Question

(3) Steel ingots (V 40 mm3; A-14 mm2) at 760 K are plunged into a bath of quench fluid at 400 K with h- 12 Ww/m2 K. Steel properties are k = 44 W/mK, density 7900 kg/m3, and heat capacity is 0.5 kJ/kgK Temperature of ingot after 45 min C Total amount of heat removed from ball up to 45 min

Answer 1

Lumped system analysis:

Check applicability of given system.

If Biot number (Bi) <=0.1 then lumped system is applicable.

Bi = (h*Lc)/k

Where, h = Convective heat transfer coefficient = 12 W/m2.K = 12 J/m2.K.sec................since 1 W = 1j/sec

Lc = Characteristic length = Volume /surface area = 40mm/14mm = 2.85 mm = 2.85 * 10^-3 m

k = thermal conductivity of steel = 44 W/mK

Bi = (12 W/m2K)*(2.85*10^-3m)/(44 W/mK) = 0.00077 <0.1

therefore Lumped system analysis is applicable,

For lumped system,

T(t) - Tm e-but - 7m

Where, T(t) = temperature of ingot after time t sec

Tm = temperature of medium = 400 K

Ti = Initial temperature of ingot = 760K

t = time = 45 min = 2700 sec

b= h* A p* V * Cp

A = surface area = 14 mm2

= 7900 kg/m3

p= densityofstelingot

V = volume = 40 mm3

Cp = specific heat = 0.5 kJ/KgK

Therefore,

b = 12m2. K Sec * 14mm²* (10 7900kg/m3* 40mm310mm *0.5 = 0.001063sec-1 100

T(t) - Tm Ti-Tm -0.001063Sec--2700sec = 0.0566

T(t) - Tm = (760 - 400)K  * 0.0566

T(t) = 400K+ (760 - 400)K  * 0.0566 = 420.40 K

Total amount of heat removed up to 45 min = Q = m*Cp*(Ti-T(t))

m = mass pf steel ingot = Density * Volume = 7900 kg/m3 * 40 mm3 * (10^-3)^3 (m3/mm3) = 0.000316 Kg

Q = 0.000316K g * 0.5 * 10– — *(760 – 420.40) K = 53.65J ) ko. k* (760