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SCORPION-xisa [38]
3 years ago
10

What will happen in a wire drawing operation when the cross-sectional area has a reduction of 60% in a single pass?

Engineering
1 answer:
Fofino [41]3 years ago
4 0

Answer:

DRAWING LOAD IS  3.67 A_{O}\sigma

Explanation:

wire drawing is a method of obtaining wire of bigger large diameter from iron rod . it is cold process which need die to obtain wire

drawing load for wire drawing is given as P = A_{F}*\sigma*ln(\frac{A_{O}}{A_{F}})

Where A f is initial area, Ao is original area, σ is yield stress

as given in question sectional area reduce 60%, therefore

A_{f} = A_{O}- 0.6A_{O}

    = 0.4 A_{O}

Due to change in area ,drawing load p is

p = 0.4A_{O}*\sigma*ln(\frac{A_{O}}{0.4A_{O}})

p = 3.67 A_{O}\sigma

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Evaporation in Double-Effect Reverse-Feed Evaporators. A feed containing 2 wt % dissolved organic solids in water is fed to a do
Ne4ueva [31]

Answer:

472,826 lb/hr

Explanation:

As per the given data:

Solids in feed= 2%

Solids in concentrate= 25%

HTA1 = HTA2 = 1000 ft^2

U1 = 500 Btu/ h ft^2 F & U2 = 700 Btu/ h ft^2 F

Overall material balance: Feed= Distillate + concentrate ----> Eq-1

Component balance: Feed * 0.02 = Distillate * 0 + concentrate * 0.25

Feed = 12.5 * concentrate ---> Eq-2

Boiling point rise = negligible, so solution & solvent vapor temperature will be same.

Assumed that the 1st effect is operating under atmospheric pressure (Boiling point - 212F).

As per the data:

Latent heat 212F = 300 Btu/lb

Latent heat 100F = 320 Btu/lb

As per material balance:

Vapor flowrate * latent heat = Overall HT coefficient * HTA * DT

1st effect: M-1= (500 * 1000 * (326-212)) / 300 = 190,000 lb/hr

2nd effect: M-2= (700*1000 * (212-100)) / 320 = 245,000 lb/hr

Distillate = M-1 + M-2 = 190,000+245,000 = 435,000 lb/hr

Substituting the above in Eq-1

Feed = 435,000 + concentrate

Substitute Eq-2 in the above

12.5 * concentrate = 435,000 + concentrate

Concentrate, L1 = 435,000/11.5 = 37826 lb /hr

Feed, F = 435,000 + 37826 = 472,826 lb/hr

1st effect operating pressure is not given, That may be the reason we are not getting the given answer. But procedure is right.

[ Find the figure in the attachment ]

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A hanging wire made of an alloy of nickel with diameter 0.19 cm is initially 2.8 m long. When a 59 kg mass is hung from it, the
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Answer:

Young's modulus for this alloy of nickel is 1.997×10^11 N/m^2

Explanation:

Young's modulus = stress/strain

Stress = Force (F)/Area (A)

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Stress = 578.2/2.836×10^-6 = 2.039×10^8 N/m^2

Strain = extension/length = 2.86×10^-3 m/2.8 m = 1.021×10^-3

Young's modulus = 2.039×10^8/1.021×10^-3 = 1.997×10^11 N/m^2

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