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2 years ago

12

A joining process in which a filler metal is melted and distributed by capillary action between faying surfaces, the base metals

does not melt, and in which the filler metal melts at a temperature less than 450ᵉC is called?a.An arc welding processb. A soldering processc. An adhesive joining processd. A brazing processe. None of the above

1 answer:

PolarNik [594]2 years ago

4 0

Answer:

A soldering process

Explanation:

Given that ,The filler metal's melting point temperature is less than 450 ° C.

Usually, the brazing material has the liquid temperature of the melting point, the full melting point of the filler material approaching 450 degrees centigrade, while the filler material is less than 450 degrees centigrade in the case of soldering.

Therefore the answer is "A soldering process".

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Bananas are to be cooled from 28°C to 12°C at a rate of 1140 kg/h by a refrigerator that operates on a vapor-compression refrige

Lera25 [3.4K]

Answer:

A) $COP = \frac{16.97}{9.8} = 1.731$

B) $P_{IN} = 0.4763$

C) Second law efficiency 4.85%

exergy destruction for the cycle = 9.3237 kW

Explanation:

Given data:

$T_1 = 28$ degree celcius

$T_2 = 12$ degree celcius

$\dot m = 1140 kg/h$

Power to refrigerator = 9.8 kW

Cp = 3.35 kJ/kg degree C

$A) Q = \dot m Cp \Delta T$

$= 1140 \times 3.35\times (28-12) = 61,104 kJ/h$

$Q_{abs} = 61,104 kJ/h = 16.97 kJ/sec$

$COP = \frac{16.97}{9.8} = 1.731$

b)

$COP ∝ \frac{1}{P_{in}}$

$P_{in}$ wil be max when COP maximum

taking surrounding temperature T_H = 20 degree celcius

$COP_{max} = \frac{T_L}{T_H- T_L} = \frac{285}{293 - 285} = 35.625$

we know that

$COP = \frac{heat\ obsorbed}{P_{in}}$

$P_{IN} = \frac{16.97}{35.62} = 0.4763$

c) second law efficiency

$\eta_{11} = \frac{COP_R}{(COP)_max} = \frac{1.731}{35.625} = 4.85\%$

exergy destruction os given as $X = W_{IN} - X_{Q2}$

= 9.8 - 0.473 = 9.3237 kW

8 0

2 years ago

The current drawn by fluorescent lighting has a high total harmonic distortion. For this case, THD is calculated to be 88%. The

Alona [7]

Answer:

displacement power factor is 0.959087

Explanation:

given data

THD = 88%

true power factor = 0.72

solution

we get here total harmonic distribution THD is express as here

THD = $\sqrt{\frac{1}{g^2}-1}$ ..............1

her g is distortion factor

so put here value and we will get g that is

0.88² = $\frac{1}{g^2} -1$

solve it we get

g = 0.750714

and

displacement power factor is express as

DPF = $\frac{PF}{g}$ .................2

put here value and we will get

DPF = $\frac{0.72}{0.750714}$

DPF = 0.959087

3 0

2 years ago

A heat pump operates on a Carnot heat pump cycle with a COP of 12.5. It keeps a space at 24°C by consuming 2.15 kW of power. Det

Vinil7 [7]

Answer:

a) $T_{L} = 273.378\,K\,(0.228\,^{\textdegree}C)$ , b) $\dot Q_{H} = 26.875\,kW$

Explanation:

a) The Coefficient of Performance of the Carnot Heat Pump is:

$COP_{HP} = \frac{T_{H}}{T_{H}-T_{L}}$

After some algebraic handling, the temperature of the cold reservoir is determined:

$T_{H}-T_{L} = \frac{T_{H}}{COP_{HP}}$

$T_{L} = T_{H}\cdot \left(1-\frac{1}{COP_{HP}} \right)$

$T_{L} = (297.15\,K)\cdot \left(1-\frac{1}{12.5}\right)$

$T_{L} = 273.378\,K\,(0.228\,^{\textdegree}C)$

b) The heating load provided by the heat pump is:

$\dot Q_{H} = COP_{HP}\cdot \dot W$

$\dot Q_{H} = (12.5)\cdot (2.15\,kW)$

$\dot Q_{H} = 26.875\,kW$

4 0

2 years ago

Write SQL queries to answer the following questions: What are the names of the course(s) that student Altvater took during the s

amm1812

Answer:

See explanations

Explanation:

Consider the following Venn diagram to retrieve the name of course that student Altvater took during semester I-2015.

• In above Venn diagram inner Ellipse represent the subquery part, this subquery part select the Student ID from STUDENT table.

• Second sub query is used to determine the SectionNo of all student whose studentID retrived in the first subquery

• Finally the main query displays the CourseName from COURSE table.

6 0

2 years ago

A 10-mm drill rod was heat-treated and ground. The measured hardness He was found to be 300 Brinell. Estimate the endurance stre

lesya [120]

Answer:

The endurance strength for the rod is 434.6 MPa

Explanation:

Since the rod is used in rotating bending, we need to use Marin equation given by

$S=k_ak_bS'_e$

Here S stands for the endurance strength for rotating bending, $S'_e$ is the endurance strength, and $k_a \text{ and } k_b$ are the parameters for Marin surface modification factor.

Endurance strength.

We can start finding the endurance strength, from the directions we know that the hardness $H_e$ was found to be 300 Brinell, thus for such value we can find the ultimate tensile strength using

$S_{ut}=3.41H_e$

Replacing the hardness we get

$S_{ut}=3.41(300) MPa \\ S_{ut}=1023 MPa$

Now since the ultimate tensile strength has a value less than 1400 MPa, we can find the endurance strength using

$S'_e =0.5S_{ut}$

Replacing the tensile strength we get

$S'_e=0.5(1023) MPa \\ S'_e = 511.5 MPa$

Parameters for Marin surface modification factor.

From the directions we know that the drill rod has a ground surface finish, so then from tables we get

$a=1.58 \text{ and } b = -0.085$

Thus the surface factor will be

$k_a=a(S_{ut})^b$

Replacing values and the ultimate tensile strength

$k_a=(1.58)1023^{-0.085}\\k_a=0.8766$

Then we can find the rotating shaft factor, for a diameter of 10 mm, we can use the equation

$k_b=1.24d^{-0.107}$

Replacing the diameter we get

$k_b=1.24(10)^{-0.107}\\k_b=0.9692$

Estimating endurance strength for rotating shaft.

We can replace now all values we have found in Marin equation.

$S=k_ak_bS'_e$

$S=(0.8766)(0.9692)(511.5) MPa$

$S=434.6 MPa$

Thus the endurance strength for the rod is 434.6 MPa

3 0

2 years ago