Question

Pipe Diameter (cm) D 10 d 5 Pipe Length (m) L1 6.57 L2 3.34 8.93 4.47 Ls 5.78 L 5.26 Lr 6.02 h = 18 m H = 20 m Note: all fittings are Threaded L. Ball Valve (closed) L1.D 45 Globe Valve (open) H) = 11.5 m 45% Ls, Led L7, D The two water tanks are connected with a pipe system made of galvanized iron. Water flows from the higher tank to the lower tank. The tanks have a much larger diameter than the pipe. Hence, change in their levels as a result of the flow is negligible. The ball valve in the purge branch remains closed, while the globe valve communicating the tanks is fully open. Considering steady flow with energy dissipation in the above figure, calculate the flow rate.

Answer 1

Let's divide the pipe into two categories. a) D=10 cm=0.1m b) d=5 cm=0.05m For d=5 cm=0.05m Length of the pipe, L=L2+ L3+ L4 +Ls +L=3.34+8.93+4.47+5.78+5.26=27.78 m Properties of water taken at 25 C Velocity of the flow is given by, Q = AV V= V= Ix0.05 4 V = 509.295Q m/s Reynolds number is given by, Re = PVd Vd Re = V 509.295Qx0.05 Re= 0.8927x10-6

Re = 25375916Q Relative roughness is given by, Esteel = 0.15mm E 0.15 mm = 0.003 d 50 mm To find friction factor, 1.11 6.9 =-1.8log 10 &/d 3.7 + Re 1.11 + -1.8log 10 0.003 3.7 + 6.9 25375916Q √f 6.9 -1.8log10 0.0008108+ 25375916Q Note: Using moody diagram also f can be calculated. Both uses the same equation. Minor loss coefficient Gate valve fully open, K=0.15 2*45 long radius 45, flanged, K=0.2 Sudden contraction

A _ doi? 0.052 = 0.25 0.12 AC From graph, K=0.38 Ball valve (closed), K= infinite (neglect the term) Minor loss coefficient K=0.15+2*0.2+0.38=0.93 Total head loss = Major loss + minor loss h = h, +h LV2 h; 2gd 2g fL hi + ΣΚ d 2g hy = (555.6fi+0.93)13220.23Q? For d=10 cm=0.1m Length of the pipe, L=Li+L7 =6.57+6.02=12.59 m Properties of water taken at 25 C Velocity of the flow is given by, А

Q V = 2x0.12 4 V =127.32Q m/s Reynolds number is given by, Vd Re= V 127.32Qx0.1 Re= 0.8927 x 10-6 Re=126879589 Relative roughness is given by, Esteel = 0.15mm E 0.15 mm = 0.0015 d 100 mm To find friction factor, 1.11 1 -1.8log 10 &ld 6.9 + Re 3.7 1-11 6.9 =-1.8log 10 0.0015 3.7 + 126879589

6.9 to =-1-8108,060-001716+ 126879589 Minor loss coefficient Reentrant entrance, K=0.8 Sharp edged exit, K=1 Sudden expansion 4_0.052 -= 0.25 A 0.12 = From graph, K=0.5 Minor loss coefficient K=1+0.8+0.25=2.05 Total head loss = Major loss + minor loss h; = hy + hm fL h; + ΣΚ d 2g h = (125.9f2 +2.05)826.26Q? By applying modified Bernoulli's theorem for water surface (1) and tank (2)

+ P. ² P V2 +Z+hp -+2 +Z2 +h; Pg 2 g Pg 28 P1 = P2 = Part Vi =V2 =V Z -Z, = h=18m;h, = 0 18=(555.6f4+0.93)13220.23Q2 + (125.982 +2.05)82626Q Note: the unit of Q is m3/s in above equation. From iterations performed in excel is attached here., From graph, the operating point is 30.175 m3/hr (0.00838 m3/s), f1=0.0326, f2=0.0235.

Q (m3/hr) Q (m3/s) 1/(f)^0.5 f1 1/(f)^0.5 f2 hp 1 0.0002778 4.94500391 0.04089 4.80915 0.04324 0.0246 3 0.0008333 5.29956679 0.03561 5.55115 0.03245 0.19368 5 0.0013889 5.39487603 0.03436 5.84887 0.02923 0.51968 7 0.0019444 5.43957094 0.0338 6.022 0.02758 1.0023 9 0.0025 5.46555201 0.03348 6.13783 0.02654 1.64147 11 0.0030556 5.48254663 0.03327 6.22162 0.02583 2.43718 13 0.0036111 5.49453237 0.03312 6.28538 0.02531 3.3894 15 0.0041667 5.5034402 0.03302 6.33568 0.02491 4.49812 17 0.0047222 5.51032123 0.03293 6.37646 0.02459 5.76335 19 0.0052778 5.51579673 0.03287 6.41022 0.02434 7.18509 21 0.0058333 5.52025754 0.03282 6.43867 0.02412 8.76332 23 0.0063889 5.52396186 0.03277 6.46297 0.02394 10.498 25 0.0069444 5.5270871 0.03273 6.48399 0.02379 12.3893 27 0.0075 5.52975924 0.0327 6.50235 0.02365 14.437 29 0.0080556 5.53207016 0.03268 6.51853 0.02353 16.6412 30 0.0083333 5.53311231 0.03266 6.52592 0.02348 17.802 30.175 0.0083819 5.53328772 0.03266 6.52717 0.02347 18.0092