Question

The total length of the piping system made of cast iron providing hot water to a house is considered to have a 42m length with a constant diameter. The entrance gage pressure is 85 kPa and exit at 15 kPa gaged. Consider the water density equal 980 Kg/m' and the viscosity 1.138x10 Kg/ms. a) Compute the piping system diameter if the volumetric flow rate provided by this system is equal to 0.1 L/s. There is no pump present in this part. Pgage = 85 kPa Supply tank Pgage = 15 kPa a House tank Transport line, 42 m b) This volumetric flow rate is considered not enough for the house, instead we use a 1 Ls flow rate, using a 3 cm diameter cast iron pipe and adding a pump. The system contains 2 gate valves fully open, 2 Manged smooth 90° bends, I sharp-edged inlet, 1 reentrant pipe exit and 4 threaded unions. The considered pump efficiency is 75%. Compute the required electric input power for the pump. Consider the velocity correction factor equal to 1.05. FOR PDF SLFTFS_4e Chapl pdf SLFTES_4e_Chapi pdf SLFTFS_4e_Chap1.pdf

Answer 1

a) Data's given The required flow rate is given by, Q=0.1 lit/s Q=0.1*10- m/s = 0.0001m/s L = 42 m Velocity of the flow is given by, Q= AV V = A 0.000lm/s IxD2 4 0.000127 V -m/s D? Reynolds number is given by, pVd Re=

0.000127 980x -XD Re= D2 1.138x10- 109.65 Re: D Relative roughness is given by, Esteel = 0.26 mm € 0.00026 = D D To find friction factor using Colebrook equation, 1-11 6.9 -1.8log10 + 3.7 Re 1.11 ST- -1.810g10 0.00026 D 3.7 + 6.9 109.65 D 2-454x10-5 -15 = -1.810g 10 + 6.9D 109.65 01:11 Note: f can be calculated by using moody diagram also. Total head loss = Major loss + minor loss

h₂ = hgthm fLV? +0 2gd 0.000127 fx42 D4 D 2x 9.81 3.45269x108xf h; PzV +2, +h; By applying Bernoulli's theorem P. ² + +Z2 +hp pg 2g Pg 2g P1 = 85 kPa ; P = 15kPa ; V1 = V2 = V Z2 -22 = Z; hp = 0 PR-P2 = 11 Pg (85-15) 103 3.45269x108xf 980x 9.81 D 3.45269x108xf 7.281 = DS Model calculation for D=12mm=0.012 m Velocity of the flow for corresponding diameter is given by,

0.000127 V= (0.012) V = 0.882 m/s To find friction factor, = --1.8log 10 2.454x10-5 01:11 + 6.9D 109.65 -1.8log 10 2.454x10-5 6.9x0.012 + (0.012)!!! 109.65 1 -1.8-2.3892 ✓f = 4.3 ✓f f = 0.054 To calculate the head supplied is given by, 3.45269x108x f 7.281= DS 3.45269x10-8 <0.054 7.281= D D=0.012 m

D=12 mm S.NO d (in mm) d (in m) V (m/s) 1/(f)^0.5 f D D (in mm) 1 11 0.011 1.049587 4.249646 0.055373 0.01213 12.12976 2 11.2 0.0112 1.012436 4.260482 0.055091 0.012117 12.11741 3 11.4 0.0114 0.977224 4.270975 0.054821 0.012105 12.10549 4 11.6 0.0116 0.943817 4.281137 0.054561 0.012094 12.09399 5 11.8 0.0118 0.912094 4.290977 0.054311 0.012083 12.08289 6 12 0.012 0.881944 4.300506 0.054071 0.012072 12.07217 7 12.2 0.0122 0.853265 4.309732 0.053839 0.012062 12.06183 8 12.4 0.0124 0.825963 4.318666 0.053617 0.012052 12.05184 9 12.6 0.0126 0.79995 4.327315 0.053403 0.012042 12.0422 10 12.8 0.0128 0.775146 4.335688 0.053197 0.012033 12.03289 11 13 0.013 0.751479 4.343793 0.052998 0.012024 12.02391

b) The required flow rate is given by, Q=1 lit/s Q=1x10- m/s = 0.001m/s L = 42 m Velocity of the flow is given by, Q = AV V= V = 0.001m/s x(0-03) 4. V =1.415 m/s Reynolds number is given by, pVd Re= 980x1.415x0.03 Re= 1.138x10-3 Re= 36548.7

Relative roughness is given by, Esteel = 0.26 mm E 0.00026 = 0.0086 D 0.03 To find friction factor using Colebrook equation, 1-11 6.9 -1.8log10 ED 3.7 + Re 1.11 0.0086 6.9 -1.810910 + 3.7 36548.68 = 5.141 f = 0.038 Note: f can be calculated by using moody diagram also. Minor loss coefficient 2*Gate valve fully open, K=0.15 2*flanged smooth 90 bends, K=0.3 1*sharp edged inlet, K=0.5 1*reentrant pipe exit, K=1

4*Threaded unions, K=0.08 Note: The loss coefficient values will differ according to various textbooks. Minor loss coefficient K=2*(0.15) + 2*(0.3) + 0.5 +1 +4*(0.08) = 2.72 Total head loss = Major loss + minor loss hệ = h +7 hi fLV2 2gd fL V2 h, +EK d 28 0.038 x 42 )(1.415) h +2.72 0.03 2x9.81 h; = 5.7043 m +E** 2g -10.05.2018 By applying Bernoulli's theorem + a = pg P, P, V? +2 +hp ta +Z2 +h; 2g pg 2g P1 = 85 kPa ; P = 15kPa ; Vi = 0; V2 =V Z2 -2 = Z;hp = ? ; h; = 5.7043m P-P V2 +h, = a 2 +h; pg 2g

(85–15)x103 (1.415) +h; = 1.05 +5.7043m 980x 9.81 2 x 9.81 7.281+hp = 5.8114m hp = 1.4695 m Pump Power is given by, P PgQh, n 980x9.81x0.001x1.4695 P= 0.75 P=18.836W