1. The following questions are about Al−Cu which is the origin of duralumin. - Select and answer the following terms tha

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1 The Following Questions Are About Al Cu Which Is The Origin Of Duralumin Select And Answer The Following Terms Tha 1 (381.68 KiB) Viewed 33 times

1. The following questions are about Al−Cu which is the origin of duralumin. - Select and answer the following terms that apply to [a] to [w] from the following choices or the temperature symbol on the vertical axis in Fig. 1. - You can use the same choices/symbols as many times if necessary. - For {A}−{E}, answer the appropriate numeric value or equation. - For {V}−{Y}, select the appropriate crystal structure shown in Fig. 2 and answer. Many aluminum alloys for aircraft such as duralumin mainly contain Al−Cu. Consider heat treatment of Al−Cu alloys. According to the binary equilibrium diagram of Al−Cu alloy (Fig. 1), in equilibrium state, Al−4mass%Cu is composed of two phases of [a ] and at room temperature (TH​). [a ] is a compound in which the molar ratio of Al to Cu is {A}:{B}, has a crystal structure of {V}, and contains {C} mass %Al. (The Fig.1 Al-Cu binary equilibrium phase diagram atomic weights of Al and Cu are 27 and 64 , respectively. A and B are simple integers. Show the intermediate formula for the calculation of C.) The mass ratio of to [a] at room temperature (TH​) is calculated to be {D}:{E} by [c] rule. By conducting [f] treatment of Al−4massCCu in the temperature range from [d] to [e], all [g] atoms are dissolved in . After that, when [h] process of the alloy is conducted from the [f] treatment temperature to room temperature, the [g] atoms remain in and form supersaturated . In this state, the hardness shows a value [j] than that before [f] treatment. This is because the ratio of [a] that increases hardness is [k]. Furthermore, when the alloy is held in the temperature range of 100∘C to 200∘C, the hardness becomes [1] over time. Such heat treatment is called [m]. This is because the [g ] atoms diffuse and form precipitates in the following order over time; [n], G.P.2 zone \{W\} and θ′ phase {X}. Among them, in [o] and [p], the molar ratio of Al and Cu is the same as in [a]. The matrix (crystal structure {Y} ) and the interface structure are [q] in [n],[r] in the G.P.2 zone, and [s] in the θ ' phase. Therefore, the mechanism of dislocation movement is [t] in [n], in the G.P.2 zone, and [v] in the θ′ phase. Due to this mechanism, the highest hardness 4. is exhibited when [w] starts to precipitate. Fig.2 Crystal structure of each phase [Choices] You can use the same choices/symbols as many times if necessary. 1: ferrite, 2: cementite, 3: austenite, 4: martensite, 5: G.P.1 zone, 6: G.P.2 zone, 7: θ phase, 8:θ′ phase, 9: LPSO phase, 10:δ ' phase, 11: equiaxed, 12: needle-shaped, 13: supercooled liquid, 14: amorphous solid, 15: eutectoid steel, 16: peritectic, 17: eutectic phase, 18: solid solution, 19: single crystal, 20: work hardening, 21: α solid solution, 22: β solid solution, 23: γ phase, 24: γ′ phase, 25 : coherent, 26: partly coherent (semi-coherent), 27: incoherent, 28: solidification, 29: precipitation, 30: quenching, 31: cooling while maintaining equilibrium, 32: slow cooling, 33: artificial aging, 34: natural aging, 35: overaging, 36: solution heat, 37: higher, 38: lower, 39: α+β process, 40: β process, 41: β transus, 42: particle shearing mechanism, 43: Orowan mechanism (accumulation of dislocations around precipitates), 44: Bausinger, 45: lever, 46: Fick, 47: dislocation, 48: superlattice dislocation pair, 49: stacking fault, 50: antiphase boundary (APB), 51: cross-slip, 52: Hole-Petch, 53: low temperature embrittlement, 54: stress corrosion cracking, 55: creep, 56: stress concentration, 57:C,58:Cu,59:Zn,60:Al,61:Mg,62:Cr,63:O,64;Ni,65:N,66:Fe,67:Li,68:Si Al