Dependence of oxidation on the temperature 10 points total The process temperature is an important parameter for the oxide growth process. To get an estimate for the temperature dependence of the Deal-Groove formula, the following considerations can be used. All processes (transport, diffusion and reaction) that are used in the Deal-Groove model are thermally activated. Thus, the dependence on the growth parameters on the temperature T can be estimated by: X(T) = Xpe Xe{A, B, B/A} (5) Here kB = 8.617 x 10-5 eV/K is the Boltzmann Constant and EX is the activation energy for the parameter. X, is the value for the corresponding parameter at T =0 K. Please note that this is not a 'physical activation energy but rather an average/fit parameter for all the physical and chemical thermally activated processes relevant for the process. In the original publication by Deal and Groove, they gave the following parameter sets for dry oxidation: Oxidation temperature in 'C A in um Bin um²/h 1000 0.165 0.0117 1200 0.040 0.045 a)EXct and X, for A, B and B/A
a) Use the given values for the parameters A and B in the table above, along with equation ?? to derive the activation energies EXd and zero temperature values X(T = 0 K) for all three parameters A, B and B/A. Then use this calculate the corresponding values of A,B, B/A for the following temperatures: 1000°C, 1200°C, 800°C and 900 °C.
b) Plot estimates for other temperatures 2 points Use the obtained values from the previous sub-task along with the full Deal-Groove model equation for Z:(t) to plot the oxide thickness over time (0-100h) using at least 20 data points each for the following temperatures: 1000°C, 1200°C, 800°C and 900 °C. Assume Zo = Om.

Answer:

In both cases, reynolds number is greater than 2000,so the flows can't be laminar.

Explanation:

A) For flow in a tube of uniform diameter, the reynolds number is defined as;

Re = 2ρvr/η

where;

ρ is the fluid density,

v its speed,

η is viscosity

r is the tube radius.

In this question,

We are given;

r = 0.25cm = 0.25 × 10^(-2) m

η of water has a standard value of 1.005 × 10^(-3)

ρ of water has a standard value of 1000 kg/m³

In the reynolds equation, we don't know the velocity. So let's calculate it from;

Q' = vA

Where; Q' is flow rate = 0.5 L/s = 0.0005 m³/s

Area = πr² = π × (0.25 × 10^(-2))²

Area = 1.963 × 10^(-5) m²

So, v = Q/A = 0.0005/(1.963 × 10^(-5)) = 25.5 m/s

So, Re = 2ρvr/η = (2*1000*25.5*0.25 × 10^(-2))/(1.005 × 10^(-3))

Re = 126865.67

Re > 2000 and so the flow is not laminar

B) Now,

radius = 0.9cm = 0.9 × 10^(-2) m

So, A = πr² = π × (0.9 × 10^(-2))²

A = 2.5447 × 10^(-4) m²

v = Q/A = 0.0005/(2.5447 × 10^(-4))

v = 1.965 m/s

Re = 2ρvr/η = (2*1000*1.965*0.9 × 10^(-2))/(1.005 × 10^(-3))

Re = 35194.03

Re > 2000. So flow is not laminar.

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 [tex]dp[/tex]= 0.003 m

particle density  = 1300 kg/m³

the bed void fraction [tex]\in =[/tex]  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 [tex]\rho[/tex] = 0.15 mol/dm ⁻³

viscosity of methane gas [tex]\mu[/tex] = 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 =  [tex]0.1 5 *\dfrac{16}{0.1^3}[/tex]

Density =  2400

Density [tex]\rho_f[/tex] =  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

[tex]Re = \dfrac{dV \rho}{\mu}[/tex]

[tex]Re = \dfrac{0.025*0.5421430*2.4}{1.429*10^5}[/tex]

[tex]Re=2276.317705[/tex]

For Re > 1000

[tex]\dfrac{\Delta P}{L}=\dfrac{1.75 \rho_f(1- \in)v_o}{\phi_sdp \in^3}[/tex]

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

[tex]\Delta P=8575.755212*2.5[/tex]

[tex]\Delta = 21439.38803 \ Pa[/tex]

To atm ; we have

[tex]\Delta P = \dfrac{21439.38803 }{101325}[/tex]

[tex]\Delta P =0.2115903087 \ atm[/tex]

ΔP  ≅  0.21159 atm

Thus; the  pressure drop  is 0.21159 atm

Answer: A

Original length = 1.60 m

Explanation: given that due to loading, a line segment has length 2 with constant normal strain 0.25

Strain is the ratio of extension to original length. That is,

Strain = e/L

If a line segment has length 2, that means:

e + L = 2

e = 2 - L

And given that the strain = 0.25

Substitute all the parameters into the formula

0.25 = ( 2 - L ) / L

Cross multiply

0.25L = 2 - L

Collect the like terms

0.25L + L = 2

1.25L = 2

L = 2/ 1.25

L = 1.6 m

Therefore, original length is 1.6 metres

Answer:

219kJ

Explanation:

The work done (W) on a gas in an isothermal process is given by;

W = -P₁V₁ ln[tex]\frac{V_{2}}{V_1}[/tex]      -----------------(i)

Where;

P₁ = initial pressure of the gas

V₁ = initial volume of the gas

V₂ = final volume of the gas

From the question;

P₁ = 600kPa = 6 x 10⁵Pa

V₁ = 0.45m³

V₂ = 0.2m³

Substitute these values into equation (i) as follows;

W = -6 x 10⁵ x 0.45 x ln [tex]\frac{0.2}{0.45}[/tex]

W = -6 x 10⁵ x 0.45 x ln (0.444)

W = -6 x 10⁵ x 0.45 x -0.811

W = 2.19 x 10⁵

W = 219 x 10³

W = 219kJ

Therefore, the work done on the gas during the compression process is 219kJ

A) The linear density expressions for BCC [110] and [111] directions in terms of the atomic radius R are;

i) LD_110 = √3/(4R√2)

ii) LD_111 = 1/(2R)

B) The linear density values for these same two directions for iron (Fe) are;

i) LD_110 = 2.4 × 10^(9) m^(-1)

ii) LD_111 = 4 × 10^(9) m^(-1)

A) i) To find linear density expression for BCC 110, first of all we will calculate the length of the vector using the length of the unit cell which is 4R/√3 and the cell edge length which is 4R. Thus, the vector length can now be calculated from this expression;

√((4R)² - (4R/√3)²)

This reduces to; 4R√(1 - 1/3) = 4R√(2/3)

Now, the expression for the linear density of this direction is;

LD_110 =

Number of atoms centered on (110) direction/vector length of 110 direction

In this case, there is only one atom centered on the 110 direction. Thus;

LD_110 = 1/(4R√(2/3))

LD_110 = √3/(4R√2)

ii)  The length of the vector for the direction 111 is equal to 4R, since

all of the atoms whose centers the vector passes through touch one another. In addition, the vector passes through an equivalent of 2 complete atoms. Thus, the linear density is;

LD_111 = 2/(4R) = 1/(2R)

B)i) From tables, the atomic radius for iron is 0.124 nm or 0.124 x 10^(-9) m. Therefore, the linear density for the [110] direction is;

LD_110 = √3/(4R√2) = √3/(4*0.124*10^(-9)(√2))

LD_110 = 2.4 × 10^(9) m^(-1)

ii) for the 111 direction, we have;

LD_111 = 1/(2R) = 1/(2*0.124*10^(-9))

LD_111 = 4 × 10^(9) m^(-1)

Read more about Linear Density of Crystalline Structures at; https://brainly.com/question/14831455

Answer:

Exit temperature = 32°C

Explanation:

We are given;

Initial Pressure;P1 = 100 KPa

Cp =1000 J/kg.K = 1 KJ/kg.k

R = 500 J/kg.K = 0.5 Kj/Kg.k

Initial temperature;T1 = 27°C = 273 + 27K = 300 K

volume flow rate;V' = 15 m³/s

W = 130 Kw

Q = 80 Kw

Using ideal gas equation,

PV' = m'RT

Where m' is mass flow rate.

Thus;making m' the subject, we have;

m' = PV'/RT

So at inlet,

m' = P1•V1'/(R•T1)

m' = (100 × 15)/(0.5 × 300)

m' = 10 kg/s

From steady flow energy equation, we know that;

m'•h1 + Q = m'h2 + W

Dividing through by m', we have;

h1 + Q/m' = h2 + W/m'

h = Cp•T

Thus,

Cp•T1 + Q/m' = Cp•T2 + W/m'

Plugging in the relevant values, we have;

(1*300) - (80/10) = (1*T2) - (130/10)

Q and M negative because heat is being lost.

300 - 8 + 13 = T2

T2 = 305 K = 305 - 273 °C = 32 °C

Answer:

0.5 kW

Explanation:

The given parameters are;

Volume of tank = 1 m³

Pressure of air entering tank = 1 bar

Temperature of air = 27°C = 300.15 K

Temperature after heating  = 477 °C = 750.15 K

V₂ = 1 m³

P₁V₁/T₁ = P₂V₂/T₂

P₁ = P₂

V₁ = T₁×V₂/T₂ = 300.15 * 1 /750.15 = 0.4 m³

[tex]dQ = m \times c_p \times (T_2 -T_1)[/tex]

For ideal gas, [tex]c_p[/tex] = 5/2×R = 5/2*0.287 = 0.7175 kJ

PV = NKT

N = PV/(KT) = 100000×1/(750.15×1.38×10⁻²³)

N = 9.66×10²⁴

Number of moles of air = 9.66×10²⁴/(6.02×10²³) = 16.05 moles

The average mass of one mole of air = 28.8 g

Therefore, the total mass = 28.8*16.05 = 462.135 g = 0.46 kg

∴ dQ = 0.46*0.7175*(750.15 - 300.15) = 149.211 kJ

The power input required = The rate of heat transfer = 149.211/(60*5)

The power input required = 0.49737 kW ≈ 0.5 kW.

Answer:

a. Compromise the fire rating by one hour.

Explanation:

One hour fire rating is given to materials that can resist the fire exposure. The Acoustical panels controls reverberation and they are used for echo controls. The Fire rating is the passive fire protection which can resist standard fire. The test for fire rating also consider normal functioning of the material.

Answer:

a) 2.45 KN

b) 0.0375 m

Explanation:

[tex](a) \quad \sigma_{v}=400 \times 10^{6} \mathrm{Pa} \quad A=\frac{\pi}{4} d^{2}=\frac{\pi}{4}(5)^{2}=19.635 \mathrm{mm}^{2}=19.635 \times 10^{-6} \mathrm{m}^{2}[/tex]

[tex]P_{U}=\sigma_{U} A=\left(400 \times 10^{6}\right)\left(19.635 \times 10^{-6}\right)=7854 \mathrm{N}[/tex]

[tex]P_{\text {al }}=\frac{P_{U}}{F S}=\frac{7854}{3.2}=2454 \mathrm{N}[/tex]

(b) [tex]\quad \delta=\frac{P L}{A E}=\frac{(2454)(60)}{\left(19.635 \times 10^{-6}\right)\left(200 \times 10^{9}\right)}=37.5 \times 10^{-3} \mathrm{m}[/tex]

Answer:

the magnitude of the stress necessary to cause slip to occur on the (111) plane in the [tex][1 0 \overline 1][/tex] direction is 4.85 MPa

Explanation:

From the given information;

To determine the angle [tex]\phi[/tex] between the direction [100] and [111]; we have:

[tex]\phi = cos ^{-1} [\dfrac{u_1u_2+v_1v_2+w_1w_2}{ \sqrt{(u_1^2+v_1^2+w_1^2) (u_2^2+v_2^2+w_2^2) }}}[/tex]

where;

[u₁,v₁,w₁] and [u₂, v₂, w₂] are directional indices.

replacing 1 for u₁ , 0 for v₁ and 0 for w₁  ;

also replacing 1 for u₂, 1 for v₂ and  1 for w₂ ; we have :

[tex]\phi = cos ^{-1} [\dfrac{1*1+0*1+0*1}{ \sqrt{(1^2+0^2+0^2) (1^2+1^2+1^2) }}}[/tex]

[tex]\phi = 54.7^0[/tex]

To determine the angle [tex]\lambda[/tex] for the slip direction [tex][1 0 \overline 1][/tex]

[tex]\lambda= cos ^{-1} [\dfrac{u_1u_2+v_1v_2+w_1w_2}{ \sqrt{(u_1^2+v_1^2+w_1^2) (u_2^2+v_2^2+w_2^2) }}}[/tex]

replacing 1 for u₁ , 0 for v₁ and 0 for w₁  ;

also replacing 1 for u₂, 1 for v₂ and  -1 for w₂ ; we have :

[tex]\lambda = cos ^{-1} [\dfrac{1*1+0*0+(0*-1)}{ \sqrt{(1^2+0^2+0^2) (1^2+0^2+(-1^2)) }}}[/tex]

[tex]\phi = cos ^{-1} [\dfrac{1}{\sqrt{2}}][/tex]

[tex]\lambda = 45^0[/tex]

The yield strength  for the slip process [tex][1 0 \overline 1][/tex] can now be calculated as:

[tex]\sigma_x = \dfrac{t_{xr}}{cos \phi \ \ cos \lambda}[/tex]

where

[tex]t_{xr}[/tex] = 1.98 MPa

[tex]\sigma_x = \dfrac{1.98}{cos 54.7^0 \ \ cos 45}[/tex]

[tex]\sigma_x = \dfrac{1.98}{0.5779 *0.7071}[/tex]

[tex]\mathbf{\sigma _x = 4.85 \ MPa}[/tex]

Hence, the magnitude of the stress necessary to cause slip to occur on the (111) plane in the [tex][1 0 \overline 1][/tex] direction is 4.85 MPa

Answer:

Electricity cost = $9.36

Explanation:

Given:

Electric power = 400 W = 0.4 KW

Unit cost of electricity = $0.13/kWh

Overall time = 1/4 (30 days) (24 hours) = 180 hours

Find:

Electricity cost

Computation:

Electricity cost = Electric power  x Unit cost of electricity x Overall time

Electricity cost = 0.4 x $0.13 x 180

Electricity cost = $9.36

Given:

Electric power = 400 W = 0.4 KW

Over all Time  = 30(1/4) = 7.5 days

Unit cost of electricity = $0.13/kWh

Find:

Electricity cost.

Computation:

Electricity cost = Electric power x Unit cost of electricity x Over all Time

Electricity cost = 0.4 x 0.13 x 7.5

Electricity cost = $

Answer:

1. B. False

2.  B. False

3. A. True

4. B. False

5. A. True

6. A. True

7. A. True

Explanation:

1. B. False

The relation of Reynolds' number, Reₓ to boundary layer thickness δ at a point x is given by the relation

[tex]\delta = \dfrac{x \times C}{\sqrt{Re_x} }[/tex]

That is the boundary layer thickness is inversely proportional to the square root of the Reynolds' number so that if the Reynolds' number were to increase, the boundary layer thickness would decrease

Therefore, the correct option is B. False

2.  B. False

From the relation

[tex]Re_x = \dfrac{U_o \times x}{v}[/tex]

As the outer flow velocity increases, the boundary layer thickness diminishes

3. A. True

As the viscous force is increased the boundary layer thickness increases

4. B. False

Boundary layer thickness is inversely proportional to velocity

5. A. True

The boundary layer model developed by Ludwig Prandtl is a special case of the Navier-Stokes equation

6. A. True

Given a definite boundary layer thickness, the curve representing the boundary layer thickness is a streamline

7. A. True

The boundary layer approximation by Prandtl Euler bridges the gap between the Euler (slip boundary conditions) and Navier-Stokes (no slip boundary conditions) equations.

Answer is given below

Explanation:

Answer:

Waterfall model

Explanation:

The waterfall model is amenable to the projects. It focused on the data structure. The software architecture and detail about the procedure. It will interfere with the procedure. It interfaces with the characterization of the objects. The waterfall model is the first model that is introduced first. This model also called a linear sequential life cycle model.

The waterfall model is very easy to use. This is the earliest approach of the SDLC.

There are different phase of the waterfall:

Answer:

Velocity = 4.73 m/s.

Explanation:

Work done by friction is;

W_f = frictional force × displacement

So; W_f = Ff * Δs = (μF_n)*Δs

where; magnitude of the normal force F_n is equal to the component of the weight perpendicular to the ramp i.e; F_n = mg*cos 24

Over the distance ab, Potential Energy change mgΔh transforms into a change in Kinetic energy and the work of friction, so;

mg(3 sin 24) = ΔKE1 + (0.22)*(mg cos 24) *(3).

Similarly, Over the distance bc, potential energy mg(2 sin 24) transforms to;

ΔKE2 + (0.16)(mg cos 24)(2).

Plugging in the relevant values, we have;

1.22mg = ΔKE1 + 0.603mg

ΔKE1 = 1.22mg - 0.603mg

ΔKE1 = 0.617mg

Also,

0.813mg = ΔKE2 + 0.292mg

ΔKE2 = 0.813mg - 0.292mg

ΔKE2 = 0.521mg

Now total increase in Kinetic Energy is ΔKE1 + ΔKE2

Thus,

Total increase in kinetic energy = 0.617mg + 0.521m = 1.138mg

Putting 9.81 for g to give;

Total increase in kinetic energy = 11.164m

Finally, if v = 0 m/s at point a, then at point c, KE = ½mv² = 11.164m

m cancels out to give; ½v² = 11.164

v² = 2 × 11.164

v² = 22.328

v = √22.328

v = 4.73 m/s.

Answer:

Fifty (50) percent. [50%]

Explanation:

Water heater is a home appliance that comprises of an electric or gas heating unit as well as a water-tank where water is heated and stored for use.

When using an alternative method of sizing with two vent connectors for draft hood-equipped water heaters, the effective area of the common vent connector or vent manifold and all junction fittings shall not be less than the area of the larger vent connector plus fifty (50) percent of the areas of smaller flue collar outlets.

A water heater is primarily vented with an approved and standardized plastic or metallic pipe such as flue or chimney, which allows gas to flow out of the water heater into the surrounding environment.

For a draft hood-equipped water heater, both the water heater and the barometric draft regulators must be installed in the same room. Also, the technician should ensure that the vent is through a concealed space such as conduit and should be labeled as Type L or Type B.

The minimum capacity of a water heater should be calculated based on the number of bathrooms, bedrooms and its first hour rating.

Answer:

False

Explanation:

The given statement is False. In real scenario the throttling valve not replaced by an isentropic turbine in the ideal vapor-compression refrigeration cycle. It is done so that the ideal vapor-compression refrigeration cycle to make the ideal vapor-compression refrigeration cycle more closely approximate the actual cycle.

Answer:

The exit temperature of the gas = 32° C

Explanation:

Solution

Given that:

Inlet temperature T₁ = 27°C ≈ 300.15 K

Inlet pressure P₁ = 100 KPa = 100 * 10^3 Pa

Volume flow rate , V = 15 m/s³

Diameter of the deduct, D = 500 mm = 0.5 m

Electric heater power, W heater = 130 kW = 130 * 10^3 W

The heat lost Q = 80 kW =  80 * 10^3 W

Now,

From the ideal gas law, density of the air at the inlet is given as :

ρ₁ = P₁/RT₁ = 100 * 10^3/500 * 300

=0.6667 kg/m³

The mass flow rate through the duct is computed below:

m = ρ₁ V = 0.6667 * 15 = 10 kg/s

Thus

Applying the first law of thermodynamics to the process is shown below:

Q + m (h₁ + V₁²/2 + gz₁) = m (h₂ + V₂²/2 + gz₂) + W (Conservation energy)

So,

If we neglect the potential and kinetic energy changes of the air, the above equation can be written again as:

Q + m (h₁) = m (h₂) + W

or

Q - W heater =m (h₂ - h₁) or Q - W heater =m (T₂ - T₁)

Thus

h₂ - h₁ = Cp T₂ - T₁

Now by method of substitution the known values are:

(- 80 *10^3) - (-130 * 10^3) = 10 * 100 * (T₂ -27)

Note: The heat transfer is  taken as negative because the heat is lost by the gas and work done is also taken as negative because the work is done on the gas

So,

Solving for T₂,

T₂ = 32° C

Therefore the exit temperature of the gas = 32° C

Answer:

There are [tex]6.624 \times 10^{21}\,atoms[/tex] of Germanium in a germanium-silicon alloy that contains 15 wt% Ge and 85 wt% Si.

Explanation:

The masses of silicon and germanium contained in a cubic centimeter of the germanium-silicon alloy by apply the concepts of mass ([tex]m[/tex]), density ([tex]\rho[/tex]) and volume ([tex]V[/tex]), as well as the mass-mass proportion of Germanium ([tex]x[/tex]):

[tex]m_{Ge} = x \cdot \rho_{Ge}\cdot V_{sample}[/tex]

[tex]m_{Ge} = 0.15\cdot \left(5.32\,\frac{g}{cm^{3}} \right)\cdot (1\,cm^{3})[/tex]

[tex]m_{Ge} = 0.798\,g[/tex]

The amount of moles of Germanium is obtained after dividing previous outcome by its atomic weight. That is to say:

[tex]n = \frac{m_{Ge}}{M_{Ge}}[/tex]

[tex]n = \frac{0.798\,g}{72.64\,\frac{g}{mol} }[/tex]

[tex]n = 0.011\,mol[/tex]

There are 0.011 moles in a cubic centimeter of the germanium-silicon alloy. According to the Law of Avogadro, there are [tex]6.022 \times 10^{23}\,atoms[/tex] in a mole of Germanium. The quantity of atoms in a cubic centimeter is therefore found by simple rule of three:

[tex]y = \frac{0.011\,mol}{1\,mol}\times \left(6.022\times 10^{23}\,\frac{atoms}{mole} \right)[/tex]

[tex]y = 6.624 \times 10^{21}\,atoms[/tex]

There are [tex]6.624 \times 10^{21}\,atoms[/tex] of Germanium in a germanium-silicon alloy that contains 15 wt% Ge and 85 wt% Si.

Answer:

The correct answer to the following question will be "[tex]a_{x}[/tex] or [tex]a_{y}[/tex]".

Explanation:

So that the above is the appropriate choice.

Answer:

Explanation:

The formula for critical stress is

[tex]\sigma_c=\frac{K}{Y\sqrt{\pi a} }[/tex]

[tex]\sigma_c =\texttt{critical stress}[/tex]

K is the plane strain fracture toughness

Y is dimensionless parameters

We are to Determine the Critical stress

Now replacing the critical stress with 54.8

a with 0.2mm = 0.2 x 10⁻³

Y with 1

[tex]\sigma_c=\frac{54.8}{1\sqrt{\pi \times 0.2\times10^{-3}} } \\\\=\frac{54.8}{\sqrt{6.283\times10^{-4}} } \\\\=\frac{54.8}{0.025} \\\\=2186.20Mpa[/tex]

The fracture will not occur because this material can handle a stress of 2186.20Mpa  before fracture. it is obvious that is greater than 2023Mpa

Therefore, the specimen does not failure for surface crack of 0.2mm

Answer:

54.52 MPa

Explanation:

power P = 46 kW = 46000 W

speed  N = 760 rpm

outer diameter of shaft D = 40 mm = 0.04 m

inner diameter of shaft d = 25 mm = 0.025 m

torque T = P/Ω

where Ω = angular speed in rad/s

Ω = 2πN/60 = (2 x 3.142 x 760)/60

Ω = 79.59 rad/s

from this,

torque T = 46000/79.59 = 577.96 N-m

the relationship between torque T, maximum shear stress τmax, and shaft diameters D and d is stated as

T = (π / 16) τmax ([tex]D^{4}[/tex] - [tex]d^{4}[/tex])/D

imputing the values, we have

577.96 = (3.142/16) x τmax x ([tex]0.04^{4}[/tex] - [tex]0.025^{4}[/tex])/0.04

577.96 = 0.196 x τmax x (5.4 x [tex]10^{-5}[/tex])

577.96 = 1.06 x [tex]10^{-5}[/tex] x τmax

τmax  ≅ 54.52 MPa

KINDLY NOTE that there is a picture in the question. Check the picture below for the picture.

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Answer:

(1). 1.2 metres.

(2). There is going to be the same pressure.

Explanation:

From the question above we can take hold of the statement Below because it is going to assist or help us in solving this particular Question or problem;

" Measurements indicate that the material of tank B will fail and the tank will burst if the air pressure in tank B exceeds 18 kPa."

=> Also, the density of oil = 930

That is if Pressure, P in B > 18kpa there will surely be a burst.

The height, h the can waste oil be poured into tank A is;

The maximum pressure  = height × acceleration due to gravity × density) + ( acceleration due to gravity × density × height, j).

18 × 10^3 = (height, h ×  10 × 930) + 10 × (2 - 1.25) × 1000.

When we make height, h the Subject of the formula then;

Approximately, Height, h = 1.2 metres.

(2). If air is accidentally trapped in the manometer line, what will be the error in the calculation of the height we will have the same pressure.

Answer:

a) [tex]a=-28m/s^{2}[/tex]

b) [tex]\frac{dT}{dx}=-5 ^{o}C/m[/tex]

Explanation:

a)

In order to solve this problem, we need to start by remembering how the acceleration is related to the velocity of a particle. We have the following relation:

[tex]a=\frac{dv}{dt}[/tex]

in other words, the acceleration is defined to be the derivative of the velocity function with respect to time. So let's take our speed function:

u=20-2x

if we take its derivative we get:

du=-2dx

this is the same as writting:

[tex]\frac{du}{dt}=-2\frac{dx}{dt}[/tex]

we also know that velocity is defined to be:

[tex]u=\frac{dx}{dt}[/tex]

so we get that:

a=-2u

when substituting we get that:

a=-2(20-2x)

when expanding we get:

a=-40+4x

and now we can use this equation to find our acceleration at x=3, so:

a=-40+4(3)

a=-40+12

[tex]a=-28 m/s^{2}[/tex]

b)

the same applies to this problem with the difference that this will be the rate of change of the temperature per m. So we proceed and take the derivative of the temperature function:

T=200-5x

[tex]\frac{dT}{dx}=-5[/tex]

so the rate of change is [tex] -5^{o}C/m [/tex]

Answer:

Explanation:

Given that

Mass of 1 = [tex]m_1[/tex]

Mass of 2 = [tex]m_2[/tex]

Temperature in  1 = [tex]T_1[/tex]

Temperature in 2 = [tex]T_2[/tex]

Pressure remains i  the group apartment

The closed system and energy balance is

[tex]E_{in}-E_{out}=\Delta E_{system}[/tex]

The kinetic energy and potential energy are negligible

since it is insulated tank ,there wont be eat transfer from the system

And there is no work involved

[tex]\Delta U = 0[/tex]

Let the final temperature be final temperature

[tex]m_1+c_v(T_3-T_1)+m_2c_v(T_3+T_2)=0\\\\m_1+c_v(T_3-T_1)=m_2c_v(T_2-T_3)---(i)[/tex]

Using mass balance

[tex]m_3+m_2+m_1[/tex]

from eqn i

[tex]m_1+c_v(T_3-T_1)=m_2c_v(T_2-T_3)m_1T_3-m_1T_1=m_2T_2-m_2T_3\\\\m_1T_3+m_2T_3=m_2T_2+m_1T_1\\\\(m_1+m_2)T_3=m_1T_1+m_2T_2\\\\(m_1)T_3=m_1T_1+m_2T_2\\\\T_3=\frac{m_1T_1_m_2T_2}{m_3}[/tex]

Therefore the final temperature can be express as

[tex]\large \boxed {T_3=\frac{m_1}{m_3} T_1+\frac{m_2}{m_3}T_2 }[/tex]

Answer:

hello attached is the free body diagram of the missing figure

Fr = [tex]\frac{\pi }{4} D^2 [ ( P1 - P2) - pV^2 ][/tex]

Explanation:

Average velocity is constant i.e V1 = V2 = V

The momentum equation for the flow in the Z - direction can be expressed as

-Fr + P1 Ac - P2 Ac = mB2V2 - mB1V1 ------- equation 1

Fr = horizontal force on the bolts

P1 = pressure of fluid at entrance

V1 = velocity of fluid at entrance

Ac = cross section area of the pipe

P2 and V2 = pressure and velocity of fluid at some distance

m = mass flow rate of fluid

B1 = momentum flux at entrance , B2 = momentum flux correction factor

Note; average velocity is constant hence substitute V for V1 and V2

equation 1 becomes

Fr = ( P1 - P2 ) Ac + mV ( 1 - 2 )

Fr = ( P1 - P2 ) Ac - mV ---------------- equation 2

equation for mass flow rate

m = pAcV  

p = density of the fluid

insert this into equation 2 EQUATION 2 BECOMES

Fr = ( P1 - P2) Ac - pAcV^2

    = Ac [ (P1 - P2) - pV^2 ]  ---------- equation 3

Note Ac = [tex]\frac{\pi }{4} D^2[/tex]

Equation 3 becomes

Fr = [tex]\frac{\pi }{4} D^2[/tex] [ (P1 -P2 ) - pV^2 ] ------- relation for the horizontal force acting on the bolts      

Answer is given below

Explanation:

Agreement is a official contract. It is written form or even in orally form. The Agreement can be written in formal or informal terms or we can use purely verbal language.

Agreement made between two or more party that allow the court to decide.  

The main 6 elements are:  

1. Offer

2. acceptance

3. consideration

4. intention to create legal relation

5. certainty

6. capacity  

1. The first elements of the contract herein are ABC Sanctions Inc., without the offer, as it is not valid under the Contract Act 1950, Contract Act.

2. Once the offer is made in the contract, acceptance must take place. The agreement must be approved by Northern Inc. When Northern Inc is clear with the offer, it will accept it once the terms and conditions of the agreement are clear.

3. Contrarification is the most important aspect of a contract, when considering a contract, the other person will give something in return. It is considered an exchange between ABC Construction Inc and Northern Inc.

4. It is necessary to have these elements in the contract. Contract law 1950 is one of the requirements of a valid contract, although there is silence about the need for a legal relationship.

5. Another important aspect of the contract is of course. The Contract Agreement sets out the terms and conditions that must be clearly understood by both ABC Contracts Inc and Northern Inc.

6. The ability of a contract to have a legal capacity on either side of the contract is more than eighteen years, since the age of 18 years is specified as the age at which the contract is entered into.

Answer:

the average normal stress in the bar is : [tex]\mathbf{\sigma = (32.5 - 20x )MPa}[/tex]

Explanation:

The free body flow of the missing diagram is attached to the answer below.

From the information given:

Let consider the sum of forces along horizontal direction to be equal to zero.

[tex]\begin{array}{l}\\\sum {{F_x}} = 0\\\\3 + 8\left( {1.25 - x} \right) - N = 0\\\\N = \left( {13 - 8x} \right){\rm{ kN}}\\\end{array}[/tex]

The average normal stress in the bar can be calculated by the formula:

[tex]\sigma = \dfrac{N}{A}[/tex]

where;

[tex]\sigma =[/tex] average normal stress in the bar

A = cross sectional area in the bar and it is given by: [tex]400*10^{-6 } m^2[/tex]

​N = (13- 8x) kN

∴ [tex]\sigma = \dfrac{(13-8)x}{400*10^{-6} m^2}[/tex]

[tex]\sigma = (32.5 - 20x )*10^3 kPa[/tex]

[tex]\sigma = (32.5 - 20x )MPa[/tex]

Thus; the average normal stress in the bar is : [tex]\mathbf{\sigma = (32.5 - 20x )MPa}[/tex]

The average normal stress in the bar as a function of x is equal to 32.5 - 20x MPa.

Given the following data:

To calculate the average normal stress in the bar as a function of x.

First of all, we would determine the sum of the forces acting on the bar in the horizontal direction;

[tex]\sum F_x=0\\\\3+8(1.25-x)-N=0\\\\3+10-8x-N=0\\\\13-8x-N=0\\\\N=(13-8x)\; kN[/tex]

For the average normal stress:

Mathematically, the average normal stress is given by this formula:

[tex]\sigma = \frac{N}{A}[/tex]

Where:

Substituting the parameters into the formula, we have;

[tex]\sigma = \frac{(13-8x) \times 10^3}{400 \times 10^{-6}}\\\\\sigma = \frac{(13-8x) \times 10^{3+6}}{400}\\\\\sigma =(32.5-20x) \times 10^{9}[/tex]

Note: 1 MPa = [tex]1\times 10^9\;Pa[/tex]

Average normal stress = 32.5 - 20x MPa.

Read more on average normal stress here: https://brainly.com/question/7958709

Answer:

Fluid viscosity, [tex]\mu = 2.57 * 10^{-3} kgm^{-1}s^{-1}[/tex]

Explanation:

Container depth, D = 12 cm = 0.12 m

Tube length, l = 30 cm = 0.3 m

Specific Gravity, [tex]\rho[/tex] = 0.95

Tube diameter, d = 2 mm = 0.002 m

Rate of flow, Q = 1.9 cm³/s = 1.9 * 10⁻⁶ m³/s

Calculate the velocity at point 2 ( check the diagram attached)

Rate of flow at section 2, [tex]Q = A_2 v_2[/tex]

[tex]Area, A_2 = \pi d^{2} /4\\A_2 = \pi/4 * 0.002^2\\A_2 = 3.14159 * 10^{-6} m^2[/tex]

[tex]v_{2} = Q/A_{2} \\v_{2} =\frac{1.9 * 10^{-6}}{3.14 * 10^{-6}} \\v_{2} = 0.605 m/s[/tex]

Applying the Bernoulli (energy flow) equation between Point 1 and point 2 to calculate the head loss:

[tex]\frac{p_{1} }{\rho g} + \frac{v_{1}^2 }{2 g} + z_1 = \frac{p_{2} }{\rho g} + \frac{v_{2}^2 }{2 g} + z_2 + h_f\\ z_1 = L + l = 0.12 + 0.3\\z_1 = 0.42\\p_1 = p_{atm}\\v_1 = 0\\z_2 = 0\\\frac{p_{atm} }{\rho g} + \frac{0^2 }{2 g} + 0.42= \frac{p_{atm} }{\rho g} + \frac{0.605^2 }{2 *9.8} +0 + h_f\\h_f = 0.401 m[/tex]

For laminar flow, the head loss is given by the formula:

[tex]h_f = \frac{128 Q \mu l}{\pi \rho g d^4} \\\\0.401 = \frac{128 * 1.9 * 10^{-6} * 0.3 \mu}{\pi *0.95* 9.8* 0.002^4}\\\\\\\mu = \frac{0.401 * \pi *0.95* 9.8* 0.002^4}{128 * 1.9 * 10^{-6} * 0.3} \\\\\mu = 2.57 * 10^{-3} kgm^{-1}s^{-1}[/tex]

Answer:

0.00257 kg / m.s

Explanation:

Given:-

- The specific gravity of a liquid, S.G = 0.95

- The depth of fluid in free container, h = 12 cm

- The length of the vertical tube , L = 30 cm

- The diameter of the tube, D = 2 mm

- The flow-rate of the fluid out of the tube into atmosphere, Q = 1.9 cm^3 / s

Find:-

To determine the viscosity of a liquid

Solution:-

- We will consider the exit point of the fluid through the vertical tube of length ( L ), where the flow rate is measured to be Q = 1.9 cm^3 / s

- The exit velocity ( V2 ) is determined from the relation between flow rate ( Q ) and the velocity at that point.

                              [tex]Q = A*V_2[/tex]

Where,

             A: The cross sectional area of the tube

- The cross sectional area of the tube ( A ) is expressed as:

                                [tex]A = \pi \frac{D^2}{4} \\\\A = \pi \frac{0.002^2}{4} \\\\A = 3.14159 * 10^-^6 m^2[/tex]

- The velocity at the exit can be determined from the flow rate equation:

                              [tex]V_2 = \frac{Q}{A} \\\\V_2 = \frac{1.9*10^-^6}{3.14159*10^-^6} \\\\V_2 = 0.605 \frac{m}{s}[/tex]

- We will apply the energy balance ( head ) between the points of top-surface ( free surface ) and the exit of the vertical tube.

                    [tex]\frac{P_1}{p*g} + \frac{V^2_1}{2*g} + z_t_o_p = \frac{P_2}{p*g} + \frac{V^2_2}{2*g} + z_d_a_t_u_m + h_L[/tex]

- The free surface conditions apply at atmospheric pressure and still ( V1 = 0 ). Similarly, the exit of the fluid is also to atmospheric pressure. Where, z_top is the total change in elevation from free surface to exit of vertical tube.

- The major head losses in a circular pipe are accounted using Poiessel Law:

                           [tex]h_L = \frac{32*u*L*V}{S.G*p*g*D^2}[/tex]

Where,

                  μ: The dynamic viscosity of fluid

                  L: the length of tube

                  V: the average velocity of fluid in tube

                  ρ: The density of water

- The average velocity of the fluid in the tube remains the same as the exit velocity ( V2 ) because the cross sectional area ( A ) of the tube remains constant throughout the tube. Hence, the velocity also remains constant.

- The energy balance becomes:

                      [tex]h + L = \frac{V_2^2}{2*g} + \frac{32*u*L*V_2}{S.G*p*g*D^2} \\\\0.42 = \frac{0.605^2}{2*9.81} + \frac{32*u*(0.3)*(0.605)}{0.95*998*9.81*0.002^2} \\\\u = 0.00257 \frac{kg}{m.s}[/tex]

- Lets check the validity of the Laminar Flow assumption to calculate the major losses:

                     [tex]Re = \frac{S.G*p*V_2*D}{u} \\\\Re = \frac{0.95*998*0.605*0.002}{0.00257} \\\\Re = 446 < 2100[/tex]( Laminar Flow )

Answer:

hello the required diagram is missing attached to the answer is the required diagram

7.9954 kip.ft

Explanation:

AB = 1550-Ib ( weight acting on AB )

BCD = 190 - Ib ( weight of cage )

169-Ib = weight of man inside cage

Attached is the free hand diagram of the question

calculate distance [tex]x![/tex]

= cos 75⁰ = [tex]\frac{x^!}{10ft}[/tex]

    [tex]x! = 10 * cos 75^{o}[/tex] = 2.59 ft

calculate distance x

= cos 75⁰ = [tex]\frac{x}{30ft}[/tex]

x = 30 * cos 75⁰ = 7.765 ft

The resultant moment  produced by all the weights about point A

∑ Ma = 0

Ma = 1550 * [tex]x![/tex] + 190 ( x + 2.5 ) + 169 ( x + 2.5 + 1.75 )

Ma = 1550 * 2.59 + 190 ( 7.765 + 2.5 ) + 169 ( 7.765 + 2.5 + 1.75 )

      = 4014.5 + 1950.35 + 2030.535

      = 7995.385 ft. Ib ≈ 7.9954 kip.ft