Question

One of the industrial process is to heat a vegetable oil to a specific temperature
by passing the oil through a metal coil that is placed in a warm water path. The
oil flows in the coil with a mass flow rate of 0.05 kg/s at 20◦C. The coil
consists of ten turns with a diameter of 50 cm that is made copper tubing (350
W/m. K) of 1 cm inside diameter with a 2 mm wall thickness. The total length of
tubing is 20 m. The coil is immersed in warm water bath of 50◦C and is stirred
in such a way as to give an external heat transfer coefficient of 4000 W/m2 K.
(Oil properties: ρ = 910 kg/ m3; k = 0.168 W/m. K; Pr = 949.82; μ = 0.081 kg/m.s;
cp = 1.97 kJ/kg. K).
(a) Determine the outlet temperature of the oil.
(b) Calculate the rate of heat transfer through the process?

Answer 1

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the = 50°C - Page 1 * Given details in problems dator m = 0.05 kg/s ta = 2016 the = 50°c te= 99 €=350 W/mk h = 4000 W/m²K d; = 1cm = 10 mm = 0.0lom h; 299 do = 14 mm = 0.014m ts 20m A=0.7540m² 7 Area of Heat transfer Azad L=(0.04 toole jx20 A= 0, 7540 m² → Velocity of vegetable oil in copper tube -'tas _Qos vos Arte 910x #xo.ol² = 0.7m/ 7 Reynold number for fluid flow in tube Sudiqloxo7x0.0L = 79 - Hout 0.081 => Average inside heat transfex co-efficient for Vegetable oid flowing through tube, 7:= 0,6646% rey (M -=0.664(2.168.) (7942 (949,82) hi - 11,1552 *8.89x9.83 hi = 974.82 W/m²K

Page 2 * Over all heat transfer co-efficient is given by U= (2) 7 + 7 ln ( loft) tt 2. do 03) 3974.877 (207) An (.com) 4000 - 0.001436 + 0.000006729 +0.00025 U = 590.76 W/m2.c => Comin = mxty = 0.05X1970 - 98.5 Ila Where min = Heat capacity of oil. for mim - R e, effectiveness of He is given by E=1 - exp CNTO) We know NT U can be caleulated, NTU= UA - 590.76x0.7540 -hes 98.5 €= 1- exp(4.522) E=1-0.01086 E = 0.9891 Cinim

Page-3 → Effectiveness of H. E. Is also given by -to-ta to-20 ax the tea 50-20 = 0.9891 ity = 49.67 °C
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