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bija089 [108]
3 years ago
12

I have a molten Ni-Cu alloy with 60 wt%Ni. (ii) I cool it down to a temperature where I have both solid and liquid phases. At th

is temperature the solid to liquid ratio is 1:1. (iii) Then I was able to remove all the solid particles out of this alloy at this stage. If I cool the remaining liquid alloy to 13000 C, how much what percentage of the remaining allow would become solid
Engineering
1 answer:
SIZIF [17.4K]3 years ago
4 0

Answer:

The percentage of the remaining alloy would become solid is 20%

Explanation:

Melting point of Cu = 1085°C

Melting point of Ni = 1455°C

At 1200°C, there is a 30% liquid and 70% solid, the weight percentage of Ni in alloy is the same that percentage of solid, then, that weight percentage is 70%.

The Ni-Cu alloy with 60% Ni and 40% Cu, and if we have the temperature of alloy > temperature of Ni > temperature of Cu, we have the follow:

60% Ni (liquid) and 40% Cu (liquid) at temperature of alloy

At solid phase with a temperature of alloy and 50% solid Cu and 50% liquid Ni, we have the follow:

40% Cu + 10% Ni in liquid phase and 50% of Ni is in solid phase.

The percentage of remaining alloy in solid is equal to

Solid = (10/50) * 100 = 20%

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Answer:

Work done for the adiabatic process = -247873.6 J/kg = - 247.9 KJ/kg

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For the first State,

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T₁ = 300 K

V₁ = ?

Second state

P₂ = 15 bar = 15 × 10⁵ Pa

T₂ = ?

V₂ = 0.1227 m³/kg

Third state

P₃ = ?

T₃ = 300 K

V₃ = ?

We require the workdone for step 1-2 (which is adiabatic)

And heat transferred for steps 2-3 (which is isochoric/constant volume)

Work done for an adiabatic process is given by

W = K(V₂¹⁻ʸ - V₁¹⁻ʸ)/(1 - γ)

where γ = ratio of specific heats = 1.4 for air since air is mostly diatomic

K = PVʸ

Using state 2 to calculate for k

K = P₂V₂ʸ = (15 × 10⁵)(0.1227)¹•⁴ = 79519.5

We also need V₁

For an adiabatic process

P₁V₁ʸ = P₂V₂ʸ = K

P₁V₁ʸ = K

(10⁵) (V₁¹•⁴) = 79519.5

V₁ = 0.849 m³/kg

W = K(V₂¹⁻ʸ - V₁¹⁻ʸ)/(1 - γ)

W = 79519.5 [(0.1227)⁻⁰•⁴ - (0.849)⁻⁰•⁴]/(1 - 1.4)

W = (79519.5 × 1.247)/(-0.4) = - 247873.6 J/kg = - 247.9 KJ/kg

To calculate the heat transferred for the constant volume process

Heat transferred = Cᵥ (ΔT)

where Cᵥ = specific heat capacity at constant volume for air = 0.718 KJ/kgK

ΔT = T₃ - T₂

We need to calculate for T₂

Assuming air is an ideal gas,

PV = mRT

T = PV/mR

At state 2,

V/m = 0.1227 m³/kg

P₂ = 15 bar = 15 × 10⁵ Pa

R = gas constant for air = 287.1 J/kgK

T₂ = 15 × 10⁵ × 0.1227/287.1 = 641.1 K

Q = 0.718 (300 - 641.1) = - 244.91 KJ/kg

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