The wall shear stress in a fully developed flow portion of a 12-in.-diameter pipe is 1.85 lb/ft^2.
Determine the pressure gradient ∂p/∂x, wherexis the flow direction when (a) the pipe is horizontal, (b) the pipe is vertical with the flow up, and (c) the pipe is vertical with the flowdown.

To solve this problem it is necessary to apply the concepts related to density in relation to mass and volume for each of the states presented.

Density can be defined as

[tex]\rho = \frac{m}{V}[/tex]

Where

m = Mass

V = Volume

For state one we know that

[tex]\rho_1 = \frac{m_1}{V}[/tex]

[tex]m_1 = \rho_1 V[/tex]

[tex]m_1 = 1.18*1[/tex]

[tex]m_1 = 1.18Kg[/tex]

For state two we have to

[tex]\rho_2 = \frac{m_2}{V}[/tex]

[tex]m_2 = \rho_2 V[/tex]

[tex]m_1 = 7.2*1[/tex]

[tex]m_1 = 7.2Kg[/tex]

Therefore the total change of mass would be

[tex]\Delta m = m_2-m_1[/tex]

[tex]\Delta m = 7.2-1.18[/tex]

[tex]\Delta m = 6.02Kg[/tex]

Therefore the mass of air that has entered to the tank is 6.02Kg

Answer:

FCF = 14.8998 N   (Compression)

FBE = 52.7342 N   (Compression)

Explanation:

We have to find the mass center of ABCD which is at G.

XCG = (0.45 m /2) = 0.225 m

YCG = ∑mi*yi / ∑mi  

⇒  YCG = (3 Kg*0.225m+3 Kg*0.225m+3 Kg*0m) / (3*3Kg) = 0.15 m

Now, we can apply  ∑F = m*a

where a is the tangential acceleration of the links

M = 3*m = 3*3 Kg = 9 Kg

∑F = m*a  ⇒  M*g*Sin 40º = M*a  ⇒  a = g*Sin 40º

⇒  a = (9.81 m/s²)*Sin 40º = 6.3057 m/s²

We apply ∑MB as follows

∑MB = (0.45)*(FCF*Sin 50º) - M*g*(0.225) = - (M*a*Sin 40º)*(0.225) - (M*a*Cos 40º)*(0.15)

⇒   (0.45)*(FCF*Sin 50º) - 9*9.81*(0.225) = - (9*6.3057*Sin 40º)*(0.225) - (9*6.3057*Cos 40º)*(0.15)

⇒  FCF = 14.8998 N   (Compression)

Then

∑Fy = m*ay

⇒   FCF*Sin 50º + FBE*Sin 50º - M*g = - M*a*Sin 40º

⇒   14.8998*Sin 50º + FBE*Sin 50º - 9*9.81 = - 9*6.3057*Sin 40º

⇒   FBE = 52.7342 N   (Compression)

We can see the system in the pic shown.

The question informs a statics problem in physics regarding finding the forces in supporting links of a system of welded bars in static equilibrium. Newton's laws and free body diagrams are typically used to solve such a problem.

The student's question pertains to determining the forces in each link that supports a system of welded bars, each with a mass of 3 kg, immediately after being released from rest. To solve this problem, one would typically employ the principles of static equilibrium for rigid bodies, ensuring that the sum of forces and the sum of moments about any pivot point equals zero. This would involve drawing free body diagrams and applying Newton's second law of motion. As the links and the bars form a static structure in this scenario, the forces in the links can be found by considering the geometry of the setup and the forces due to gravity acting on the masses of the bars. However, without a detailed figure or further information about the geometry of the setup, we cannot provide a specific answer to this problem.

Answer:

The answers are on the attachment.

Explanation:

To solve the problem, we can use the Maxwell model equation:

σ(t) = σ(0)exp(-t/τ) + Eε

where σ(t) is the stress at time t, σ(0) is the initial stress, τ is the relaxation time, E is the instantaneous modulus, and ε is the strain. using this equation the time before the seal begins to leak is approximately 1304 days (or about 3.6 years).

The Maxwell model assumes that the material is composed of a spring and a dashpot in parallel. The spring represents the elastic component of the material and the dashpot represents the viscous component. The model assumes that the total strain is the sum of the elastic and viscous strains and that the stress is proportional to the total strain.

The seal is held at a fixed compressive strain of 0.2, so ε = 0.2. We are also given that the short-term modulus E is 3 MPa and the relaxation time τ is 300 days. Finally, we are asked to find the time before the seal begins to leak under an internal fluid pressure of 0.3 MPa, so σ(t) = 0.3 MPa.

Substituting these values into the equation, we get:

0.3 MPa = σ(0)exp(-t/300 days) + (3 MPa)(0.2)

Solving for σ(0), we get:

σ(0) = 0.3 MPa - (3 MPa)(0.2)exp(t/300 days)

At the point when the seal begins to leak, the stress will have reached the tensile strength of the material. Let's assume that the tensile strength of the material is 10 MPa. Then, we can set σ(0) = 10 MPa and solve for t:

10 MPa = 0.3 MPa - (3 MPa)(0.2)exp(t/300 days)

9.7 MPa = (3 MPa)(0.2)exp(t/300 days)

Taking the natural logarithm of both sides, we get:

ln(9.7 MPa / (3 MPa)(0.2)) = -t/300 days

t ≈ 1304 days

Therefore, the time before the seal begins to leak is approximately 1304 days (or about 3.6 years).

To know more about the Maxwell model equation:

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Answer:

A (81 N)

Explanation:

P = F/A, where P is the pressure on an object exerted by the force F over the area A. Using "D" to represent drag force specifically, this equation can be rearranged to D = CPA, where C is a constant of proportionality that represents the friction coefficient. The pressure exerted on an object undergoing motion through a fluid can be expressed as (1/2) ρv2, where ρ is the density of the fluid and v is the velocity by which the object is moving. Therefore, D = (1/2)(C)(ρ)(v²)(A).

V = 150 m/s, C(d) = 0.002, A = 3m², ρ(density  of air) = 1.2 kg/m3

Fd = (1.2 x 3 x 0.002 x 150²)/2 = 81 kg.m/s²

The drag force exerted on the airfoil is 81 N.

The output DC voltage after using a full-wave rectifier on an AC source with a Vrms of 60V, considering a forward voltage drop of 0.7V and negligible ripple, is approximately 53.7V.

To convert an AC voltage to DC voltage using a full-wave rectifier, you first need to understand the relationship between the root mean square (RMS) value of the AC voltage and its peak value. The RMS value is given as Vrms = 60V. For a full-wave rectified sine wave, the peak voltage (Vp) is Vrms × √2. Therefore, Vp = 60V × √2 ≈ 84.85V. However, because of the rectifier's forward voltage drop (Vf), we have to subtract this from the peak voltage to get the peak DC voltage (Vdc_peak). Therefore, Vdc_peak = Vp - Vf ≈ 84.85V - 0.7V ≈ 84.15V.

However, to find the average or output DC voltage (Vdc_avg) from a full-wave rectifier, you'd multiply the peak DC voltage by 2/π (since it's a sinusoidal wave), so Vdc_avg ≈ (2/π) × 84.15V ≈ 53.7V. Thus, the output DC voltage after using a full-wave rectifier directly on the line voltage, assuming negligible ripple, is approximately 53.7V.

The student's question requires using Raoult's law and the Peng-Robinson equation to calculate vapor pressures in order to determine the liquid fraction and phase compositions in a flash drum separation of a 50 mol% propane and n-butane mixture at 37°C and 0.6 MPa.

The flash drum separation involves determining the fraction of liquid that exits the drum and the composition of the exiting phases when a 50 mol% mixture of propane and n-butane is subjected to isothermal conditions at 37°C and 0.6 MPa pressure. This calculation requires the use of Raoult's law and the Peng-Robinson equation to calculate vapor pressures for the given conditions. Two scenarios are considered:

To answer the specific question we need to perform these calculations, but it is important to note that without the actual Peng-Robinson parameters or computational tools to carry out these calculations, providing a numeric answer would be speculative. Generally, the approach involves setting up material balances around the drum, calculating the equilibrium constant for each component, and then solving the equations iteratively until a convergent solution is found.

Based on the calculations, the acceleration of this hollow, spherical shell is equal to 3.62 [tex]m/s^2[/tex].

Given the following data:

The torque for a hollow, spherical shell is given by this formula:

[tex]F=\frac{2}{3} ma[/tex]

From Newton's Second Law of Motion, the forces acting on the shell is given by:

[tex]mgsin\theta -F=ma\\\\mgsin\theta -\frac{2}{3} ma=ma\\\\mgsin\theta=ma+\frac{2}{3} ma\\\\mgsin\theta=\frac{5}{3} ma\\\\gsin\theta=\frac{5}{3} a\\\\a=\frac{3gsin\theta}{5} \\\\a=\frac{3 \times 9.8 \times sin38.0}{5} \\\\a = \frac{18.10}{5}[/tex]

Acceleration, a = 3.62 [tex]m/s^2[/tex]

Mathematically, the friction force acting on this shell is equal to the torque:

[tex]F=\frac{2}{3} ma\\\\F=\frac{2}{3} \times 2.00 \times 3.62[/tex]

F = 4.83 Newton.

First of all, we would determine the normal force:

[tex]N=mgcos\theta\\\\N= 2 \times 9.8 \times cos 38.0[/tex]

N = 15.45 Newton.

For the coefficient of friction:

[tex]\mu = \frac{F}{N} \\\\\mu = \frac{4.83}{15.45}\\\\\mu=0.31[/tex]

Read more on force here: brainly.com/question/1121817

Answer:

1.44 mm

Explanation:

Compute the maximum allowable surface crack length using

[tex]C=\frac {2E\gamma}{\pi \sigma_c^{2}}[/tex] where E is the modulus  of elasticity, [tex]\gamma[/tex] is surface energy and [tex]\sigma_c[/tex] is tensile stress

Substituting the given values

[tex]C=\frac {2\times 393\times 10^{9}\times 0.9}{\pi\times (16\times 10^{6})^{2}= 0.001441103 m\approx 1.44mm[/tex]

The maximum allowable surface crack is 1.44 mm

To calculate the downstream velocity of air in a pipe, use the principle of conservation of mass and the continuity equation, which relates the product of cross-sectional area and velocity at two points in a system. The cross-sectional areas of the pipe at both points are computed, and using the continuity equation, the downstream velocity can be found. Convert diameters to meters, calculate areas, and apply the equation.

The student's question involves calculating the velocity of air at a downstream location in a pipe system, using the principles of fluid dynamics. First, to find the velocity of gas at the downstream point, we can invoke the principle of conservation of mass, specifically the continuity equation for an incompressible fluid. This equation states that the product of the cross-sectional area (A) and velocity (V) at one point must equal the product of A and V at any other point in the system, assuming the density of the fluid remains constant.

First, we must calculate the cross-sectional areas at both points:

Using the given velocities and the continuity equation A1 x V1 = A2 x V2, we can solve for the downstream velocity, V2.

Given:

We can convert these diameters to meters, find the areas, and then apply the continuity equation to find the downstream velocity. However, it seems the student is not required to solve for changes in air density due to temperature and pressure variations, hence we're treating the air as incompressible.

The velocity of the gas at the downstream point is approximately 57.3 m/s, considering the conservation of mass flow rate and the given conditions of temperature and pressure.

To find the velocity of the gas at the downstream point, we need to apply the principle of conservation of mass for a compressible flow, which states that the mass flow rate must be constant throughout the pipe. The mass flow rate [tex](\(\dot{m}\))[/tex] can be expressed using the following equation:

[tex]\[\dot{m} = \rho \cdot A \cdot v\][/tex]

where:

- [tex]\(\rho\)[/tex] is the density of the air,

- A is the cross-sectional area of the pipe,

- v is the velocity of the air.

First, we'll calculate the density of the air at both the upstream and downstream points using the ideal gas law:

[tex]\[\rho = \frac{P}{R \cdot T}\][/tex]

where:

- P is the absolute pressure,

- R is the specific gas constant for air [tex](\(R = 287.05 \, \text{J/(kg·K)}\))[/tex],

- T is the temperature in Kelvin.

Let's start by converting the gauge pressures to absolute pressures. Since gauge pressure is the pressure relative to atmospheric pressure, we need to add atmospheric pressure (1.01325 bar) to each gauge pressure:

Upstream Conditions:

- Gauge pressure: [tex]\(1.80 \, \text{bar}\)[/tex]

- Absolute pressure: [tex]\(P_1 = 1.80 \, \text{bar} + 1.01325 \, \text{bar} = 2.81325 \, \text{bar} = 281325 \, \text{Pa}\)[/tex]

- Temperature: [tex]\(27.0^\circ \text{C} = 27.0 + 273.15 = 300.15 \, \text{K}\)[/tex]

- Pipe diameter: [tex]\(7.00 \, \text{cm} = 0.07 \, \text{m}\)[/tex]

- Velocity: [tex]\(v_1 = 30.0 \, \text{m/s}\)[/tex]

Downstream Conditions:

- Gauge pressure: [tex]\(1.63 \, \text{bar}\)[/tex]

- Absolute pressure: [tex]\(P_2 = 1.63 \, \text{bar} + 1.01325 \, \text{bar} = 2.64325 \, \text{bar} = 264325 \, \text{Pa}\)[/tex]

- Temperature: [tex]\(60.0^\circ \text{C} = 60.0 + 273.15 = 333.15 \, \text{K}\)[/tex]

- Pipe diameter: [tex]\(5.50 \, \text{cm} = 0.055 \, \text{m}\)[/tex]

Calculate the density at both points:

[tex]\[\rho_1 = \frac{P_1}{R \cdot T_1} = \frac{281325 \, \text{Pa}}{287.05 \, \text{J/(kg·K)} \times 300.15 \, \text{K}} \approx 3.268 \, \text{kg/m}^3\][/tex]

[tex]\[\rho_2 = \frac{P_2}{R \cdot T_2} = \frac{264325 \, \text{Pa}}{287.05 \, \text{J/(kg·K)} \times 333.15 \, \text{K}} \approx 2.769 \, \text{kg/m}^3\][/tex]

Calculate the cross-sectional area of the pipes:

[tex]\[A_1 = \pi \left(\frac{0.07 \, \text{m}}{2}\right)^2 \approx 0.00385 \, \text{m}^2\][/tex]

[tex]\[A_2 = \pi \left(\frac{0.055 \, \text{m}}{2}\right)^2 \approx 0.00238 \, \text{m}^2\][/tex]

Mass flow rate at the upstream point:

[tex]\[\dot{m} = \rho_1 \cdot A_1 \cdot v_1 = 3.268 \, \text{kg/m}^3 \times 0.00385 \, \text{m}^2 \times 30.0 \, \text{m/s} \approx 0.378 \, \text{kg/s}\][/tex]

Velocity at the downstream point:

Using the conservation of mass flow rate:

[tex]\[\dot{m} = \rho_2 \cdot A_2 \cdot v_2\][/tex]

Solving for [tex]\(v_2\):[/tex]

[tex]\[v_2 = \frac{\dot{m}}{\rho_2 \cdot A_2} = \frac{0.378 \, \text{kg/s}}{2.769 \, \text{kg/m}^3 \times 0.00238 \, \text{m}^2} \approx 57.3 \, \text{m/s}\][/tex]

So, the velocity of the gas at the downstream point is approximately [tex]\(57.3 \, \text{m/s}\).[/tex]

To solve this problem it is necessary to apply the concepts related to Arc Length.

From practical terms we know that the Angle can be calculated based on the arc length and the radius of the circle, in other words

[tex]\theta = \frac{S}{r}[/tex] or

[tex]S = \theta r[/tex]

Where,

S = Length of the arc

r = Radius

From the information given, the object to bending has a millimeter thick and a radius of 12 mm. This way your net radio is given by

[tex]R_1 = r_B+\frac{t}{2}[/tex]

After Springback the radius increases and the thickness ratio is therefore maintained.

[tex]R_2 = r_S+\frac{t}{2}[/tex]

For both cases we have different angles but that maintains the electron-magnetic proportions, therefore the arc length is maintained:

[tex]S_B = S_S[/tex]

[tex]\theta_B R_1 = \theta_S R_2[/tex]

[tex]\theta_B (r_B+\frac{t}{2})=\theta_S(r_S+\frac{t}{2})[/tex]

Re-arrange to find [tex]\theta_S[/tex]

[tex]\theta_S = \frac{\theta_B (r_B+\frac{t}{2})}{(r_S+\frac{t}{2})}[/tex]

[tex]\theta_S= \frac{90\°*(12+\frac{1}{2})}{13+\frac{1}{2}}[/tex]

[tex]\theta_S= 83.33\°[/tex]

Therefore the angle that should be bent before springback is 83.33°

Answer:

Load to fracture = 17212.5 N

Explanation:

Firstly we need to find the flexural strength of the material. We do this by using the date obtained by the test of the circular sample and the equation shown below

σ (flexural strength) = F x L/ (π x r³), where F is the load at fracture, L is the distance between the two load points, r is the radius of the circular cross section. Substituting the values we get

σ (flexural strength) = 3000 x 40 x 10⁻³/ (π x 5 x 10⁻³)³ = 306 x 10⁶ N/m²

Now, the flexural stress for a square sample is written as

σ (flexural strength) = 3 x F x L/ 2 x b x d² , and since in a square sample breadth = width the equation becomes

σ (flexural strength) = 3 x F x L/ 2 x d³

Solving for F, we get

F = σ (flexural strength) x 2 x d³ /3 x L

For same material we will use the σ (flexural strength) as calculated above, furthermore L remains the same and d = 15 x 10⁻³. Solving for F

F = 306 x 10⁶ x 2 x (15 x 10⁻³)³ /3 x 40 x 10⁻³ = 17212.5 N

The load required for this specimen to fracture would be 17212.5 N.

Answer:

a. Park should be adjacent to 48th Street, b. Difference in noise level = 707dBa, c. Yes

Explanation:

Data given for Avenue A

Cars = 496, Medium Truck = 52, Heavy Truck = 19, Buses = 10

Data given for 48th Street

Cars = 822, Medium Truck = 22, Heavy Truck = 8, Buses = 3

Consider the PCEs to be Cars = 1, Medium Truck = 13, Heavy Truck = 47, Buses = 18

a. Noise Level = Number of vehicles x PCE

For Avenue A

Noise level = (496 x 1) + (52 x 13) + (19 x 47) + (10 x 18) = 2245dBa

For 48th Street

Noise level = (822 x 1) + (22 x 13  + (8 x 47) + (3 x 18) = 1538dBa

The park should adjacent to 48th street as it is quieter than Avenue A

b. Let the Setback be 50ft. We know that the reduction of noise for 100ft = 5-8 dBa, hence

For Avenue A Noise Reduction due to 50 ft = (8/100) x 50 = 4dBa

Noise at Setback distance = 2245 - 4 = 2241dBa

Considering the same setback the noise at 48th street would be = 1538 - 4 = 1534 dBa

The difference is noise level between the two sides would be = 2241 - 1534 = 707 dBa

c. Yes the developer concerns are valid because there is a clear difference in noise levels of the two sites. This can be seen even after setting the same Setback. Locating the park next to Avenue A will cause serious noise problems.

Answer:

[tex]R_{min} = 4.84\times 10^{-3} m[/tex]

Explanation:

Given data:

Applied force 750 N

Flexural strength is 105 MPa

separation is 50 mm = 0.05 m

flexural strength is given as

[tex]\sigma_f = \frac{FL}{\pi R_{min}^3}[/tex]

solving for R so we have

[tex]R_{min} = [\frac{FL}{\pi \sigma_f}]^{1/3}[/tex]

plugging all value to get minimum radius

[tex]R_{min} = [\frac{750 \times 0.05}{\pi 105 \times 10^6}]^{1/3}[/tex]

[tex]R_{min} = 4.84\times 10^{-3} m[/tex]

Answer:

[tex]\tau_a = 22.7 ksi[/tex]

Explanation:

Given data:

[tex]\alpha  = 12.2 \times 10^{-6} [/tex]degree F

Diameter is d = 0.4375

[tex]L_1 = 19 inch[/tex]

[tex]A_1 = 0.50 in^2[/tex]

[tex]L_2 = 27 inch[/tex]

[tex]A_2 = 0.65 in^2[/tex]

force in titanium and bronze will be equal

from equilibrium condition we have

F_T = F_B = p

From the information given as bar is tightened from ends thus net deformation is assummed to be zero

so we have

[tex]\Delta_T +\Delta_B = \Delta  = 0[/tex]

[tex]\frac{PL}{AE}_T + \alpha \Delta T L = \frac{PL}{AE}_B + (\alpha \Delta T L)_B[/tex]

[tex]\frac{19\times P}{0.5\times 16500} + 19 \times 5.3\times 10^{-6} \times 40 = \frac{27P}{0.65 \times 15200} + 27\times 12.2\times 10^{-6} \times 40 = 0 [/tex]

solving for P we get

P = 341 kips

average shear at B

[tex]\tau_a = \frac{P}{A}[/tex]

[tex]\tau_a = \frac{341}{\frac{\pi}{4} \times 0.4375^2}[/tex]

[tex]\tau_a = 22.7 ksi[/tex]

Answer:

t = 25.10 sec

Explanation:

we know that Avrami equation

[tex]Y = 1 - e^{-kt^n}[/tex]

here Y is percentage of completion  of reaction = 50%

t  is duration of reaction = 146 sec

so,

[tex]0.50 = 1 - e^{-k^146^2.1}[/tex]

[tex]0.50 = e^{-k306.6}[/tex]

taking natural log on both side

ln(0.5) = -k(306.6)

[tex]k = 2.26\times 10^{-3}[/tex]

for 86 % completion

[tex]0.86 = 1 - e^{-2.26\times 10^{-3} \times t^{2.1}}[/tex]

[tex]e^{-2.26\times 10^{-3} \times t^{2.1}} = 0.14[/tex]

[tex]-2.26\times 10^{-3} \times t^{2.1} = ln(0.14)[/tex]

[tex]t^{2.1} = 869.96[/tex]

t = 25.10 sec

Answer:

The area at the base should be 2.91cm²

Explanation:

The separation of the molten metal may occur when the pressure of the flowing liquid falls below the atmospheric pressure and the air entrapped inside the mold tries to escape but rather gets rapped inside the casting (which is solidifying) causing the liquid to separate and the casting to be porous.

The velocity at the bottom of the sprue is written as V = √(2gh), where g is the acceleration due to gravity and h is the height of the sprue

V = √(2 x 9.81 x 100 x 15) = 171.6 cm² (Multiplying by 100 to convert 9.81 m/s to cm/s)

The expression for flow rate is Q = VA where V is the velocity as calculated above and A is the area

500 = 171.6 x A, solving for A we get A = 500/171.6 = 2.91cm²

The cross - sectional Area at the base of the sprue should be 2.91cm² to avoid separation of the molten metal.

Answer:

a. Volume = 13.36 x 10^-3 m³ Pressure = 29.17 bar  b. Volume = 14.06 x 10^-3 m³ Pressure = 22.5 bar

Explanation:

Mass of O₂ = 1kg, Pressure (P1) = 35bar, T1= 180K, T2= 150k Molecular weight of O₂ = 32kg/Kmol

Volume of tank and final pressure using a)Ideal Gas Equation and b) Redlich - Kwong Equation

a. PV=mRT

V = {1 x (8314/32) x 180}/(35 x 10⁵) = 13.36 x 10^-3

Since it is a rigid tank the volume of the tank must remain constant and hnece we can say

T2/T1 = P2/P1, solving for P2

P2 = (150/180) x 35 = 29.17bar

b. P1 = {RT1/(v1-b)} - {a/v1(v1+b)(√T1)}

where R, a and b are constants with the values of, R = 0.08314bar.m³/kmol.K, a = 17.22(m³/kmol)√k, b = 0.02197m³/kmol

solving for v1

35 = {(0.08314 x 180)/(v1 - 0.02197)} - {17.22/(v1)(v1 + 0.02197)(√180)}

35 = {14.96542/(v1-0.02197)} - {1.2835/v1(v1 + 0.02197)}

Using Trial method to find v1

for v1 = 0.5

Right hand side becomes =  {14.96542/(0.5-0.02197)} - {1.2835/0.5(0.5 + 0.02197)} = 31.30 ≠ Left hand side

for v1 = 0.4

Right hand side becomes =  {14.96542/(0.4-0.02197)} - {1.2835/0.4(0.4 + 0.02197)} = 39.58 ≠ Left hand side

for v1 = 0.45

Right hand side becomes =  {14.96542/(0.45-0.02197)} - {1.2835/0.45(0.45 + 0.02197)} = 34.96 ≅ 35

Specific Volume = 35 m³/kmol

V = m x Vspecific/M = (1 x 0.45)/32 = 14.06 x 10^-3 m³

For Pressure P2, we know that v2= v1

P2 = {RT2/(v2-b)} - {a/v2(v2+b)(√T2)} = {(0.08314 x 150)/(0.45 - 0.02197)} - {17.22/(0.45)(0.45 + 0.02197)(√150)} = 22.5 bar

Answer:

speed of water flow in the test is 27 m/s

Explanation:

given data

cross section radius r1 = 1.8 m

radius constricts r2 = 0.60 m

speed of water flow = 3.0 m/s

to find out

speed of water flow in the test section

solution

we using the continuity equation here

A1 × V1 = A2 × V2     ...............1

put here value we get speed

[tex](\frac{\pi }{4} d1^2) v1 = (\frac{\pi }{4} d2^2) v2[/tex]

1.8²×3 = 0.6² × v2

v2 = 27 m/s

speed of water flow in the test is 27 m/s

Answer: No, not in favor of the idea.

This is because, an increase in the refrigerator's work output is equal to work input. In real live application, the work input of the refrigerator is always greater than the additional work output. This will result to reducing the efficiency of the power plant.

Refrigerating cooling water before the condenser in a power plant is impractical due to high energy costs, added complexity, and inefficiencies, outweighing the potential efficiency gains.

While the idea of refrigerating the cooling water before it enters the condenser in a power plant to increase efficiency might seem theoretically sound, in practice it is not advisable due to several reasons:

1.Energy Cost of Refrigeration: Refrigeration itself is an energy-intensive process. The energy required to cool the water to a significantly lower temperature would likely be greater than the energy savings achieved by the increased efficiency of the heat engine. This means that the net energy gain would be negative.

2.Second Law of Thermodynamics: According to the second law of thermodynamics, any energy conversion process, including refrigeration, involves irreversibilities and losses. The additional step of refrigeration would introduce more inefficiencies and would not be 100% efficient. Thus, the overall efficiency of the system would decrease.

3.Complexity and Cost: Adding a refrigeration system to cool the condenser water would significantly increase the complexity and cost of the power plant. This includes the initial capital cost, operational costs, and maintenance costs. The benefits gained from a marginal increase in thermal efficiency would likely not justify these additional expenses.

4. Practical Alternatives: There are more practical and cost-effective ways to improve the efficiency of a power plant. These include optimizing the thermodynamic cycle, improving the insulation, using better quality fuels, and upgrading to more efficient machinery and technology. Improving the heat exchange process itself, rather than cooling the water, is generally more practical.

5.Environmental Considerations: The energy used for refrigeration would typically come from the power plant itself or the grid, potentially leading to increased fuel consumption and higher greenhouse gas emissions. This is counterproductive to the goal of improving efficiency and sustainability.

In summary, while lowering the temperature at which heat is rejected can theoretically improve the thermal efficiency of a heat engine, refrigerating the cooling water is not a practical or economical solution due to the high energy cost, increased complexity, and overall inefficiency of the refrigeration process itself. Instead, other methods to enhance efficiency should be explored.

The question asks for writing a function named 'payOut' that calculates the player's winnings based on the outcomes of rolling a die 10 times. A Python function is provided as a solution, which counts the occurrences of numbers 5 and 6 in the input array and determines the payout according to the specified rules.

The question involves writing a function called payOut that calculates the winnings based on the outcomes of rolling a die 10 times, with specific payouts for different totals of rolling a 5 or a 6. The function takes as input an array of 10 integers, which represent the results of the die rolls, and outputs the payout based on the criteria provided.

To address this, we create a function in Python since it's a commonly used programming language for such tasks. The function will iterate through the input array, count the occurrences of the numbers 5 and 6, and then determine the payout based on the specified rules. Here's a basic implementation of this logic:

This function works by first counting how many times a 5 or a 6 appears in the input array using a generator expression inside the sum() function. It then uses conditional statements to determine the payout based on the count, according to the game's rules.

Answer:

generalizing

Explanation:

We all have a generalization system that operates as an autopilot, allowing us to be fast and consistent with our own identity. Thanks to these we package and label all the information with which we are bombed every second, to immediately think and act.

Otherwise, if we pay attention to each data individually, however tiny, every minute of our lives would become an exhausting and extremely slow process of analyzing and digesting, leaving us so overloaded to the point of collapse and not being able to function more mentally.

Answer:

0.024 m = 24.07 mm

Explanation:

1) Notation

[tex]\sigma_c[/tex] = tensile stress = 200 Mpa

[tex]K[/tex] = plane strain fracture toughness= 55 Mpa[tex]\sqrt{m}[/tex]

[tex]\lambda[/tex]= length of a surface crack (Variable of interest)

2) Definition and Formulas

The Tensile strength is the ability of a material to withstand a pulling force. It is customarily measured in units (F/A), like the pressure. Is an important concept in engineering, especially in the fields of materials and structural engineering.

By definition we have the following formula for the tensile stress:

[tex]\sigma_c=\frac{K}{Y\sqrt{\pi\lambda}} [/tex]   (1)

We are interested on the minimum length of a surface that will lead to a fracture, so we need to solve for [tex]\lambda[/tex]

Multiplying both sides of equation (1) by [tex]Y\sqrt{\pi\lambda}[/tex]

[tex]\sigma_c Y\sqrt{\pi\lambda}=K[/tex]   (2)

Sequaring both sides of equation (2):

[tex](\sigma_c Y\sqrt{\pi\lambda})^2=(K)^2[/tex]  

[tex]\sigma^2_c Y^2 \pi\lambda=K^2[/tex]   (3)

Dividing both sides by [tex]\sigma^2_c Y^2 \pi[/tex] we got:

[tex]\lambda=\frac{1}{\pi}[\frac{K}{Y\sigma_c}]^2[/tex]   (4)

Replacing the values into equation (4) we got:

[tex]\lambda=\frac{1}{\pi}[\frac{55 Mpa\sqrt{m}}{1.0(200Mpa)}]^2 =0.02407m[/tex]

3) Final solution

So the minimum length of a surface crack that will lead to fracture, would be 24.07 mm or more.

Answer:

Visual Basic program

Sub cosApprox()

' Variable declaration

Dim x, seriesSum, exactValue As Double

' Variable for holding result at each iteration

seriesSum = 0

' Initializing x value

x = CDbl(InputBox("Enter x value: ", "Cos Approx", "0"))

n = CInt(InputBox("Enter n value: ", "Cos Approx", "0"))

' Iterating for n terms

For term = 0 To n

' Calculating current value and accumulating total result

seriesSum = seriesSum + (((-1) ^ term) ((x ^ (term 2)) / (WorksheetFunction.Fact((term * 2)))))

' Incrementing loop variable

Next term

' Storing exact value

exactValue = Cos(x)

' Displaying result

MsgBox ("Approximation Value: cos(" & x & "): " & seriesSum & vbNewLine & vbNewLine & "Exact Value: cos(" & x & "): " & exactValue)

End Sub

' Factorial function

Public Function factorial(ByVal n As Integer) As Long

' Base Case

If ((n = 0) Or (n = 1)) Then

factorial = 1

Else

' Recursive case

factorial = factorial(n - 1) * n

End If

End Function

Answer:

The area or sproket is A 0.8 m

Explanation:

Answer:

Time of submersion in years = 7.71 years

Explanation:

Area of plate (A)= 16in²

Mass corroded away = Weight Loss (W) = 3.2 kg = 3.2 x 106

Corrosion Penetration Rate (CPR) = 200mpy

Density of steel (D) = 7.9g/cm³

Constant = 534

The expression for the corrosion penetration rate is

Corrosion Penetration Rate = Constant x Total Weight Loss/Time taken for Weight Loss x Exposed Surface Area x Density of the Metal

Re- arrange the equation for time taken

T = k x W/ A x CPR x D

T = (534 x 3.2 x 106)/(16 x 7.9 x 200)

T = 67594.93 hours

Convert hours into years by

T = 67594.93 x (1year/365 days x 24 hours x 1 day)

T = 7.71 years

Answer:

def extract_word_with_given_letter(sentence, letter):

   words = sentence.split()

   for word in words:

       if letter in word.lower():

           return word

   return ""

# Testing the function here. ignore/remove the code below if not required

print(extract_word_with_given_letter('hello HOW are you?', 'w'))

print(extract_word_with_given_letter('hello how are you?', 'w'))

Answer:

% od sidelap is [tex]P_w = 0.251\ or\ 25%[/tex]

Explanation:

Given data:

flying height is 7000 ft

focal dimension is 9× 9

focal length f = 88.90 mm

side width is x = 13,500 ft

we know side width is given as

[tex]x = (1- P_w) \frac{w}{s}[/tex]

where s is scale and given as [tex]s = \frac{f}{h}[/tex]

[tex]13,500 = (1- P_w) \times \frac{\frac{9}{12}}{\frac{0.29}{7000}}[/tex]

[tex](1- P_w) = 0.748[/tex]

[tex]P_w = 0.251\ or\ 25%[/tex]

The power required to drive the heat pump is 2 MW. For the same COP, a Carnot heat pump would achieve a high temperature of 324°C, starting from a low temperature of 85°C. The calculations involve both the COP formula and the Carnot efficiency formula.

To determine the power required to drive a heat pump with a given Coefficient of Performance (COP), we first use the formula:

COP = QH / W

Where QH is the heat delivered and W is the work input. Given that the heat delivered QH is 5 MW and the COP is 2.5, we can solve for the work W as follows:

[tex]2.5 = 5 MW / W[/tex]

Solving for W gives:

[tex]W = 5 MW / 2.5 = 2 MW[/tex]

To find the maximum temperature that a Carnot heat pump would achieve with the same low temperature of 85°C and the same COP, we use the Carnot heat pump efficiency formula:

[tex]COP_{Carnot[/tex] [tex]= TH / (TH - TC)[/tex]

We know that:

[tex]2.5 = TH / (TH - 358 K)[/tex]

Rewriting it:

[tex]2.5(TH - 358) = TH[/tex]

Solving for TH, we get:

[tex]2.5TH - 895 = TH[/tex]

[tex]1.5TH = 895[/tex]

[tex]TH = 597 K[/tex]

Converting 597 K to Celsius, [tex]TH = 597 - 273 =[/tex] 324°C

Thus, the required power to drive the unit is 2 MW, and for the same COP, a Carnot heat pump would have a high temperature of 324°C, assuming the same low temperature of 85°C.

Answer:

a. Please see attachment .

b. 612.377

c. 24.22%

Explanation:

Please see attachment.