# DESIGN AND CONSTRUCTION OF AN INDUCTION FURNACE (COOLING SYSTEM) Presented by MG THANT ZIN WIN Roll No: Ph.D-M-7 Supervisors : Dr Mi Sanda Mon Daw Khin.

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DESIGN AND CONSTRUCTION OF AN INDUCTION FURNACE (COOLING SYSTEM) Presented by MG THANT ZIN WIN Roll No: Ph.D-M-7 Supervisors : Dr Mi Sanda Mon Daw Khin War Oo 21th Seminar 24.11.2004

Evaporative Cooling System Cooling Pond System Spray Pond System Cooling Tower System 1.Forced-draft cooling tower 2.Natural-draft cooling tower

Cooling Ponds Satisfactory method of removing heat from water Availability in large ground areas Small investment by pushing up an earth dike 1.8 to 3.1 m (6 to 10 ft) high Involving four principal heat transfer process _ Heat loss : evaporation, convection, radiation Heat gain : solar radiation, water supply

Mainly Consideration in Cooling Ponds Required pond area: it depends on the number of degrees of cooling required and the net heat loss from each square foot of pond surface. Equilibrium temperature: it is designed as E, at which heat loss would equal heat gain, is greatly affected by the amount of solar radiation, which is usually not known very accurately and which varies through the day.

Nomograph for Determining Cooling-Pond Performance and Size

Fig – Nomograph of cooling-pond preformance [Langhaar, Chem. Eng., 60(8), 194 (1953).]

where, D 1 & D 2 = the approaches to equilibrium for the entering and leaving water, ºF V w = the wind velocity, mi/h PQ = the area of the pond surface, ft 2 /(gal.min) of flow to the pond The P factor assumes a pond with uniform flow, with turbulence, and with the water warmer than the air.

For Example : Determine the required size of a cooling pond operating at the following conditions. Relative humidity, percent = 50 Wind velocity, mi/h = 5 Dry-bulb air temperature, ºF = 68 Solar heat gain, Btu/(h.ft 2 ) = 100 Water quantity, gal/min = 10,000 Water inlet temperature, ºF = 110 Water outlet temperature, ºF = 90

Calculation By using the nomograph, P = 68, Q = 1.07 Required Area, A = PQ = 73 ft 2 /(gal.min) Area for 10,000 gal/min = 730,000 ft 2 If we use a depth of 5 ft of pond, total volume of the pond would amount to a 45.5 - h holdup, which is more than the 24 - h holdup required to maintain a fairly constant discharge temperature throughout the day.

Table : Maximum Expected Solar Radiation at Various North Latitudes 24-hr. avg. at north latitudeNoon value at north latitude 30º35º40º45º30º35º40º45º Jan. 1------------ Feb. 1------------ Mar. 1----------- Apr. 1----------- May. 1----------- June. 1----------- July. 1----------- Aug. 1----------- Sept. 1----------- Oct. 1------------ Nov. 1----------- Dec. 1----------- 65 75 90 110 120 130 125 115 100 80 65 50 65 80 105 120 130 125 110 90 70 55 40 55 75 95 120 130 125 105 80 60 45 30 45 65 90 115 130 120 100 75 50 35 240 270 305 340 360 365 360 350 315 270 240 205 240 285 320 350 360 350 335 295 245 210 170 210 255 300 335 345 350 340 315 270 215 175 135 175 230 280 320 335 340 325 300 245 185 140 Source - [Langhaar, Chem. Eng., 60(8), 194 (1953).]

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