Question

Asteelcompanyisproducingsteelshaftsforaxisofturbines by  forgingprocess. The forgingtemperatureis1200  °C . After theforgingprocesstheshafthastobecooleddown . This isdonebysprayingwaterhomogeniouslyontothesurfaceoftheshaft.

I.-Describethecoolingprocesswiththehelpof dimensionlessnumbers. Makethefollowingassumptions:

- The shaftisa cylinderwithdiamterD andlenghtl

- The physical propertiesofthe steelare:

    - density ofthesteel:  ρ

    - thermal diffusity: a

-homogenoustemperatureat thebeginning: T_0,

-constantsurfacetemperatureduringcooling: T_1

-meantemperatureinside: T(t)

II.- How will the cooling time change  when the mass of the shaft will be halved (shape of the shaft will be unchanged: L/D=const)?

Answer 1

Diameter of shaft (d) = D. Length of shaft (2) = 1 Forging temperature (fo) = 1200°C - Density of steel = p. Thermal diffusivity - a. This is a case of transient Conduction. The dimensionless heat numbers. involved are - Fourier number = last Where, at a time and he is the Characteristic dimension. Biot number = (h*Lel IK. where, ha convective heat transfer Coefficient K = thermal Conductivity L = (Volume surface area) If Biot number cool, we can assume lumped system where there will a

be no Variation position. In this of Temperature with case we can. write, -has (T-To) = 4 me (IT) Solving we get (T-Tool - e- pva hast Here, t = time; v= volume; As - Surface area; Cp - Specific heat Capacity, = denpity; To = Ambient temperature I P

D Let Initial mass = m . Final mass = m/ (b) = constant = c (say) - 0 we know, we mass = (Density & Volume) If we are not changing the material of shaft, then density will remain constant Hence, mass & volume.. If, initial volume a Va a deli. then final volume a 2 = 1/2 = 7 de la So, dily = 2*£ dzdy -> di? l = 2* d * lz - we know, la at 'Fourier Number = z). a will remain same as it is a material Property. =) **, ) or, La (1/2 + C).

2 + c) from equation (1) t, (% + c)² ta (1/2 + c)2 - se, Also from equation ① and on lix Cils) = 2* (1/)24 17 - l = 2 & 3 From or, lille = (2) ( and = +,*(;) or it, = 0.629**, So, change in Cooling time = (t , - t 2 x 100 = (- 2x 100 = 3 71. decrease So Cooling time will decrease by 37% (Ans--2)

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