A cylinder of length L would be made to carry a torque T with an angle of twist ɸ. There are two options considered: 1) hollowed cylinder with an inner radius that is equal to 0.9 of the outer radius, and 2) solid cylinder (with a different radius). If both options would be made from the same material and must have the same angle of twist ɸ under the torque T, find the ratio of the weight between the cylinder designed for option 1 and option 2.

Answer:

15 atm

Explanation:

Solution

Given that:

The first step is to convert mass flow rate unit

m = 87.6 lbm/s = 2.72  slug/sec

Now,

We express the mass flow rate, which is

m = p₀ A* /√T₀√y/R ( 2/ y+1)^(y+1) (y-1)

so,

2.72 = p₀ ^(0.5)/√6000√1.21/2400 (2/2.21)^10.52

Thus,

p₀ =  31736 lb/ft^2

Then,

We convert the pressure in terms of atmosphere

p₀ = 31736/ 2116 = 15 atm

Answer:

Explanation:

The description of all the pieces to the equations

(a)[tex]\( T_2 = 87.92°C \), (b) \( W = 8089.91 \text{ kJ/kg} \)[/tex]  for the isentropic compression of air in a piston-cylinder assembly.

To solve this problem, we'll use the isentropic compression process for an ideal gas. The process being isentropic means that entropy remains constant during the compression, which implies [tex]\( S_1 = S_2 \). Given that it's an ideal gas, we can use the specific gas constant \( R \) and the specific heats \( C_{p} \) and \( C_{v} \) to solve the problem. The relationship between these variables is \( C_{p} - C_{v} = R \).[/tex]

First, let's find

[tex]\( T_2 \) using the given information:\( T_1 = 35°C \)\( v_2 = \frac{1}{10} v_1 \)Since it's an isentropic process, we have:\[ \frac{T_2}{T_1} = \left( \frac{v_1}{v_2} \right)^{k-1} \][/tex]

[tex]where \( k = \frac{C_p}{C_v} \) is the ratio of specific heats. We need to express \( T_1 \) and \( v_2 \) in terms of \( v_1 \) to solve for \( T_2 \):\[ v_2 = \frac{1}{10} v_1 \]\[ v_1 = \frac{RT_1}{P_1} \][/tex]

Since it's an ideal gas, we can use the ideal gas law [tex]\( Pv = RT \):[/tex]

[tex]\[ P_1v_1 = RT_1 \]\[ P_1 = \frac{RT_1}{v_1} \]Now substitute \( v_1 \) into the equation for \( P_1 \):\[ P_1 = \frac{RT_1}{v_1} = \frac{RT_1}{\frac{RT_1}{P_1}} = P_1 \]So, \( P_1 \) cancels out. This means pressure remains constant during the process. Now, let's substitute \( v_1 \) and \( v_2 \) into the equation for \( T_2 \):[/tex]

[tex]\[ T_2 = T_1 \left( \frac{v_1}{v_2} \right)^{k-1} \]Now, calculate \( T_2 \):\[ T_2 = 35 \left( \frac{RT_1}{\frac{RT_1}{10}} \right)^{k-1} \]\[ T_2 = 35 \left( 10 \right)^{k-1} \]We know \( k = \frac{C_p}{C_v} \), and since \( C_p - C_v = R \), we can rewrite \( k \) as \( k = 1 + \frac{R}{C_v} \). Substituting this into the equation:\[ T_2 = 35 \left( 10 \right)^{\left( 1 + \frac{R}{C_v} \right) - 1} \]\[ T_2 = 35 \left( 10 \right)^{\frac{R}{C_v}} \][/tex]

Now, let's find the specific heat ratio[tex]\( k \):\[ k = 1 + \frac{R}{C_v} \]\[ k - 1 = \frac{R}{C_v} \]\[ C_v = \frac{R}{k - 1} \]Given that \( C_p - C_v = R \), we can also express \( C_p \) in terms of \( k \):\[ C_p = C_v + R = \frac{R}{k - 1} + R = \frac{Rk}{k - 1} \][/tex]

Now substitute

[tex]\( C_v \) and \( C_p \) into the equation for \( T_2 \):\[ T_2 = 35 \left( 10 \right)^{\frac{R}{\frac{R}{k - 1}}} \]\[ T_2 = 35 \left( 10 \right)^{k - 1} \]Since \( k = \frac{C_p}{C_v} \):\[ T_2 = 35 \left( 10 \right)^{\frac{C_p}{C_v} - 1} \]\[ T_2 = 35 \left( 10 \right)^{\frac{\frac{Rk}{k - 1}}{\frac{R}{k - 1}} - 1} \]\[ T_2 = 35 \left( 10 \right)^{k - 1} \][/tex]

Now we can calculate

[tex]\( T_2 \):\[ T_2 = 35 \times 10^{0.4} \]\[ T_2 = 35 \times 2.5119 \]\[ T_2 = 87.9185°C \][/tex]

Now that we have [tex]\( T_2 \), we can calculate the work done during the process using the first law of thermodynamics:\[ W = C_v (T_1 - T_2) \]Given that \( C_v = \frac{R}{k - 1} \), we can substitute this expression for \( C_v \):\[ W = \frac{R}{k - 1} (T_1 - T_2) \][/tex]

Substitute the known values:

[tex]\[ W = \frac{8.314 \text{ kJ/kg-K}}{\frac{1.4 - 1}{1.4}} (308.15 - 87.9185) \]\[ W = \frac{8.314 \times 1.4 \text{ kJ/kg-K}}{0.4} \times 220.2315 \]\[ W = 36.779 \times 220.2315 \text{ kJ/kg} \]\[ W = 8089.91 \text{ kJ/kg} \]So, the final answers are:(a) \( T_2 = 87.9185°C \)(b) \( W = 8089.91 \text{ kJ/kg} \)[/tex]

Answer:

4. The velocity gradient at the wall is greater for a laminar boundary layer than a turbulent boundary layer.

Explanation:

This is false

Answer:

Explanation:

The solutions to this question can be seen in the screenshot taken from the solution manual.

Answer:

(a). 42.1 ft/s, (b). 3366.66 ft^3/s, (c). 0.235, (d). 18.2 ft, (e). 3.8 ft.

Explanation:

The following parameters are given in the question above and they are;

Induced hydraulic jump, j = 80 ft wide channel, and the water depths on either side of the jump are 1 ft and 10 ft. Let k1 and k2 represent each side of the jump respectively.

(a). The velocity of the faster moving flow can be calculated using the formula below;

k1/k2 = 1/2 [ √ (1 + 8g1^2) - 1 ].

Substituting the values into the equation above a s solving it, we have;

g1 = 7.416.

Hence, g1 = V1/ √(L × k1).

Therefore, making V1 the subject of the formula, we have;

V1 = 7.416× √ ( 32.2 × 1).

V1 = 42.1 ft/s.

(b). R = V1 × j × k1.

R = 42.1 × 80 × 1.

R = 3366.66 ft^3/s.

(c). Recall that R = V2 × A.

Where A = 80 × 10.

Therefore, V2 = 3366.66/ 80 × 10.

V2 = 4.21 ft/s.

Hence,

g2 = V1/ √(L × k2).

g2 = 4.21/ √ (32.2 × 10).

g2 = 0.235.

(d). (k2 - k1)^3/ 4 × k1k2.

= (10 - 1)^3/ 4 × 1 × 10.

= 18.2 ft.

(e).The critical depth;

[ (3366.66/80)^2 / 32.2]^ 1/3.

The The critical depth = 3.80 ft.

Answer:

a) 42.08 ft/sec

b) 3366.33 ft³/sec

c) 0.235

d) 18.225 ft

e) 3.80 ft

Explanation:

Given:

b = 80ft

y1 = 1 ft

y2 = 10ft

a) Let's take the formula:

[tex] \frac{y2}{y1} = \frac{1}{5} * \sqrt{1 + 8f^2 - 1} [/tex]

[tex] 10*2 = \sqrt{1 + 8f^2 - 1[/tex]

1 + 8f² = (20+1)²

= 8f² = 440

f² = 55

f = 7.416

For velocity of the faster moving flow, we have :

[tex] \frac{V_1}{\sqrt{g*y_1}} = 7.416 [/tex]

[tex] V_1 = 7.416 *\sqrt{32.2*1}[/tex]

V1 = 42.08 ft/sec

b) the flow rate will be calculated as

Q = VA

VA = V1 * b *y1

= 42.08 * 80 * 1

= 3366.66 ft³/sec

c) The Froude number of the sub-critical flow.

V2.A2 = 3366.66

Where A2 = 80ft * 10ft

Solving for V2, we have:

[tex] V_2 = \frac{3666.66}{80*10} [/tex]

= 4.208 ft/sec

Froude number, F2 =

[tex] \frac{V_2}{g*y_2} = \frac{4.208}{32.2*10} [/tex]

F2 = 0.235

d) [tex]El = \frac{(y_2 - y_1)^3}{4*y_1*y_2}[/tex]

[tex] El = \frac{(10-1)^3}{4*1*10} [/tex]

[tex] = \frac{9^3}{40} [/tex]

= 18.225ft

e) for critical depth, we use :

[tex] y_c = [\frac{(\frac{3366.66}{80})^2}{32.2}]^1^/^3 [/tex]

= 3.80 ft

Answer:

Explanation:

Part c:

(Q)out = m(u1 - u2) = 5(2506-193.67) = 11562KJ

Total Entropy change = Δs + (Qout/Tsurrounding) = -33.36 + (11562/393) = 5.44KJ/K

To estimate the uncertainty of a tachometer with 1% accuracy, you calculate 1% of the rpm readings at 50, 500, and 5,000 rpm, resulting in uncertainties of 0.5 rpm, 5 rpm, and 50 rpm respectively.

To estimate the design-stage uncertainty in a tachometer reading with an accuracy of 1% of reading, we calculate the uncertainty for different values of revolutions per minute (rpm). Given readings at 50 rpm, 500 rpm, and 5,000 rpm, the uncertainty can be calculated by taking 1% of each reading:

This means that the display of the tachometer can vary by the calculated uncertainty for each of these readings, which represents the design-stage accuracy of the instrument.

The question involves calculations within AC circuits, focusing on impedance, phase angles, power in RLC circuits, and the effects of frequency on circuit behavior.

The student has presented a scenario involving alternating current (AC) circuits with various components like coils, resistors, inductors, and capacitors. These types of problems require an understanding of impedance, phase angles, power calculations, and resonance in RLC circuits. Calculating these values involves the use of formulas derived from AC circuit theory which take into account the sinusoidal nature of the voltage and current.

To illustrate, in an AC circuit with a resistor (R) and an inductor (L), the total impedance (Z) can be found using the formula Z = \(\sqrt{R^2 + (\omega L - 1/\omega C)^2}\), where \(\omega\) is the angular frequency of the source. The phase angle (\(\phi\)) between the voltage and the current can be determined using the arctangent of the reactive component over the resistive component, \(\phi = arctan((\omega L - 1/\omega C)/R)\). Power calculations involve both the voltage and current, with the power factor being the cosine of the phase angle.

Answer:

a) 9.995 psi

b) 1.3725 psi

Explanation:

Given:-

- The diameter of the pipe at inlet, d1 = 1 in

- The diameter of the pipe at exit, de = 0.5 in

- The exit velocity, Ve = 30 ft/s

- The exit discharge pressure Pe = 0 ( gauge )

- The density of water ρ = 1.940 slugs/ft3

Find:-

Calculate the minimum gage pressure required at the section (1)

Solution:-

- The mass flow rate m ( flow ) for the fluid remains constant via the continuity equation applies for all steady state fluid conditions.

                 m  ( flow ) = ρ*An*Vn = constant

Where,

           An: the area of nth section

           Vn: the velocity at nth section

- Consider the point ( 1 ) and exit point. Determine the velocity at point ( 1 ) via continuity equation.

- The cross sectional area of the pipe at nth point is given by:

                 An = π*dn^2 / 4

- The continuity equation becomes:

                 ρ*A1*V1 = ρ*Ae*Ve

Note: Water is assumed as incompressible fluid; hence, density remains constant.

                V1 = ( Ae / A1 ) * Ve

                V1 = ( (π*de^2 / 4 ) / (π*d1^2 / 4) ) *Ve

                V1 = ( de / d1 ) ^2 * Ve

                V1 = ( 0.5 / 1 )^2 * 30

                V1 =7.5 ft/s

- The required velocity at section ( 1 ) is V1 = 7.5 ft/s.

- Apply the bernoulli's principle for the point ( 1 ) and exit. Assuming the frictional losses are minimal.

        P1 + 0.5*ρ*V1^2 + ρ*g*h1 = Pe + 0.5*ρ*Ve^2 + ρ*g*he

- We will set " h1 " as datum; hence, h1 = 0. The elevation of exit nozzle from point (1) is at he = 10 ft.

- The bernoulli's equation expressed above is in " gauge pressure ". So the gauge pressure of exit Pe = 0 ( Patm ).

Therefore the simplified equation becomes:

        P1 + 0.5*ρ*V1^2  =  0.5*ρ*Ve^2 + ρ*g*he

        P1 = 0.5*ρ* ( Ve^2 - V1^2 ) + ρ*g*he

        P1  = 0.5*1.940*( 30^2 - 7.5^2 ) + 1.940*32*10

        P1 = 818.4375 + 620.8

        P1 = 1439.2375 lbf / ft^2 = 9.995 psi

- If the device is inverted then the velocity at the inlet " V1 " would remain same as there is no change in continuity equation - ( Diameters at each section remains same ).

- The only thing that changes is the application of bernoulli's equation as follows:

         P1 + 0.5*ρ*V1^2 + ρ*g*h1 = Pe + 0.5*ρ*Ve^2 + ρ*g*he

- We will set " he " as datum; hence, he = 0. The elevation of point ( 1 ) from exit nozzle is at h1 = 10 ft.

- The bernoulli's equation expressed above is in " gauge pressure ". So the gauge pressure of exit Pe = 0 ( Patm ).

Therefore the simplified equation becomes:

        P1 + 0.5*ρ*V1^2 + ρ*g*h1 =  0.5*ρ*Ve^2

        P1 = 0.5*ρ* ( Ve^2 - V1^2 ) - ρ*g*h1

        P1  = 0.5*1.940*( 30^2 - 7.5^2 ) - 1.940*32*10

        P1 = 818.4375 - 620.8

        P1 = 197.6375 lbf / ft^2 = 1.3725 psi

Answer:

[tex]L = 0.319\,m[/tex]

Explanation:

Let suppose that temperature of air is 15°C. The heating process of the solar concentrator is modelled after the First Law of Thermodynamics:

[tex]\dot Q = h\cdot \pi\cdot D\cdot L\cdot (\bar T-T_{\infty})[/tex]

The required length is:

[tex]L = \frac{\dot Q}{h\cdot \pi\cdot D\cdot (\bar T-T_{\infty})}[/tex]

But,

[tex]\dot Q = \dot m \cdot c_{w}\cdot (T_{o}-T_{i})[/tex]

[tex]\dot Q = \left(0.015\,\frac{kg}{s}\right)\cdot \left(4180\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (85^{\circ}C-15^{\circ}C)[/tex]

[tex]\dot Q = 4389\,W[/tex]

[tex]\bar T = \frac{15^{\circ}C + 85^{\circ}C}{2}[/tex]

[tex]\bar T = 50^{\circ}C[/tex]

[tex]L = \frac{4389\,W}{\left(2500\,\frac{W}{m^{2}} \right)\cdot \pi \cdot (0.05\,m)\cdot (50^{\circ}C-15^{\circ}C)}[/tex]

[tex]L = 0.319\,m[/tex]

Answer:

The shortest distance d to the edge of the plate is 66.67 mm

Concepts and reason

Moment of a force:

Moment of a force refers to the propensity of the force to cause rotation on the body it acts upon. The magnitude of the moment can be determined from the product of force’s magnitude and the perpendicular distance to the force.

Moment(M) = Force(F)×distance(d)

Moment of inertia ( I )

It is the product of area and the square of the moment arm for a section about a reference. It is also called as second moment of inertia.

First prepare the free body diagram of sectioned plate and apply moment equilibrium condition and also obtain area and moment of inertia of rectangular cross section. Finally, calculate the shortest distance using the formula of compressive stress (σ) in combination of axial and bending stress

Solution and Explanation:

[Find the given attachments]

The shortest distance (d) to the edge of the rectangular-plate is equal to 66.65 mm.

Given the followin data:

Thickness (length) of plate = 10 mm to m = 0.001 m.

Width of plate = 200 mm to m = 0.2 m.

c = [tex]\frac{0.2}{2}[/tex] = 0.1 m.

First of all, we would determine the area of the rectangular-plate and its moment of inertia.

For area:

[tex]A=LW\\\\A=0.001 \times 0.2\\\\A= 0.002 \;m^2[/tex]

For moment of inertia:

Mathematically, the moment of inertia of a rectangular-plate is given by this formula:

[tex]I=\frac{b^3d}{12} \\\\I=\frac{0.2^3 \times 0.01}{12} \\\\I=\frac{0.008 \times 0.01}{12}\\\\I=\frac{0.0008 }{12}\\\\I=6.67 \times 10^{-6}\;m^4[/tex]

The compressive stress of a rectangular-plate with respect to axial and bending stress is given by this formula:

[tex]\sigma = \frac{P}{A} -\frac{Mc}{I} \\\\\sigma = \frac{P}{0.002} -\frac{P(0.1-d)\times 0.1}{6.67 \times 10^{-6}} \\\\\sigma = \frac{P}{0.002} -\frac{0.01P-0.1Pd}{6.67 \times 10^{-6}} \\\\\frac{P}{0.002}=\frac{0.01P-0.1Pd}{6.67 \times 10^{-6}}\\\\500P=\frac{0.01P-0.1Pd}{6.67 \times 10^{-6}}\\\\3.335 \times 10^{-3}P=0.01P-0.1Pd\\\\[/tex]

[tex]0=0.01P-0.1Pd-3.335 \times 10^{-3}P\\\\0=(0.01-0.1d-3.335 \times 10^{-3})P\\\\0.01-0.1d-3.335 \times 10^{-3}=0\\\\0.1d=0.01-3.335 \times 10^{-3}\\\\d=\frac{6.665\times 10^{-3}}{0.1} \\\\d=6.665\times 10^{-2}\;m\\\\[/tex]

d = 66.65 mm.

Read more on moment of inertia here: https://brainly.com/question/3406242

Answer:

Explanation:

The solutions to this question can be seen in the following screenshots taken from the solution manual.

Answer:

Check the explanation

Explanation:

Kindly check the attached images below to see the step by step explanation to the question above.

Answer:

transfer function from R(s) to E(s) and determine the steady-state error (ess) for a unit-step reference input (c) Select the system parameters (k, kP, kI) such that the closed-loop system has damping coefficient ζ = 0.707 Figure Control system diagram.

Explanation:

hope this helps

Answer:

The correct answer is 0.004382 kW/K, the second law is a=satisfied and established because it is a positive value.

Explanation:

Solution

From the question given we recall that,

The transferred heat rate is = 2kW

A reservoir cold at = 300K

The next step is to find the rate  at which the entropy of the two reservoirs changes is kW/K

Given that:

Δs =  Q/T This is the entropy formula,

Thus

Δs₁ = 2/ 300 = 0.006667 kW/K

Δs₂ = 2 / 875 =0.002285

Therefore,

Δs = 0.006667 - 0.002285

= 0.004382 kW/K

Yes, the second law is satisfied, because it is seen as positive.

The student's question regarding heat loss by convection requires the convective heat transfer coefficient, which is not provided, making it impossible to calculate the rate of heat loss without additional data or empirical formulas.

The question pertains to heat loss from the wall of a house by convection on a cold winter day when the wind is blowing parallel to the wall. However, to answer the question accurately, we need additional information related to the convective heat transfer coefficient for the scenario described, which typically requires the use of empirical relationships that take into account the flow of air over the surface (such as the wind speed) and the properties of the air (temperature, viscosity, etc.). Unfortunately, the student's question does not provide the necessary details or equations to calculate the convective heat transfer coefficient, which is crucial to determining the rate of heat loss.

To calculate the rate of heat loss through convection, the formula Q = hA(T_s - T_{∞}) is typically used, where Q is the heat transfer rate, h is the convective heat transfer coefficient, A is the surface area, T_s is the surface temperature, and T_{∞} is the ambient temperature. Without the convective heat transfer coefficient, we cannot complete this calculation.

Answer:

See explaination

Explanation:

Given that

the cross section at right is mode 2024-T3

Ftu=62 ksi

E=10.5 Msi

Ec=10.7 Msi

the angle b1=1.25"

b2=1.75"

t1=0.050"

t2=0.080"

Kindly check attachment for the step by step solution of the given problem.

Answer: Both Technicians

Explanation:

When testing a spring break it is advisable to Step on and off the brake, with the engine off, the parking brake knob is expected to pop out when air pressure falls between 20-40 psi.

Go under the vehicle and pull the spring brakes.

Turn on the engine back and pump the brake pedal down to the floor. To test the spring breaks

Answer:

surface temperature = 128.74⁰c

Explanation:

Given data

diameter of cable = 5 mm = 0.005 m

length of cable = 4 m

T∞ ( surrounding temperature ) = 20⁰c

voltage drop across cable ( dv )= 60 V

current across cable = 1.5 A

attached to this answer is the comprehensive analysis and solution to the problem.

The assumption made is not a good one since the calculated Ts ( surface temperature ) is very much different from the assumed Ts

Given Information:

Inductance = L = 5 mH = 0.005 H

Time = t = 2 seconds

Required Information:

Current at t = 2 seconds = i(t) = ?

Energy at t = 2 seconds = W = ?

Answer:

Current at t = 2 seconds = i(t) = 735.75 A

Energy at t = 2 seconds = W = 1353.32 J

Explanation:

The voltage across an inductor is given as

[tex]V(t) = 5(1-e^{-0.5t})[/tex]

The current flowing through the inductor is given by

[tex]i(t) = \frac{1}{L} \int_0^t \mathrm{V(t)}\,\mathrm{d}t \,+ i(0)[/tex]

Where L is the inductance and i(0) is the initial current in the inductor which we will assume to be zero since it is not given.

[tex]i(t) = \frac{1}{0.005} \int_0^t \mathrm{5(1-e^{-0.5t}}) \,\mathrm{d}t \,+ 0\\\\i(t) = 200 \int_0^t \mathrm{5(1-e^{-0.5t}}) \,\mathrm{d}t \\\\i(t) = 200 \: [ {5\: (t + \frac{e^{-0.5t}}{0.5})]_0^t \\i(t) = 200\times5\: \: [ { (t + 2e^{-0.5t} + 2 )] \\[/tex]

[tex]i(t) = 1000t +2000e^{-0.5t} -2000\\[/tex]

So the current at t = 2 seconds is

[tex]i(t) = 1000(2) +2000e^{-0.5(2)} -2000\\\\i(t) = 735.75 \: A[/tex]

The energy stored in the inductor at t = 2 seconds is

[tex]W = \frac{1}{2}Li(t)^{2}\\\\W = \frac{1}{2}0.005(735.75)^{2}\\\\W = 1353.32 \:J[/tex]

Answer:

See explaination

Explanation:

Please kindly check attachment for the step by step solution of the given problem.

The attached file gave a detailed solution of the problem.

The acceleration of point A on the hoop can be calculated by considering both the tangential acceleration due to the hoop's angular acceleration and the radial (centripetal) acceleration. These values are vectorially added along with the deceleration of the hoop's center.

To determine the acceleration of point A, we must consider both the tangential and radial (centripetal) accelerations since the hoop is experiencing angular acceleration and the center of the hoop has a deceleration. The tangential acceleration (at) is due to the angular acceleration and is calculated by multiplying the angular acceleration (
a) by the radius (r) of the hoop. The radial (centripetal) acceleration (ar) depends on the angular velocity (
w) and the radius of the hoop. The equation for centripetal acceleration is ar = w2
r. The total acceleration (aA) of point A can be found by vectorially adding the tangential and radial accelerations, considering the direction of deceleration of the center as well.

Answer:

a) [tex]\dot W = 0.978\,kW[/tex], b) [tex]I = \left(50000\,\frac{kJ}{h} \right)\cdot \left(\frac{1}{3600}\,\frac{h}{s}\right)\cdot \left(\frac{1}{COP_{real}} \right) - 0.978\,kW[/tex]

Explanation:

a) The ideal Coefficient of Performance for the heat pump is:

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

[tex]COP_{HP} = \frac{298.15\,K}{298.15\,K - 277.15\,K}[/tex]

[tex]COP_{HP} = 14.198[/tex]

The reversible work input is:

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

[tex]\dot W = \left(\frac{50000\,\frac{kJ}{h} }{14.198} \right)\cdot \left(\frac{1}{3600}\,\frac{h}{s} \right)[/tex]

[tex]\dot W = 0.978\,kW[/tex]

b) The irreversibility is given by the difference between real work and ideal work inputs:

[tex]I = \dot W_{real} - \dot W_{ideal}[/tex]

[tex]I = \left(50000\,\frac{kJ}{h} \right)\cdot \left(\frac{1}{3600}\,\frac{h}{s}\right)\cdot \left(\frac{1}{COP_{real}} \right) - 0.978\,kW[/tex]

Answer:

Given:

P₁ = 500 kPa

T₁ = 860 K

P₂= 100 kPa

T₂ = 460 K

Let's take entropy properties of T1 and T2 from ideal properties of air,

at T = 860K, s(T₁) = 2.79783 kJ/kg.K

at T = 460K, s(T₂) = 2.13407 kJ/kg.K

using entropy balance equation:

[tex] \frac{\sigma _cv}{m} = s(T_2)- s(T_1) - R In [\frac{P_2}{P_1}] [/tex]

[tex] \frac{\sigma _cv}{m} = 2.79783 - 2.13407 - 0.287 In [\frac{100}{500}] [/tex]

= - 0.2018 kJ/kg. K

In this case the entropy is negative, which means the value of exit temperature is not correct, beacause entropy should always be positive(>0).

Answer:

The answer 1 is correct.

Explanation:

When a junction called the p-n is forward biased, the capacitance of diffusion will form or structured across depletion layers.

When at a higher frequency or frequencies, the dopants, are vibrated across the metallurgical junction and the doping profiles smooths out.

Answer:

Explanation:

Solution:-

- To categorize the flow conditions of any fluid we utilize a dimensionless number, called Reynold's number ( Re ) to study the behaviour of the fluid.

- Reynold's number is proportional to the ratio of inertial forces ( forces that resist any change in motion of a unit mass ) and viscous forces ( forces that resist any iner-plane deformations between layers of fluid ).

- Considering 2-Dimensional viscous flow over a flat plate, the Reynold number (Re) is mathematically expressed as a function of distance "x" denoted from leading edge and along the length of the plate:

                             [tex]Re_x = \frac{U*x}{v}[/tex]

Where,

               U: The free stream velocity of the fluid

               ν: The kinematic viscosity of the fluid

- The distance "x" along the length of the plate is substituted in the above formula and the corresponding Reynold number is evaluated. This gives a highly localized value about "x".

- The purpose of the Reynold number is the substitution of dynamically similar fluids i.e Fluid with the same Reynold's number when testing models to see how they would behave in a specific environment.

- The Reynolds number has a set of ranges above and below the critical range defined by the critical length "xc" along the plate.  The range below the critical has laminar flow characteristics, whereas the range above the critical has turbulent flow. The laminar region has flow along smooth streamlines, while turbulent region is characterized by 3 - dimensional random eddy.

- The critical length " xc " is determined from the critical Reynold number i.e ( 5 x 10^5 ) which is a small region that has mixed characteristics of laminar and turbulent conditions.

- The flow at the boundaries has zero velocity, there is a steep velocity gradient from the boundary into the flow. This velocity gradient in a real fluid sets up shear forces near the boundary that reduce the flow speed to that of the boundary. That fluid layer which has had its velocity affected by the boundary shear is called the "boundary layer"

-  For smooth upstream boundaries, " x << xc " or " Re_x < 5 x 10^5 " , the boundary layer starts out as a laminar boundary layer in which the fluid particles move in smooth layer.

- As the laminar boundary layer increases in thickness, it becomes unstable as changes in motion become more predominant ( inertial ) than viscous effect of fluid layers. This leads to a transformation of laminar boundary layer into turbulent boundary layer in which fluid particles move in haphazard paths.  " x > xc " or " Re_x > 5 x 10^5 "

- The boundary layer thickness/height d increases as x increases. The relationships for laminar and turbulent regions of boundary layer are given as follows:

                  [tex]\frac{d}{x} = \left \{ {{\frac{5}{\sqrt{Re_x} } , 10^3 < Re_x < 10^6} \\\\\atop {\frac{0.16}{\frac{1}{7} \sqrt{Re_x} } , 10^6 < Re_x}} \right.[/tex]

- To construct a function of boundary layer thickness " d " and length from leading edge of the plate " x ". We use the Re_x relation and substitute, we get the following proportionalities for our sketch:

                  d ∝ √x  .... Laminar region

                  d ∝ [tex]x^\frac{6}{7}[/tex]  .... Turbulent region

- Use the above relation to develop sketch for the boundary layer along the length "x" from leading edge.

- The sketch is given as an attachment.

Answer:

See explaination

Explanation:

We can describe Aspiration Effect as a phenomenon of providing an allowance for the release of air from the mold cavity during the metal pouring.

See the attached file for detailed solution of the given problem.

Answer:

The diameter at the bottom of the sprue is =0.725 in, and the sprue will not occur when 0.021<0.447.

Explanation:

Solution

The first step to take is to define the Bernoulli's eqaution

h effective = v²top/2g + ptop /ρg = hbottom + v² bottom/2g + p bottom/ ρg

h effective  + 0 +0= 0 +v² bottom/2g + 0

Thus,

v bottom = √ 2gh total

=√ 2 (32. 6 ft/ s²) + (12/12 ft)

Which is = 8.074 ft/s

We now, express the relation for flow rate.

Q =π/4 D² bottom v bottom

= 40 in 3/s = π/4 D²₃ ( 8.074 ft/s) (12  in/ ft)

so,

D bottom = 0.725 in.

Then,

We express the relation to avoid aspiration

A₃/A₂ < √ h top /h total

= π/4 D²₃/ π/4 D²₂ < √3/15

= 0.725²/5² < √3/15

=0.021<0.447

Therefore, the aspiration will not happen or occur

The dimensions of the specific resistance of the filter medium, Rf, are F/L (force per unit length), and the dimensions of the filtrate "viscosity", alpha, are T/L^2 (time per area). This ensures the dimensional consistency of the vacuum filtration equation.

To answer the question regarding vacuum filtration, we need to determine the dimensions of Rf and alpha. The equation provided is Q = Delta pA2 alpha(VRw + ARf), where each symbol has an associated dimension. To keep the equation dimensionally consistent, the dimensions of the terms added within the parenthesis must be the same, i.e., the dimensions of VRw must equal the dimensions of ARf.

Let's break down the dimensions:

Hence, the dimensions of Rf are F/L (force per unit length), and the dimensions of alpha are T/L2 (time per area).

Answer:

i) Heat transfer coefficient (h) = 7 w/m²k

ii) Heat transfer per meter width of wall

      = h x L x 1 x (Ts - T₆₀)

        = 7 x 0.3048 x (505.4 - 322) = 414.747 w/m

Explanation:

see attached image