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vekshin1
3 years ago
7

) Please label the following statements as either True (T) or False (F). (a) In general, the greater the % of cold work, the sma

ller the recrystallization grain size. (b) The higher the annealing temperature, the smaller the recrystallization grain size. (c) The greater the % of cold work, the lower the recrystallization temperature.
Engineering
1 answer:
Alexandra [31]3 years ago
6 0

Answer:

A. This option is true

B. This option is false

C. This option is true

Explanation:

A. Generally speaking, the greater percentage of cold, the recrystallization grain size would turn out to be smaller. Therefore this true.

B. A higher annealing temperature does not result in smaller recrystallization grain size. Therefore this is false.

C. As the percentage of cold work is greater, the recrystallization temperature would tend to be lower. Therefore this is true.

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A permanent magnet DC motor has an armature resistance of 0.5 Ω and when a voltage of 120 V is applied to the motor it reaches a
BaLLatris [955]

Answer:

12 N-m

Explanation:

The dc motor is operating at 24 V that is its terminal voltage V =24 V

Armature resistance  = 0.2 ohm

No load speed = 240 radian /sec

For motor we know that  as the motor is on no load so  so

Power developed in the motor

Now we know that power = torque× angular speed

So

3 0
4 years ago
Refrigerant 134a vapor in a piston-cylinder assembly undergoes a process at constant pressure from an initial state at 8 bar and
jonny [76]

Answer:

- Work done is 2.39 kJ

- heat transfer is 20.23 kJ/kg

Explanation:    

Given the data in the question;

First we obtain for specific volumes and specific enthalpy from "Table Properties Refrigerant 134a;

Specific Volume v₁ =  0.02547 m³/kg

Specific enthalpy u₁ = 243.78 kJ/kg

Specific Volume V₂ = 0.02846 m³/kg

Specific enthalpy u₂ = 261.62 kJ/kg

p = 8 bar = 800 kPa

Any changes in kinetic and potential energy are negligible.

So we determine the work done by using the equation at constant pressure

]Work done W = p( v₂ - v₁ )

we substitute

W = 800 kPa( 0.02846 m³/kg - 0.02547 m³/kg )

W = 800 kPa( 0.00299 m³/kg )

W = 2.39 kJ

Therefore, Work done is 2.39 kJ

Heat transfer;

using equation at constant pressure

Heat transfer Q = W + ( u₂ - u₁ )

so we substitute

Q = 2.392 kJ + ( 261.62 kJ/kg - 243.78 kJ/kg )

Q = 2.392 kJ +  17.84 kJ/kg )

Q = 20.23 kJ/kg

Therefore, heat transfer is 20.23 kJ/kg

3 0
3 years ago
Consider a very long, cylindrical fin. The temperature of the fin at the tip and base are 25 °C and 50 °C, respectively. The dia
Mrrafil [7]

Answer:

The fin temperature in °C at a distance of 10 cm from the base = 33.78°C

Explanation:

The following assumptions will be made to solve this problem

- The heat transfer coefficient does not change with the time or distance.

- The temperature of the fins varies just in only one direction.

The temperature of the fin at x = 10 cm = 0.10 m from the base can be calculated from the temperature variation with distance formula for a very long fin.

(T - T∞) = (T₀ - T∞)e⁻ᵐˣ

T = T(x) = temperature at any point along the fin

T∞ = temperature at the tip of the fin = ambient temperature = 25°C

T₀ = temperature at the base of thw fin = 50°C

x = any distance along the length of the fin from the base of the fin = 0.1 m

m = √(hP/KA)

h = Heat transfer coefficient = 123 W/m².K

P = perimeter in contact with the base = πD = π × 0.03 = 0.0943 m

K = thermal conductivity = 150 W/m.K

A = surface area in contact with the base = πD²/4 = π(0.03)²/4 = 0.0007071 m²

m = √(123 × 0.0943)/(150 × 0.0007071)

m = 10.46

mx = 10.46 × 0.1 = 1.046

(T - 25) = (50 - 25) e⁻¹•⁰⁴⁶

T = 25 + 25 e⁻¹•⁰⁴⁶ = 25 + 8.78 = 33.78°C

8 0
3 years ago
A thick aluminum block initially at 26.5°C is subjected to constant heat flux of 4000 W/m2 by an electric resistance heater whos
Yanka [14]

Given Information:

Initial temperature of aluminum block = 26.5°C

Heat flux = 4000 w/m²

Time = 2112 seconds

Time = 30 minutes = 30*60 = 1800 seconds

Required Information:

Rise in surface temperature = ?

Answer:

Rise in surface temperature = 8.6 °C after 2112 seconds

Rise in surface temperature = 8 °C after 30 minutes

Explanation:

The surface temperature of the aluminum block is given by

T_{surface} = T_{initial} + \frac{q}{k} \sqrt{\frac{4\alpha t}{\pi} }

Where q is the heat flux supplied to aluminum block, k is the conductivity of pure aluminum and α is the diffusivity of pure aluminum.

After t = 2112 sec:

T_{surface} = 26.5 + \frac{4000}{237} \sqrt{\frac{4(9.71\times 10^{-5}) (2112)}{\pi} }\\\\T_{surface} = 26.5 + \frac{4000}{237} (0.51098)\\\\T_{surface} = 26.5 + 8.6\\\\T_{surface} = 35.1\\\\

The rise in the surface temperature is

Rise = 35.1 - 26.5 = 8.6 °C

Therefore, the surface temperature of the block will rise by 8.6 °C after 2112 seconds.

After t = 30 mins:

T_{surface} = 26.5 + \frac{4000}{237} \sqrt{\frac{4(9.71\times 10^{-5}) (1800)}{\pi} }\\\\T_{surface} = 26.5 + \frac{4000}{237} (0.4717)\\\\T_{surface} = 26.5 + 7.96\\\\T_{surface} = 34.5\\\\

The rise in the surface temperature is

Rise = 34.5 - 26.5 = 8 °C

Therefore, the surface temperature of the block will rise by 8 °C after 30 minutes.

5 0
3 years ago
A body is moving with simple harmonic motion. It's velocity is recorded as being 3.5m/s when it is at 150mm from the mid-positio
natima [27]

Answer:

1) A=282.6 mm

2)a_{max}=60.35\ m/s^2

3)T=0.42 sec

4)f= 2.24 Hz

Explanation:

Given that

V=3.5 m/s at x=150 mm     ------------1

V=2.5 m/s at x=225 mm   ------------2

Where x measured  from mid position.

We know that velocity in simple harmonic given as

V=\omega \sqrt{A^2-x^2}

Where A is the amplitude and ω is the natural frequency of simple harmonic motion.

From equation 1 and 2

3.5=\omega \sqrt{A^2-0.15^2}    ------3

2.5=\omega \sqrt{A^2-0.225^2}   --------4

Now by dividing equation 3 by 4

\dfrac{3.5}{2.5}=\dfrac {\sqrt{A^2-0.15^2}}{\sqrt{A^2-0.225^2}}

1.96=\dfrac {{A^2-0.15^2}}{{A^2-0.225^2}}

So    A=0.2826 m

A=282.6 mm

Now by putting the values of A in the equation 3

3.5=\omega \sqrt{A^2-0.15^2}

3.5=\omega \sqrt{0.2826^2-0.15^2}

ω=14.609 rad/s

Frequency

ω= 2πf

14.609= 2 x π x f

f= 2.24 Hz

Maximum acceleration

a_{max}=\omega ^2A

a_{max}=14.61 ^2\times 0.2826\ m/s^2

a_{max}=60.35\ m/s^2

Time period T

T=\dfrac{2\pi}{\omega}

T=\dfrac{2\pi}{14.609}

T=0.42 sec

8 0
4 years ago
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