Gasoline at 20°C is pumped through a smooth 12-cm-diameter pipe 10 km long, at a flow rate of 75 m^3/h (330 gal/min). The inlet is fed by a pump at an absolute pressure of 24 atm. The exit is at standard atmospheric pressure and is 150 m higher. Estimate the frictional head loss hf, and compare it

Question

Gasoline at 20°C is pumped through a smooth 12-cm-diameter pipe 10 km long, at a flow rate of 75 [latex]m^3[/latex]/h (330 gal/min). The inlet is fed by a pump at an absolute pressure of 24 atm. The exit is at standard atmospheric pressure and is 150 m higher. Estimate the frictional head loss [latex]h_f[/latex], and compare it to the velocity head of the flow [latex]V^2[/latex]/(2g). (These numbers are quite realistic for liquid flow through long pipelines.)

Accepted Answer

• Property values: From Table A.3 for gasoline at 20°C, ρ = 680 kg/[latex]m^3[/latex], or [latex]\gamma[/latex] = (680)(9.81) = 6670 N/[latex]m^3[/latex].

Table A.3 Properties of Common Liquids at 1 atm and 20°C (68°F)
Liquid	ρ, kg/[latex]m^3[/latex]	µ, kg/(m·s)	Y, N/[latex]m^*[/latex]	[latex]p_{ u}, N/m^2[/latex]	Bulk modulus K, N/[latex]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- [latex]1^{\ddagger}[/latex]	3.6 E-2	–	1.31 E+9	15.7
SAE 10W30 oil	876	1.7 E-[latex]1^{\ddagger}[/latex]	–	–	–	14.0
SAE 30W oil	891	2.9 E-[latex]1^{\ddagger}[/latex]	3.5 E-2	–	1.38 E+9	18.3
SAE 50W oil	902	8.6 E-[latex]1^{\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]	[latex]\mu ,N\cdot s/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 [latex]0\leq T\leq 100°C[/latex]:
[latex] ho (kg/m^{3})\approx 1000-0.0178\mid T°C-4°C\mid ^{1.7}\pm 0.2\% [/latex]
[latex]\ln \frac{\mu }{\mu _{0}} \approx -1.704-5.306z+7.003z^{2}[/latex]
[latex]z=\frac{273K}{TK} \mu _{0}=1.788E-3kg/(m\cdot s)[/latex]

• Assumptions: Steady flow. No shaft work, thus [latex]h_p = h_t = 0[/latex]. If [latex]z_1 = 0[/latex], then [latex]z_2 = 150[/latex] m.

• Approach: Find the velocity and the velocity head. These are needed for comparison. Then evaluate the friction loss from Eq. (3.73).

[latex]\left(\frac{p}{\gamma} + \frac{V^2}{2g} + z ight)_{in} = \left(\frac{p}{\gamma} + \frac{V^2}{2g} + z ight)_{out} + h_{friction} - h_{pump} + h_{turbine}[/latex] (3.73)

• Solution steps: Since the pipe diameter is constant, the average velocity is the same everywhere:

[latex]V_{in} = V_{out} = \frac{Q}{A} = \frac{Q}{(\pi/4)D^2} = \frac{(75 m^3/h)/(3600 s/h)}{(\pi/4)(0.12 m)^2} \approx 1.84 \frac{m}{s}[/latex]

Velocity head = [latex]\frac{V^2}{2g} = \frac{(1.84 m/s)^2}{2(9.81 m/s^2)} \approx 0.173 m[/latex]

Substitute into Eq. (3.73) and solve for the friction head loss. Use pascals for the pressures and note that the velocity heads cancel because of the constant-area pipe.

[latex]\frac{p_{in}}{\gamma} + \frac{V^2_{in}}{2g} + z_{in} = \frac{p_{out}}{\gamma} + \frac{V^2_{out}}{2g} + z_{out} + h_f[/latex]

[latex]\frac{(24)(101,350 N/m^2)}{6670 N/m^3} + 0.173 m + 0 m = \frac{101,350 N/m^2}{6670 N/m^3} + 0.173 m + 150 m + h_f[/latex]

or [latex]h_f = 364.7 - 15.2 - 150 \approx 199 m[/latex]

The friction head is larger than the elevation change Δz, and the pump must drive the flow against both changes, hence the high inlet pressure. The ratio of friction to velocity head is

[latex]\frac{h_f}{V^2/(2g)} \approx \frac{199 m}{0.173 m} \approx 1150[/latex]

• Comments: This high ratio is typical of long pipelines. (Note that we did not make direct use of the 10,000-m pipe length, whose effect is hidden within [latex]h_f[/latex].) In Chap. 6 we can state this problem in a more direct fashion: Given the flow rate, fluid, and pipe size, what inlet pressure is needed? Our correlations for [latex]h_f[/latex] will lead to the estimate [latex]p_{inlet}[/latex] ≈ 24 atm, as stated here.

Question 3.21: Gasoline at 20°C is pumped through a smooth 12-cm-diameter p...

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