Question

Problem #2 Water tank is supplied with water at a flow rate Q = 0.005 m/s. A 5 cm diameter, 15 m long pipe is attached to the tank at the bottom and discharges to the atmosphere as shown in the figure. The pipe has roughness e = 0.0015 mm and the water is at 18°C. Answer the following: a) If the depth of water in the tank is initially 7 m, does the amount of water in the tank increase or decrease with time? b) What would be the steady state depth of water in the tank? c) If a valve is fitted at the end of the drainage pipe, find the loss coefficient K of the valve that would maintain the depth of water in the tank at 7 m. d-5 cm

Answer 1

. .. Solution : - . solutionGiven that? d=15cm K 15m G1 = 0.0508) ; d = 5CM ; L=151 .. Daughness, e = 0.015 M. (a) Min = Pa =.60.005 XP). LL . properties of water need to be taken at 18c. P=998.68 kg led; it = 0.00105+ kgle-s So, nin=10.005 X99866) = 4.9934 kg/s M = 0.001054 kg /4-5, Apply Teapoulli prin ciped between land 9;- Il tvtzi = P2 + v + Zg thei - 0 eg. 29 : eg 29. point 1& 2 are exposed to atmosphere, 30, Po=P9 =0 ( Grouge press wre)

Also, at the surface of large tanks velocity of water to. Take point 2 as the daton level. so, 22=0 and 2,=hm e hafi - Head loss due to friction : ... Assuming flow through pipe as turbulent Then fine mody's chart 15 Relatie soughness. (cl) = (0 C houdhui 3x105.. .. we, get friction factor (f) = 0,009 I hfiz fave 10.009x15xv2 ago 12x9.81 x(5/100) J +0.137649" @ ..fwcy the egin , : 0+o+h=+ vo tö +0.13.76V ? a) so, vý= 5.393.h. - ® - Also; Min = Mout = (dm) =74.9934 - PAQ VA = 1 dage) . Tag

74.9937 - 995.65 x ( 56)* x 75.306h - Come) 14.9934 - 4.5.15.14 =don) - ca. at time t=0; ih = 7M ( Given) :50) dec.6.954 kg/s since this value is coming out negative it means that the tank negatine is' feasing water at the sente, of 6:954kgls. so the level of water in the tank decuases with time. by At steady state, there will be no variation in flow parameters with respect to the tione. ..so, das ) -0 fracy Equation © we see, .4.9934 - 4,515h2=0 So, 1 ha). 293MA -Simg. This will be the depth of water in the tack at steady state,

29 (e) of a valve is filled at the end of . drainge then we need to account Minor loss as well Minoy head loss due to valve, hf2 - Kvinne So, Equation @° is modified as 942 4 79 the thelle so, ototh = oth to 10 +0.1376 va three => h=(0.1855 0.051K) - © Min - Mout = (dm) 27 4.9934 - PAR "= Colori > 48934-995-18** *

Now in quals tion , we need to maintain the depth of water in tack to M at Steady state. Sos C) = 0; heart 30, 4.988-98.15 xÅ MED! One solving this we can get, ah . fk = 17,97 tis This is the loss coefficient of valve.