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Mrrafil [7]
3 years ago
6

Q7. A cylindrical rod of 1040 steel originally 15.2 mm (0.60 in.) in diameter is to be cold worked by drawing; the circular cros

s section will be maintained during deformation. A cold-worked tensile strength in excess of 840 MPa (122,000 psi) and a ductility of at least 12%EL are desired. Furthermore, the final diameter must be 10 mm (0.40 in.). Explain how this may be accomplished. (10 points)

Engineering
1 answer:
umka2103 [35]3 years ago
5 0

Answer:

11.2mm or 0.45in

Explanation:

The percent cold work, attendant tensile strength and ductility if drawing is carried out without interruption is given by the equation you will find in the attached file.

Please go through the attached file for a step by step solution to this question.

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A petrol engine produces 20 hp using 35 kW of heat transfer from burning fuel. What is its thermal efficiency, and how much powe
hoa [83]

Answer:

efficiency =42.62%

AMOUNT OF POWER REJECTED IS 20.080 kW

Explanation:

given data:

power 20 hp

heat energy = 35kW

power production = 20 hp = 20* 746 W = 14920 Watt   [1 hp =746 watt]

efficiency = \frac{power}{heat\ required}

efficiency = \frac{14920}{35*10^3}

                = 0.4262*10^100

                 =42.62%

b) heat\ rejected = heat\ required - amount\ of\ power\ generated

                           = 35*10^3 - 14920

                           = 20.080 kW

AMOUNT OF POWER REJECTED IS 20.080 kW

5 0
3 years ago
An experimental arrangement for measuring the thermal conductivity of solid materials involves the use of two long rods that are
Kamila [148]

Answer:

Explanation:

Given:

The two rods could be approximated as a fins of infinite length.

TA = 75 0C θA = (TA - T∞) = 75 - 25 = 50 0C

TB = 55 0C     θB = (TB - T∞) = 55 - 25 = 30 0C

Tb = 100 0C   θb = (Tb - T∞) = (100 - 25) = 75 0C

KA = 200 W/m · K

T∞ = 25 0C

Solution:

The temperature distribution for the infinite fins are given by

θ/θb=e⁻mx

θA/θb= e-√(hp/A.kA) x1    ....................(1)

 θB/θb = e-√(hp/A.kB) x1.......................(2)

Taking natural log on both sides we get,

Ln(θA/θb) = -√(hp/A.kA) x1 ...................(3)

Ln(θB/θb) = -√(hp/A.kB) x1 .....................(4)

Dicving (3) and (4) we get

[ Ln(θA/θb) /Ln(θB/θb)] = √(KB/KA)

 [ Ln(50/75) /Ln(30/75)] = √(KB/200)

4 0
2 years ago
The smallest crystal lattice defects is a) cracks b) point defects c) planar defects d) dislocations.
ss7ja [257]

Answer:b) Point defects

Explanation: The point defect is the tiny defect that occurs in the lattice. It usually occurs when there is the atoms or ions missing in the lattice structure that creates a irregularity in the structure.The name point defect itself describes that the occurring defect is having a size of point thus is the smallest defect. Therefore option(b) is the correct option.

8 0
3 years ago
A flame ionization detector, which is often used in gas chromatography, responds to a change in
Lena [83]

Answer:

Option A

Explanation:

We know that ions are present in hydrogen-air flame and when the burning of an organic compound takes place in this flame more ions are produced in the flame.  

Thus when we apply a voltage across this flame, the ion collector plate attracts the all the ions in the flame.

The presence of organic compounds increases the voltage across the hydrogen ion flame produced at the ion collector increases and as the voltage increases, the detection of the organic compound can be made in turn.

Thus flame ionization detector clearly responds to the variation in the collection of ions or electrons in a flame.

3 0
3 years ago
ASAP correct answer plss When you are driving, if you see this traffic sign it means
vlabodo [156]

Answer:

C

Explanation:

5 0
3 years ago
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