Concept explainers
(1)
The transformation product after quenching and then after cool at room temperature.
(1)
Explanation of Solution
Introduction:
The process of rapid cooling of a metal in water, or air or oil in order to alter the
(a)
The steel has eutectoid composition that is initially equilibrated at
(b)
When the product formed in part (a) is cooled to room temperature, then the transformation product that is formed will be the
Conclusion:
Thus, the product formed after quenching is
(2)
The transformation product after quenching and then after cool at room temperature.
(2)
Explanation of Solution
Introduction:
The process of rapid cooling of a metal in water, or air or oil in order to alter the mechanical properties of the metal is called quenching. The process of quenching helps in the prevention of undesired low-temperature processes like phase transformation. Under normal conditions quenching is performed for few seconds.
(a)
The steel has eutectoid composition that is initially equilibrated at
(b)
When the product formed in part (a) is cooled to room temperature, then the transformation product that is formed will be the
Conclusion:
Thus, the product formed after quenching is
(3)
The transformation product after quenching and then after cool at room temperature.
(3)
Explanation of Solution
Introduction:
The process of rapid cooling of a metal in water, or air or oil in order to alter the mechanical properties of the metal is called quenching. The process of quenching helps in the prevention of undesired low-temperature processes like phase transformation. Under normal conditions quenching is performed for few seconds.
(a)
The steel has eutectoid composition that is initially equilibrated at
(b)
When the product formed in part (a) is cooled to room temperature, then the transformation product that is formed will be the
Conclusion:
Thus, the product formed after quenching is
(4)
The transformation product after quenching and then after cool at room temperature.
(4)
Explanation of Solution
Introduction:
The process of rapid cooling of a metal in water, or air or oil in order to alter the mechanical properties of the metal is called quenching. The process of quenching helps in the prevention of undesired low-temperature processes like phase transformation. Under normal conditions quenching is performed for few seconds.
(a)
The steel has eutectoid composition that is initially equilibrated at
(b)
When the product formed in part (a) is cooled to room temperature, then the transformation product that is formed will be the
Conclusion:
Thus, the product formed after quenching is
(5)
The transformation product after quenching and then after cool at room temperature.
(5)
Explanation of Solution
Introduction:
The process of rapid cooling of a metal in water, or air or oil in order to alter the mechanical properties of the metal is called quenching. The process of quenching helps in the prevention of undesired low-temperature processes like phase transformation. Under normal conditions quenching is performed for few seconds.
(a)
The steel has eutectoid composition that is initially equilibrated at
(b)
When the product formed in part (a) is cooled to room temperature, then the transformation product that is formed will be the
Conclusion:
Thus, the product formed after quenching is
(6)
The transformation product after quenching and then after cool at room temperature.
(6)
Explanation of Solution
Introduction:
The process of rapid cooling of a metal in water, or air or oil in order to alter the mechanical properties of the metal is called quenching. The process of quenching helps in the prevention of undesired low-temperature processes like phase transformation. Under normal conditions quenching is performed for few seconds.
(a)
The steel has eutectoid composition that is initially equilibrated at
(b)
When the product formed in part (a) is cooled to room temperature, then the transformation product that is formed will be the
Conclusion:
Thus, the product formed after quenching is
(7)
The transformation product after quenching and then after heating.
(7)
Explanation of Solution
Introduction:
The process of rapid cooling of a metal in water, or air or oil in order to alter the mechanical properties of the metal is called quenching. The process of quenching helps in the prevention of undesired low-temperature processes like phase transformation. Under normal conditions quenching is performed for few seconds.
(a)
The steel has eutectoid composition that is initially equilibrated at
(b)
When the product formed in part (a) is heated at the temperature of
Conclusion:
Thus, the product formed after quenching is
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Chapter 5 Solutions
Materials Science And Engineering Properties
- The following information is given for n-pentane at 1 atm:boiling point = 36.20°C ΔHvap (36.20°C) = 357.6 J/g melting point = − 129.7°C ΔHfus (−129.7°C) = 116.7 J/g specific heat gas = 1.650 J/g ⋅ °C specific heat liquid = 2.280 J/g ⋅ °C A 26.10-g sample of liquid n-pentane is initially at −51.40°C. If the sample is heated at constant pressure (P = 1 atm), how many kJ of energy are needed to raise the temperature of the sample to 65.30°C?arrow_forwardSHOW SOLUTION AND DRAW THE ILLUSTRATIONarrow_forwardView Policies Current Attempt in Progress Using the Animated Figure 10.40, the isothermal transformation diagram for a 0.45 wt% C steel alloy, specify the nature of the final microstructure (in terms of the microconstituents present) of a small specimen that has been subjected to the following temperature treatments. In each case assume that the specimen begins at 845 °C and that it has been held at this temperature long enough to have achieved a complete and homogeneous austenitic structure. a) Rapidly cool to 700 degrees C, hold for 100,000 s, then quench to room temperature. b) Rapidly cool to 450 degrees C, hold for 10 s, then quench to room temperature. proeutectoid ferrite + pearlite proeutectoid ferrite + martensite proeutectoid ferrite + pearlite + martensite eT proeutectoid ferrite + pearlite + bainite + martensite all spheroidite Save all bainite Attempts: 0 of 5 used Submit Answer all martensite bainite + martensitearrow_forward
- A sufficient amount of pure copper is(4 to be heated for casting a large plate in an open mold. The plate has dimensions: length = 20 in, width=D10 in, and thickness =3 in. Compute the amount of heat that must be added to the metal to heat it to a temperature of 21500F for pouring. Assume that the amount of metal heated will be 10% more than what is needed to fill the mold cavity. Properties of the metal are: density=0.324 Ibm/in3, melting point = 19810F, specific heat of the %3D metal=0.093Btu/lbm-F in the solid state and 0.090 Btu/lbm-F in the liquid * .state, and heat of fusion = 80 Btu/lbmarrow_forwardPlease answer correctly and box the final answer.arrow_forwardUsing the isothermal transformation diagram for a 0.45 wt% C steel alloy (Figure 10.40), determine the final microstructure (in terms of just the microconstituents present) AND approximate percentages of the microconstituents that form in a small specimen that has been subjected to the following time-temperature treatments. In each case assume that the specimen begins at 845°C (1550°F), and that it has been held at this temperature long enough to have achieved a complete and homogeneous austenitic structure. (a) Rapidly cool to 250°C (480°F), hold for 10^3 s, then quench to room temperature. (b) Rapidly cool to 700°C (1290°F), hold for 30 s, then quench to room temperature. (c) Rapidly cool to 700°C (1290°F), hold at this temperature for 10^5 s, then quench to room temperature. (d) Rapidly cool to 400°C (750°F), hold for 500 s, then quench to room temperature.arrow_forward
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- 2. A 2-m long aluminum pipe at 27°C is heated until it is 2.0024 m at 77°C. What is the coefficient of linear expansion of aluminum? Complete FBD, FORMULA, AND SOLUTIONarrow_forward6 A steel pipe with a 50 mm outside diameter and a 43.75 'mm inside diameter surrounds a solid brass rod 37.5 mm in diameter as shown. Both materials are joined to a rigid cover plate at each end. The assembly is free to expand longitudinally. The assembly is stress free at a temperature of 27°C. For steel, Young's modulus is 200 GN/m², and the coefficient of linear thermal expansion is 11.7x10-6 1/°C. For brass, Young's modulus is 93.33 GN/m², and the coefficient of linear thermal expansion is 1.872 x 10-5 1/°C. What is the stress in the steel tube when the temperature is raised to 121°C? brass rod (solid) 37.5 mm diameter steel tube 50 mm outside diameter 43.75 mm inside diameterarrow_forwardIn shown figure the aluminum is subjected to a thermal stress where T1 is 21.957 deg. C and T2 is 87.860 deg. C. Determine the required diameter (mm) of the steel rods to limit the expansion of the aluminum just 3.238 mm if the thermal coefficient of aluminum is 24.650 x10^-6 Where: H1 = 2000.392 mm H2 = 2047.231 mm H3 = 1999.692 mm Aal = 2288 mm^2 Eal = 70.018 GPa Est = 202.930 GPaarrow_forward
Materials Science And Engineering PropertiesCivil EngineeringISBN:9781111988609Author:Charles GilmorePublisher:Cengage Learning