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10.53: PROBLEM DEFINITION Situation: Water is draining out of a tank through a galvanized iron pipe. ks = 0.006 in = 5 × 10−4 ft, L = 10 ft, H = 4 ft. The pipe is 1-in schedule 40 NPS, D = 1.049 in = 0.08742 ft. Find: Velocity in the pipe (ft/s). Flow rate (cfs) Assumptions: Steady flow. Component head loss is zero. Turbulent flow. Also, α2 = 1.0. Properties: Water (70 ◦ F), Table A.5, ρ = 1.94 slug/ ft3 , γ = 62.3 lbf/ ft3 , ν = 1.06 × 10−5 ft2 / s. PLAN Classify this problem as case 2 (V is unknown), then 1. Write the energy eqn., the Darcy-Weisbach eqn., etc. to produce a set of 4 equations with 4 unknowns. 2. Solve the set of equations using a computer program (we used TK Solver). 3. Find the flow rate with Q = V A. SOLUTION 1. Governing equations: • Energy equation (section 1 on water surface, section 2 at exit plane) p1 V2 V2 p2 + α1 1 + z1 + hp = + α2 2 + z2 + ht + hL γ 2g γ 2g V22 0 + 0 + (H + L) + 0 = 0 + + 0 + hf 2g • Darcy-Weisbach: hf = f 78 L V2 D 2g (1) (2) • Swamee-Jain: • Reynolds number: 0.25 f=£ ¡ ks log10 3.7D + Re = 5.74 Re0.9 ¢¤2 VD ν (3) (4) 2. Solution of Eqs. (1) to (4): hf = 11.04 ft Re = 113800 f = 0.0326 V = 13.8 ft/ s 3. Flow rate equation: πD2 V Q = ! Ã4 π (0.08742 ft)2 (13.8 ft/ s) = 8. 283 1 × 10−2 ft3 / s = 4 Q = 0.0828 cfs REVIEW • Notice that the turbulent flow assumption is valid because Re > 2300. • An easy way to solve case 2 and case 3 problems is to acquire a computer program that can solve coupled, non-linear equations. • Notice that most of the elevation head (14 ft) is converted to head loss (11 ft) . 79