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3 years ago

The temperature of an electric welding arc is about?

Engineering

1 answer:

N76 [4]3 years ago

7 0

Answer:

The heat of the arc melts the surface of the base metal and the end of the electrode. The electric arc has a temperature that ranges from 3,000 to 20,000 °C

Explanation:

Welding fumes are complex mixtures of particles and ionized gases.

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Air enters the compressor of an ideal gas refrigeration cycle at 7∘C and 35 kPa and the turbine at 37∘C and 160 kPa. The mass fl

amid [387]

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5 0

4 years ago

Consider a steam turbine, with inflow at 500oC and 7.9 MPa. The machine has a total-to-static efficiency ofηts=0.91, and the pre

sergiy2304 [10]

Answer: $\dot m_{in} = 23.942 \frac{kg}{s}$ , $\dot H_{out} = 39632.62 kW$

Explanation:

Since there is no information related to volume flow to and from turbine, let is assume that volume flow at inlet equals to $\dot V = 1 \frac{m^{3}}{s}$ . Turbine is a steady-flow system modelled by using Principle of Mass Conservation and First Law of Thermodynamics:

Principle of Mass Conservation

$\dot m_{in} - \dot m_{out} = 0$

First Law of Thermodynamics

$- \dot W_{out} + \eta\cdot (\dot m_{in} \dot h_{in} - \dot m_{out} \dot h_{out}) = 0$

This 2 x 2 System can be reduced into one equation as follows:

$-\dot W_{out} + \eta \cdot \dot m \cdot ( h_{in}- h_{out})=0$

The water goes to the turbine as Superheated steam and goes out as saturated vapor or a liquid-vapor mix. Specific volume and specific enthalpy at inflow are required to determine specific enthalpy at outflow and mass flow rate, respectively. Property tables are a practical form to get information:

Inflow (Superheated Steam)

$\nu_{in} = 0.041767 \frac{m^{3}}{kg} \\h_{in} = 3399.5 \frac{kJ}{kg}$

The mass flow rate can be calculated by using this expression:

$\dot m_{in} =\frac{\dot V_{in}}{\nu_{in}}$

$\dot m_{in} = 23.942 \frac{kg}{s}$

Afterwards, the specific enthalpy at outflow is determined by isolating it from energy balance:

$h_{out} =h_{in}-\frac{\dot W_{out}}{\eta \cdot \dot m}$

$h_{out} = 1655.36 \frac{kJ}{kg}$

The enthalpy rate at outflow is:

$\dot H_{out} = \dot m \cdot h_{out}$

$\dot H_{out} = 39632.62 kW$

3 0

3 years ago

Compression is maintained during combustion because on top of the motor is a _____

stealth61 [152]

Answer:

fan

Explanation:

7 0

3 years ago

A pin must be inserted into a collar of the same steel using an expansion fit. The coefficient of thermal expansion of the metal

nirvana33 [79]

Answer:

a) the temperature to which the pin must be cooled for assembly is $T_2 = -101.89^ \ ^0}C$

b) the radial pressure at room temperature after assembly is $P_f = 62.8 \ MPa$

c) the safety factor in the resulting assembly = 6.4

Explanation:

Coefficient of thermal expansion $\alpha = 12.3*10^{-6} \ ^0 C$

Yield strength $\sigma_y$ = 400 MPa

Modulus of elasticity (E) = 209 GPa

Room Temperature $T_1$ = 20°C

outer diameter of the collar $D_o = 95 \ mm$

inner diameter of the collar $D_i = 60 \ mm$

pin diameter $D_p$ = $60.03 \ mm$

Clearance c = 0.06 mm

The temperature to which the pin must be cooled for assembly can be calculated by using the formula:

$(D_i - c )-D_p = \alpha * D_p(T_2-T_1)$

$(60-0.06)-60.03=12.3*10^{-6}*60.03(T_{2}-20^{0}C)$

-0.09 = $7.38369*10^{-4}$ $(T_{2}-20^{0}C)$

-0.09 = $7.38369*10^{-4}T_2$ $\ \ - \ \ 0.01476738$

$-0.09 + 0.01476738$ = $7.38369*10^{-4}T_2$

−0.07523262 = $7.38369*10^{-4}T_2$

$T_2 = \frac{-0.07523262}{7.38369*10^{-4}}$

$T_2 = -101.89^ \ ^0}C$

To determine the radial pressure at room temperature after assembly ;we have:

$P_f = \frac{E * (D_p-D_i)(D_o^2-D_1^2)}{D_i*D_o} \\ \\ \\ P_f = \frac{209*10^9* 0.03(95^2-60^2)}{60*95^2} \\ \\ P_f = 62815789.47 \ Pa \\ \\ P_f = 62.8 \ MPa$