Question

A secondary hot-water service system has 55m of 28mm circulation pipework and 40m of 15mm circulation pipework. Water leaves the cylinder at 65°C and returns at 60°C. Air temperature around the pipes is 15°C. Pressure loss rate in the pipework is 260N/m'. The frictional resistance of the pipe fittings is equivalent to 25% of measured pipe length. Specify the head and flowrate required for the secondary circulation pump and choose a suitable pump. Use Table 4.3 and Fig. 4.16. Table 43 Flow of water in copper pipes of various diameters 360 Apel(im) Water flow rate (kes) for diameters of 6 mm 15 mm 22m 28 mm 35 mm 200 0.013 0.065 0.174 0.381 0.656 0.015 0.075 0.202 0.441 0.760 0.016 0.081 0.219 0.478 0.823 0.018 0.090 0242 0.529 0.910 0.019 0.096 0.257 0561 0.965 0.0200.104 0.278 0.607 040 500 v0.50 0.021 0.109 0.291 0.635 1.090 0.023 0.116 0310 0.677 L160 0.024 0.120 0.323 0.703 1.210 0.025 0.127 0.340 0.741 1.270 700 0.026 0,131 0.352 0.766 1.320 760 0.027 0.138 0.368 0.802 1.380 0028 0.14210379 0.825 1420 42 mm 1.060 v= 1.0 1.230 1.330 1.470 1.560 1.680 v 1.50 1.760 1.880 1.950 2.050 2.120 2220 2280 800 15 Wher low rate Source Reproduced from OBSEGA (CIRSE. 1906) by permission of the Chartered In tion of Being Services Engineert Note Www Wlocky (m/ ap/EL. rate of pressure out due to friction 16 e performanse curves

Answer 1

sater m Solution Given that; A Secondary hot-water Service System * for 55m of 28mm from table 14.3. - At 2600/m3 pressure loss rate alkals) = 0.491 Kgls. * For yom of 15mm. from table 4.3 → At 260 N/m2 altegls) = 0.075 kgls. The fructional resistance of the pipe fittings is Equivalent to 25% of measured pipe length EL = 1.25855 EL = 68.75m. relation- DPH E , X9807 N/m=260 EL 260 x 68.75 Head (H) = 9807 | Hi= 1.82.267 89 < 0.441

Elg = 1.25 X40 [Elz = som Head (H) = 260x50 - 9807 14 = 1.325m 89 = 0.075. 1 and head * As less flow rate (8) is less than than 4. - is suitable from A so pump 'C' would be 1. fig. 4.16