Question

= View Policies Show Attempt History A novel heat exchanger concept consists of a large number of extruded polypropylene sheets (k = 0.17 W/m-K), each having a fin- like geometry, that are subsequently stacked and melted together to form the heat exchanger core. Besides being inexpensive, the heat exchanger can be easily recycled at the end of its life. Carbon dioxide at a mean temperature of 10°C and pressure of 2 atm flows in the cool channels at a mean velocity of um = 0.1 m/s. Air at 30°C and 2 atm flows at 0.2 m/s in the warm channels. The channel pitch is d - 12 mm. CO2 Welded interface a = 4 mm Air Polypropylene Sheet N N+1 N12 b= 4 mm C4 mm Neglecting the thermal contact resistance at the welded interface, determine the product of the overall heat transfer coefficient and heat transfer area, U A, for a heat exchanger core consisting of 400 cool channels and 400 warm channels. Physical Properties Mathematical Functions

Answer 1

a=4mm CO2 Welded Air interface + Polypropylene Sheet N Ntl N+2 b=4mm C=4mm 2 * Given Data:- Kp=0.17 w/m-k Tmc = 10°c. Pc = a atm. Umc d=12mm. = 12x10 m a=4mm. = 4x10 m. b = 4mm = 4x10 mi (= 4 mm. 4x10 m. - 0.1 m/s. Tm, n = 30°c. 2 atm. Nc- 300 Nh=300. Ph Um, h= 0.2 m/s *Now, carbondioxide properties at (loc=283k) and a atm, Pc = 3.764 kg/mo, CBC=0-833x10°J/kg-k, Mc = 141.3 x 103 Nslim, ke=15-4X103 W/mK, P = 0.765

'* Now, Air properties at (30°C =303k) and zatm, la = 2.303 kg/m, CP,a = 1.007 X10 J/kg-k, Ma = 186-01X1õ' Nsim, ka= 26-52x10 w/mk, Pg=0.706. - Now, calculate the cross-sectional area, A c= [(d-c)ka • [[12x102) – (4x1õp)]X[4x103) | Ac= 32x100 m² - Now, calculate the perimeter, P-2 [ld-c) + a) 2 [[12x1o°)-(44173)]+[C4x153)]]. 3 pe 24X10 ! -> Now, the value of hydraulic diameter is, Dh= 4 Ac P. 4X32X10 24x10 Dh= 5.33x103 Dhe 0.00533 m.

* NOW, CO2 Reynolds number is Rebic pa - Um, c Dh uc 3.764 x 0.1 x 5.33x103 141-3 X107 d-C = 12-4 4 Rep, C = 142.06 m <230om [Laminar flow] d-c From the observation table 8.7", Nusselt number for fully developed "Laminar flow, ASSU ming uniform wall temperature, by interpolation NUD, = 3.39 For Nub, c = 3.9 hn Dh = 3.39 ka hna 3.39xka Dh hc=3.399kc Db -3-39x15 4x10? 5.629x10 3.39X26'52x103 5.629x103 : hc = 9.274 w/mk kn = 15.97 W/mk * Now, consider mid point of sheet is adiabatic fin length is (L) = doc d. - (12x109) -(4x153) -3 4x10 m. 2 4x10 m

* Now, surface area of the fin with unit width (W=1) is, Af=alw = 2L (1) +2x4x10 Af=8x103 m2 * Now, total area of the fin with unit width (Wel) is, Az Aft Ab QLWtaw =(2 x 4x10x1)+(4x103) (1) ::JA= 12x103 m2 → For CO2, single fin efficiency is, Mc- 165.15 ml 2hc 2x 9.274 Kpb 0-17 X 4X103 Inc=165-15 ml Now, Efficiency is, Of, c Tanhimch) Ta n h (165-45 x 4 x 10) 165.15 X4 X103 mch Hf, c = 0.7461 Now, overall efficiency of fin for coz is No; c = 1 Af (1-16,c). 1- 8X10 12X103 (1-0-7461) - 0-8307 hoca 0.8307)

* Now, for air Efficiency of single fion is mna ahn kpb 2x 16.97 223.40 0-17X4X103 mn=223.40 Of, ha Tanh (MnL) MIL Ta nh (22 3x + x^o^) 223:4X4 X 153 nf, h= 0.802 * overall effieiency - I- Af (1-nfin). А 1-6 (1-0-802) 12 Noch=0.867 * Now total area for 400 warm channels is, An: 400x1axio 4.8 m² * Total area for 400 cool channels is, AC= 400 X 12 X10%: 4,8 m?

* Now, welded interface area for unit width (W=1 is, Aw- CathwN = (4x103)+(4x103) 0) (400) Aw = 3.2 m² + UA No, che AC f Kp Aw No,h hnan - 0.004 f + 0-8307x9.274x5:4 0.17 X3.2 0.867X16-97X5:4 UA = 22.83 WIK
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