction factor, f must remain a constant between the model and the prototype (which is a direct result of maintaining a constant ɛ/D and a constant R between the model and the prototype) as follows:

[latex] \underbrace{\left[\frac{\frac{h_{f}}{v^{2}L} }{2gD} ight]_{p} }_{f_{p}} = \underbrace{\left[\frac{\frac{h_{f}}{v^{2}L} }{2gD} ight]_{m} }_{f_{m}} [/latex]
[latex] \gamma _{p}: = ho _{p}.g =0.074 \frac{lb}{ft^{3}} [/latex]

Guess value: [latex] h_{fp}: = 1 ft [/latex] [latex] \Delta p_{p}: = 1 \frac{lb}{ft^{2}} [/latex] [latex] f_{p}: = 0.1 [/latex]

Given
[latex] f_{p} = \frac{h_{fp}}{\left(\frac{V^{2}_{p}. L_{p}}{2.g. D_{p}} ight) } [/latex] [latex] \Delta p_{p} = h_{fp}. \gamma _{p} [/latex]

[latex] f_{p} = f_{m} [/latex]
[latex] \left ( \begin{matrix} h_{fp} \ \Delta p_{p} \ f_{p} \end{matrix} ight ) : = Find (h_{fp},\Delta p_{p}, f_{p}) [/latex]

[latex] h_{fp} = 0.015 ft [/latex] [latex] \Delta p_{p} = 1.139 imes 10^{-3} \frac{lb}{ft^{2}} [/latex] [latex] f_{p}: = 0.049 [/latex]

Therefore, although the similarity requirements regarding the independent π term, [latex] \varepsilon /D ((\varepsilon /D)_{p} = (\varepsilon /D)_{m} = 0.01)[/latex] and the independent π term, R (“viscosity model”) ([latex] R_{p} = R_{m} = 3.966 imes 10^{3} [/latex]) are theoretically satisfied, the dependent π term (i.e., the friction factor, f ) will actually/practically remain a constant between the model and its prototype ([latex] f_{p} = f_{m} = 0.049 [/latex]) only if it is practical to maintain/attain the model velocity, pressure, fluid, scale, and cost.

Table A.2
Physical Properties for Water at Standard Sea-Level Atmospheric Pressure as a Function of Temperature
Temperature (θ) [latex]^{\circ } F[/latex]	Density (ρ) [latex]slug/ft^{3} [/latex]	Specific Weight (γ) [latex]Ib/ft^{3}[/latex]	Absolute (Dynamic) Viscosity (μ) [latex]10^{-6} Ib-sec/ft^{3}[/latex]	Kinematic Viscosity (ν) [latex]10^{-6} ft^{2}/sec[/latex]	Surface Tension (σ) lb=ft	Vapor Pressure [latex]( ho _{ u } )[/latex] psia	Bulk Modulus of Elasticity [latex](E_{\upsilon } )[/latex] psi
32	1.940	62.42	37.46	19.31	0.00518	0.0885	293,000
40	1.940	62.43	32.29	16.64	0.00514	0.1220	294,000
50	1.940	62.41	27.35	14.10	0.00509	0.1780	305,000
60	1.938	62.37	23.59	12.17	0.00504	0.2560	311,000
70	1.936	62.30	20.50	10.59	0.00498	0.3630	320,000
80	1.934	62.22	17.99	9.30	0.00492	0.5070	322,000
90	1.931	62.11	15.95	8.26	0.00486	0.6980	323,000
100	1.927	62.00	14.24	7.39	0.00480	0.9490	327,000
110	1.923	61.86	12.84	6.67	0.00473	1.2750	331,000
120	1.918	61.71	11.68	6.09	0.00467	1.6920	333,000
130	1.913	61.55	10.69	5.58	0.00460	2.2200	334,000
140	1.908	61.38	9.81	5.14	0.00454	2.8900	330,000
150	1.902	61.20	9.05	4.76	0.00447	3.7200	328,000
160	1.896	61.00	8.38	4.42	0.00441	4.7400	326,000
170	1.890	60.80	7.80	4.13	0.00434	5.9900	322,000
180	1.883	60.58	7.26	3.85	0.00427	7.5100	318,000
190	1.876	60.36	6.78	3.62	0.00420	9.3400	313,000
200	1.868	60.12	6.37	3.41	0.00413	11.5200	308,000
212	1.860	59.83	5.93	3.19	0.00404	14.6900	300,000
[latex]^{\circ } C[/latex]	[latex]kg/m^{3} [/latex]	[latex]KN/m^{3} [/latex]	[latex]N-sec/m^{2} [/latex]	[latex]10^{-6} m^{2} /sec[/latex]	[latex]N/m[/latex]	[latex]KN/m^{2} abs[/latex]	[latex]10^{6} KN/m^{2} [/latex]
0	999.8	9.805	0.001781	1.785	0.0756	0.611	2.02
5	1000.0	9.807	0.001518	1.519	0.0749	0.872	2.06
10	999.7	9.804	0.001307	1.306	0.0742	1.230	2.10
15	999.1	9.798	0.001139	1.139	0.0735	1.710	2.14
20	998.2	9.789	0.001002	1.003	0.0728	2.340	2.18
25	997.0	9.777	0.000890	0.893	0.0720	3.170	2.22
30	995.7	9.765	0.000798	0.800	0.0712	4.240	2.25
40	992.2	9.731	0.000653	0.658	0.0696	7.380	2.28
50	988.0	9.690	0.000547	0.553	0.0679	12.330	2.29
60	983.2	9.642	0.000466	0.474	0.0662	19.920	2.28
70	977.8	9.589	0.000404	0.413	0.0644	31.160	2.25
80	971.8	9.530	0.000354	0.364	0.0626	47.340	2.20
90	965.3	9.467	0.000315	0.326	0.0608	70.100	2.14
100	958.4	9.399	0.000282	0.294	0.0589	101.330	2.07

Table A.5
Physical Properties for Some Common Gases at Standard Sea-Level Atmospheric Pressure at Room Temperature [latex](68^{\circ }F or 20^{\circ }C )[/latex]
Gas at [latex]68^{\circ }[/latex]F	Chemical Formula	Molar Mass (m) slug=slug- mol	Density (ρ) [latex]slug/ft^{3} [/latex]	Absolute (Dynamic) Viscosity (μ) [latex]10^{-6} Ib-sec/ft^{2} [/latex]	Gas Constant (R) [latex]ft-Ib/(slug-^{\circ }R )=ft^{2}/(sec^{2} -^{\circ }R )[/latex]	Specific Heat		Specific Heat Ratio, [latex]K=C_{ ho }/C_{\upsilon }[/latex]
						[latex]C_{ ho }[/latex]	[latex]C_{\upsilon } [/latex]
						[latex]ft-Ib/(slug-^{\circ }R )=ft^{2}/(sec^{2} -^{\circ }R )[/latex]
Air		28.960	0.002310	0.376	1715	6000	4285	1.40
Carbon dioxide	[latex]CO_{2} [/latex]	44.010	0.003540	0.310	1123	5132	4009	1.28
Carbon monoxide	CO	28.010	0.002260	0.380	1778	6218	4440	1.40
Helium	He	4.003	0.000323	0.411	12,420	13,230	18,810	1.66
Hydrogen	[latex]H_{2}[/latex]	2.016	0.000162	0.189	24,680	86,390	61,710	1.40
Methane	[latex]CH_{2}[/latex]	16.040	0.001290	0.280	3100	13,400	10,300	1.30
Nitrogen	[latex]N_{2}[/latex]	28.020	0.002260	0.368	1773	6210	4437	1.40
Oxygen	[latex]O_{2}[/latex]	32.000	0.002580	0.418	1554	5437	3883	1.40
Water vapor	[latex]H_{2}O[/latex]	18.020	0.001450	0.212	2760	11,110	8350	1.33
at [latex]20^{\circ } C[/latex]		kg/kg-mol	[latex]kg/m^{3} [/latex]	[latex]10^{-6} N-sec/m^{2} [/latex]	[latex]N-m/(kg-^{\circ}K )=m^{2} /(sec^{2}-^{\circ}K )[/latex]	[latex]N-m/(kg-^{\circ}K )=m^{2} /(sec^{2}-^{\circ}K )[/latex]
Air		28.960	1.2050	18.0	287	1003	716	1.40
Carbon dioxide	[latex]CO_{2} [/latex]	44.010	1.8400	14.8	188	858	670	1.28
Carbon monoxide	CO	28.010	1.1600	18.2	297	1040	743	1.40
Helium	He	4.003	0.1660	19.7	2077	5220	3143	1.66
Hydrogen	[latex]H_{2}[/latex]	2.016	0.0839	9.0	4120	14,450	10,330	1.40
Methane	[latex]CH_{2}[/latex]	16.040	0.6680	13.4	520	2250	1730	1.30
Nitrogen	[latex]N_{2}[/latex]	28.020	1.1600	17.6	297	1040	743	1.40
Oxygen	[latex]O_{2}[/latex]	32.000	1.3300	20.0	260	909	649	1.40
Water vapor	[latex]H_{2}O[/latex]	18.020	0.7470	10.1	462	1862	1400	1.33

Question 11.12: (Refer to Example Problem 11.11.) Air at 68^◦ F flows in a p...

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