Drill Rod Quenching A steel drill rod that is 8 mm in diameter and 15 cm long is being heat treated in an oil bath. The 850°C rod is quenched and held at 600°C and then quenched a second time to 20°C. Calculate the quantities of heat that must be removed at the two quenching stages. Approach We

Question

Drill Rod Quenching

A steel drill rod that is 8 mm in diameter and 15 cm long is being heat treated in an oil bath. The 850°C rod is quenched and held at [latex] 600°C [/latex]and then quenched a second time to [latex]20°C[/latex]. Calculate the quantities of heat that must be removed at the two quenching stages.

Approach

We will calculate the quantities of heat that fl ow from the rod during the two quenching stages by applying Equation (7.7).

[latex]Q = mc(T - T _{0})[/latex] (7.7)

Table 5.2

Table 5.2 Elastic Modulus and Weight Density of Selected Metals*

	Elastic Modulus, E			Weight Density, [latex]E/p_ω (m)[/latex]
Material	Mpsi	GPa	Poisson’s Ratio, [latex]v[/latex]	[latex]lb/ft^3[/latex]	[latex]kN/m^3[/latex]
Aluminum alloys	10	72	0.32	172	27
Copper alloys	16	110	0.33	536	84
Steel alloys	30	207	0.30	483	76
Stainless steels	28	190	0.30	483	76
Titanium alloys	16	114	0.33	276	43
* The numerical values given are representative, and values for specific materials could vary with composition and processing.

lists the weight density of steel as [latex]\rho _w = 76 kN/m^3[/latex] ,which we will use in calculating the rod’s
mass. The specific heat of steel is listed in Table 7.4

Table 7.4
Specific Heat of Certain Materials

		Specific Heat, c
Type	Substance	[latex]kJ/(kg. °C)[/latex]	[latex]Btu/(lbm .°F)[/latex]
Liquid	Oil Water	1.9 4.2	0.45 1.0
Solid	Aluminum Copper Steel Glass	0.90 0.39 0.50 0.84	0.21 0.093 0.11 0.20

as [latex]0.50 kJ/(kg .°C).[/latex]

Accepted Answer

The volume V of the rod is calculated based on its length L and cross sectional area A

[latex]A=\pi \frac{(0.008m)^2}{4} \longleftarrow \left[A=\pi \frac{d^2}{4} ight] [/latex]

[latex]= 5.027 imes 10^{-5} m^2[/latex]

The volume becomes

[latex]V=(5.027 imes 10^{-5} m^2)(0.15m) \longleftarrow \left[V=AL ight] [/latex]

[latex]= 7.540 imes 10^{-6} m^3 [/latex]

The rod’s weight is therefore

[latex]w=\left(76 imes 10^3\frac{N}{m^3} ight) (7.540 imes 10^{-6} m^3) \longleftarrow \left[w= ho _{w}V ight] [/latex]

[latex]=0.5730\left(\frac{N}{\cancel{m^3}} ight) (\cancel{m^3})[/latex]

[latex]=0.5730 N [/latex]

and its mass is

[latex]m=\frac{0.5730N}{9.81m/s^2} \longleftarrow \left[w=mg ight] [/latex]

[latex]= 5.841 imes 10^{-2} (N)\left(\frac{s^2}{m} ight) [/latex]

[latex]= 5.841 imes 10^{-2}\left(\frac{kg.\cancel{m}}{\cancel{s^2}} ight) \left(\frac{\cancel{s^2}}{\cancel{m}} ight) [/latex]

[latex]= 5.841 imes 10^{-2} kg [/latex]

In the last step, we expanded the derived unit newton in terms of the base units meter, kilogram, and second. The quantities of heat removed during the two quenching stages become

[latex]Q_1=( 5.841 imes 10^{-2} kg )\left(0.50\frac{kJ}{kg.^\omicron C } ight) (850^\omicron C - 600^\omicron C) \longleftarrow \left[Q=mc(T-T_0) ight] [/latex]

[latex]= 7.301(\cancel{kg})\left(\frac{kJ}{\cancel{kg}.^\omicron \cancel{C}} ight)( \cancel{^ºC }) [/latex]

[latex]= 7.301 kJ[/latex]

and

[latex]Q_2=( 5.841 imes 10^{-3} kg )\left(0.50\frac{kJ}{kg.^\omicron C } ight) (600^\omicron C - 20^\omicron C)) \longleftarrow \left[Q=mc(T-T_0) ight] [/latex]

[latex]= 16.94(\cancel{kg})\left(\frac{kJ}{\cancel{kg}. \cancel{^\omicron C}} ight)( \cancel{^\omicron C }) [/latex]

[latex]= 16.94 kJ[/latex]

Discussion
The heat flows from the drill rod into the oil bath. Following the two quenching stages, the temperature of the oil bath rises by amounts that can be calculated by using Equation (7.7). That the second quenching removes more heat makes sense because the second temperature reduction is significantly more than the reduction in the first quenching.

[latex]Q_1= 7.301 kJ[/latex]
[latex]Q_2= 16.94 kJ[/latex]

Question 7.7: Drill Rod Quenching A steel drill rod that is 8 mm in diamet...

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