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kirill115 [55]
3 years ago
14

Determine whether or not it is possible to cold work steel so as to give a minimum Brinell hardness of 225 and at the same time

have a ductility of at least 12%EL. Justify your decision
Engineering
1 answer:
rodikova [14]3 years ago
7 0

Answer:

First we determine the tensile strength using the equation;

Tₓ (MPa) = 3.45 × HB

{ Tₓ is tensile strength, HB is Brinell hardness = 225 }

therefore

Tₓ = 3.45 × 225

Tₓ = 775 Mpa

From Conclusions, It is stated that in order to achieve a tensile strength of 775 MPa for a steel, the percentage of the cold work should be 10

When the percentage of cold work for steel is up to 10,the ductility is 16% EL.

And 16% EL is greater than 12% EL

Therefore, it is possible to cold work steel to a given minimum Brinell hardness of 225 and at the same time a ductility of at least 12% EL

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Answer:

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b) the LEDs drew 0.0555 Amp current

Explanation:

Given the data in the question;

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Therefore the LEDs drew 0.0555 Amp current

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3 years ago
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Answer:

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3 years ago
Consider a Carnot refrigeration cycle executed in a closed system in the saturated liquid-vapor mixture region using 0.96 kg of
Eduardwww [97]

Answer:

P_m_i_n= 356.9 KPa

Explanation:

Coefficient of performance (COP) of refrigeration cycle is given by:

COP=\frac{1}{\frac{T_H}{T_L}-1 }

We are given:

T_H=1.2T_L

COP=\frac{1}{1.2-1 }

COP= 5

We can also write Coefficient of performance (COP) of refrigeration cycle  as:

COP_R=\frac{Q_L}{W_i_n}

Amount of heat absorbed by low temperature reservoir can be found as:

Q_L=COP_R * W_i_n

Q_L=5 * 22 KJ

Q_L=110 KJ

According to first law of thermodynamics amount of heat rejected by hot reservoir is given by:

Q_H=Q_L + W_i_n

Q_H=110 KJ + 22 KJ

Q_H=132 KJ

We are given the mass of 0.96 kg. So,

q_H=\frac{Q_H}{m}

q_H=\frac{132 KJ}{0.96Kg}

q_H=137.5 KJ/Kg

Since it is a saturated liquid-vapour mixture q_H=h_f_g.

q_H=h_f_g=137.5 KJ/Kg

From Refrigerant 134-a tables T_H at h_f_g=137.5 KJ/Kg is 61.3 C. (We calculated this by interpolation)

Converting T_H from Celsius to Kelvin:

61.3^{o} C+273 = 334.3^{o} K

T_H= 334.3^{o} K

We are given:

T_H=1.2T_L

T_L=\frac{T_H}{1.2}

T_L=\frac{334.3}{1.2}

T_L=278.58^{o} K

Converting T_L from Kelvin to Celsius:

278.58^{o} K-273 = 5.58^{o} C

T_L= 5.58^{o} C

From Refrigerant 134-a tables P_m_i_n at T_L=5.58^{o} C is 356.9 KPa. (We calculated this by interpolation).

P_m_i_n= 356.9 KPa

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