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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.

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Presentation on theme: "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."— Presentation transcript:

1 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

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

3 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

4 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.

5 Nomograph for Determining Cooling-Pond Performance and Size

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

7 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.

8 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

9 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 h holdup, which is more than the 24 - h holdup required to maintain a fairly constant discharge temperature throughout the day.

10 Table : Maximum Expected Solar Radiation at Various North Latitudes 24-hr. avg. at north latitudeNoon value at north latitude 30º35º40º45º30º35º40º45º Jan Feb Mar Apr May June July Aug Sept Oct Nov Dec Source - [Langhaar, Chem. Eng., 60(8), 194 (1953).]

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