Question

Quenching of a Steel Billet. A cylindrical steel billet 1 ft in diameter and 3 ft long, initiallyat 1000°F is quenched in oil. Assume that the surface of the billet is at 200°F through out the quenching. The steel has the following properties, assumed to be independent of temperature:k = 25 Btu/(hr×ft×°F), r= 7.7 g/cm3, CP= 0.12 cal/g×°C

Estimate the temperature of the hottest point in the billet after five minutes of quenching.Neglect end effects, that is , make the calculation for a cylinder of the given diameter but of infinite length.

There is a solution for it all data is given or can be found on different sources, please need a solution

Answer 1

Given data

Thermal conductivity k = 25 x 4.1365 x 10^-3

= 0.10341 cal/s-cm-C

Density r= 7.7 g/cm3

Specific heat capacity Cp= 0.12 cal/g°C

Calculate the thermal diffusivity = k/(r*Cp)

= (0.10341 cal/s cm-C) / (7.7 g/cm3 x 0.12 cal/g°C)

= 0.112 cm2/s

Time parameter (dimensionless)

= thermal diffusivity x t / R²

= ( 0.112 cm2/s x 5x60 s) / (0.5 x 30.48 cm)²

= 0.145

From the below graph, center line temperature = 0.31

The hottest point would be at the center

(Tc - To) /(T1 - To) = 0.31

Tc = 0.31(T1 - To) + To

= 0.31(200-1000) + 1000

= 750 °F