Is there anyone who can help me with welding?

A one-dimensional slab without heat generation has a thickness of 20 mm with surfaces main- tained at temperatures of 275 K and

vlada-n [284]

Answer:

a) 512.5 KW/m2

b) 40.75 KW/m2

c) 2 KW/m2

Explanation:

Given data;

T_2 = 325 K

T_1 = 275 K

dx = 0.20 mm

a) for aluminium K = 205 W/m k

heat flux $= k \frac{dt}{dx}$

$= 205 \frac{325 - 275}{0.02}$

= 512.5 KW/m2

b) for AISI 316 stainless steel

k = 16.3 W/ m k

heat flux $= k \frac{dt}{dx}$

$= 16.3 \frac{325 - 275}{0.02}$

= 40.75 KW/m2

C) for Concrete

k = 0.8 W/ m k

heat flux $= k \frac{dt}{dx}$

$= 0.8 \times \frac{325 - 275}{0.02}$

= 2 KW/m2

6 0

4 years ago

Five kg of nitrogen gas (N2) in a rigid, insulated container fitted with a paddle wheel is initially at 300 K, 150 kPa. The N2 g

andrew-mc [135]

Answer:

A) attached below

B) 743 KJ

C) 1.8983 KJ/K

Explanation:

A) Diagram of system schematic and set up states

attached below

<u>B) Calculate the amount of work received from the paddle wheel </u>

assuming ideal gas situation

v1 = v2 ( for a constant volume process )

work generated by paddle wheel = system internal energy

dw = mCv dT . where ; Cv = 0.743 KJ/kgk

= 5 * 0.743 * ( 500 - 300 )

= 3.715 * 200 = 743 KJ

<u>C) calculate the amount of entropy generated ( KJ/K )</u>

S2 - S1 = 1.8983 KJ/K

attached below is the detailed solution

4 0

3 years ago

For better thermal control it is common to make catalytic reactors that have many tubes packed with catalysts inside a larger sh

Paul [167]

Answer:

the pressure drop is 0.21159 atm

Explanation:

Given that:

length of the reactor L = 2.5 m

inside diameter of the reactor d= 0.025 m

diameter of alumina sphere $dp$ = 0.003 m

particle density = 1300 kg/m³

the bed void fraction $\in =$ 0.38

superficial mass flux m = 4684 kg/m²hr

The Feed is methane with pressure P = 5 bar and temperature T = 400 K

Density of the methane gas $\rho$ = 0.15 mol/dm ⁻³

viscosity of methane gas $\mu$ = 1.429 x 10⁻⁵ Pas

The objective is to determine the pressure drop.

Let first convert the Density of the methane gas from 0.15 mol/dm ⁻³ to kg/m³

SO; we have :

Density = 0.15 mol/dm ⁻³

Molar mass of methane gas (CH₄) = (12 + (1×4) ) = 16 mol

Density = $0.1 5 *\dfrac{16}{0.1^3}$

Density = 2400

Density $\rho_f$ = 2.4 kg/m³

Density = mass /volume

Thus;

Volume = mass/density

Volume of the methane gas = 4684 kg/m²hr / 2.4 kg/m³

Volume of the methane gas = 1951.666 m/hr

To m/sec; we have :

Volume of the methane gas = 1951.666 * 1/3600 m/sec = 0.542130 m/sec

$Re = \dfrac{dV \rho}{\mu}$

$Re = \dfrac{0.025*0.5421430*2.4}{1.429*10^5}$

$Re=2276.317705$

For Re > 1000

$\dfrac{\Delta P}{L}=\dfrac{1.75 \rho_f(1- \in)v_o}{\phi_sdp \in^3}$

$\dfrac{\Delta P}{2.5}=\dfrac{(1.75 *2.4)(1- 0.38)*0.542130}{1*0.003 (0.38)^3}$

$\Delta P=8575.755212*2.5$

$\Delta = 21439.38803 \ Pa$

To atm ; we have

$\Delta P = \dfrac{21439.38803 }{101325}$

$\Delta P =0.2115903087 \ atm$

ΔP ≅ 0.21159 atm

Thus; the pressure drop is 0.21159 atm

4 0

3 years ago

Iron has a BCC crystal structure, an atomic radius of 0.124 nm, and an atomic weight of 55.85 g/mol. Compute and compare its the

Andre45 [30]

Answer:

Explanation:

The density of the unit cell of a material, Iron in this case, has to be approximately equal with its experimental value of 7.87 g/cm³.

The density d = m/v, so what we need to do is calculate the volume of the unit cell and its mass and perform the calculation.

For a BCC crystal structure the length of the side of the cube is given by:

a = 4r/√3

where a is the atomic radius of Iron

first we will convert this radius to cm since we want the density in g/cm³:

0.124 nm x 1 x 10⁻⁷ cm / nm = 1.24 x 10⁻⁸ cm

a = 4 x 1.24 x 10⁻⁸ cm /√3 = 2.86 x 10⁻⁸ cm

the volume of the cubic cell is:

v = a³ = ( 2.86 x 10⁻⁸ cm )³ =2.35 x 10⁻²³ cm³

The mass of iron in the body centered cubic cell is obtained from the mass of the atoms in it:

BCC = 2 atoms / unit cell ( 1/8 from the 8 corners + 1 in the center)

m = 2 atoms/unit cell x 1 mol/ 6.022 x 10²³ atoms x 55.85 g/mol

= 1.85 x 10⁻²² g

Therefore,

d = m/v = 1.85 x 10⁻²² g / 2.35 x 10⁻²³cm³ = 7.88 g/cm³

An excelent agreement which confirms that the density of the BCC unit cell agrees with the experimental value.

4 0

4 years ago

Steel grade AISI 1045 contains 4.5% carbon. a)-True b)- False

yarga [219]

Answer:

False

Explanation:

According to the AISI (SAE) statements the last two numbers of the designation indicates the porcentage of carbon in the steel. And the firs numbers indicate the most important alloy alements.

In the case of the AISI 1045 indicates that the composition of this steel is 0,45% of carbon.

4 0

3 years ago