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2 years ago

What happens to the duty cycle for a GMAW Gun when 75Ar/25COzgas

Engineering

1 answer:

skad [1K]2 years ago

6 0

So what happens is the host will not kill the y no se que hacer para no one can see it in

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Disc brake rotors that are too thin cannot handle as much heat and will experience ___________.

konstantin123 [22]

Answer:

brake fade

Explanation:

Among the most frequent problems in the brake disc is their overheating. When there is overheating, the temperature of the disc rises to critical values, thus the brake pad starts to slide with respect to the disc, and the efficiency of the brake system decreases to a minimum.

Overheating of the brake disc is detected when inspecting the parts. Steel is the most common base material in the creation of brake discs. When heated, the material becomes its color. A disc of steel at critical temperature, turns bright orange and later on cooling becomes purple.

If a color change is seen on the brake discs when inspecting, a service station must be contacted without delay for an in-depth examination. After detecting the problem, the brake discs and pads will be changed in a mandatory way.

6 0

3 years ago

g A food department is kept at -12oC by a refrigerator in an environment at 30oC. The total heat gain to the food department is

boyakko [2]

Answer:

a) $\dot W = 0.417\,kW$ , b) $COP_{R} = 2.198$ , c) Irreversible.

Explanation:

a) The power input required by the refrigerator is:

$\dot W = \dot Q_{H} - \dot Q_{L}$

$\dot W = \left(4800\,\frac{kJ}{h} - 3300\,\frac{kJ}{h}\right)\cdot \left(\frac{1}{3600} \,\frac{h}{s} \right)$

$\dot W = 0.417\,kW$

b) The Coefficient of Performance of the refrigerator is:

$COP_{R} = \frac{\dot Q_{L}}{\dot W}$

$COP_{R} = \frac{3300\,\frac{kJ}{h} }{(0.417\,kW)\cdot \left(3600\,\frac{s}{h} \right)}$

$COP_{R} = 2.198$

c) The maximum ideal Coefficient of Performance of the refrigeration is given by the inverse Carnot's Cycle:

$COP_{R,ideal} = \frac{T_{L}}{T_{H}-T_{L}}$

$COP_{R,ideal} = \frac{261.15\,K}{303.15\,K - 261.15\,K}$

$COP_{R,ideal} = 6.218$

The refrigeration cycle is irreversible, as $COP_{R} < COP_{R,ideal}$ .

3 0

3 years ago

The rate at which velocity changes is called?

Travka [436]

The rate at which velocity changes is called acceleration

7 0

3 years ago

Read 2 more answers

WILL MARK BRAINLEST PLEASE HELP

TEA [102]

I put

People who pursue a career in the creative imaging fields have qualities like a good imagination, creativity, open minds, good with ideas, and handling situations. If you enter that field, you need imagination to create things and an open mind to be open to all creations. You need good ideas to make good thing that will work.

please don't plagiarise tho, re-word it.

6 0

3 years ago

Find the thickness of the material that will allow heat transfer of 6706.8 *10^6 kcal during the 5 months through the rectangle

Vinvika [58]

Answer:

The thickness of the material is 6.23 cm

Explanation:

Given;

quantity of heat, Q = 6706.8 *10⁶ kcal

duration of the heat transfer, t = 5 months

thermal conductivity of copper, k = 385 W/mk

outside temperature of the heater, T₁ = 30° C

inside temperature of the heater, T₂ = 50° C

dimension of the rectangular heater = 450 cm by 384 cm

1 kcal = 1.163000 Watt-hour

6706.8 *10⁶ kcal = 7800008400 watt-hour

I month = 730 hours

5 months = 3650 hours

Rate of heat transfer, P = $\frac{7800008400 \ Watt-Hour}{3650 \ Hours} = 2136988.6 \ W$

Rate of heat transfer, $P = \frac{K*A *\delta T}{L}$

where;

P is the rate of heat transfer (W)

k si the thermal conductivity (W/mk)

ΔT is change in temperature (K)

A is area of the heater (m²)

L is thickness of the heater (m)

$P = \frac{KA(T_2-T_1)}{L} \\\\L = \frac{KA(T_2-T_1)}{P}\\\\L = \frac{385(4.5*3.84)(50-30)}{2136988.6}\\\\L = 0.0623 \ m$

L = 6.23 cm

Therefore, the thickness of the material is 6.23 cm

8 0

3 years ago