Question

A shell and tube heat exchanger with one shell pass and two tube passes is used to heat 8.82 kg/s of fluid from 15.6 °C to 60 °C by using saturated steam at 150 kPa. The steam is condensing on the outside of the tubes with h= 15 kW/m2.K. There are 50 tubes with an outside diameter of 1.91 cm and a wall thickness of 0.211 cm. If the fouling coefficient on the inside of the tubes is 5678 W/m2.K, determine: A. The length of heat exchanger per one pass B. The pressure drop through the tubes C. The pumping power required when the pump efficiency is 60% E. The wall temperature at each end of the exchanger if the wall thickness of the tube is neglected For turbulent flow, the friction factor is given as: vin = -1.8 log .8 log(+)") The properties of the fluid are given as: p=790 kg/m), K = 0.182 W/m.K, p = 12 x 104 Ns/m2, Cp = 2470 J/kg.K, Pr = 16.29.

Answer 1

1 liven as mpa Sol" A shell 2 tube heat exchanges with one shell pass and two tube passes. The data for heat exchanger given fluid 8.82Kgls, Tp;= 15-6°C, TPo= 60°c 3=790kg 1203, K=0:18288 Whook, pe=12*10" , (p= 2470J/kgk , Pg=16:29 Saturated steam Ps= 150kpa infy = 2226:2K.J|Kg Data taken for saturated stomat 150kPa) ho=15kwhook ; N = 50, d=1-91cm . Up=5678Whom?k, di-do-24t=488com Ts=111.4°C tt = 0.211cm / saturated steam Inlet fluid outlet fluid I caturated liquid out One shell & two tubes heat exchanger (21, 72) To calculate heat transfer coefficient for fluid flowing through tube. Red = sxvxdi m-SxVolume 3xAXVXN SAN 3 adxn xd con

Red= 4m Naredi = 4x 8.82 2.** X12X10 4 X 0.01488 Red = 31.458104 turbulent flow [Re>4000(Pipe flow (6:29)004 Note: Here, N= No. of tubes are taken as 2, because the mass flow rate of fluid is given for one shell and two tube heat exchanger. -> The Nusselt muomber relation for turbulent water flow through tube. Nud = 0.023 Red 8x 20.4 = 0.023x(31-45x104)058 Hud=1756.095 The heat transfer coefficient for turbulanst fleild flow through tube Nuda hixdi hizdkf di 0.01488 bi = 21479.11 work to high value of hi due to rate high mass flow through tubes) The heat transfer equation for heat exchanger for saturated steam to fluidis, Q mpxCp% of TRO-Tri) = mixhing 8.82% 2470 x ( 60-15.6) mg x 22263x102 967, 271.76 Watts, Ims= 0.4345 kg/ 19756-095X0.182 9

le A} The length of heat exchanger per ome pase ~ The heat transfer equation for heat exchanger can be written as, @ = Vox As X (AT) CMTD where, Uo= overall heat transfer coefficient ㅗ chila ho + but tt ho U 21479:11 5678 uf (hi) == 4490.86 Whock! + 1 + I 15 000 U PLINTO difference Tso TS: Third 4490.86 [Vo = 20 115612. Whoek Logarithmic mean temperature ne for heat exchanger. . (AT), = Ts; -Tfo=1114-60 = 51.4°C (AT = Toom Tf i=111-4-15-6=958°C - COD (A7) LMTO (@I), - (Ana ein ( 1 ) (AT) LMTD = 51.4-95.8 en (5104/958) AD LMTD = 71.311° c :: Q=UA COD).MTO -> 96727 1.76 = 3456-12% Asx H.311 As=3.925m2 but, Ag=27xLx do aço [L=32.7oom =) L=3.925 2X TXo.0191

90 By the pressure drop through tubes (AP) The pressure doops due to friction im is calculated App=SxLxV2 x Sf for one tube di x2 The friction factor for tube flow is calculated CEldi jf Re but surface roughness Eis mot given considered the tube surface is smooth 18 log (6.a) rf 5x104) Vf If=0.01422 AP= 2x FxLXV2x5f di x2 2x790X /4x6.014889 =-1.8 -1.8xlog (3.45x10" 8.82 -0.01422X 32.703x 0.01488 BP = 25441.508x 1 value to pascale high mass flow nate. high

(P5 The pumping power required whom promp efficiency is mp=607. Power, P=DPx Of mp sex mf sf me P= 25441.508x103 (8.82 790 P=472331-807 Watts
Note:Plzz don't give dislike ...plzzz comment if you have any problem i will try to solve your problem.....plzzz give thumbs up i am in need.......