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Leona [35]
3 years ago
14

Bainite has finer grains because the transformation takes place at a lower temperature where the nucleation rate is high relativ

e to the growth rate. True or False
Engineering
1 answer:
kkurt [141]3 years ago
8 0

Answer:

False

Explanation:

Bainite is a type of steel. It is formed by the decomposition of austenite. This happens at a temperature above MS but given that MS temperature is below the one where fine pearlite is formed.

In other words, when iron goes through a metastable crystallization phase Bainite is formed. It takes the rapid cooling, or quenching, of austenite for that to happen.

Metastability is the phenomenon where matter remains in an apparent state of equilibrium or stability even though it can achieve a more stable state. It can also be referred to as the condition of remaining in that pseudo unstable state for a very protracted period of time.

Bainite steel is used for the construction of components that require frequent usage and where plasticity, or tendency for deformity, as well as wear, cannot be tolerated.

Cheers

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Nitrogen at an initial state of 300 K, 150 kPa and 0.2 m3 is compressed slowly in an isothermal process to a final pressure of 8
s344n2d4d5 [400]

Answer:

Work = - 4175.23 J

Heat Transfer = -4175.23 J

Explanation:

The work done in an isothermal process, is given by the following formula:

W = RT ln (P1/P2)

where,

W = Work done

R = Universal Gas Constant = 8.314 J/mol.k

T = Constant Temperature = 300 K

P1 = initial pressure = 150 KPa

P2 = final pressure = 800 KPa

using these values, we get:

W = (8.314 J/mol.K)(300 k) ln (150/800)

<u>W = - 4175.23 J</u>

Here, negative sign shows that work is done on the system.

In isothermal process, from 1st law of thermodynamics:

Heat Transfer = Q = W     (Since change in internal energy is zero for isothermal processes)

<u>W = - 4175.23 J</u>

Here, negative sign shows that heat is transferred from the system to surrounding.

6 0
3 years ago
Read 2 more answers
Which process must first take place in order for the proteins in foods to be used by body cells?
barxatty [35]

Answer:

Digestion

1 is correct since you need to eat something before anything else.

2 is incorrect since if this was first, what are you storing?

3 is incorrect since that is the second thing you need to do.

4 is incorrect since that is the last thing to do.

5 0
3 years ago
An ideal Otto cycle has a compression ratio of 8. At the beginning of the compression process, air is at 95 kPa and 278C, and 75
Inessa [10]

Answer:

(a). The value of temperature at the end of heat addition process            T_{3} = 2042.56 K

(b). The value of pressure at the end of heat addition process                    P_{3} = 1555.46 k pa

(c). The thermal efficiency of an Otto cycle   E_{otto} = 0.4478

(d). The value of mean effective pressure of the cycle P_{m} = 1506.41 \frac{k pa}{kg}

Explanation:

Compression ratio r_{p} = 8

Initial pressure P_{1} = 95 k pa

Initial temperature T_{1} = 278 °c = 551 K

Final pressure P_{2} = 8 × P_{1} = 8 × 95 = 760 k pa

Final temperature T_{2} = T_{1} × r_{p} ^{\frac{\gamma - 1}{\gamma} }

Final temperature T_{2} = 551 × 8 ^{\frac{1.4 - 1}{1.4} }

Final temperature T_{2} = 998 K

Heat transferred at constant volume Q = 750 \frac{KJ}{kg}

(a). We know that Heat transferred at constant volume Q_{S} = m C_{v} ( T_{3} - T_{2}  )

⇒ 1 × 0.718 × ( T_{3} - 998 ) = 750

⇒ T_{3} = 2042.56 K

This is the value of temperature at the end of heat addition process.

Since heat addition is constant volume process. so for that process pressure is directly proportional to the temperature.

⇒ P ∝ T

⇒ \frac{P_{3} }{P_{2} } = \frac{T_{3} }{T_{2} }

⇒ P_{3} = \frac{2042.56}{998} × 760

⇒ P_{3} = 1555.46 k pa

This is the value of pressure at the end of heat addition process.

(b). Heat rejected from the cycle Q_{R} = m C_{v} ( T_{4} - T_{1}  )

For the compression and expansion process,

⇒ \frac{T_{3} }{T_{2} } = \frac{T_{4} }{T_{1} }

⇒ \frac{2042.56}{998} = \frac{T_{4} }{551}

⇒ T_{4} = 1127.7 K

Heat rejected Q_{R} = 1 × 0.718 × ( 1127.7 - 551)

⇒ Q_{R} = 414.07 \frac{KJ}{kg}

Net heat interaction from the cycle Q_{net} = Q_{S} - Q_{R}

Put the values of Q_{S} & Q_{R}  we get,

⇒ Q_{net} = 750 - 414.07

⇒ Q_{net} = 335.93 \frac{KJ}{kg}

We know that for a cyclic process net heat interaction is equal to net work transfer.

⇒ Q_{net} = W_{net}

⇒ W_{net} = 335.93 \frac{KJ}{kg}

This is the net work output from the cycle.

(c). Thermal efficiency of an Otto cycle is given by

E_{otto} = 1- \frac{T_{1} }{T_{2} }

Put the values of T_{1} & T_{2} in the above formula we get,

E_{otto} = 1- \frac{551 }{998 }

⇒ E_{otto} = 0.4478

This is the thermal efficiency of an Otto cycle.

(d). Mean effective pressure P_{m} :-

We know that mean effective pressure of  the Otto cycle is  given by

P_{m} = \frac{W_{net} }{V_{s} } ---------- (1)

where V_{s} is the swept volume.

V_{s} = V_{1}  - V_{2} ---------- ( 2 )

From ideal gas equation P_{1} V_{1} = m × R × T_{1}

Put all the values in above formula we get,

⇒ 95 × V_{1} = 1 × 0.287 × 551

⇒ V_{1} = 0.6 m^{3}

From the same ideal gas equation

P_{2} V_{2} = m × R × T_{2}

⇒ 760 × V_{2} = 1 × 0.287 × 998

⇒ V_{2} = 0.377 m^{3}

Thus swept volume V_{s} = 0.6 - 0.377

⇒ V_{s} = 0.223 m^{3}

Thus from equation 1 the mean effective pressure

⇒ P_{m} = \frac{335.93}{0.223}

⇒ P_{m} = 1506.41 \frac{k pa}{kg}

This is the value of mean effective pressure of the cycle.

4 0
3 years ago
Consider a refrigerator that consumes 320 W of electric power when it is running. If the refrigerator runs only one quarter of t
Semenov [28]

Answer:

B. $5.18

Explanation:

Cost of electricity per kWh = $0.09

Power consumption of refrigerator = 320W = 320/1000 = 0.32kW

In a month (30 days) the refrigerator works 1/4 × 30 days = 7.5 days = 7.5 × 24 hours = 180 hours

Energy consumed in 180 hours = 0.32kW × 180h = 57.6kWh

Cost of electricity of 57.6kWh energy consumed by the refrigerator = 57.6 × $0.09 = $5.18

3 0
4 years ago
Read 2 more answers
Determine the enthalpy, volume and density of 1.0 kg of steam at a pressure of 0.5 MN/m2 and with a dryness fraction of 0.96
Viktor [21]

Answer:

Enthalpy, hsteam = 2663.7 kJ/kg

Volume, Vsteam = 0.3598613 m^3 / kg

Density = 2.67 kg/ m^3

Explanation:

Mass of steam, m = 1 kg

Pressure of the steam, P = 0.5 MN/m^2

Dryness fraction, x = 0.96

At P = 0.5 MPa:

Tsat = 151.831°C

Vf = 0.00109255 m^3 / kg

Vg = 0.37481 m^3 / kg

hf = 640.09 kJ/kg

hg = 2748.1 kJ/kg

hfg = 2108 kJ/kg

The enthalpy can be given by the formula:

hsteam = hf + x * hfg

hsteam = 640.09 + ( 0.96 * 2108)

hsteam = 2663.7 kJ/kg

The volume of the steam can be given as:

Vsteam = Vf + x(Vg - Vf)

Vsteam = 0.00109255 + 0.96(0.37481 - 640.09)

Vsteam = 0.3598613 m^3 / kg

From the steam table, the density of the steam at a pressure of 0.5 MPa is 2.67 kg/ m^3

5 0
3 years ago
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