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dexar [7]
3 years ago
14

In DC electrode positive, how much power is at the work clamp?

Engineering
1 answer:
Korolek [52]3 years ago
7 0

Answer:

1/3 power

Explanation:

I'm just a smart guy

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A nozzle receives an ideal gas flow with a velocity of 25 m/s, and the exit at 100 kPa, 300 K velocity is 250 m/s. Determine the
Margaret [11]

Given Information:

Inlet velocity = Vin = 25 m/s

Exit velocity = Vout = 250 m/s

Exit Temperature = Tout = 300K

Exit Pressure = Pout = 100 kPa

Required Information:

Inlet Temperature of argon = ?

Inlet Temperature of helium = ?

Inlet Temperature of nitrogen = ?

Answer:

Inlet Temperature of argon = 360K

Inlet Temperature of helium = 306K

Inlet Temperature of nitrogen = 330K

Explanation:

Recall that the energy equation is given by

$ C_p(T_{in} - T_{out}) = \frac{1}{2} \times (V_{out}^2 - V_{in}^2) $

Where Cp is the specific heat constant of the gas.

Re-arranging the equation for inlet temperature

$ T_{in}  = \frac{1}{2} \times \frac{(V_{out}^2 - V_{in}^2)}{C_p}  + T_{out}$

For Argon Gas:

The specific heat constant of argon is given by (from ideal gas properties table)

C_p = 520 \:\: J/kg.K

So, the inlet temperature of argon is

$ T_{in}  = \frac{1}{2} \times \frac{(250^2 - 25^2)}{520}  + 300$

$ T_{in}  = \frac{1}{2} \times 119  + 300$

$ T_{in}  = 360K $

For Helium Gas:

The specific heat constant of helium is given by (from ideal gas properties table)

C_p = 5193 \:\: J/kg.K

So, the inlet temperature of helium is

$ T_{in}  = \frac{1}{2} \times \frac{(250^2 - 25^2)}{5193}  + 300$

$ T_{in}  = \frac{1}{2} \times 12  + 300$

$ T_{in}  = 306K $

For Nitrogen Gas:

The specific heat constant of nitrogen is given by (from ideal gas properties table)

C_p = 1039 \:\: J/kg.K

So, the inlet temperature of nitrogen is

$ T_{in}  = \frac{1}{2} \times \frac{(250^2 - 25^2)}{1039}  + 300$

$ T_{in}  = \frac{1}{2} \times 60  + 300$

$ T_{in}  = 330K $

Note: Answers are rounded to the nearest whole numbers.

5 0
3 years ago
The term _______________refers to the science of using fluids to perform work.
Alla [95]
The term is hydraulics.
5 0
3 years ago
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un contenedor de 0.01m∧3 se llena con 2kg de nitrogeno a una presion de 15mpa ¿cual es la temperatura del nitrogeno?resolver uti
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Example – a 100 kW, 60 Hz, 1175 rpm motor is coupled to a flywheel through a gearbox • the kinetic energy of the revolving compo
rjkz [21]

Answer:

1200KJ

Explanation:

The heat dissipated in the rotor while coming down from its running speed to zero, is equal to three times its running kinetic energy.

P (rotor-loss) = 3 x K.E

P = 3 x 300 = 900 KJ

After coming to zero, the motor again goes back to running speed of 1175 rpm but in opposite direction. The KE in this case would be;

KE = 300 KJ

Since it is in opposite direction, it will also add up to rotor loss

P ( rotor loss ) = 900 + 300 = 1200 KJ

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3 years ago
A video inspection snake is use
LekaFEV [45]

Answer:

very good thx

Explanation:

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