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Vinil7 [7]3 years ago

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Additional scals apply to the

REY [17]

Location of the class depends on satiation

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2 years ago

What is the difference between filler and electrode in Welding? Can a filler be an electrode? Can an electrode be a filler? Why?

Vlada [557]

Explanation:

<u>Filler:</u>

Filler is the material rod is used when we are joining two material by using welding process.If thickness of work piece is more so it will become compulsory to provide some filler material for making the welding join to withstand high stresses.

<u>Electrode:</u>

Electrode is the element which is used to complete the electric circuit in welding .Some time electrode is connected with positive terminal and some time with negative terminal ,it depends on the requirement of welding process.In Tungsten inert gas welding electrode is connected negative terminal but on the other hand Metal inert gas welding electrode is connected with positive terminal.Electrode can be consumable non-consumable depends on the condition.

Yes electrode can be work as filler material ,in Metal inert gas welding wire is used as electrode as well as filler material.In Metal inert gas welding consumable electrode is used on the other hand Tungsten inert gas welding non-consumable electrode is used.In Tungsten inert gas welding if thickness of work pieces is less than 5 mm then no need to used any filler material but if thickness is more than 5 mm then we have to use filler material.

8 0

3 years ago

Shows a closed tank holding air and oil to which is connected a U-tube mercury manometer and a pressure gage. Determine the read

damaskus [11]

Answer:

$P_2-P_1=27209h$

Explanation:

For pressure gage we can determine this by saying:

The closed tank with oil and air has a pressure of P₁ and the pressure of oil at a certain height in the U-tube on mercury is p₁gh₁. The pressure of mercury on the air in pressure gauge is p₂gh₂. The pressure of the gage is P₂.

$P_1+p_1gh_1=p_2_gh_2+P_2$

We want to work out P₁-P₂: Heights aren't given so we can solve it in terms of height: assuming h₁=h₂=h

$P_1-P_2=p_1gh_1-p_2gh_2=(55)\cdot{32.2}h-845\cdot{32.2}h$

$P_2-P_1=27209h$

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3 years ago

A 03-series cylindrical roller bearing with inner ring rotating is required for an application in which the life requirement is

-BARSIC- [3]

Answer:

$\mathbf{C_{10} = 137.611 \ kN}$

Explanation:

From the information given:

Life requirement = 40 kh = 40 $40 \times 10^{3} \ h$

Speed (N) = 520 rev/min

Reliability goal $(R_D)$ = 0.9

Radial load $(F_D)$ = 2600 lbf

To find C10 value by using the formula:

$C_{10}=F_D\times \pmatrix \dfrac{x_D}{x_o +(\theta-x_o) \bigg(In(\dfrac{1}{R_o}) \bigg)^{\dfrac{1}{b}}} \end {pmatrix} ^{^{^{\dfrac{1}{a}}$

where;

$x_D = \text{bearing life in million revolution} \\ \\ x_D = \dfrac{60 \times L_h \times N}{10^6} \\ \\ x_D = \dfrac{60 \times 40 \times 10^3 \times 520}{10^6}\\ \\ x_D = 1248 \text{ million revolutions}$

$\text{The cyclindrical roller bearing (a)}= \dfrac{10}{3}$

The Weibull parameters include:

$x_o = 0.02$

$(\theta - x_o) = 4.439$

$b= 1.483$

∴

Using the above formula:

$C_{10}=1.4\times 2600 \times \pmatrix \dfrac{1248}{0.02+(4.439) \bigg(In(\dfrac{1}{0.9}) \bigg)^{\dfrac{1}{1.483}}} \end {pmatrix} ^{^{^{\dfrac{1}{\dfrac{10}{3}}}$

$C_{10}=3640 \times \pmatrix \dfrac{1248}{0.02+(4.439) \bigg(In(\dfrac{1}{0.9}) \bigg)^{\dfrac{1}{1.483}}} \end {pmatrix} ^{^{^{\dfrac{3}{10}}$

$C_{10} = 3640 \times \bigg[\dfrac{1248}{0.9933481582}\bigg]^{\dfrac{3}{10}}$

$C_{10} = 30962.449 \ lbf$

Recall that:

1 kN = 225 lbf

∴

$C_{10} = \dfrac{30962.449}{225}$

$\mathbf{C_{10} = 137.611 \ kN}$

7 0

3 years ago

The unit for volume flow rate is gallons per minute, but cubic feet per second is preferred. Use the conversion factor tables in

amm1812

Answer:

The conversion factor is 0.00223 ( 1 gallon per minute equals 0.00223 cubic feet per second)

Explanation:

Since the given volume flow rate is gallons per minute.

We know that 1 gallon = 3.785 liters and

1 minute = 60 seconds

Let the flow rate be $Q\frac{gallons}{minute}$

Now replacing the gallon and the minute by the above values we get

$Q'=Q\frac{gallon}{minute}\times \frac{3.785liters}{gallon}\times \frac{1minute}{60seconds}$

Thus $Q'=0.631Q\frac{liters}{second}$

Now since we know that 1 liter = $0.0353ft^{3}$

Using this in above relation we get

$Q'=0.631Q\frac{liters}{second}\times \frac{0.0353ft^3}{liters}\\\\\therefore Q'=0.00223Q$

From the above relation we can see that flow rate of 1 gallons per minute equals flow rate of 0.00223 cubic feet per second. Thus the conversion factor is 0.00223.

3 0

3 years ago