This set of Engineering Materials Science "Numerical Problems Questions & Answers" are based on "PHASE DIAGRAM"
PHASE DIAGRAM
NUMERICAL PROBLEMS
Hello reader, here you can find some Numerical Problems with answers. But before reading this I will suggest you take a practice test on these questions given below. so you can understand your strength on this topic.
Click here to go Practice Test section on this topic namely Phase Diagram.
But if you want to read this topic first and then practice it then read it carefully. The Numerical Problems based on the Phase Diagram are given below:
PART - II
1) Consider the sugar–water phase diagram is given below: How much sugar will dissolve in 500 g of water at 40°C?
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Correct Answer is "A"
At 40°C maximum solubility of sugar in water is 67%
i.e. 66 g sugar dissolve in (100-67) = 33 g water
Or in other words, in 33 g water dissolve sugar = 67 g
Hence in 500 g water dissolve sugar = 500 × (67/33) = 1015 g
i.e. 66 g sugar dissolve in (100-67) = 33 g water
Or in other words, in 33 g water dissolve sugar = 67 g
Hence in 500 g water dissolve sugar = 500 × (67/33) = 1015 g
2) At a pressure of 0.1 atm, the melting temperature for ice and the boiling temperature for water are respectively
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Correct Answer is "C"
A horizontal line from 0.1 atm pressure parallel to the temperature axis cuts the OB and OC curve at two points.
The temperature corresponding to the intersection of the horizontal line and OB curves gives the melting temperature at 0°C = 273K
The temperature corresponding to the intersection of the horizontal line and OC curves gives boiling temperature at 70°C = (273 + 70)K = 343K
The temperature corresponding to the intersection of the horizontal line and OB curves gives the melting temperature at 0°C = 273K
The temperature corresponding to the intersection of the horizontal line and OC curves gives boiling temperature at 70°C = (273 + 70)K = 343K
3) The degrees of freedom for a system which has an equal number of components and phases is
Correct Answer is "C"
D.O.F = C – C + 2 = 2 (No of components =no of phases = C)
4) At 30°C, hot chocolate (liquid) with 35% chocolate and 65% vanilla transform to chocolate ripple (eutectic mixture of vanilla containing 10% chocolate and chocolate containing 5% vanilla). Just below 30°C, the fraction of chocolate ripple in composition with 45% chocolate is
Correct Answer is "A"
With respect to vanilla, the eutectic reaction is,
Liquid (65%) → Chocolate rich region (5%) + Vanilla rich region (90%)
The fraction of chocolate ripple below 30°C in a composition with 45% chocolate i.e. 55% vanilla is calculated from the justify side of the eutectic point (65% vanilla)
= (55-5)/(65-5) = 50/60 = 0.83
Liquid (65%) → Chocolate rich region (5%) + Vanilla rich region (90%)
The fraction of chocolate ripple below 30°C in a composition with 45% chocolate i.e. 55% vanilla is calculated from the justify side of the eutectic point (65% vanilla)
= (55-5)/(65-5) = 50/60 = 0.83
5) The fraction of eutectoid in a 0.43% C steel is
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Correct Answer is "B"
Fraction of eutectoid α = fraction of total α – fraction of pro-eutectoid α
= (6.7-0.43)/(6.7-0.022) - (0.76-0.43)/(0.76-0.022)
= 0.94 – 0.45
= 0.49
= (6.7-0.43)/(6.7-0.022) - (0.76-0.43)/(0.76-0.022)
= 0.94 – 0.45
= 0.49
6) How many kilograms of nickel must be added to 2.8 kg of copper to yield a liquidus temperature of 1250°C?
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Correct Answer is "A"
The horizontal line from 1250°C parallel to the composition axis cuts the liquidus line at a point from which by the vertical movement we get the composition, in that case, is 33% Ni.
So the amount of Ni must be added = 2.8 × 33/(100-33) = 1.38 Kg
So the amount of Ni must be added = 2.8 × 33/(100-33) = 1.38 Kg
7) For a 99.72 wt.% Fe–0.28 wt.% C alloy at a temperature just below the eutectoid, determine the fraction of pro eutectoid pearlite.
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Correct Answer is "C"
Fraction of proeutectoid pearlite = (0.28-0.022)/(0.76-0.022) = 0.349
8) If alpha of 72% B and liquid of 33% B are in equilibrium in an alloy of 51% B, the fraction of liquid is
Correct Answer is "B"
The fraction of liquid = (72-51)/(72-33) = 0.538
9) The fraction of pearlite in a 0.65% C steel is
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Correct Answer is "C"
Fraction of Pearlite = (0.65-0.022)/(0.76-0.022) = 0.85
10) For the 40 wt.% Ag – 60 wt.% Cu copper-silver alloy the relative amount of α and β phase present in terms of volume fraction at 700°C is respectively ____.
Given the densities of Cu and Ag are 8.96 and 10.49 g/cm3, respectively
Correct Answer is "D"
Mass fraction of α phase = (95-40)/(95-5 ) = 0.61
Mass fraction of β phase = (40-5)/(95-5 ) = 0.39
Density of α phase = 100/(5/10.49+ 95/8.96 ) = 9.03
Density of β phase = 100/(95/10.49+ 5/8.96 ) = 10.40
Volume fraction of α phase = (0.61/9.03)/(0.61/9.03+ 0.39/10.40) = 0.643
Volume fraction of β phase = (0.39/10.40)/(0.61/9.03+ 0.39/10.40) = 0.357
Mass fraction of β phase = (40-5)/(95-5 ) = 0.39
Density of α phase = 100/(5/10.49+ 95/8.96 ) = 9.03
Density of β phase = 100/(95/10.49+ 5/8.96 ) = 10.40
Volume fraction of α phase = (0.61/9.03)/(0.61/9.03+ 0.39/10.40) = 0.643
Volume fraction of β phase = (0.39/10.40)/(0.61/9.03+ 0.39/10.40) = 0.357
Do you read all the questions and answers given above carefully? let's make a practice on all the questions given above.
Go to: Practice Test on Phase Diagram (Numerical Problems) PART - II
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