Question

Question 4 Liquid water enters a feedwater heater at inlet 1 with inlet conditions 15 MPa, 40°C at 60 kg/s. Another stream of water in saturated mixture enters the heater at inlet 2 at temperature of 175 °C. The heater operates at steady state and heat transfer to surrounding can be neglected. At exit 3, saturated liquid flows out at 275 kPa. Select two (2) different values of mixture quality at inlet 2 between 0.5 and 0.8, and subsequently plot the mass flow rate in kg/s at inlet 2 for these two cases as a function of the chosen quality values. 1 Question 5 Heat is to be transferred to Refrigerant 134a by flowing it through a long horizontal pipe at steady state. The pipe has an inner diameter of 5 cm. When refrigerant saturated vapour flows in at 20 kg/min and -6 °C, while exits at 800 the heat transfer rate to the refrigerant was found to be 3.5 kW. Assume an exit temperature in the range of 80 - 100 °C and determine the percentage error between that assumed value and that from calculation

Answer 1

Answer-4: Given mi 1 15MPa, 40°c m₂ = ? 3 Iseal Feed water Heater (FWH) in = ? Saturated liquid 275kPa Saturated mixture 175°C. Apply mass Balance to the FWH in ting=ing Applying energy Balance to F WHE As there is no heat transfer and thus - enthalpy at inlet - enthalpy at exit m, h + m₂ hz + m, ha = mghz from steam tables - at 15 MPa, Tsat = 342-16°C We have T=40°c <342-16°c which means I is sub- cooled, there fore from compressed liquid tables- h = 180.78 kg 1 kg At Inlet 2 - we have saturated mixture of quality say and hf (1-2) + x properties at esaturated state to T=175c - 44115 (1-x) + x (2772.7) - 74.15 + x (20 31.55) aging at exit 3 - Saturated liquid reated liquid - P = 275 kPa from saturation tables, nee get hz = (hg) @ 275 kPa = 548.87 kg 1kg x, and therefore - where ha a hg are enthalpy corresponding

Putting the values in eq (0) 60 (180.78) + m, {141.5 + x (2031055)} = (60+^) (548-87)-(110) taking different values for x between 0.5 to 0.8 - Now i) x=0.5 5. 60 (180-78) + mi, { 741.5 + 0.5 (2031.55)} = (60+ min ) (548.87) 60 (180.78) – 66 (548.87) 548.87 - {741.5 +0.5 (2031.95)} +22085.4 + 1208.405 18.276 6 kg/sec î 18.3 11) X=0.6 ว, Go (180.78) - 60 (548087) + 22085-4 548.87-{(741.5 +0.612031-55)} & 1411.56 in = 150646 46 kg | sec 2 15.7 111) X=0.7 22085.4 74105+0.7 (2031,55) -548,87 = 13.677 22085.4 1614.715 13.67767/sec ł 13.7 it) X=0.8 n. 22085.4 • 12.149 1817.87 ł 12.15 kg/see

18.3 17- Mass 16 flow 157 Rate ma (kg/sec) 14 13.7 13- 12-15 12 0.5 0.6 0.6 0.7 On 8 Quality (x)