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alina1380 [7]
2 years ago
7

Dampness or moisture introduces ____ into the weld, which causes cracking when some metals are welded.

Engineering
1 answer:
N76 [4]2 years ago
5 0

Answer: Dampness or moisture introduces hydrogen into the weld, which causes cracking when some metals are welded.

Explanation:

<em>This moisture (hydrogen) is a major cause of weld cracking and porosity. </em>

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A steady stream (1000 kg/hr) of air flows through a compressor, entering at (300 K, 0.1 MPa) and leaving at (425 K, 1 MPa). The
AleksandrR [38]

Answer:

The work furnished by the compressor is 69.77kJ/s

The minimum work required for the state to change is 55.26kW

Explanation:

The explanation to these solution is on the first, second , third and fourth uploaded image respectively

8 0
2 years ago
What is the activation energy (Q) for a vacancy formation if 10 moles of a metal have 2.3 X 10^13 vacancies at 425°C?
Yakvenalex [24]

Answer:

Activation\ Energy=2.5\times 10^{-19}\ J

Explanation:

Using the expression shown below as:

N_v=N\times e^{-\frac {Q_v}{k\times T}

Where,

N_v is the number of vacancies

N is the number of defective sites

k is Boltzmann's constant = 1.38\times 10^{-23}\ J/K

{Q_v} is the activation energy

T is the temperature

Given that:

N_v=2.3\times 10^{13}

N = 10 moles

1 mole = 6.023\times 10^{23}

So,

N = 10\times 6.023\times 10^{23}=6.023\times 10^{24}

Temperature = 425°C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (425 + 273.15) K = 698.15 K  

T = 698.15 K

Applying the values as:

2.3\times 10^{13}=6.023\times 10^{24}\times e^{-\frac {Q_v}{1.38\times 10^{-23}\times 698.15}

ln[\frac {2.3}{6.023}\times 10^{-11}]=-\frac {Q_v}{1.38\times 10^{-23}\times 698.15}

Q_v=2.5\times 10^{-19}\ J

4 0
2 years ago
Consider the expansion of a gas at a constant temperature in a water-cooled piston-cylinder system. The constant temperature is
Leona [35]

Answer:

Q_{in} = W_{out} = nRT ln (\frac{V_{2}}{V_{1}})

Explanation:

According to the first thermodynamic law, the energy must be conserved so:

dQ = dU - dW

Where Q is the heat transmitted to the system, U is the internal energy and W is the work done by the system.

This equation can be solved by integration between an initial and a final state:

(1) \int\limits^1_2 {} \, dQ = \int\limits^1_2 {} \, dU - \int\limits^1_2 {} \, dW

As per work definition:

dW = F*dr

For pressure the force F equials the pressure multiplied by the area of the piston, and considering dx as the displacement:

dW = PA*dx

Here A*dx equals the differential volume of the piston, and considering that any increment in volume is a work done by the system, the sign is negative, so:

dW = - P*dV

So the third integral in equation (1) is:

\int\limits^1_2 {- P} \, dV

Considering the gas as ideal, the pressure can be calculated as P = \frac{n*R*T}{V}, so:

\int\limits^1_2 {- P} \, dV = \int\limits^1_2 {- \frac{n*R*T}{V}} \, dV

In this particular case as the systems is closed and the temperature constant, n, R and T are constants:

\int\limits^1_2 {- \frac{n*R*T}{V}} \, dV = -nRT \int\limits^1_2 {\frac{1}{V}} \, dV

Replacion this and solving equation (1) between state 1 and 2:

\int\limits^1_2 {} \, dQ = \int\limits^1_2 {} \, dU + nRT \int\limits^1_2 {\frac{1}{V}} \, dV

Q_{2} - Q_{1} = U_{2} - U_{1} + nRT(ln V_{2} - ln V_{1})

Q_{2} - Q_{1} = U_{2} - U_{1} + nRT ln \frac{V_{2}}{V_{1}}

The internal energy depends only on the temperature of the gas, so there is no internal energy change U_{2} - U_{1} = 0, so the heat exchanged to the system equals the work done by the system:

Q_{in} = W_{out} = nRT ln (\frac{V_{2}}{V_{1}})

4 0
3 years ago
g If the rails are originally laid in contact, what is the stress in them on a summer day when their temperature is 31.0
Anettt [7]

A) The amount of space must be left between adjacent rails if they are just to touch on a summer day when their temperature is 31.0 °C is; 0.6048 cm

B) The stress in the rails on a summer day when their temperature is 31.0 °C is; 86.4 × 10⁶ Pa

<h3>Linear Thermal Expansion</h3>

We are given;

Length; L = 14 m

Initial Temperature; T_i = −5 °C

Final Temperature; T_f = 31 °C

The formula for Linear Thermal Expansion is;

ΔL = L_i * α * ΔT

where;

L_i is initial length

α is thermal expansion

ΔL is change in length

ΔT is change in temperature

Now, the thermal expansion of steel from online tables is α = 1.2 × 10⁻⁵ C⁻¹

Thus;

ΔL = 14 * 1.2 × 10⁻⁵  * (31 - (-5))

ΔL = 6.048 × 10⁻³ m = 0.6048 cm

The formula to get the stress is;

σ = Y * α  * ΔT

where;

Y is young's modulus of steel = 20 × 10¹⁰ Pa

α is thermal expansion

ΔT is change in temperature

Thus;

σ = 20 × 10¹⁰ × 1.2 × 10⁻⁵ × (31 - (-5))

σ = 86.4 × 10⁶ Pa

The complete question is;

Steel train rails are laid in 14.0-m long segments placed end to end. The rails are laid on a winter day when their temperature is −5 °C.

(a) How much space must be left between adjacent rails if they are just to touch on a summer day when their temperature is 31.0 °C?

(b) If the rails are originally laid in contact, what is the stress in them on a summer day when their temperature is 31.0 °C?

Read more about Linear Thermal Expansion at; brainly.com/question/6985348

4 0
2 years ago
A TPMS (tire pressure monitoring system) instrument panel indicator lamp is on. Technician A says the most likely cause is low t
zheka24 [161]

Answer:

Actualmente estoy trabajando en una pregunta diferente en este momento.

Explanation:

Actualmente estoy trabajando en una pregunta diferente en este momento.

6 0
2 years ago
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