Question

[10] Design Problem As shown in the figure below, a double-pipe parallel-flow heat exchanger is used to heat cold fluid which is water (Cp = 4180 J/(kg.K), p = 1000 kg/m) from 20°C to 80°C at a rate of 0.15 kg/s. The heating is to be accomplished by hot fluid, which is geothermal water (Cp = 4310 J/(kg.K), p = 1050 kg/m²) available at 130°C at a mass flow rate of 0.25 kg/s. The inner tube has an inner diameter of 0.8 cm and outer dimeter of 1.0 cm and made of material having thermal conductivity of 45 W/(m.K). Also, the inner diameter of the shell tube is 2.0 cm. Assume the viscosity of the hot and cold fluids are u = 0.001 kg/(m.s). Design the double-pipe parallel-flow heat exchanger by determining the length of the heat exchanger required to achieve the desired heating. Thereafter, investigate the effect of the inlet temperature of the geothermal water on the length of the heat exchanger for different values that varies from 100°C to 200°C. Then, plot the length of the heat exchanger as function of inlet temperature of the geothermal water, and discuss the results. Cold out Hot in Hot out Cold in

Answer 1

1 90 :-> A double-pipe parallel flow heat exchanger, as shown in figare below has the ihlet & outlet data for bot & cold fluid as stro are, .. cold. out Hot in Hot out I di Hot Water (fluid) di=0.8cm do=1cm cold water (Plug k=ustulma cold cp=4480 J 1kg k (p=43107) kg, 3= 1050 kg/h Di= 2cm in 3= 1000 kglu min=0.25 kg/s, Thi=13002 Toi=20°c, Tco 80°c m= SAV en=0.ool pass me zo 15 kq la V-(m/sa) le=0.001 pass The outlet temperature of water hot fluid is calculated from balence egn of heat excharger, energy Q = mucph CT6 - The) = mcxcpcx (Tco-Tci) Q = 0.15* 4180x (80-20) = 37,620 Watts . Q = max cph & CThi-Tho). 37,620=0.25+ 43107(130-Tho Tho=g5.0862 The thermophysical properties of hot & cold fluid at mean temp are, Hot fluid Tm- Thit The 2 cold fluid. Tm= TcitTco_ 2018 Tm=sor k=0.644 wlmk, P2= 3.358 112.54°C 2 Tm-Bot 95.086 sagt segles k= 0.68524 luk, Pr=1.443

U= 2641-0 rulmek بي The LMTD equation for paralles flow heat exchanger is Thi. ATI= Thi-The = 130-05: 88620, AT = 11000 tt Tho 8T2= Tho- Teo 295.086-80 AT2=15.886°0 ATI 4 T2 ... LPTD = 110-15-086 4T, - AT In (AT1/4h) Teo m 110/15.086 Tci LATD=47.82 جا

4 The length of heat exchanger requires to meet the best transfer rake, Q = U Asx LMTD = Um Utd.LX LIMTD 37,620=26410 17xo.018 L x 47.8, L=9.49m So, length of heat exchanges is about L=9.4gm to to tronstes heat rate (@=37 620 watts Q = 39 620 Watts from not fluid to cold fluid.

→ The effect of not Pluid temperature on length of heat exchanges for temperature range 100-200%, for Thi= 120°c The LIPTO for this condition is, ATI= Thi - Ti = 120-20-10002 T6 - Tho-To= but Tho is calulate from energy balence ech Q = max cph + (Thi-Tho), 37620 = 0.25 x 4310X (120- Tho) Thoz 85.086 : AT2=85.086-80 i 5.086°c LPYTD=31-864°C LMTD = 100 - 5086 Inc 100/5.086) Q=37620=26418 x 0.01 X LX 31-864 L=14-23m For Thi= 160°C : Q = 37620 0.25x Prat (140- Tho) x 4610 Tho=105.086°C ... ATI = 140-20= 120°C, 4T2= 105.086-80 = 25.086°C LPITO 120- 25.086 LATD=60.64°c) In(120/25.086) 37620 = 2641*11*0.01 LX 60-64. L=7.480 (11) For Thi = 180°C Q=376202 0.25x4310 + (180- Tho). The 24. 145.086°c. : ATI = 180-200 1600 4722 145.086-86 = 65.086€. LMTD= 160-65.086 In (160/65.086) = 3631 los.52°C

6 37620 = 26418 xo.olx Lx 105.52 = 4.3m IV For Thiz 2000 : Q=37620 - 0.25 x 43lox (20o- Tho) Tho=165.0860 ATIF 200-20=1800 4T2² Tho - To= 165.086-80 = 85.086°c : LRTD= 180-85.086 In (180/85.086) 126.67°C : Q = 37620 = 2661* X0:01 XL X 126.67 L=3.58m → The graph between hot fluid inlet temperature with length of heat exchanger. 14 12 10 L 8 m 6 4 2 O 140 160 180 200 20 90 60 80 100 120 Thi (oc)