LQ 128(0.0012)(1.2 m)q πρgd π(789)(9.81)(0.002) Solve for Q 1.90E 6 m /s = m /h. Ans.

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1 6.5 For the configuration shown in Fig. P6.5, the fluid is ethyl alcohol at 0 C, and the tanks are very wide. Find the flow rate that occurs, in m /h. Is the flow laminar? Solution: For ethanol, take ρ = 789 kg/m and µ = kg/m s. Write the energy equation from upper free surface (1) to lower free surface (): Fig. P f 1 1 p V p V + + z = + + z + h, with p = p and V V 0 ρg g ρg g 18µ LQ 18(0.001)(1. m)q Then hf = z1 z = 0.9 m = = 4 4 πρgd π(789)(9.81)(0.00) Solve for Q 1.90E 6 m /s = m /h. Ans. Check the Reynolds number Re = 4ρQ/(πµd) 795 OK, laminar flow.

2 6. SAE 0 oil at 0 C flows in the - cm-diameter pipe in Fig. P6., which slopes at 7. For the pressure measurements shown, determine (a) whether the flow is up or down and (b) the flow rate in m /h. Solution: For SAE 0 oil, take ρ = 891 kg/m and µ = 0.9 kg/m s. Evaluate the hydraulic grade lines: Fig. P6. pb HGLB = + zb = + 15 = 5.6 m; HGLA = + 0 = 57. m ρg 891(9.81) 891(9.81) Since HGL > HGL the flow is up Ans. (a) The head loss is the difference between hydraulic grade levels: A B 18µ LQ 18(0.9)(5)Q h = = 1.6 m = = πρgd π (891)(9.81)(0.0) f 4 4 Solve for Q = m /s 1.86 m / h Ans. (b) Finally, check Re = 4ρQ/(πµd) 68 (OK, laminar flow).

3 6.47 The gutter and smooth drainpipe in Fig. P6.47 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 5 inches 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. P6.47 V L V gδz (9.81)(4.) Δ z= + hf, hf = f, solve V = = g d g 1 + fl/ d 1 + f (4./0.07) For water, take ρ = 998 kg/m and µ = kg/m s. Guess f 0.0 to obtain the velocity estimate V 6 m/s above. Then Red ρvd/µ (998)(6)(0.07)/(0.001) 48,000 (turbulent). Then, for a smooth pipe, f 0.015, and V is changed slightly to 6.74 m/s. After convergence, we obtain V = 6.77 m/s, Q = V( π/4)(0.07) = 0.06 m /s Ans. (a)

4 A rainfall of 5 in/h = (5/1 ft/h)(0.048 m/ft)/(600 s/h) = m/s. The required roof area is roof drain rain A = Q / V = (0.06 m /s)/ m/s 740 m Ans. (b)

5 6.55 The reservoirs in Fig. P6.55 contain water at 0 C. If the pipe is smooth with L = 4500 m and d = 4 cm, what will the flow rate in m /h be for Δz = 100 m? Solution: For water at 0 C, take ρ = 998 kg/m and µ = kg/m s. The energy equation from surface 1 to surface gives p1= p and V1= V, thus h = z z = 100 m f 1 Fig. P6.55! 4500" V Then 100 m = f $, or fv & % 0.04 ' (9.81) Iterate with an initial guess of f 0.0, calculating V and Re and improving the guess: 1/! " m 998(0.94)(0.04) V $ 0.94, Re 700, fsmooth 0.04 & % 0.0 ' s /! " m Vbetter $ 0.88, Rebetter 500, fbetter 0.06, etc... & % 0.04 ' s Alternately, one could, of course, use EES. The above process converges to f = 0.07, Re = 5000, V = m/s, Q m /s 4.0 m / h. Ans.

6 6.6 Water at 0 C is to be pumped through 000 ft of pipe from reservoir 1 to at a rate of ft /s, as shown in Fig. P6.6. If the pipe is cast iron of diameter 6 in and the pump is 75 percent efficient, what horsepower pump is needed? Solution: For water at 0 C, take ρ = 1.94 slug/ft and µ =.09E 5 slug/ft s. For cast iron, take ε ft, or ε/d = /(6/1) Compute V, Re, and f: Fig. P6.6 Q ft V = = = 15. ; A ( π/4)(6/1) s ρvd 1.94(15.)(6/1) Re = = ε/ d = , fmoody 0.07 µ.09e 5 The energy equation, with p1 = p and V1 V 0, yields an expression for pump head: L V! 000 " (15.) hpump = Δ z + f = 10 ft % & = ft d g ' 6/1 ( (.)

7 ρgqhp 1.94(.)(.0)(450) Power: P = = = hp η 0.75 Ans.

8 6.78 In Fig. P6.78 the connecting pipe is commercial steel 6 cm in diameter. Estimate the flow rate, in m /h, if the fluid is water at 0 C. Which way is the flow? Solution: For water, take ρ = 998 kg/m and µ = kg/m s. For commercial steel, take ε mm, hence ε/d = 0.046/ With p1, V1, and V all 0, the energy equation between surfaces (1) and () yields Fig. P6.78 p z z + h f, or hf = m (flow to left) ρg 998(9.81) LV 50 V Guess turbulent flow: hf = f = f = 5.4, or: fv dg 0.06(9.81) ε " 0.178# m = , guess ffully rough , V %.64, Re = d ' & ( s m fbetter 0.004, Vbetter =.50, Rebetter , frd iteration (converged) s The iteration converges to f 0.005, V.49 m/s, Q = (π/4)(0.06) (.49) = m /s = 5 m /h Ans. 1/

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ρg 998(9.81) LV 50 V. d2g 0.062(9.81)

ρg 998(9.81) LV 50 V. d2g 0.062(9.81) 6.78 In Fig. P6.78 the connecting pipe is commercial steel 6 cm in diameter. Estimate the flow rate, in m 3 /h, if the fluid is water at 0 C. Which way is the flow? Solution: For water, take ρ = 998 kg/m

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