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Chapter 9: Problem 56

Specify the number of degrees of freedom for th following alloys: (a) \(20 \mathrm{wt} \% \mathrm{Ni}-80 \mathrm{wt} \% \mathrm{Cu}\) at \(1300^{\circ} \mathrm{C}\) (b) \(71.9 \mathrm{wt} \%\) Ag-28.1 \(\mathrm{wt} \% \mathrm{Cu}\) at \(779^{\circ} \mathrm{C}\) (c) \(52.7 \mathrm{wt} \% \mathrm{Zn}-47.3 \mathrm{wt} \% \mathrm{Cu}\) at \(525^{\circ} \mathrm{C}\) (d) \(81 \mathrm{wt} \% \mathrm{~Pb}-19 \mathrm{wt} \% \mathrm{Mg}\) at \(545^{\circ} \mathrm{C}\) (e) \(1 \mathrm{wt} \% \mathrm{C}-99 \mathrm{wt} \% \mathrm{Fe}\) at \(1000^{\circ} \mathrm{C}\)

Short Answer

Expert verified
a) 20% Nickel - 80% Copper Alloy at 1300°C b) 71.9% Silver - 28.1% Copper Alloy at 779°C c) 52.7% Zinc - 47.3% Copper Alloy at 525°C d) 81% Lead - 19% Magnesium Alloy at 545°C e) 1% Carbon - 99% Iron Alloy at 1000°C Answer: a) 3 degrees of freedom b) 2 degrees of freedom c) 2 degrees of freedom d) 2 degrees of freedom e) 3 degrees of freedom

Step by step solution

01

(a) 20% Nickel - 80% Copper Alloy

First, check the phase diagram for the Nickel-Copper system using the given temperature (1300°C) and composition (20% Nickel - 80% Copper). The system is completely in the liquid phase. Thus, there is only one phase present (P=1), and the number of components (C) is 2, as there are two elements (Ni and Cu). Now, let's use the Gibbs Phase Rule: \(F = C - P + 2 = 2 - 1 + 2 = 3\) So for the alloy system with 20% Ni and 80% Cu at 1300°C, there are 3 degrees of freedom.
02

(b) 71.9% Silver - 28.1% Copper Alloy

Check the phase diagram for the Silver-Copper system using the given temperature (779°C) and composition (71.9% Silver - 28.1% Copper). The system is at the eutectic point with both solid phase α-Ag and solid phase β-Cu. Thus, there are two phases present (P=2), and the number of components (C) is 2, as there are two elements (Ag and Cu). Using the Gibbs Phase Rule: \(F = C - P + 2 = 2 - 2 + 2 = 2\) So for the alloy system with 71.9% Ag and 28.1% Cu at 779°C, there are 2 degrees of freedom.
03

(c) 52.7% Zinc - 47.3% Copper Alloy

Check the phase diagram for the Zinc-Copper system using the given temperature (525°C) and composition (52.7% Zinc - 47.3% Copper). The system is at the eutectic point with solid phase α-Zn and solid phase β-Cu. Thus, there are two phases present (P=2), and the number of components (C) is 2, as there are two elements (Zn and Cu). Using the Gibbs Phase Rule: \(F = C - P + 2 = 2 - 2 + 2 = 2\) So for the alloy system with 52.7% Zn and 47.3% Cu at 525°C, there are 2 degrees of freedom.
04

(d) 81% Lead - 19% Magnesium Alloy

Check the phase diagram for the Lead-Magnesium system using the given temperature (545°C) and composition (81% Lead - 19% Magnesium). The system is at the partial melting point with both the solid phase (Pb) and liquid phase (L). Thus, there are two phases present (P=2), and the number of components (C) is 2, as there are two elements (Pb and Mg). Using the Gibbs Phase Rule: \(F = C - P + 2 = 2 - 2 + 2 = 2\) So for the alloy system with 81% Pb and 19% Mg at 545°C, there are 2 degrees of freedom.
05

(e) 1% Carbon - 99% Iron Alloy

Check the phase diagram for the Iron-Carbon system using the given temperature (1000°C) and composition (1% Carbon - 99% Iron). The system is in the austenite single-phase region (γ-Fe). Thus, there is only one phase present (P=1), and the number of components (C) is 2, as there are two elements (C and Fe). Using the Gibbs Phase Rule: \(F = C - P + 2 = 2 - 1 + 2 = 3\) So for the alloy system with 1% C and 99% Fe at 1000°C, there are 3 degrees of freedom.

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Most popular questions from this chapter

Construct the hypothetical phase diagram for metals \(\mathrm{A}\) and \(\mathrm{B}\) between room temperature \(\left(20^{\circ} \mathrm{C}\right)\) and \(700^{\circ} \mathrm{C}\), given the following information: \- The melting temperature of metal \(\mathrm{A}\) is \(480^{\circ} \mathrm{C}\). \- The maximum solubility of \(B\) in \(A\) is 4 wt \(\%\) B, which occurs at \(420^{\circ} \mathrm{C}\). \- The solubility of \(\mathrm{B}\) in \(\mathrm{A}\) at room temperature is 0 wt \(\%\) B. \- One eutectic occurs at \(420^{\circ} \mathrm{C}\) and \(18 \mathrm{wt} \%\) B-82 wt \(\%\) A. \- A second eutectic occurs at \(475^{\circ} \mathrm{C}\) and \(42 \mathrm{wt} \%\) B- \(58 \mathrm{wt} \% \mathrm{~A}\) \- The intermetallic compound AB exists at a composition of \(30 \mathrm{wt} \% \mathrm{~B}-70 \mathrm{wt} \% \mathrm{~A}\), and melts congruently at \(525^{\circ} \mathrm{C}\). \- The melting temperature of metal B is \(600^{\circ} \mathrm{C} .\) \- The maximum solubility of \(\mathrm{A}\) in \(\mathrm{B}\) is \(13 \mathrm{wt} \% \mathrm{~A}\), which occurs at \(475^{\circ} \mathrm{C}\). \- The solubility of \(\mathrm{A}\) in \(\mathrm{B}\) at room temperature is \(3 \mathrm{wt} \% \mathrm{~A}\)

Given here are the solidus and liquidus temperatures for the copper-gold system. Construct the phase diagram for this system and label each region. $$ \begin{array}{ccc} \hline \begin{array}{c} \text { Composition } \\ \text { (wt\% Au) } \end{array} & \begin{array}{c} \text { Solidus } \\ \text { Temperature }\left({ }^{\circ} \mathrm{C}\right) \end{array} & \begin{array}{c} \text { Liquidus } \\ \text { Temperature }\left({ }^{\circ} \boldsymbol{C}\right) \\ \hline 0 \end{array} & 1085 & 1085 \\ \hline 20 & 1019 & 1042 \\ \hline 40 & 972 & 996 \\ \hline 60 & 934 & 946 \\ \hline 80 & 911 & 911 \\ \hline 90 & 928 & 942 \\ \hline 95 & 974 & 984 \\ \hline 100 & 1064 & 1064 \\ \hline \end{array} $$

A hypothetical A-B alloy of composition 40 \(\mathrm{wt} \% \mathrm{~B}-60 \mathrm{wt} \% \mathrm{~A}\) at some temperature is found to consist of mass fractions of \(0.66\) and \(0.34\) for the \(\alpha\) and \(\beta\) phases, respectively. If the composition of the \(\alpha\) phase is \(13 \mathrm{wt} \%\) B-87 wt \(\% \mathrm{~A}\), what is the composition of the \(\beta\) phase?

What thermodynamic condition must be met for a state of equilibrium to exist?

(a) What is the distinction between hypoeutectoid and hypereutectoid steels? (b) In a hypoeutectoid steel, both eutectoid and proeutectoid ferrite exist. Explain the difference between them. What will be the carbon concentration in each?
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