Why is low voltage advantageous in arc welding?

A plane wall of thickness 0.1 m and thermal conductivity 25 W/m·K having uniform volumetric heat generation of 0.3 MW/m3 is insu

Contact [7]

Answer:

T = 167 ° C

Explanation:

To solve the question we have the following known variables

Type of surface = plane wall ,

Thermal conductivity k = 25.0 W/m·K,

Thickness L = 0.1 m,

Heat generation rate q' = 0.300 MW/m³,

Heat transfer coefficient hc = 400 W/m² ·K,

Ambient temperature T∞ = 32.0 °C

We are to determine the maximum temperature in the wall

Assumptions for the calculation are as follows

Negligible heat loss through the insulation
Steady state system
One dimensional conduction across the wall

Therefore by the one dimensional conduction equation we have

$k\frac{d^{2}T }{dx^{2} } +q'_{G} = \rho c\frac{dT}{dt}$

During steady state

$\frac{dT}{dt}$ = 0 which gives $k\frac{d^{2}T }{dx^{2} } +q'_{G} = 0$

From which we have $\frac{d^{2}T }{dx^{2} } = -\frac{q'_{G}}{k}$

Considering the boundary condition at x =0 where there is no heat loss

$\frac{dT}{dt}$ = 0 also at the other end of the plane wall we have

$-k\frac{dT }{dx } =$ hc (T - T∞) at point x = L

Integrating the equation we have

$\frac{dT }{dx } = \frac{q'_{G}}{k} x+ C_{1}$ from which C₁ is evaluated from the first boundary condition thus

0 = $\frac{q'_{G}}{k} (0)+ C_{1}$ from which C₁ = 0

From the second integration we have

$T = -\frac{q'_{G}}{2k} x^{2} + C_{2}$

From which we can solve for C₂ by substituting the T and the first derivative into the second boundary condition s follows

$-k\frac{q'_{G}L}{k} = h_{c}( -\frac{q'_{G}L^{2} }{k} + C_{2}-T∞)$ → C₂ = $q'_{G}L(\frac{1}{h_{c} }+ \frac{L}{2k} } )+T∞$

T(x) = $\frac{q'_{G}}{2k} x^{2} + q'_{G}L(\frac{1}{h_{c} }+ \frac{L}{2k} } )+T∞$ and T(x) = T∞ + $\frac{q'_{G}}{2k} (L^{2}+(\frac{2kL}{h_{c} }} )-x^{2} )$

∴ Tmax → when x = 0 = T∞ + $\frac{q'_{G}}{2k} (L^{2}+(\frac{2kL}{h_{c} }} ))$

Substituting the values we get

T = 167 ° C

4 0

3 years ago

A circular bar will be subjected to an axial force (P) of 2000 lbf. The bar will be made of material that has a strength (S) of

schepotkina [342]

Answer:

$n = 2.36$

Explanation:

The stress experimented by the circular bar is:

$\sigma = \left[\frac{2000\, lbf}{\frac{\pi}{4}\cdot (0.5\,in)^{2}}\right]\cdot \left(\frac{1\,kpsi}{1000\,psi} \right)$

$\sigma = 10.186\,kpsi$

The safety factor is:

$n = \frac{24\,kpsi}{10.186\,kpsi}$

$n = 2.36$

5 0

3 years ago

The lattice constant of a simple cubic lattice is a0.

Oksi-84 [34.3K]

Answer:

A)The sketches for the required planes were drawn in the first attachment.

B)The sketches for the required directions were drawn in the second attachment.

To draw a plane in a simple cubic lattice, you have to follow these instructions:

1- the cube has 3 main directions called "a", "b" and "c" (as shown in the first attachment)

2- The coordinates of that plane are written as: π:(1/a₀ 1/b₀ 1/c₀) (if one of the coordinates is 0, for example (1 1 0), c₀ is ∞, therefore that plane never cross the direction c).

3- Identify the points a₀, b₀, and c₀ at the plane that crosses this main directions and point them in the cubic cell.

4- Join the points.

To draw a direction in a simple cubic lattice, you have to follow these instructions:

1- Identify the points a₀, b₀, and c₀ in the cubic cell.

2- Draw the direction as a vector-like (a₀ b₀ c₀).

7 0

2 years ago

A circuit contains a resistor, an inductor, and a capacitor. When an AC voltage is applied across the circuit, the impedance of

katen-ka-za [31]

Answer:

The impedance of the circuit depends on the angular frequency of the voltage source.

Explanation:

In a electric circuit, the magnitude of the impedance, is given by the following expression:

$Z = \sqrt{R^{2} + (Xl-Xc)^{2} (1)$

where R = Resistance

Xl = Inductive reactance = ω*L

Xc = Capacitive Reactance = 1/ωC

and ω = angular frequency of the voltage source.

So, it can be seen that the impedance depends on the value of the constants R,L and C, and on the angular frequency ω.

3 0

2 years ago

Will a simple circuit work without a switch? Explain why or why not.

Marina CMI [18]

yes, a simple circuit will work without a switch. If we don't have a switch in a circuit then whichever device you are using electricity for wouldn't be able to switch off and will be working continuously. An open/close switch is usually all that is needed to supply or disconnect power from the source to the circuit.So, the circuit will work but it will work continuously.

7 0

2 years ago