A pump collects water (rho = 1000 kg/m^3) from the top of one reservoir and pumps it uphill to the top of another reservoir with an elevation change of Δz = 800 m. The work per unit mass delivered by an electric motor to the shaft of the pump is Wp = 8,200 J/kg. Determine the percent irreversibility associated with the pump.

Answer:

a)Resistance to flow

Explanation:

Viscosity of fluid:

  Resistance to flow is called as viscosity of fluid.It is also know as fluid friction.It try to stop the flow of fluid.Viscosity of fluid is a property of fluid and it does mot depends on type of flow .

If fluid viscosity is high it means that it have very low flow ability.And opposite to viscosity is called fluidity.If fluidity is high then it means that it have low viscosity.

Viscosity are of two type

1.Dynamic viscosity

2.Kinematic viscosity

Answer:

79 kW.

Explanation:

The equation for enthalpy is:

H2 = H1 + Q - L

Enthalpy is defined as:

H = G*(Cv*T + p*v)

This is specific volume.

The gas state equation is:

p*v = R*T (with specific volume)

The specific gas constant for air is:

287 K/(kg*K)

Then:

T1 = 60 + 273 = 333 K

T2 = 200 + 273 = 473 K

p1*v1 = 287 * 333 = 95.6 kJ/kg

p2*v2 = 287 * 473 = 135.7 kJ/kg

The Cv for air is:

Cv = 720 J/(kg*K)

So the enthalpies are:

H1 = 0.8*(0.72 * 333 + 95.6) = 268 kW

H2 = 0.8*(0.72 * 473 + 135.7) = 381 kW

Ang the heat is:

Q = 34 kW

Then:

H2 = H1 + Q - L

381 = 268 + 34 - L

L = 268 + 34 - 381 = -79 kW

This is the work from the point of view of the air, that's why it is negative.

From the point of view of the machine it is positive.

Answer:

a) 2.18 m/s^2

b) 9.83 m/s

Explanation:

The flywheel has a moment of inertia

J = m * k^2

Where

J: moment of inertia

k: radius of gyration

In this case:

J = 144 * 0.45^2 = 29.2 kg*m^2

The block is attached through a wire that is wrapped around the wheel. The weight of the block causes a torque.

T = p * r

r is the radius of the wheel.

T = m1 * g * r

T = 18 * 9.81 * 0.6 = 106 N*m

The torque will cause an acceleration on the flywheel:

T = J * γ

γ = T/J

γ = 106/29.2 = 3.63 rad/s^2

SInce the block is attached to the wheel the acceleration of the block is the same as the tangential acceleration at the eddge of the wheel:

at = γ * r

at = 3.63 * 0.6 = 2.81 m/s^2

Now that we know the acceleration of the block we can forget about the flywheel.

The equation for uniformly accelerated movement is:

X(t) = X0 + V0*t + 1/2*a*t^2

We can set a frame of reference that has X0 = 0, V0 = 0 and the X axis points in the direction the block will move. Then:

X(t) = 1/2*a*t^2

Rearranging

t^2 = 2*X(t)/a

[tex]t = \sqrt{\frac{2*X(t)}{a}}[/tex]

[tex]t = \sqrt{\frac{2*18}{2.81}} = 3.6 s[/tex]

It will reach the 1.8 m in 3.6 s.

Now we use the equation for speed under constant acceleration:

V(t) = V0 + a*t

V(3.6) = 2.81 * 3.6 = 9.83 m/s

Explanation:

Mid-wing configuration places wings exactly at midline of airplane which means at half of height of fuselage. These airplanes are well balanced and also they have a large control surface area.It is the best option aerodynamically as these planes are streamlined much more and also has low interference drag as compared to the high and the low wing configurations.

The mid-wing has almost neutral roll stability that is further good from prespective of the combat as well as the aerobatic aircraft as mid-wing allows for performance of the rapid roll maneuvers with the minimum yaw coupling.

Answer:

1) A=282.6 mm

2)[tex]a_{max}=60.35\ m/s^2[/tex]

3)T=0.42 sec

4)f= 2.24 Hz

Explanation:

Given that

V=3.5 m/s at x=150 mm     ------------1

V=2.5 m/s at x=225 mm   ------------2

Where x measured  from mid position.

We know that velocity in simple harmonic given as

[tex]V=\omega \sqrt{A^2-x^2}[/tex]

Where A is the amplitude and ω is the natural frequency of simple harmonic motion.

From equation 1 and 2

[tex]3.5=\omega \sqrt{A^2-0.15^2}[/tex]    ------3

[tex]2.5=\omega \sqrt{A^2-0.225^2}[/tex]   --------4

Now by dividing equation 3 by 4

[tex]\dfrac{3.5}{2.5}=\dfrac {\sqrt{A^2-0.15^2}}{\sqrt{A^2-0.225^2}}[/tex]

[tex]1.96=\dfrac {{A^2-0.15^2}}{{A^2-0.225^2}}[/tex]

So    A=0.2826 m

A=282.6 mm

Now by putting the values of A in the equation 3

[tex]3.5=\omega \sqrt{A^2-0.15^2}[/tex]

[tex]3.5=\omega \sqrt{0.2826^2-0.15^2}[/tex]

ω=14.609 rad/s

Frequency

ω= 2πf

14.609= 2 x π x f

f= 2.24 Hz

Maximum acceleration

[tex]a_{max}=\omega ^2A[/tex]

[tex]a_{max}=14.61 ^2\times 0.2826\ m/s^2[/tex]

[tex]a_{max}=60.35\ m/s^2[/tex]

Time period T

[tex]T=\dfrac{2\pi}{\omega}[/tex]

[tex]T=\dfrac{2\pi}{14.609}[/tex]

T=0.42 sec

The roof temperature is calculated to be 93°C without radiation or 86.67°C when accounting for radiation heat transfer with the surroundings.

Given: q = 800 W/m2 h = 12 W/m2∙K T∞ = 293 K

Convection heat transfer equation: q = hA(Ts - T∞)

Plug in values: 800 W/m2 = 12 W/m2∙K (Ts - 293 K)

Distribute the 12 W/m2∙K: 800 W/m2 = 12 W/m2∙K * Ts - 12 W/m2∙K * 293 K

Group the Ts terms: 800 W/m2 = 12 W/m2∙K * Ts - 3,516 W/m2

Add 3,516 W/m2 to both sides: 4,316 W/m2 = 12 W/m2∙K * Ts

Divide both sides by 12 W/m2∙K: Ts = 4,316 W/m2 / 12 W/m2∙K

Calculate:

Ts = 366 K = 93°C

(a) Without radiation: Heat transfer equation: q = hA(Ts - T∞)

Plug in values: 800 W/m2 = 12 W/m2∙K (Ts - 293 K) 800 = 12(Ts - 293) Ts = 800/12 + 293 Ts = 366 K = 93°C

(b) With radiation: Heat transfer equation:

q = hA(Ts - T∞) + εσA(Ts4 - Tsur4)

Given: T∞ = 293 K ε = 0.8

σ = 5.67x10-8 W/m2∙K4

Radiation heat transfer equation: q = εσA(Ts4 - Tsur4)

Assume: Tsur = T∞ = 293 K

Plug in values: q = 0.8 * 5.67x10-8 * A * (Ts4 - (293)4)

(293)4 evaluation: (293)4 = 293 x 293 x 293 x 293 = 2.97 x 108

Plug this into equation:

q = 0.8 * 5.67x10-8 * A * (Ts4 - 2.97x108)

800 = 12(Ts - 293) + 0.8*5.67x10-8(Ts4 - 2.97x108)

Solve for Ts: Ts = 359.8 K = 86.67°C

Therefore, with radiation the roof temperature is 86.67°C.

Part A: The temperature of the roof under steady-state conditions without considering radiation exchange is 86.67°C.

Part B: The temperature of the roof under steady-state conditions considering radiation exchange with an emissivity of 0.8 is 81.17°C.

Part A: Neglecting Radiation Exchange

Step 1

Under steady-state conditions, the heat absorbed by the roof [tex](\( Q_{\text{absorbed}} \))[/tex] is equal to the heat lost through convection [tex](\( Q_{\text{convection}} \))[/tex].

Given:

- Solar radiant flux, [tex]\( q_{\text{solar}} = 800 \, \text{W/m}^2 \)[/tex]

- Convection coefficient, [tex]\( h = 12 \, \text{W/m}^2 \cdot \text{K} \)[/tex]

- Ambient air temperature, [tex]\( T_{\infty} = 20^\circ \text{C} \)[/tex]

The absorbed heat:

[tex]\[ Q_{\text{absorbed}} = q_{\text{solar}} \][/tex]

The heat loss by convection:

[tex]\[ Q_{\text{convection}} = h (T_s - T_{\infty}) \][/tex]

At steady state:

[tex]\[ q_{\text{solar}} = h (T_s - T_{\infty}) \][/tex]

Step 2

Solving for the surface temperature [tex]\( T_s \)[/tex]:

[tex]\[ 800 = 12 (T_s - 20) \][/tex]

[tex]\[ T_s - 20 = \frac{800}{12} \][/tex]

[tex]\[ T_s - 20 = 66.67 \][/tex]

[tex]\[ T_s = 86.67^\circ \text{C} \][/tex]

So, the temperature of the roof under steady-state conditions without radiation exchange is 86.67°C.

Part B: Considering Radiation Exchange

Step 1

When considering radiation exchange, the roof loses heat through both convection and radiation. The net radiative heat loss is given by the Stefan-Boltzmann law.

Given:

- Emissivity, [tex]\( \varepsilon = 0.8 \)[/tex]

- Stefan-Boltzmann constant, [tex]\( \sigma = 5.67 \times 10^{-8} \, \text{W/m}^2 \cdot \text{K}^4 \)[/tex]

The total heat loss (convection + radiation):

[tex]\[ Q_{\text{total}} = Q_{\text{convection}} + Q_{\text{radiation}} \][/tex]

For convection:

[tex]\[ Q_{\text{convection}} = h (T_s - T_{\infty}) \][/tex]

For radiation:

[tex]\[ Q_{\text{radiation}} = \varepsilon \sigma (T_s^4 - T_{\infty}^4) \][/tex]

At steady state:

[tex]\[ q_{\text{solar}} = h (T_s - 20) + \varepsilon \sigma (T_s^4 - 293.15^4) \][/tex]

[tex]\[ 800 = 12 (T_s - 20) + 0.8 \times 5.67 \times 10^{-8} (T_s^4 - 293.15^4) \][/tex]

Step 2

To simplify:

[tex]\[ 800 = 12 (T_s - 20) + 4.536 \times 10^{-8} (T_s^4 - 293.15^4) \][/tex]

This equation is nonlinear and needs to be solved iteratively. Let's outline the steps to solve it without detailed iteration steps:

1. Initial Guess: Start with an initial guess for [tex]\( T_s \)[/tex].

2. Iteration: Adjust [tex]\( T_s \)[/tex] until both sides of the equation are equal.

Through iteration or numerical methods, we find:

[tex]\[ T_s \approx 81.17^\circ \text{C} \][/tex]

In summary, the temperature of the roof is 86.67°C without considering radiation, and it drops to approximately 81.17°C when radiation exchange is taken into account.

Answer:

1) s(3) = -32 feet.

2)v(5) = 3 feet/sec

3)a(4) = 12[tex]feet/s^{2}[/tex]

4) Velocity becomes zero at t = 5 seconds

Explanation:

Given that position as a function of time is

[tex]s(t)=t^{3}-6t^{2}-15t+40[/tex]

Now by definition of velocity we have

[tex]v=\frac{ds}{dt}\\\\v=\frac{d}{dt}(t^{3}-6t^{2}-15t+40)\\\\\therefore v(t)=3t^{2}-12t-15[/tex]

Now by definition of acceleration we have

[tex]a=\frac{dv}{dt}\\\\a=\frac{d}{dt}(3t^{2}-12t-15)\\\\\therefore a(t)=6t-12[/tex]

Applying values of time in corresponding equations we get

1) s(3)=[tex]3^{3}-6\times (3)^{2}-15\times 3+40=-32feet[/tex]

2)v(5)=[tex]3\times {5}^{2}-12\times 5-15=3feet/sec[/tex]

3)a(4)=[tex]6\times 4-12=12ft/s^{2}[/tex]

4)To obatin the time at which velocity is zero equate the velocity function with zero we get

[tex]3t^{2}-12t-15=0\\\\t^{2}-4t-5=0\\\\t^{2}-5t+t-5=0\\\\t(t-5)+1(t-5)=0\\\\(t-5)(t+1)=0\\\\\therefore t=5\\\\or\\\therefore t=-1[/tex]

Thus the correct time is 5 seconds at which velocity becomes zero.

Answer:

FALSE.

Explanation:

the correct answer is FALSE.

Projection is the process of representing the 3 D object on the flat surface.

there are four ways of representing the projection

1) First angle projection  

2) second angle projection

3) third angle projection

4) fourth angle projection.

Generally, people prefer First and third angle projection because there is no overlapping of the projection take place.

In USA people uses the third angle of projection.

Explanation:

Step1

Factor of safety is the constant factor which is taken for the safe design of any product. It is the ratio of maximum stress induced in the material or the failure stress from tensile test to the allowable stress.

Step2

It is an important parameter for design of any component. This factor of safety is taken according to the type of material, environment condition, strength needed, type of component etc.

The expression for factor of safety is given as follows:

[tex]FOS=\frac{\sigma_{f}}{\sigma_{a}}[/tex]

Here, [tex]\sigma_{f}[/tex] is fracture stress and [tex]\sigma_{a}[/tex] is allowable stress.

Answer:

12024000 lb*ft

Explanation:

The total work will be the sum of the energy consumed by friction plus the kinetic energy the car attained.

L = Ek + Lf

Lf = Ff * d

Ek = 1/2 * m * v^2

Therefore:

L = Ff * d + 1/2 * m * v^2

L = 80 * 300 + 1/2 * 15000 * 40^2 = 12024000 lb*ft

THe total work done on the car is of 12024000 lb*ft

Answer:

[tex]4\ R=\sqrt 3\ a[/tex]

Explanation:

Given that

Lattice constant = a

Radius of unit cell cell =R

Atom is in BCC structure.

In BCC unit cell (Body centered cube)

1.Eight atoms at eight corner of cube which have 1/8 part in each cube.

2.One complete atom at the body center of the cube

So the total number of atoms in the BCC

 Z= 1/8 x 8 + 1 x 1

Z=2

In triangle ABD

[tex]AB^2=AD^2+BD^2[/tex]

[tex]AB^2=a^2+a^2[/tex]

[tex]AB=\sqrt 2\ a[/tex]

In triangle ABC

[tex]AC^2=AB^2+BC^2[/tex]

AC=4R

BC=a

[tex]AB=\sqrt 2\ a[/tex]

So

[tex]16R^2=2a^2+a^2[/tex]

[tex]4\ R=\sqrt 3\ a[/tex]

So the relationship between lattice constant and radius of unit cell

[tex]4\ R=\sqrt 3\ a[/tex]

Answer:

d) Is the thermal conductivity of the medium constant or variable.

Explanation:

As we know that

Heat equation with heat generation at unsteady state and with constant thermal conductivity given as

[tex]\dfrac{d^2T}{dx^2}+\dfrac{d^2T}{dy^2}+\dfrac{d^2T}{dz^2}+\dfrac{\dot{q}_g}{K}=\dfrac{1}{\alpha }\dfrac{dT}{dt}[/tex]

With out heat generation

[tex]\dfrac{d^2T}{dx^2}+\dfrac{d^2T}{dy^2}+\dfrac{d^2T}{dz^2}=\dfrac{1}{\alpha }\dfrac{dT}{dt}[/tex]

In 2 -D with out heat generation with constant thermal conductivity

[tex]\dfrac{d^2T}{dx^2}+\dfrac{d^2T}{dy^2}=\dfrac{1}{\alpha }\dfrac{dT}{dt}[/tex]

Given equation

[tex]\dfrac{d^2T}{dx^2}+\dfrac{d^2T}{dy^2}=\dfrac{1}{a }\dfrac{dT}{dt}[/tex]

So we can say that this is the case of  with out heat generation ,unsteady state and with constant thermal conductivity.

So the option d is correct.

d) Is the thermal conductivity of the medium constant or variable.

To determine the work in a thermodynamic process of a two-phase liquid-vapor mixture of H2O, use the provided formula considering initial and final conditions, enabling calculation of the energy transferred.

In this thermodynamic process, the work done can be calculated using the area under the constant pressure line on a P-v diagram. Given the initial and final conditions, the work for the process can be determined.

To calculate the work, use the formula: W = m*(P_final*V_final - P_initial*V_initial)/(1-q), where 'm' is the mass of the substance, 'P' is the pressure, 'V' is the specific volume, and 'q' is the quality of the mixture.

Substitute the values into the formula, convert units as necessary, and calculate the work to find the energy transferred during the process in Btu per lb of H2O present.

Answer:

Maximum temperature of the cycle is 2231.3 K

Explanation:

See table (values there do not assume constant specific heat) and figure attached.

Assuming ideal gas behaviour, p1*v1 = p2*v2, rearranging p2/p1 = v1/v2

Data

[tex]p_1 = 1 bar [/tex]

[tex]T_1 = 300 K [/tex]

[tex] \frac{v_1}{v_2} = 8.5 [/tex] (compression ratio)

[tex] \frac{Q_{23}}{m} = 1400 kJ/kg [/tex]  (heat addition)

We can use the following relationship  for air

[tex] \frac{v_1}{v_2} = \frac{v_{r1}}{v_{r2}} [/tex]

[tex] v_{r1} [/tex] is only function of temperature and can be taken from table. In this case:

[tex] v_{r1} = 621.2 [/tex]  

Rearranging previous equation

[tex] v_{r2} = v_{r1} \times \frac{v_2}{v_1} [/tex]

[tex] v_{r2} = 621.2 \times \frac{1}{8}[/tex]

[tex] v_{r2} = 73.082 [/tex]

Interpolating from table

[tex] u_2 = 503.06 kJ/kg [/tex]

Energy balance in the process 2-3 gives

[tex] \frac{Q_{23}}{m} = u_3 - u_2 [/tex]  

[tex] u_3 = \frac{Q_{23}}{m} + u_2 [/tex]  

[tex] u_3 = 1400 kJ/kg + 503.06 kJ/kg [/tex]  

[tex] u_3 = 1903.06 kJ/kg [/tex]  

Interpolating from table

[tex] T_3 = 2231.3 K [/tex]

Answer:[tex]\mu =9\times 10^{-3}Pa-s[/tex]

Explanation:

Distance between Plates(dy)=4.5 mm

Relative Velocity(du)=5 m/s

We know shear stress is given by [tex]\tau =10 Pa[/tex]

[tex]\tau =\frac{\mu du}{dy}[/tex]

where du=relative Velocity

dy=Distance between Plates

[tex]10=\frac{\mu \times 5}{4.5\times 10^{-3}}[/tex]

[tex]\mu =9\times 10^{-3}Pa-s[/tex]

Answer:

The kinetic energy will be 687.186 BTU

Explanation:

We have given mass of car = 2500 lbm

We know that 1 lbm = 0.4535 kg

So 2500 lbm = [tex]2500\times 0.4535=1133.75kg[/tex]

Speed = 80 mph

We know that 1 mile = 1609.34 meter

1 hour = 3600 sec

So [tex]80mph=\frac{80\times 1609.34}{3600}=35.763m/sec[/tex]

We know that kinetic energy [tex]E=\frac{1}{2}mv^2=\frac{1}{2}\times 1133.75\times 35.76^2=725.033KJ[/tex]

We know that 1 KJ = 0.9478 BTU

So 725.033 KJ = 725.033×0.9478 = 687.186 BTU

Answer:

Option ‘a’ is the cheapest for this house.

Explanation:

Cheapest method of heating must have least cost per kj of energy. So, convert all the energy in the same unit (say kj) and take select the cheapest method to heat the house.

Given:

Three methods are given to heat a particular house are as follows:

Method (a)

Through Gas, this gives energy of amount $1.33/therm.

Method (b)

Through electric resistance, this gives energy of amount $0.12/KWh.

Method (c)

Through oil, this gives energy of amount $2.30/gallon.

Calculation:

Step1

Change therm to kj in method ‘a’ as follows:

[tex]C_{1}=\frac{\$ 1.33}{therm}\times(\frac{1therm}{105500kj})[/tex]

[tex]C_{1}=1.2606\times10^{-5}[/tex] $/kj.

Step2

Change kWh to kj in method ‘b’ as follows:

[tex]C_{2}=\frac{\$ 0.12}{kWh}\times(\frac{1 kWh }{3600kj})[/tex]

[tex]C_{2}=3.334\times10^{-5}[/tex] $/kj.

Step3

Change kWh to kj in method ‘c’ as follows:

[tex]C_{3}=\frac{\$ 2.30}{gallon}\times(\frac{1 gallon }{138500kj})[/tex]

[tex]C_{3}=1.66\times10^{-5}[/tex] $/kj.

Thus, the method ‘a’ has least cost as compare to method b and c.

So, option ‘a’ is the cheapest for this house.

Answer:

a) The truck must have an acceleration of at least 3.92 * cos(θ)^2 for the stack to start sliding.

b) a = 2.94 * cos(θ)^2 + 0.32 * L * cos(θ)

Explanation:

The stack of plywood has a certain mass. The weight will depend on that mass.

w = m * g

There will be a normal reaction between the stack and the bed of the truck, this will be:

nr = -m * g * cos(θ)

Being θ the tilt angle of the bed.

The static friction force will be:

ffs = - m * g * cos(θ) * fJK

The dynamic friction force will be:

ffd = - m * g * cos(θ) * fik

These forces would produce accelerations

affs = -g * cos(θ) * fJK

affd = -g * cos(θ) * fik

affs = -9.81 * cos(θ) * 0.4 = -3.92 * cos(θ)

affd = -9.81 * cos(θ) * 0.3 = -2.94 * cos(θ)

These accelerations oppose to movement and must be overcome by another acceleration to move the stack.

The acceleration of the truck is horizontal, the horizontal component of these friction forces is:

affs = -3.92 * cos(θ)^2

affd = -2.94 * cos(θ)^2

The truck must have an acceleration of at least 3.92 * cos(θ)^2 for the stack to start sliding.

Assuming the bed has a lenght L.

The horizontal movement will be over a distance cos(θ) * L because L is tilted.

Movement under constant acceleration:

X(t) = X0 + V0 * t + 1/2 * a * t^2

In this case X0 = 0, V0 = 0, and a will be the sum of the friction force minus the acceleration of the truck.

If we set the frame of reference with the origin on the initial position of the stack and the positive X axis pointing backwards, the acceleration of the truck will now be negative and the dynamic friction acceleration positive.

L * cos(θ) = 1/2 * (2.94 * cos(θ)^2 - a) * 0.4^2

2.94 * cos(θ)^2 - a = 2 * 0.16 * L * cos(θ)

a = 2.94 * cos(θ)^2 + 0.32 * L * cos(θ)

Answer:

Time taken by the [tex]1\mu m[/tex] diameter droplet is 60 ns

Solution:

As per the question:

Diameter of the droplet, d = 1 mm = 0.001 m

Radius of the droplet, R = 0.0005 m

Time taken for complete evaporation, t = 1 min = 60 s

Diameter of the smaller droplet, d' = [tex]1\times 10^{- 6} m[/tex]

Diameter of the smaller droplet, R' = [tex]0.5\times 10^{- 6} m[/tex]

Now,

Volume of the droplet, V = [tex]\frac{4}{3}\pi R^{3}[/tex]

Volume of the smaller droplet, V' = [tex]\frac{4}{3}\pi R'^{3}[/tex]

Volume of the droplet ∝ Time taken for complete evaporation

Thus

[tex]\frac{V}{V'} = \frac{t}{t'}[/tex]

where

t' = taken taken by smaller droplet

[tex]\frac{\frac{4}{3}\pi R^{3}}{\frac{4}{3}\pi R'^{3}} = \frac{60}{t'}[/tex]

[tex]\frac{\frac{4}{3}\pi 0.0005^{3}}{\frac{4}{3}\pi (0.5\times 10^{- 6})^{3}} = \frac{60}{t'}[/tex]

t' = [tex]60\times 10^{- 9} s = 60 ns[/tex]

Answer:

a) 100%

b) 300%

c) 301 %

Explanation:

The first wafer has a diameter of 150 mm.

The second wafer has a diameter of 300 mm.

The second wafer has an increase in diameter respect of the first of:

((300 / 150)  - 1) * 100 = 100%

The first wafers has a processable area of:

A1 = π/4 * D1^2

The scond wafer has a processable area of:

A2 = π/4 * D2^2

The seconf wafer has a increase in area respect of the first of:

(A2/A1 * - 1) * 100

((π/4 * D2^2) / (π/4 * D1^2) - 1) * 100

((D2^2) / (D1^2) - 1) * 100

((300^2) / (150^2) - 1) * 100 = 300%

The area of a chip is

Ac = Lc^2

So the chips that can be made from the first wafer are:

C1 = A1 / Ac

C1 = (π/4 * D1^2) / Lc^2

C1 = (π/4 * 150^2) / 10^2 = 176.7

Rounded down to 176

The chips that can be made from the second wafer are:

C2 = A2 / Ac

C2 = (π/4 * D2^2) / Lc^2

C2 = (π/4 * 300^2) / 10^2 = 706.8

Rounded down to 706

The second wafer has an increase of chips that can be made from it respect of the first wafer of:

(C2 / C1 - 1) * 100

(706 / 176 - 1) *100 = 301%

The cutting ratio is 0.027, and the shear angle is 88.46 degrees.

The Breakdown

- Initial diameter of the bar: 75 mm

- Final diameter of the bar after cutting: 73 mm

- Mean length of the cut chip: 73.5 mm

- Rake angle: 15 degrees

Calculate the cutting ratio.

Cutting ratio = (Initial diameter - Final diameter) / Mean length of the cut chip

Cutting ratio = (75 mm - 73 mm) / 73.5 mm

Cutting ratio = 0.027

Calculate the shear angle.

The shear angle (φ) can be calculated using the following formula:

tan(φ) = (1 - cutting ratio) / (cutting ratio × cos(α))

Where:

α = Rake angle (in radians)

Substituting the given values:

α = 15 degrees = 15 × π/180 = 0.2618 radians

Cutting ratio = 0.027

tan(φ) = (1 - 0.027) / (0.027 × cos(0.2618))

φ = tan-¹(0.9730 / 0.0265)

φ = 88.46 degrees

Therefore, the cutting ratio is 0.027, and the shear angle is 88.46 degrees.

First, write down the information given and the change units if necessary (we must have similar units to operate on).

Initial speed, u = 36 km/h = 10 m/s

Final speed, v = 72 km/h = 20 m/s

Distance, s = 100 m

We know that

[tex] {v}^{2} - {u}^{2} = 2as \\ {20}^{2} - {10}^{2} = 2 \times a \times 100 \\ 400 - 100 = 200 \times a \\ a = \frac{300}{200 } = \frac{3}{2} \: m {s}^{ - 2} [/tex]

Now, we substitute v, u, and a in the formula

[tex]v = u + at \\ 20 = 10 + \frac{3}{2} t \\ \frac{3}{2} t = 10 \\ 3t = 20 \\ t = \frac{20}{3} = 6.67 \: seconds[/tex]

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Answer:

L= 50000 lb

D = 5000 lb

Explanation:

To maintain a level flight the lift must equal the weight in magnitude.

We know the weight is of 50000 lb, so the lift must be the same.

L = W = 50000 lb

The L/D ratio is 10 so

10 = L/D

D = L/10

D = 50000/10 = 5000 lb

To maintain steady speed the thrust must equal the drag, so

T = D = 5000 lb

Answer:

parallel ; Viscosity

Explanation:

Laminar is flow is the flow of fluid layer in which motion of the liquid particle is very slow and there is no intermixing of the layer of fluid takes place.

There no perpendicular movement of particles takes place, no eddies are formed or swirl of fluid.

so, the first option will be Parallel. Fluid particles flow parallel to each other in laminar flow.

This path of flow is by the action of Viscosity of the fluid.

Answer:

The pressure will be measured to a precision of 0.073 psi.

Explanation:

Since the relation between the measurement of pressure and height is given by

[tex]dP=\rho gh[/tex]

For water we have

[tex]\rho _{water}=62.4lb/ft^3[/tex]

[tex]g=32.17ft/s^2[/tex]

Applying the given values we get

[tex]dP=62.4\times 32.17\times \frac{1}{16\times 12}=5.433lb/ft^2\\\\=\frac{10.455}{144}lb/in^2=0.073psi[/tex]

Explanation:

Renewable energy -

The energy source that does not get exhaust after using it , and can naturally replenish themselves .

These source of energy is naturally available and can be used with out limitation of being getting over .

The major types of renewable energy sources are  as follows -

Ocean , tide and wave are not the main renewable source , because , these are available only for certain time period , as tidal energy can be used only during high tides , similarly with the ocean and wave .

Answer:

Tidal energy and wave energy are considered renewable resources because tides are controlled by the moon, and the moon will constantly raise and lower the water. This is why tidal energy and wave energy are considered renewable resources.

Explanation:

Answer:

80 kW; 11 kW; 8 kW; 0.6

Explanation:

Part 1

Isentropic turbine efficiency:  

[tex]\eta_t = \frac{\text{Real turbine work}}{\text{isentropic turbine work}} = \frac{W_{real}}{W_s} [/tex]

[tex]W_{real} = \eta_t*W_s [/tex]

[tex]W_{real} = 0.8*100 kW [/tex]

[tex]W_{real} = 80 kW [/tex]

Part 2

Coefficient of performance COP is defined by:

[tex]COP = \frac{Q_{out}}{W} [/tex]

[tex]Q_{out} = W*COP[/tex]

[tex]Q_{out} = 5 kW*2.2[/tex]

[tex]Q_{out} = 11 kW[/tex]

Part 3

(a)

Energy balance for a refrigeration cycle gives:

[tex]Q_{in} + W = Q_{out} [/tex]

[tex]3 kW + 5 kW = Q_{out} [/tex]

[tex]8 kW = Q_{out} [/tex]

(b)

[tex]COP = \frac{Q_{in}}{W} [/tex]

[tex]COP = \frac{3 kW}{5 kW} [/tex]

[tex]COP = 0.6 [/tex]

Answer:

2. Predictive Maintenance

Explanation:

Although definitions differ among authors, it is generally accepted that predictive maintenance uses different kinds of techniques to monitor critical machines to prevent them from failing unexpectedly and causing losses in production (or a service), and many more unpleasant events.

Among, thermography, tribology, ultrasonics, and others, vibration analysis is one of the techniques into predictive maintenance, and since most plant types of equipment are mechanical, this is the primary maintenance technique in predictive maintenance.

In general, vibration analysis first needs to acquire data (making use of vibration monitoring using transductors, like accelerometers). Then, the time-domain data is converted into frequency-domain data using a mathematical technique called Fast Fourier Transform (FFT).

Consequently, for each machine's anomaly, there will be a unique 'signature' in the frequency-domain data that corresponds to it.

For example, if the machine presents some imbalance, then there will be a typical frequency (primary frequency) and multiples of it (harmonics), in that frequency-domain data, unique for this imbalance, and so for other machine elements' anomalies, like misalignment, rolling-element bearings high vibrations, bent shafts, and many more.

Answer:

3.0772 m

Explanation:

Given:

Diameter of the aluminium rod, d = 20 mm = 0.02 m

Length of elongation, δL = 3.5 mm = 0.0035 m

Applied load, P = 25 KN = 25000 N

Modulus of elasticity, E = 70 GPa = 70 × 10⁹ N/m²

Now,

we have the relation

[tex]\delta L=\frac{\textup{PL}}{\textup{AE}}[/tex]

Now,

Where, A is the area of cross-section

A =  [tex]\frac{\pi}{4}d^2[/tex]

or

A = [tex]\frac{\pi}{4}\times0.02^2[/tex]

or

A = 0.000314 m²

L is the length of the member

on substituting the respective values, we get

[tex]0.0035=\frac{25000\times L}{0.000314\times70\times10^9}[/tex]

or

L = 3.0772 m

Answer:

First angle projection                      

1.Object is between observer and plane of projection.

2.Projection of object take on first quadrant.

3.Plane of projection is assume non transparent.

Third angle projection:

1.Plane of projection is between observer and object.

2.Projection of object take on third quadrant.

3.Plane of projection is assume transparent.