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Neporo4naja [7]

3 years ago

9

A hypereutectoid steel often presents hard and brittle cementite along the grain boundaries of pearlite. Which of the following

heat treatments can alleviate this problem by changing the microstructure?(a) Annealing(b) Normalizing(c) Austenizing(d) Spheroidizing(e) Process-annealing

1 answer:

Anvisha [2.4K]3 years ago

6 0

Answer:

(d) Spheroidizing

Explanation:

Spheroidizing

This is the heat treatment process for steel which having carbon percentage more than 0.8 %.As we know that a hard and brittle material is having carbon percentage more than 0.8 %.That is why this process is suitable for the hard materials.

In this process a hard and brittle materials convert into soft and ductile after this it improve the machine ability as well as improve the tool life.

In this process grain become spheroidal and these grains are ductile.

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Answer:

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3 0

3 years ago

A heat engine is coupled with a dynamometer. The length of the load arm is 900 mm. The spring balance reading is 16. Applied wei

miss Akunina [59]

Answer:

P = 80.922 KW

Explanation:

Given data;

Length of load arm is 900 mm = 0.9 m

Spring balanced read 16 N

Applied weight is 500 N

Rotational speed is 1774 rpm

we know that power is given as

$P = T\times \omega$

T Torque = (w -s) L = (500 - 16)0.9 = 435.6 Nm

$\omega$ angular speed $=\frac{2 \pi N}{60}$

Therefore Power is

$P =\frac{435.6 \time 2 \pi \times 1774}{60} = 80922.65 watt$

P = 80.922 KW

4 0

4 years ago

Firefighters are holding a nozzle at the end of a hose while trying to extinguish a fire. The nozzle exit diameter is 8 cm, and

ivanzaharov [21]

Question

Determine the average water exit velocity

Answer:

53.05 m/s

Explanation:

Given information

Volume flow rate, $Q=16 m^{3}/min$

Diameter d= 8cm= 0.08 m

Assumptions

The flow is jet flow hence momentum-flux correction factor is unity
Gravitational force is not considered
The flow is steady, frictionless and incompressible
Water is discharged to the atmosphere hence pressure is ignored

We know that Q=AV and making v the subject then

$V=\frac {Q}{A}$ where V is the exit velocity and A is area

Area, $A=\frac {\pi d^{2}{4}$ where d is the diameter

By substitution

$V=\frac {16\times 4}{\pi 0.08^{2}}=3183.098862 m/min$

To convert v to m/s from m/s, we simply divide it by 60 hence

$V=\frac {3183.098862 m/min}{60 s}=53.0516477 m/s\approx 53.05 m/s$

3 0

3 years ago

Discuss in detail the following methods used to redistribute income and wealth in cash grants?

aev [14]

Answer:

Redistribution of income and wealth is the transfer of income and wealth (including physical property) from some individuals to others through a social mechanism such as taxation, welfare, public services, land reform, monetary policies, confiscation, divorce or tort law.

-The term typically refers to redistribution on an economy-wide basis rather than between selected individuals.

Interpretations of the phrase vary, depending on personal perspectives, political ideologies and the selective use of statistics.

-It is frequently heard in politics, usually referring to perceived redistribution from those who have more to those who have less. Occasionally, however, it is used to describe laws or policies that cause opposite redistribution that shift monetary burdens from low-income earners to the wealthy.

-

The phrase is often coupled with the term class warfare, with high-income earners and the wealthy portrayed as victims of unfairness and discrimination.

-

Redistribution tax policy should not be confused with predistribution policies. "Predistribution" is the idea that the state should try to prevent inequalities from occurring in the first place rather than through the tax and benefits system once they have occurred. For example, a government predistribution policy might require employers to pay all employees a living wage, not just a minimum wage, as a "bottom-up" response to widespread income inequalities or high poverty rates.

Many alternate taxation proposals have been floated without the political will to alter the status quo. One example is the proposed "Buffett Rule", which is a hybrid taxation model composed of opposing systems, intended to minimize the favoritism of the special interest tax design.

7 0

3 years ago

A four-cylinder, four-stroke internal combustion engine has a bore of 3.7 in. and a stroke of 3.4 in. The clearance volume is 16

Bad White [126]

Answer:

the net work per cycle $\mathbf{W_{net} = 0.777593696}$ Btu per cycle

the power developed by the engine, W = 88.0144746 hp

Explanation:

the information given includes;

diameter of the four-cylinder bore = 3.7 in

length of the stroke = 3.4 in

The clearance volume = 16% = 0.16

The cylindrical volume $V_2 = 0.16 V_1$

the crankshaft N rotates at a speed of 2400 RPM.

At the beginning of the compression , temperature $T_1$ = 60 F = 519.67 R

and;

Otto cycle with a pressure = 14.5 lbf/in² = (14.5 × 144 ) lb/ft²

= 2088 lb/ft²

The maximum temperature in the cycle is 5200 R

From the given information; the change in volume is:

$V_1-V_2 = \dfrac{\pi}{4}D^2L$

$V_1-0.16V_1= \dfrac{\pi}{4}(3.7)^2(3.4)$

$V_1-0.16V_1= 36.55714291$

$0.84 V_1 =36.55714291$

$V_1 =\dfrac{36.55714291}{0.84 }$

$V_1 =43.52040823 \ in^3 \\ \\ V_1 = 43.52 \ in^3$

$V_1 = 0.02518 \ ft^3$

the mass in air ( lb) can be determined by using the formula:

$m = \dfrac{P_1V_1}{RT}$

where;

R = 53.3533 ft.lbf/lb.R°

$m = \dfrac{2088 \ lb/ft^2 \times 0.02518 \ ft^3}{53.3533 \ ft .lbf/lb.^0R \times 519 .67 ^0 R}$

m = 0.0018962 lb

From the tables of ideal gas properties at Temperature 519.67 R

$v_{r1} =158.58$

$u_1 = 88.62 Btu/lb$

At state of volume 2; the relative volume can be determined as:

$v_{r2} = v_{r1} \times \dfrac{V_2}{V_1}$

$v_{r2} = 158.58 \times 0.16$

$v_{r2} = 25.3728$

The specific energy $u_2$ at $v_{r2} = 25.3728$ is 184.7 Btu/lb

From the tables of ideal gas properties at maximum Temperature T = 5200 R

$v_{r3} = 0.1828$

$u_3 = 1098 \ Btu/lb$

To determine the relative volume at state 4; we have:

$v_{r4} = v_{r3} \times \dfrac{V_1}{V_2}$

$v_{r4} =0.1828 \times \dfrac{1}{0.16}$

$v_{r4} =1.1425$

The specific energy $u_4$ at $v_{r4} =1.1425$ is 591.84 Btu/lb

Now; the net work per cycle can now be calculated as by using the following formula:

$W_{net} = Heat \ supplied - Heat \ rejected$

$W_{net} = m(u_3-u_2)-m(u_4 - u_1)$

$W_{net} = m(u_3-u_2- u_4 + u_1)$

$W_{net} = m(1098-184.7- 591.84 + 88.62)$

$W_{net} = 0.0018962 \times (1098-184.7- 591.84 + 88.62)$

$W_{net} = 0.0018962 \times (410.08)$

$\mathbf{W_{net} = 0.777593696}$ Btu per cycle

the power developed by the engine, in horsepower. can be calculated as follows;

In the four-cylinder, four-stroke internal combustion engine; the power developed by the engine can be calculated by using the expression:

$W = 4 \times N' \times W_{net$

where ;

$N' = \dfrac{2400}{2}$

N' = 1200 cycles/min

N' = 1200 cycles/60 seconds

N' = 20 cycles/sec

W = 4 × 20 cycles/sec × 0.777593696

W = 62.20749568 Btu/s

W = 88.0144746 hp

8 0

3 years ago