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2 years ago

A gas metal arc welder is also known as a _____ welder.

Engineering

1 answer:

dusya [7]2 years ago

7 0

Answer:

Metal inert gas or Metal Active Gas

Explanation:

Gas Metal Arc Welder(GMAW) also is also termed as Metal Inert Gas(MIG) welder or Metal Active Gas (MAG) welder.

Gas Metal Arc welding is a process in which 2 metal pieces are melted through the use of electricity and joined together. A continuous wire is fed through the welding gun, which heats up the metal. This type of welding was introduced back in 1949 and was used for welding of Aluminium.

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Annealing is a process by which steel is reheated and then cooled to make it less brittle. Consider the reheat stage for a 100-m

kodGreya [7K]

Complete question is;

Annealing is a process by which steel is reheated and then cooled to make it less brittle. Consider the reheat stage for a 100 mm thick plate (ρ = 7830 kg/m3, Cp = 550 J/kg K, k = 48 W/m K). The plate initially is at 200 °C and is to be heated to a minimum temperature of 550 °C. Heating is effected in a gas-fired furnace where the products of combustion at T∞ = 800 °C maintain a convection heat transfer coefficient of h = 250 W/m.K on both surfaces of the plate. How long should the plate be left in the furnace?

Answer:

860 seconds

Explanation:

We are given;

Initial Temperature; Ti = 200 °C

Minimum Temperature; T_i = 550 °C

T∞ = 800 °C

convection coefficient; h = 250 W/m².K

ρ = 7830 kg/m³

Cp = 550 J/kg K

k = 48 W/m K

Plate thickness = 100mm

Thus,L = 100/2 = 50mm = 0.05 m

Let's find the biot number from the formula;

Bi = hL/K

Bi = (250 × 0.05)/48

Bi = 0.2604

Now, lowest temperature in the slab is given as;

θ_o = (T_o - T∞)/(T_i - T∞)

θ_o = (550 - 800)/(200 - 800)

θ_o = 0.4167

Now, from online tables calculation, we can find the root of the biot number.

Thus, root of the biot number Bi = 0.2604 is;

ζ1 = 0.488 rad

Also, C1 is gotten as 1.0396

Now,formula for thermal diffusivity is;

α = k/ρc

α = 48/(7830 × 550)

α = 1.115 × 10^(-5) m²/s

Also, from online tables, f(ζ1) = 0.401

Thus, we can find the time the plate should the plate be left in the furnace from;

-(ζ1)²(αt/L²) = In 0.401

-(ζ1)²(αt/L²) = -0.9138

t = (-0.9138 × 0.05²)/-(0.488² × 1.115 × 10^(-5))

t ≈ 860 s

8 0

2 years ago

A 150-lbm astronaut took his bathroom scale (a spring scale) and a beam scale (compares masses) to the moon where the local grav

kozerog [31]

Answer:

a)Wt =25.68 lbf

b)Wt = 150 lbf

F= 899.59 N

Explanation:

Given that

$g = 5.48 ft/s^2.$

m= 150 lbm

Weight on the spring scale(Wt) = m g

We know that

$1\ lbf=32.17 \ lmb.ft/s^2$

Wt = 150 x 5.48/32 lbf

Wt =25.68 lbf

On the beam scale

This is scale which does not affects by gravitational acceleration.So the wight on the beam scale will be 150 lbf.

Wt = 150 lbf

If the plane is moving upward with acceleration 6 g's then the for F

F = m a

We know that

$1\ ft/s^2= 0.304\ m/s^2$

$5.48\ ft/s^2= 1.66\ m/s^2$

a=6 g's

$a=9.99\ m/s^2$

F = 90 x 9.99 N

F= 899.59 N

3 0

2 years ago

True or False:A retractable service pit cover can help other shop employees from falling into the pit while another technician i

coldgirl [10]

Answer:

Explanation:

7 0

2 years ago

Read 2 more answers

Consider air entering a heated duct at P1 = 1 atm and T1 = 288 K. Ignore the effect of friction. Calculate the amount of heat pe

Ne4ueva [31]

Answer:

The solution for the given problem is done below.

Explanation:

M1 = 2.0

$\frac{p1}{p*}$ = 0.3636

$\frac{T1}{T*}$ = 0.5289

$\frac{T01}{T0*}$ = 0.7934

Isentropic Flow Chart: M1 = 2.0 , $\frac{T01}{T1}$ = 1.8

T1 = $\frac{1}{0.7934}$ (1.8)(288K) = 653.4 K.

In order to choke the flow at the exit (M2=1), the above T0* must be stagnation temperature at the exit.

At the inlet,

T02= $\frac{T02}{T1}T1$ = (1.8)(288K) = 518.4 K.

Q= Cp(T02-T01) = $\frac{1.4(287 J / (Kg.K)}{1.4-1}(653.4-518.4)K$ = 135.7* $10^{3}$ J/Kg.

5 0

2 years ago

Read 2 more answers

A 20 mm 3 20 mm silicon chip is mounted such thatthe edges are flush in a substrate. The substrate provides anunheated starting

sasho [114]

Answer:

q= 1.77 W

Explanation:

Given data, Dimensions = 20mm x 20mm, Unheated starting length (ε) = 20 mm, Air flow temperature = 25 C, Velocity of flow = 25m/s, Surface Temperature = 75 C

To find maximum allowed power dissipated use the formula Q = hA(ΔT), where Q = Maximum allowed power dissipated from surface , h = heat transfer coefficiant, A = Area of Surface, ΔT = Temperature difference between surface and surrounding

we need to find h, which is given by (Nu x K)/x, where Nu is the Nusselt's Number, k is the thermal conductivity at film temperature and x is the length of the substrate.

For Nu use the Churchill and Ozoe relation used for parallel flow over the flat plate , Nu = (Nux (ε=0))/[1-(ε/x)^3/4]^1/3

Nux (ε=0) = 0.453 x Re^0.5 x Pr^1/3, where Re is the Reynolds number calculated by [Rex = Vx/v, where V is the velocity and v is the kinematic velocity, x being the length of the substrate] and Pr being the Prandtl Number

The constants, namel Pr, k and v are temperature dependent, so we need to find the film temperature

Tfilm = (Tsubstrate + Tmax)/2 = (25 + 75)/2 = 50 C

At 50 C, Pr = 0.7228, k = 0.02735w/m.k , v = 1.798 x 10^-5 m2/s

First find Rex and keep using the value in the subsequent formulas until we reach Q

Rex = Vx/v = 25 x (0.02 + 0.02)/ 1.798 x 10^-5 = 55617.35 < 5 x 10^5, thus flow is laminar, (x = L + ε)

Nux = (Nux (ε=0))/[1-(ε/x)^3/4]^1/3 = 0.453 x Re^0.5 x Pr^1/3/[1-(ε/x)^3/4]^1/3

= 0.453 x 55617.35 ^0.5 x 0.7228^1/3/ [1 - (0.02/0.04)^3/4]^1/3 = 129.54

h = (Nu x K)/x = (129.54 x 0.02735)/0.04 = 88.75W/m2.k

Q = 88.75 x (0.02)^2 x (75-25) = 1.77 W

7 0

2 years ago