An iron-carbon alloy initially containing 0.258 wt% C is exposed to an oxygen-rich and virtually carbon-free atmosphere at 1120°C. Under these circumstances the carbon diffuses from the alloy and reacts at the surface with the oxygen in the atmosphere; that is, the carbon concentration at the surface position is maintained essentially at 0.0 wt% C. At what position will the carbon concentration be 0.194 wt% after a 6 h treatment? The value of D at 1120°C is 6.9 × 10-11 m2/s.

Answer:

Explanation:

As per Amdahl's law :

[tex]\text {Speedup} = {\frac{\text{Old Execution time}}{\text {New Execution time} }[/tex]

[tex]\text {Speedup} = \frac{1}{( (1- \text {FractionEnhanced}) + (\text {FractionEnhanced} / \text {SpeedupEnhanced}) )}[/tex]

Here :

Design 1:

FractionEnhanced = 0.5 (50% of computation )

[ Note that I/O wait has nothing to do with speed ]

SpeedupEnhanced = 15 times

[tex]\text {Overall speedup} =\frac{1}{( ( 1- 0.5) + (0.5/ 15) )}[/tex]

[tex]\text {overall Speedup} = \frac{1}{(0.5 + 0.033)}[/tex]

[tex]\text {overall Speedup} = \frac{1}{ 0.533} = 1.876[/tex]

========

Design 2:

FractionEnhanced = 0.3 (30% of computation )

SpeedupEnhanced = 20 times

[tex]\text {overall speedup} = 1 / ( ( 1- 0.3) + (0.3/ 20) )\\\\\text {overall speedup} = 1/ (0.7 + 0.015)\\\\\text {overall speedup} = 1/ 0.715 \\\\\text {overall speedup}= 1.398[/tex]

========

So as we can see Design 1 is better with overall speedup of 1.876 times the original processor.

Answer:

a) 0.697*10³ lb.in

b) 6.352 ksi

Explanation:

a)

For cylinder AB:

Let Length of AB = 12 in

[tex]c=\frac{1}{2}d=\frac{1}{2} *1.1=0.55in\\ J=\frac{\pi c^4}{2}=\frac{\pi}{2}0.55^4=0.1437\ in^4\\[/tex]

[tex]\phi_B=\frac{T_{AB}L}{GJ}=\frac{T_{AB}*12}{3.3*10^6*0.1437} =2.53*10^{-5}T_{AB}[/tex]

For cylinder BC:

Let Length of BC = 18 in

[tex]c=\frac{1}{2}d=\frac{1}{2} *2.2=1.1in\\ J=\frac{\pi c^4}{2}=\frac{\pi}{2}1.1^4=2.2998\ in^4\\[/tex]

[tex]\phi_B=\frac{T_{BC}L}{GJ}=\frac{T_{BC}*18}{5.9*10^6*2.2998} =1.3266*10^{-6}T_{BC}[/tex]

[tex]2.53*10^{-5}T_{AB}=1.3266*10^{-6}T_{BC}\\T_{BC}=19.0717T_{AB}[/tex]

[tex]T_{AB}+T_{BC}-T=0\\T_{AB}+T_{BC}=T\\T_{AB}+T_{BC}=14*10^3\ lb.in\\but\ T_{BC}=19.0717T_{AB}\\T_{AB}+19.0717T_{AB}=14*10^3\\20.0717T_{AB}=14*10^3\\T_{AB}=0.697*10^3\ lb.in\\T_{BC}=13.302*10^3\ lb.in[/tex]

b) Maximum shear stress in BC

[tex]\tau_{BC}=\frac{T_{BC}}{J}c=13.302*10^3*1.1/2.2998=6.352\ ksi[/tex]

Maximum shear stress in AB

[tex]\tau_{AB}=\frac{T_{AB}}{J}c=0.697*10^3*0.55/0.1437=2.667\ ksi[/tex]

The reaction Torque at A and the maximum shear stress in BC are respectively; T_ab = 697 lb.in; T_bc = 13302 lb.in and τ_bc = 6.352 Ksi

The torques in cylinders AB and BC are statically indeterminate. Matching the rotation φ_b for each cylinder.

For Cylinder AB; c = ¹/₂d = ¹/₂ * 1.1 = 0.55 in

Polar Moment of Inertia; J = ¹/₂ *πc⁴ = ¹/₂ * π * 0.55⁴ = 0.1437 in⁴

Rotation; φ_b = (T_ab * L)/GJ

We are given;

G = 3.3 * 10⁶

L = 12 in

Thus;

φ_b = (12/(3.3 * 10⁶ * 12))T_ab

φ_b = 2.53 × 10⁻⁵ T_ab

For Cylinder BC; c = ¹/₂d = ¹/₂ * 2.2 = 1.1 in

Polar Moment of Inertia; J = ¹/₂ *πc⁴ = ¹/₂ * π * 1.1⁴ = 2.2998 in⁴

Rotation; φ_b = (T_bc * L)/GJ

We are given;

G = 5.9 * 10⁶

L = 18 in

Thus;

φ_b = (18/(5.9 * 10⁶ * 2.2998))T_bc

φ_b = 1.3266 × 10⁻⁶ T_bc

Matching expressions for φ_b, we have;

2.53 × 10⁻⁵ T_ab = 1.3266 × 10⁻⁶ T_bc

Thus; T_bc = 19.0717T_ab    -----(eq 1)

Equilibrium of connection at B;

T_ab + T_bc - T = 0

We are given T = 14 kip·in = 14 × 10³ lb.in

Thus;

T_ab + T_bc - (14 × 10³) = 0

T_ab + T_bc = 14000     ----(eq 2)

Combining equation 1 and 2, we have;

T_ab = 697 lb.in

T_bc = 13302 lb.in

B) Formula for maximum shear stress is;

τ = Tc/J

Maximum shear stress in BC is;

τ_bc = (13302 * 1.1)/2.2998

τ_bc = 6.352 Ksi

Read more about Torque and Maximum Shear stress at; https://brainly.com/question/13507543

Answer:

Entropy generation rate of the two reservoirs is approximately zero ([tex]\dot S_{gen} = 9.318 \times 10^{-4}\,\frac{kW}{K}[/tex]) and system satisfies the Second Law of Thermodynamics.

Explanation:

Reversible heat pumps can be modelled by Inverse Carnot's Cycle, whose key indicator is the cooling Coefficient of Performance, which is the ratio of heat supplied to hot reservoir to input work to keep the system working. That is:

[tex]COP_{H} = \frac{\dot Q_{H}}{\dot W}[/tex]

The following simplification can be used in the case of reversible heat pumps:

[tex]COP_{H,rev} = \frac{T_{H}}{T_{H} - T_{L}}[/tex]

Where temperature must written at absolute scale, that is, Kelvin scale for SI Units:

[tex]COP_{H, rev} = \frac{297.15\,K}{297.15\,K-280.15\,K}[/tex]

[tex]COP_{H, rev} = 17.479[/tex]

Then, input power needed for the heat pump is:

[tex]\dot W = \frac{\dot Q}{COP_{H,rev}}[/tex]

[tex]\dot W = \frac{300\,kW}{17.749}[/tex]

[tex]\dot W = 16.902\,kW[/tex]

By the First Law of Thermodynamics, heat pump works at steady state and likewise, the heat released from cold reservoir is now computed:

[tex]-\dot Q_{H} + \dot W + \dot Q_{L} = 0[/tex]

[tex]\dot Q_{L} = \dot Q_{H} - \dot W[/tex]

[tex]\dot Q_{L} = 300\,kW - 16.902\,kW[/tex]

[tex]\dot Q_{L} = 283.098\,kW[/tex]

According to the Second Law of Thermodynamics, a reversible heat pump should have an entropy generation rate equal to zero. The Second-Law model for the system is:

[tex]\dot S_{in} - \dot S_{out} - \dot S_{gen} = 0[/tex]

[tex]\dot S_{gen} = \dot S_{in} - \dot S_{out}[/tex]

[tex]\dot S_{gen} = \frac{\dot Q_{L}}{T_{L}} - \frac{\dot Q_{H}}{T_{H}}[/tex]

[tex]\dot S_{gen} = \frac{283.098\,kW}{280.15\,K} - \frac{300\,kW}{297.15\,K}[/tex]

[tex]\dot S_{gen} = 9.318 \times 10^{-4}\,\frac{kW}{K}[/tex]

Albeit entropy generation rate is positive, it is also really insignificant and therefore means that such heat pump satisfies the Second Law of Thermodynamics. Furthermore, [tex]\dot S_{in} = \dot S_{out}[/tex].

The rate of entropy change of the two reservoirs is; 9.318 * 10⁻⁴ kW/K and it satisfies second law of thermodynamics

The formula for Coefficient of Performance is;

COP = T_H/(T_H - T_L)

Where;

T_H = 24°C = 297.15 K

T_L = 7°C = 280.15 K

Thus;

COP = 297.15/(297.15 - 280.15)

COP = 17.479

Input power is;

Input power needed for the heat pump is:

W' = Q'/COP

We are given; Q' = 300 kW

Thus;

W' = 300/17.479

W' = 16.902 kW

From first law of thermodynamics, we can deduce that;

Q_L = Q_H - W'

Thus;

Q_L = 300 - 16.902

Q_L = 283.098 kW

From second law of thermodynamics, the rate of entropy generation is;

S_gen = (Q_L/T_L) - (Q_H/T_H)

S_gen = (283.098/280.15) - (300/297.15)

S_gen = 9.318 * 10⁻⁴ kW/K

Read more about Entropy at; https://brainly.com/question/15022152

Answer:

Data Governance

Explanation:

Data governance is a government organization which allows the users to access the shared data.

This is achieved as the Data governance involves the management, integrity, usability, availability and security.

Data governance plays an important role in IT industries and business practices so that business can work more efficiently. The governance involves selecting a team, discover data quality and data security.

Thus, Data governance is the correct answer.

Answer:

ate/kuya poca please

21. What is the best explanation of ATP utilization and production?

A. ATP is being recycled from the process glycolysis, undergo oxidation process and later it is being remade during

Krebs cycle

B. ATP is used to reduced carbon molecule to produce oxaloacetate and used for the bulk reproduction of ATP

during electron transport chain

C. ATP is used during energy investment phase to harvest the energy stored in the molecule 2,3 – carbon molecule

during the energy harvesting phase

D. ATP is adenosine triphosphate that has 3 phosphate groups attached and it is being reduced into Adenosine

diphosphate after losing one phosphate group in its molecule

22. The best description on the role of oxygen during respiration is ________.

A. Oxygen is used to phosphorylate ADP to ATP

B. Oxygen is used to oxidize carbon dioxide to form glucose

C. Oxygen is utilized to add carbon in carbon dioxide as a by-product of cellular respiration

D. Oxygen is used to oxidize glucose molecule to form ATP—the energy needed by the body

23. Where are the 36 ATP come from during cellular respiration?

A. 2 from glycolysis, 2 from oxidation of pyruvate and 32 from Krebs cycle

B. 2 from glycolysis, 2 from Krebs cycle and 32 from electron transport chain

C. 2 from glycolysis, 2 from oxidation of pyruvate and 32 from electron transport chain

D. 2 from oxidation of pyruvate, 2 from Krebs cycle and 32 from electron transport cha

The question is incomplete! Complete question along with answer and step by step explanation is provided below.

Question:

In the given circuit, V(t)=12cos(2000t+45)V, R1=R2=2Ω, L1=L2=L3=3mH and C1=250μF. You are required to find the Thevenin equivalent of this circuit using phasors.

a. If you write the Thevenin voltage, Vth, in phasor form, what is the magnitude of this phasor? Put your answer in the box below without units.

b. What is the value of the angle associated with the phasor Vth, in degrees?

c. Now, calculate the Thevenin impedance, Zth. What is the magnitude of this phasor?

d. What is the angle associated with the phasor Zth, in degrees?

Answer:

Vth = 6 < 45° V

Zth = 1.414 < 45°

a. The magnitude of the Thevenin voltage is 6 V

b. The phase angle of the Thevenin voltage is 45°

c. The magnitude of the Thevenin impedance is 1.414 V

d. The phase angle of the Thevenin impedance is 45°

Explanation:

The given voltage is

V(t)=12cos(2000t+45)

In phasor form,

V(t) = 12 < 45° V

So the magnitude of voltage is 12 V and the phase angle is 45°

Also the frequency ω = 2000

then the inductance is

L₁ = L₂ = L₃ = jωL = j×2000×0.003 = j6 Ω

and the capacitance is

C₁ = 1/jωC = 1/(j×2000×250x10⁻⁶) = -j2 Ω

and the resistance is

R₁ = R₂ = 2 Ω

Thevenin voltage:

The Thevenin voltage is the voltage that appears across the open-circuited terminals a-b (after removing L₃)

The Thevenin voltage is given by

Vth = V(t) × [ (R₂ + L₂) / (R₁ + L₁) + (R₂ + L₂) ]

Please note that there is no current flow in the capacitor due to open-circuited terminal a-b

Vth = 12 < 45°  × [ (2 + j6) / (2 + j6) + (2 + j6) ]

Vth = 12 < 45°  × [ (2 + j6) / (4 + j12) ]

Vth = 4.24264 + j4.24264 V

In phasor form,

Vth = 6 < 45° V

a. The magnitude of the Thevenin voltage is 6 V

b. The phase angle of the Thevenin voltage is 45°

Thevenin Impedance:

The Thevenin Impedance is the impedance of the circuit calculated when looking from the terminal a-b

Zth = [ (R₁ + L₁) × (R₂ + L₂) / (R₁ + L₁) + (R₂ + L₂) ] + (-j2)

Zth = [ (2 + j6) × (2 + j6) / (2 + j6) + (2 + j6) ] - j2

Zth = [ (2 + j6) × (2 + j6) / (2 + j6) + (2 + j6) ] - j2

Zth = [ (-32 + j24) / (4 + j12) ] -j2

Zth = [ (1 +j3) ] - j2

Zth = 1 + j Ω

In phasor form,

Zth = 1.414 < 45°

c. The magnitude of the Thevenin impedance is 1.414 V

d. The phase angle of the Thevenin impedance is 45°

Answer:

The overview of the given question is described in the explanation segment below.

Explanation: