You are asked to design a small wind turbine (D-x+1.25 ft, where x is the last two digits of your student ID). Assume the wind speed is 15 mph at T-10°C and p-0.9 bar. The efficiency of the turbine is n-25%, meaning that 25% of the kinetic energy in the wind can be extracted. Calculate the power in watts that can be produced by your turbine. Scan the solution of the problem and upload in the VUWS before closing the vUWS or moving to other question.
x = 85

The flow field is described in cylindrical coordinates by the stream function y = AO+Br sin o where A and B are positive constants and the corresponding velocity potential is calculated as follows:As per the continuity equation,The velocity potential is given by the following equation:

Where vr is the radial velocity and vo is the tangential velocity. The velocity vector is then given by the gradient of the velocity potential. Thus, The angle θ is given by This equation shows that the velocity vector makes an angle of π/2 with the x-axis when r = B/A, that is, at the surface of the cylinder. Stagnation points occur where the velocity vector is zero,

which is the case for vr = vo = 0. Thus, Setting each factor to zero, we obtain the following equations: The equation A = 0 is not a physical solution since it corresponds to zero velocity, thus, the stagnation point occurs at (r,θ) = (B,π/2).

The magnitude of the velocity vector is 2.236 m/s, and the angle it makes with the x-axis is 63.4°. Stagnation points occur where the velocity vector is zero, which is the case for Vx = Vy = 0. Since Vx = -4x, the stagnation point occurs at x = 0.

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A 0.5 m long vertical plate is at a temperature of 70°C. The air around it is at 30°C and 1 atm. At 10 m/s, the air comes into the plate from the blower.

The answers to the given queries are as follows:

1) Grashof Number of Flow Grashof Number is calculated using the following formula:

Gr = (gβΔTl³) / (ν²) Here, g is acceleration due to gravity, β is coefficient of thermal expansion, ΔT is temperature difference between the two surfaces, l is the length of the plate, and ν is the kinematic viscosity of the fluid.The values of the constants can be found in the following way:g = 9.81 m/s²β = 1/T where T is the average temperature between the two surfacesν = μ / ρ, where μ is dynamic viscosity, and ρ is density.

Now, we can use these formulas to find the values of the constants, and then use the Grashof Number equation to solve for Gr.Gr = 4.15 x 10^9

The Reynolds number is used to determine whether the flow is laminar or turbulent. It is defined as:

Re = (ρvl) / μ Here, ρ is the density of the fluid, v is the velocity of the fluid, l is the length of the plate, and μ is the dynamic viscosity of the fluid.

The value of the constants can be found in the following way:

ρ = 1.18 kg/m³

μ = 1.85 x 10^-5 Ns/m²

Re = 31,783

Since the value of Re is greater than 2300, the flow is turbulent.

3) The type of flow is mixed convection flow because it is influenced by both natural and forced convection.

4) The most accurate estimate for the average heat transfer coefficient can be found using the following equation:

Nu = (0.60 + 0.387(Gr Pr)^(1/6)) / (1 + (0.559 / Pr)^(9/16))

Here, Nu is the Nusselt number, Gr is the Grashof number, and Pr is the Prandtl number.

We already know the value of Gr, and we can find the value of Pr using the following formula:

Pr = ν / αwhere α is the thermal diffusivity of the fluid. α = k / (ρ cp), where k is the thermal conductivity of the fluid, and cp is the specific heat at constant pressure.

Now we can use these equations to find the value of Nu, which will help us solve for h, using the following formula:

Nu = h l / k

The value of h is found to be 88.8 W/m²K.5)

The rate of convection heat transfer from the plate is given by the following formula:

q = h A ΔTwhere A is the area of the plate, and ΔT is the temperature difference between the two surfaces.

Now, the width of the plate is 1m, so the area of the plate is 0.5 m x 1 m = 0.5 m².

Now, we can use the equation to find the value of q:

q = 88.8 x 0.5 x (70-30)q = 2220 W6)

The thickness of the thermal boundary at the top of the plate can be found using the following equation:

δ = 5 x ((x / l) + 0.015(Re x / l)^(4/5))^(1/6)

Here, δ is the thermal boundary layer thickness, l is the length of the plate, and x is the distance from the leading edge of the plate.

The value of Re x / l can be found using the following formula:

Re x / l = (ρ v x) / μ

Now, we can use these equations to find the value of δ, when x = 0.5 m.

In conclusion, the Grashof number is 4.15 x 10^9, and the flow is turbulent because the Reynolds number is 31,783. The type of flow is mixed convection flow because it is influenced by both natural and forced convection. The most accurate estimate for the average heat transfer coefficient is 88.8 W/m²K. The rate of convection heat transfer from the plate is 2220 W. Finally, the thickness of the thermal boundary at the top of the plate is 0.0063 m.

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(a) The acceleration of block A is 5.2 m/s².

(b) The speed of block A after it has moved 5 ft is approximately 3.82 m/s.

To determine the acceleration of block A, we need to consider the forces acting on the system. The weight of the flywheel exerts a downward force, which is balanced by the tension in the wire attached to block A. The tension in the wire causes block A to accelerate.

The equation for the net force acting on block A is given by:

Net Force = Mass * Acceleration

We can calculate the mass of block A using its weight:

Weight = Mass * Gravity

Rearranging the equation, we have:

Mass = Weight / Gravity

Substituting the given values, we find the mass of block A to be approximately 112.9 kg.

Now, using the equation for torque, we can determine the tension in the wire:

Torque = Moment of Inertia * Angular Acceleration

The moment of inertia of the flywheel is given by:

Moment of Inertia = Mass of Flywheel * Radius of Gyration²

Substituting the given values, we find the moment of inertia to be approximately 60.94 kg·m².

Rearranging the torque equation, we have:

Angular Acceleration = Torque / Moment of Inertia

Since the torque is equal to the tension in the wire multiplied by the radius of the flywheel, we can write:

Torque = Tension * Radius of Flywheel

Substituting the given values, we have:

Tension * 0.5m = Tension * 0.375m = 60.94 kg·m² * Angular Acceleration

Simplifying the equation, we find:

Tension = 162.51 N

Finally, we can calculate the acceleration of block A:

Net Force = Tension - Weight of A

Mass of A * Acceleration = Tension - Weight of A

Substituting the given values, we have:

112.9 kg * Acceleration = 162.51 N - 1110 N

Solving for acceleration, we find:

Acceleration = (162.51 N - 1110 N) / 112.9 kg ≈ 5.2 m/s²

To determine the speed of block A after it has moved 5 ft, we need to convert the distance to meters:

5 ft = 5 ft * 0.3048 m/ft ≈ 1.524 m

We can use the equation of motion to find the final speed of block A:

Final Speed² = Initial Speed² + 2 * Acceleration * Distance

Assuming the system starts from rest, the initial speed is zero. Substituting the given values, we have:

Final Speed² = 2 * 5.2 m/s² * 1.524 m

Simplifying the equation, we find:

Final Speed ≈ √(15.96 m²/s²) ≈ 3.82 m/s

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(a) The force required to produce an elongation in a rod of a given length
[tex]F = A (σ − σY)[/tex]where F is the force required to produce an elongation in the rod, A is the cross-sectional area of the rod, σ is the final stress in the rod, and σY is the yield stress of the material.

For a given elongation, the final stress in the rod is given by the following equation;[tex]σ = Eε[/tex]
[tex]Δd/d = Δl/[/tex]
[tex]Δl = (π/4) (d² - d'²)[/tex] lf
[tex]ε = (Δd/d) / (1 - (Δd/d'))[/tex]
[tex]σ = E (Δd/d) / (1 - (Δd/d'))[/tex]
[tex]F = A (E (Δd/d) / (1 - (Δd/d')) - σY[/tex])
[tex]Δd = 1 mm, d' = 4 mm, A = π (5 mm)² / 4, E = 8005 ksi = 55.2 GPa,[/tex]
[tex]F = (π/4) (5 mm)² (55.2 GPa) (1 mm / (1 - (1 mm / 4 mm))) - σYF = 1537 kN - σY[/tex]

(b) The yield strength of the produced rod can be calculated from the force required to produce the given elongation and the original cross-sectional area of the rod using the following equation
[tex]σY = F / A[/tex]
[tex]σY = 1537 kN / [(π/4) (5 mm)²][/tex]
[tex]σY = 39.1 MPa[/tex]

The force needed to perform this process is 1537 kN and the yield strength of the produced rods is 39.1 MPa.

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The production order quantity (POQ) is 1830 units.

Given information:

Annual demand, D = 37,085 units

Holding cost, H = $22.8 per unit per year

Ordering cost, S = $4,591

Replenishment rate, R = 705 units/day

Working days per year, W = 365 days/year

To calculate the production order quantity (POQ), use the following formula:

POQ = √((2DS)/H(1-D/RW))

Put the given values in the above formula:

POQ = √((2 × 37,085 × 4,591)/(22.8 × (1 - 37,085/705 × 365)))

POQ = √(3,346,733.34)

POQ = 1829.80

≈ 1830

Therefore, the production order quantity (POQ) is 1830 units.

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Given data:Width of tapered segment (w1) at the bottom end = 2 inWidth of tapered segment (w2) at the top end = 5 inThickness of the bar (t) = 0.50 in Length of the bar (L) = 5 ftLoad applied (P) = 30 kips = 30,000 lbYoung's modulus of steel (E) = 30,000 ksi = 30,000,000 psi

Area of uniform-width segment = A1 = w1 * t = 2 * 0.5 = 1 in²Area of tapered segment at the bottom end = A2 = w1 * t = 2 * 0.5 = 1 in²

Area of tapered segment at the top end = A3 = w2 * t = 5 * 0.5 = 2.5 in²

Area of the bar = A = A1 + A2 + A3 = 1 + 1 + 2.5 = 4.5 in²

Stress produced by the load applied,P/A = 30000/4.5 = 6666.67 psi

Deflection of the uniform-width segment = [tex]Δ1 = PL1/(AE) = 30000*12*60/(1*30,000,000*1) = 0.24[/tex] in

Deflection of the tapered segment = Δ2 = PL2/(AE) ... (1)Here, [tex]L2 = L - L1 = 60 - 12 = 48[/tex] in,

since the tapered segment starts at 12 in from the bottom end and extends up to the top end.

Plug in the values,[tex]Δ2 = (30,000 x 48 x 0.50²) / (30,000,000 x (5/2) x (2² + 2(2.5)²)) = 0.37[/tex]

inTotal deflection of the bar,[tex]Δ = Δ1 + Δ2 = 0.24 + 0.37 = 0.61[/tex]in

The elongation of the bar = [tex]Δ x L = 0.61 x 12 = 7.32[/tex] The elongation of the bar resulting from the application of the 30 kip load is 7.32 in.

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By using Euler's method of solving O.D.E., with the step size of h = 1.5, an approximate value of \int_5^8 6x^3 dx can be found.

Euler's method is given as:by_{i+1} = y_i +hf(x_i,y_i)Let us consider the integral, \int_{5}^{8}6x^3dxHere,a=5, b=8, h=1.5$and ]f(x,y)=6x^3]. x_0 = We can find y_1 by using the formula of Euler's method, y_{i+1} = y_i +hf(x_i,y_i)where i=0.So,y_1 = y_0 + hf(x_0,y_0)Substitute x_0=5 and y_0=0, we get,y_1 = 0 + 1.5*6*5^3 = 2250Next, find y_2,y_2 = y_1 + hf(x_1,y_1)where$x_1 = 5+1.5 = 6.5. Substituting the values, we get,y_2 = 2250 + 1.5*6*6.5^3 = 7031.25Similarly,y_3 = y_2 + hf(x_2,y_2)\implies y_3 = 7031.25 + 1.5*6*8^3 = 149560.5Now, we can approximate the integral using the formula of the definite integral,\int_a^b f(x)dx = [F(b)-F(a)]\implies \int_{5}^{8}6x^3dx = \left[ \frac{1}{4}x^4\right]_{5}^{8} \implies \int_{5}^{8}6x^3dx \ approx 3179$$Therefore, the approximate value of \int_{5}^{8}6x^3dx$using Euler's method of solving O.D.E. with a step size of h = 1.5 is 3179.

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(i). Sketch the free workspace and trapezoidate it (using the sweeping trapezoidation algorithm):The sketch of the free workspace and the trapezoidal partition using the sweeping trapezoidal algorithm are as follows: Fig. 2: Problem 3(ii). Sketch the dual graph for the trapezoidal partition and the roadmap:

The dual graph for the trapezoidal partition and the roadmap can be shown as follows: Fig. 2: Problem 3(iii). Sketch a path from start point to goal point in the dual graph and an associated path in the workspace that a robot can follow.A path from the start point to the goal point in the dual graph is shown below. The solid lines indicate the chosen path from the start to the goal node in the dual graph. The associated path in the workspace is indicated by the dashed line. Fig. 2: Problem 3

To summarize, the given problem is related to Partitions and roadmaps, and the solution of the problem is given in three parts. In the first part, we sketched the free workspace and trapezoidated it using the sweeping trapezoidal algorithm. In the second part, we sketched the dual graph for the trapezoidal partition and the roadmap. Finally, we sketched a path from the start point to the goal point in the dual graph and an associated path in the workspace that a robot can follow.

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A hot upset-forging operation is carried out in an open die at a high temperature of 1200°C. As a result, the material behavior can be assumed to be perfectly plastic. The work metal at this temperature yields at 90 MPa. The original cylindrical work part has a diameter of 30 mm and a height of 50 mm.

The part is upset to a final diameter of 45 mm, and the coefficient of friction at the die-work interface is 0.40, while the impact of barreling can be overlooked. The ultimate height of the part and the true strain and corresponding flow stress at the end of the stroke are the parameters to be determined.a) Calculation of final height:

Initial volume of work metal, V1=π(30/2)²×50=70685 mm³

Final volume of work metal, V2=π(45/2)²×h = (π(22.5)²)h

Final height of work metal, h= V1/ (π(22.5)²)= 25.68 mm

Therefore, the final height of the work part is 25.68 mm.

b) Calculation of True strain and the corresponding flow stress at the end of the stroke:True strain, εt= ln (h1/h2)=ln (50/25.68)=0.6524Flow stress, σf = Kεtⁿσf = 90 MPa, εt = 0.6524MPa = K(0.6524)^n90 = K(0.6524)^n...equation (i)Now, σf = Kεtⁿσf = 90 MPa, εt = 0.6524, n = ln σ2/ ln σ1σ2/σ1 = (ln (σ2) − ln(90))/(ln(90)− ln(90)) = 1.1σ2 = σ1(e^n)σ2 = 90e^1.1 = 249.21 MPaTherefore, the true strain at the end of the stroke is 0.6524, while the corresponding flow stress is 249.21 MPa.

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A -  v = 24.08 km/s  To determine the arrival excess velocity when reaching Mars' sphere of influence following a Hohmann transfer trajectory, we can use the vis-viva equation  v^2 = GM*(2/r - 1/a)

where v is the velocity, G is the gravitational constant, M is the mass of Mars, r is the distance from Mars' center, and a is the semi-major axis of the spacecraft's transfer orbit.

For a Hohmann transfer, the semi-major axis of the transfer orbit is the sum of the radii of the departure and arrival orbits. The departure orbit is the Earth's orbit and the arrival orbit is the Mars' orbit.

Let's assume the radius of Earth's orbit is 1 AU (149.6 million km) and the radius of Mars' orbit is 1.52 AU (227.9 million km). We can calculate the semi-major axis of the transfer orbit:

a = (149.6 + 227.9) / 2 = 188.75 million km

Next, we can calculate the velocity at Mars' orbit:

v = sqrt(GM*(2/r - 1/a))

v = sqrt(6.674e-11 * 6.39e23 * (2/(227.9e6 * 1000) - 1/(188.75e6 * 1000)))

v = 24.08 km/s

To calculate the arrival excess velocity, we subtract the velocity of Mars in its orbit around the Sun (24.08 km/s) from the velocity of the spacecraft:

Arrival excess velocity = v - 24.08 km/s

Arrival excess velocity = 0 km/s

Therefore, the arrival excess velocity is 0 km/s.

B - Since the arrival excess velocity is 0 km/s, the spacecraft is on a parabolic trajectory when entering Mars' sphere of influence with a periapsis altitude of 400 km.

C - To circularize the orbit, we need to change the velocity of the spacecraft at periapsis to match the orbital velocity required for a circular orbit at the given altitude. The delta-v required to circularize the orbit can be calculated using the vis-viva equation:

v_circular = sqrt(GM/r)

where v_circular is the circular orbital velocity, G is the gravitational constant, M is the mass of Mars, and r is the periapsis altitude.

Let's assume the periapsis altitude is 400 km (400,000 meters). We can calculate the delta-v required to circularize the orbit:

Delta-v = v_circular - v_periapsis

Delta-v = sqrt(GM/r) - v_periapsis

Using the known values:

Delta-v = sqrt(6.674e-11 * 6.39e23 / (3389e3 + 400e3)) - v_periapsis

Delta-v = 2.65 km/s - v_periapsis

The magnitude of the delta-v is given in absolute value, so the answer is:

Delta-v = |2.65 km/s - v_periapsis|

D - The delta-v required to circularize the orbit should be oriented tangentially to the spacecraft's orbit at periapsis. This means the delta-v vector should be perpendicular to the radius vector at periapsis.

E - If the spacecraft circularized the orbit before entering Mars' sphere of influence, it would be on a circular orbit around Mars with a radius equal to the periapsis altitude (400 km).

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To determine the volume of the tank, temperature at the final state, and the heat transfer, we need to consider the principles of thermodynamics and the properties of water.

First, let's calculate the mass of water in the tank. Given that there are 0.8 lb of saturated vapor and 6.0 lb of saturated liquid, the total mass of water in the tank is:

Mass of water = Mass of vapor + Mass of liquid

= 0.8 lb + 6.0 lb

= 6.8 lb

Next, we need to determine the specific volume of water at the initial state. The specific volume of saturated liquid water at 212°F is approximately 0.01605 ft³/lb. Assuming the water in the tank is incompressible, we can approximate the specific volume of the water in the tank as:

Specific volume of water = Volume of tank / Mass of water

Rearranging the equation, we have:

Volume of tank = Specific volume of water x Mass of water

Plugging in the values, we get:

Volume of tank = 0.01605 ft³/lb x 6.8 lb

= 0.10926 ft³

So, the volume of the tank is approximately 0.10926 ft³.

Since the tank is closed and rigid, the specific volume remains constant during the heating process. Therefore, the specific volume of the water at the final state is still 0.01605 ft³/lb.

To find the temperature at the final state, we can use the steam tables or properties of water. The saturation temperature corresponding to saturated vapor at atmospheric pressure (since the tank is closed) is approximately 212°F. Thus, the temperature at the final state is 212°F.

Lastly, to determine the heat transfer, we can use the principle of conservation of energy:

Heat transfer = Change in internal energy of water

Since the system is closed and there are no changes in kinetic or potential energy, the heat transfer will be equal to the change in enthalpy:

Heat transfer = Mass of water x Specific heat capacity x Change in temperature

The specific heat capacity of water is approximately 1 Btu/lb·°F. The change in temperature is the final temperature (212°F) minus the initial temperature (212°F).

Plugging in the values, we get:

Heat transfer = 6.8 lb x 1 Btu/lb·°F x (212°F - 212°F)

= 0 Btu

Therefore, the heat transfer in this process is 0 Btu.

In summary, the volume of the tank is approximately 0.10926 ft³, the temperature at the final state is 212°F, and the heat transfer is 0 Btu.

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1. A field analogue outcrop refers to a rock exposure in the field that resembles a subsurface petroleum reservoir. They are used as geological models for studying subsurface reservoirs, and they are known to be an important tool for petroleum engineers in training.

2. Field analogues are useful for a petroleum engineer in various ways. One of the benefits is that they enable petroleum engineers to determine reservoir properties such as porosity, permeability, and capillary pressure. The rock formations exposed on the surface are analogous to the subsurface reservoirs, and the data obtained from the field analogues can be extrapolated to subsurface reservoirs to make predictions of the petroleum production.

The information obtained from the field analogues allows engineers to make important decisions on the drilling and completion of a well. They can also help in determining which reservoir model is most appropriate. Finally, they are also useful in verifying subsurface data acquired from well logs.

A field analogue outcrop is a rock exposure in the field that mimics a subsurface petroleum reservoir. They are useful as geological models for studying subsurface reservoirs, and they are an essential tool for petroleum engineers in training. Field analogues enable petroleum engineers to determine reservoir properties such as porosity, permeability, and capillary pressure.

The rock formations exposed on the surface are analogous to the subsurface reservoirs, and the data obtained from the field analogues can be extrapolated to subsurface reservoirs to make predictions of the petroleum production.Field analogues are helpful in many ways to a petroleum engineer. One of the benefits is that they allow engineers to make important decisions on the drilling and completion of a well. They can also help in determining which reservoir model is most appropriate. Finally, they are also useful in verifying subsurface data acquired from well logs.

Field analogue outcrops refer to rock exposures in the field that mimic a subsurface petroleum reservoir. The geological models obtained from field analogues are beneficial for petroleum engineers in training. Field analogues are valuable tools in determining reservoir properties such as porosity, permeability, and capillary pressure.

Field analogues are essential for petroleum engineers, and they offer many benefits. For instance, field analogues allow engineers to make important decisions on the drilling and completion of a well. Petroleum engineers can determine which reservoir model is most suitable based on the data obtained from field analogues.Field analogues are also useful in verifying subsurface data acquired from well logs.

The data from field analogues is similar to the subsurface reservoirs, and it can be extrapolated to make predictions about petroleum production

Field analogue outcrops are crucial geological models for studying subsurface petroleum reservoirs. Petroleum engineers use field analogues to determine reservoir properties such as porosity, permeability, and capillary pressure. Field analogues are beneficial for petroleum engineers, as they allow them to make informed decisions on the drilling and completion of a well. Furthermore, they assist in verifying subsurface data obtained from well logs.

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The design margin of this link is -23.5 dB.

The formula for optical power is given as follows: Optical Power

= P0 × 10 ^ (- L/10)

where, P0 is the power transmitted, L is the total losses and optical power is the power received. The given total loss is 0.25 dB/km, so the total loss for 120 km is calculated as follows:

L = 0.25 dB/km × 120 km

= 30 dB

The total loss includes 2 dB of loss from splices and connector and it includes a dispersion compensator with a loss of 3 dB.

Therefore, the fiber loss is 30 dB + 2 dB + 3 dB

= 35 dB.

The receiver's sensitivity is -33 dBm,

so the power received at the end of the link is:

Optical Power = P0 × 10 ^ (- L/10)

Optical Power = +5 dBm × 10 ^ (-35/10)

Optical Power = 2.238 × 10 ^ -4 mW or -56.48 dBm

The design margin of the link is the difference between the power received at the end of the link and the minimum sensitivity of the receiver. Design Margin = Optical Power - Receiver Sensitivity Design Margin

= (-56.48 dBm) - (-33 dBm)

Design Margin = -23.48 dB

The design margin of this link is -23.5 dB.

The design margin of this link is -23.5 dB.

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In assembly language programming, the XOR instruction (exclusive OR) is used to take the exclusive OR of the two operands, with the resulting value being written to the first operand (also known as the destination operand).

The XOR R1,R2 instruction, for example, will XOR the value of R2 with the value of R1, resulting in the result being saved to R1.The control word for the instruction XOR R1,R2 is CW=45B3. The control word is a set of bits that govern the behavior of the processor.

Each operation has its control word, which specifies the type of operation, the number of operands, and the size of the operands (in terms of bits).The control word 45B3 is the hexadecimal representation of the 16-bit binary number 0100 0101 1011 0011. The most significant nibble (4 bits) specifies the type of operation (in this case, XOR).

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a) For horizontal flow of a liquid in a rectangular duct between parallel plates, the pressure varies linearly both in the direction of flow and in the direction normal to the plates. This is a true statement.

b) For flows occurring between r= 0 and r= a in cylindrical coordinates, the term In r may appear in the final expression for one of the velocity components. This statement is also true.

c) For flows in ducts and pipes, the volumetric flow rate can be obtained by differentiating the velocity profile. This is a true statement

d) If, in three dimensions, the pressure obeys the equation Op/ dy = -pg, and both Op/ Ox and op/ öz are nonzero, then integration of this equation gives the pressure as p = -ogy+c, where c is a constant. This statement is true.

a) For horizontal flow of a liquid in a rectangular duct between parallel plates, the pressure varies linearly both in the direction of flow and in the direction normal to the plates. This is a true statement. For horizontal flow of a liquid in a rectangular duct between parallel plates, the pressure varies linearly both in the direction of flow and in the direction normal to the plates.

b) For flows occurring between r= 0 and r= a in cylindrical coordinates, the term In r may appear in the final expression for one of the velocity components. This statement is also true. In r may appear in the final expression for one of the velocity components in flows occurring between r= 0 and r= a in cylindrical coordinates.

c) For flows in ducts and pipes, the volumetric flow rate can be obtained by differentiating the velocity profile. This is a true statement as well. For flows in ducts and pipes, the volumetric flow rate can be obtained by differentiating the velocity profile.

d) If, in three dimensions, the pressure obeys the equation

Op/ dy = -pg,

and both Op/ Ox and op/ öz are nonzero, then integration of this equation gives the pressure as

p = -ogy+c,

where c is a constant. This statement is true. If, in three dimensions, the pressure obeys the equation

Op/ dy = -pg,

and both Op/ Ox and op/ öz are nonzero, then integration of this equation gives the pressure as

p = -ogy+c,

where c is a constant.

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The power of the induction motor is 3.03 kW.

The formula for calculating the power is given below: Power = (3 * V * I * cosφ) / (10^3)

where 3 is the square root of 3, V is the line-to-line voltage, I is the current, cosφ is the power factor, and 10^3 is used to convert the power to kW.

Let us find out the required parameters. We are given the rated voltage V = 440 V and rated frequency f = 60 Hz.The synchronous speed of the motor is given by the formula: Ns = (120 * f) / P

where P is the number of poles of the motor. The value of Ns is: Ns = (120 * 60) / 4 = 1800 rpm The slip of the motor is given by the formula: s = (Ns - n) / Ns

where n is the actual speed of the motor in rpm. The value of s is: s = (1800 - 1746) / 1800 = 0.03

The rotor resistance is r = 2.256 ΩThe rotor reactance is given by the formula:X2 = (s * Xs) / R

where Xs is the synchronous reactance. The value of Xs is: Xs = 2 * π * f * Lm where Lm is the magnetizing inductance. The value of Xs is:

Xs = 2 * π * 60 * 572 / 1000 = 216 ΩThe value of X2 is: X2 = (0.03 * 216) / 2.256 = 2.88 Ω

The equivalent rotor resistance is: R2' = r + X2 = 2.256 + 2.88 = 5.136 Ω The equivalent rotor leakage reactance is:

Xlr' = (s * Llr) / (Xs + R2')The value of Xlr' is: Xlr' = (0.03 * 32) / (216 + 5.136) = 0.45 Ω

The value of X1 is: X1 = Xls + Xlr' The value of X1 is:

X1 = 32 + 0.45 = 32.45 ΩThe equivalent stator resistance is:R1' = rs + R2'The value of R1' is:R1'

= 1 + 5.136 = 6.136 ΩThe equivalent stator leakage reactance is:Xls' = X1 - Xlr'The value of Xls' is: Xls' = 32.45 - 0.45 = 32 ΩThe impedance of the motor is given by: Z = R1' + jXls'

The value of Z is: Z = 6.136 + j32 = 32.27 ∠81.39° ΩThe current drawn by the motor is given by:

I = V / Z The value of I is:I = 440 / 32.27 ∠81.39°

= 13.62 ∠-81.39° A

The power factor is given by:cosφ = R1' / ZThe value of cosφ is:cosφ = 6.136 / 32.27 = 0.1902The power can be calculated as follows:

Power = (3 * V * I * cosφ) / (10^3)

Substituting the given values, the value of power is:

Power = (3 * 440 * 13.62 * 0.1902) / 1000 = 3.03 kW

Therefore, the power of the induction motor is 3.03 kW.

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a) i) Fatigue loading Fatigue loading refers to the type of loading that develops due to cyclic stress conditions. Fatigue loading, unlike static loading, can occur when the same loading is repeatedly applied on a material that is already under stress.

This fatigue loading effect can result in a material experiencing different amounts of stress at different times during its lifespan, ultimately leading to failure if the stress levels exceed the endurance limit of the material. ii) Endurance limit. The endurance limit is defined as the maximum amount of stress that a material can endure before it starts to experience fatigue failure.

This means that if the material is subjected to stresses below its endurance limit, it can withstand an infinite number of stress cycles without undergoing fatigue failure. The fatigue strength of a material is typically determined by subjecting the material to a series of cyclic loading conditions at different stress levels.

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Given equation is e^x = 10(x^2 - 1).

By arranging the given equation, we get x = g(x).

Let us consider x1 as the negative root of the given equation.

First case, using x = ln(10(x² - 1)),

the iteration formula is given as

Xn + 1 = ln (10 (Xn^2 - 1))

The initial approximation is

x0 = -0.5

The iteration procedure is shown below in the table.

For n = 4, the value of Xn+1 = -1.48 is closer to the real root -1.49.

Case 2, x = (ln⁡(10x² - 1))/x iteration formula is given as Xn + 1 = (ln⁡(10Xn^2 - 1))/Xn

The initial approximation is x0 = 1.5

The iteration procedure is shown below in the table. For n = 4, the value of Xn+1 = 1.28 is closer to the real root 1.28.Case 3, x = √(ln⁡10(x² - 1)) / √10

iteration formula is given as Xn + 1 = √(ln⁡10(Xn^2 - 1))/√10

The initial approximation is x0 = 0.5

The iteration procedure is shown below in the table. For n = 4, the value of Xn+1 = 0.88 is closer to the real root 0.89.

Therefore, the three roots of the equation are x = -1.49, 1.28, and 0.89, respectively.

The sequences of approximation for each case are shown above.

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The estimated endurance strength for a cold-rolled 1040 steel 100 mm square bar is the stress level below which the metal can sustain indefinitely without experiencing fatigue failure. The operating temperature is 100 C and a reliability of 99% will be required, and the bar will be loaded axially. The estimated endurance strength for a cold-rolled 1040 steel 100 mm square bar is 279.3 MPa.

An endurance limit is given by a graph of stress amplitude against the number of cycles. If a specimen is subjected to cyclic loading below its endurance limit, it will withstand an infinite number of cycles without experiencing fatigue failure. The fatigue limit, sometimes known as the endurance limit, is the stress level below which the metal can endure an infinite number of stress cycles without failure.

According to the given terms, the estimated endurance strength for a cold-rolled 1040 steel 100 mm square bar can be calculated as follows:The endurance strength can be estimated using the equation:

Endurance strength= K × (ultimate tensile strength)^a

Where:K = Fatigue strength reduction factor (related to reliability)

α = Exponent in the S-N diagram

N = Number of cycles to failure

Therefore,

Endurance strength= K × (ultimate tensile strength)^a

Here, for the cold-rolled 1040 steel, the value of K and α will be determined based on the type of loading, surface condition, and other factors. For a rough estimate, we can assume that the value of K is 0.8 for reliability of 99%.Thus,

Endurance strength= K × (ultimate tensile strength)^a

= 0.8 × (590 MPa)^0.1

= 279.3 MPa

The estimated endurance strength for a cold-rolled 1040 steel 100 mm square bar is 279.3 MPa.

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There are 16 quantization levels available for the ADC and the quantization interval for this ADC is 2V.

To find the number of quantization levels and the quantization interval for a 4-bit analog-to-digital converter (ADC) with a maximum detection voltage of 32V, we need to consider the resolution of the ADC.

i) The number of quantization levels (N) can be determined using the formula:

N = 2^B

where B is the number of bits. In this case, B = 4, so the number of quantization levels is:

N = 2^4 = 16

ii) The quantization interval (Q) represents the difference between two adjacent quantization levels and can be calculated by dividing the maximum detection voltage by the number of quantization levels. In this case, the maximum detection voltage is 32V, and the number of quantization levels is 16:

Q = Maximum detection voltage / Number of quantization levels

= 32V / 16

= 2V

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The natural frequency (ω) of the second-order system is 5 rad/s, and the damping ratio (ζ) is 1.

To find the natural frequency (ω) and damping ratio (ζ), we need to analyze the transfer function of the second-order system.

Given the transfer function is G(s) = (2 + as + b) / s^2, where a = -8 and b = 25.

The natural frequency (ω) can be determined by finding the square root of the coefficient of the s^2 term. In this case, the coefficient is 1. Therefore, ω = √1 = 1 rad/s.

The damping ratio (ζ) can be calculated by dividing the coefficient of the s term (a) by twice the square root of the product of the coefficient of the s^2 term (1) and the constant term (b). In this case, ζ = -8 / (2 * √(1 * 25)) = -8 / (2 * 5) = -8 / 10 = -0.8.

Since the damping ratio (ζ) cannot be negative, we take the absolute value of -0.8, resulting in ζ = 0.8.

In summary, the natural frequency (ω) is 1 rad/s, and the damping ratio (ζ) is 0.8.

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The bare cost of providing and installing 12 sliding windows 5ft x 4ft including screen assuming vinyl clad, premium with double insulating glass in Miami, Florida is approximately $9,120.38.

The calculation of the bare cost of providing and installing 12 sliding windows 5ft x 4ft including screen, assuming vinyl clad, premium with double insulating glass in Miami, Florida can be done in the following way. The material cost is found by calculating the area of one window which is 5ft x 4ft = 20 sq.ft. For 12 windows, the area would be 240 sq.ft. The vinyl material cost is assumed to be $15 per sq.ft. So, the material cost would be 240 x 15 = $3,600.

The labor cost is calculated by taking 69.5% of the material cost. Then, the total cost is found by adding the material and labor costs. The total cost is equal to 90.7% of the sum of the material and labor costs.

Let X be the cost of materials. Therefore; Labor cost = 69.5/100 × X Total cost = 90.7/100 × (X + 69.5/100 × X) Total cost = 90.7/100 × (1 + 69.5/100) × X Total cost = 90.7/100 × 1.695 × X Total cost = 153.8125/100 × X

Using this formula, the bare cost of providing and installing 12 sliding windows 5ft x 4ft including screen assuming vinyl clad, premium with double insulating glass in Miami, Florida is approximately $9,120.38.

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Part (a):Let us first calculate the resolved shear stress (RSS) along a (110) plane. To do so, we must first draw the tensile stress vector () and the normal to the (110) plane () to which the stress vector is applied, and then rotate both vectors by 45° as shown below.

The direction of the shear stress () will be perpendicular to both the and vectors in this case, so we can use the cross-product rule. The following is the solution to the problem:

Resolving shear stress in a (110) planeRSS = ( sin )/sin = (52 sin 45° )/sin 35.3°= 45.85 MPa

Therefore, the resolved shear stress along the (110) plane is 45.85 MPa.

To calculate the resolved shear stress in a [111] direction, we must first draw the tensile stress vector () and the normal to the [111] direction () to which the stress vector is applied, as shown below. Since the tensile stress vector is already parallel to the [010] direction, which is perpendicular to the [111] direction, the normal vector () to which the tensile stress is applied must be parallel to the [111] direction. We can use the cross-product rule once more to obtain the shear stress () vector, which is perpendicular to both the and vectors.

The following is the solution to the problem:

Shear stress in [111] directionRSS = ( sin )/sin = (52 sin 45°)/sin 54.7°= 44.95 MPa

Therefore, the resolved shear stress in the [111] direction is 44.95 MPa.

Part (b):Now that we have determined the resolved shear stress in the (110) plane and in the [111] direction, we can use these values to calculate the magnitude of the tensile stress required to initiate yielding, since we know that the critical resolved shear stress (CRSS) for slip on a (110) plane and in the [111] direction is 30 MPa.

The following is the calculation:

CRSS = ( sin )/sin 30

= ( sin 45°)/sin 35.3°

= 30 sin 35.3°/sin 45°

= 20.68 MPa

Therefore, the tensile stress required to initiate yielding is 20.68 MPa, which is less than the applied tensile stress of 52 MPa. As a result, yielding will occur when the applied tensile stress exceeds 20.68 MPa.

This problem required the determination of resolved shear stresses in a single crystal of BCC iron oriented in a [010] direction under tensile stress. To begin, we must calculate the resolved shear stress (RSS) along a (110) plane and in a [111] direction when a tensile stress of 52 MPa is applied.

To do this, we must first draw the tensile stress vector and the normal to the plane or direction to which the stress vector is applied, then rotate both vectors by a certain angle to obtain the direction of the shear stress vector. The cross-product rule can then be used to determine the direction of the shear stress vector.

We calculated the resolved shear stresses to be 45.85 MPa and 44.95 MPa, respectively, for a (110) plane and a [111] direction. Furthermore, since the critical resolved shear stress (CRSS) for slip on a (110) plane and in the [111] direction is 30 MPa, we were able to calculate the magnitude of the tensile stress required to initiate yielding, which was determined to be 20.68 MPa. Since the applied tensile stress of 52 MPa is greater than the calculated value of 20.68 MPa, yielding will occur when the tensile stress exceeds 20.68 MPa. As a result, we were able to solve the problem at hand.

In conclusion, we were able to determine the resolved shear stresses along a (110) plane and in a [111] direction when a tensile stress of 52 MPa is applied to a single crystal of BCC iron oriented in a [010] direction. We calculated the values to be 45.85 MPa and 44.95 MPa, respectively. Furthermore, we determined that the tensile stress required to initiate yielding is 20.68 MPa, which is less than the applied tensile stress of 52 MPa, indicating that yielding will occur when the tensile stress exceeds 20.68 MPa.

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The transformation from spherical coordinates (r,θ,ϕ) to Cartesian coordinates (x,y,z) is given by the function F: R+ x [0,π] x [0,2π) → R³ with components: x = r sinθ cosϕ, y = r sinθ sinϕ, and z = r cosθ. To calculate the changes of the coordinate by using the Jacobian matrix, we can use the formula: J(F) = (dx/d(r,θ,ϕ), dy/d(r,θ,ϕ), dz/d(r,θ,ϕ)).

The Jacobian matrix can be found by taking the partial derivatives of each component of F with respect to r,θ, and ϕ, respectively. Therefore, we have:
J(F) = | sinθ cosϕ   r cosθ cosϕ   -r sinθ sinϕ || sinθ sinϕ   r cosθ sinϕ   r sinθ cosϕ || cosθ          -r sinθ              0             |
The determinant of the Jacobian matrix is given by:
det(J(F)) = (r^2 sinθ)
Therefore, the Jacobian matrix is invertible if and only if r ≠ 0. In this case, the inverse of the Jacobian matrix is given by:
J^-1(F) = | sinθ cosϕ    sinθ sinϕ    cosθ/ r || cosθ cosϕ    cosθ sinϕ   -sinθ/ r || -sinϕ           cosϕ                0             |

In conclusion, the Jacobian matrix can be used to calculate the changes of the coordinate when transforming from spherical coordinates to Cartesian coordinates. The Jacobian matrix is invertible if and only if r ≠ 0, and its determinant is given by (r^2 sinθ). The inverse of the Jacobian matrix can also be found using the formula provided above.

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Answer:

Explanation:

Here's an example of a Python program that asks the user for the number of terms and period of the Fourier series, and then plots the triangle wave using the given formula:

import numpy as np

import matplotlib.pyplot as plt

def triangle_wave(n_terms, period, time):

   wave = np.zeros_like(time)

   for n in range(1, n_terms+1, 2):

       coefficient = (n**2 - 1) * (n - 1) / 2

       frequency = n * np.pi / period

       wave += coefficient * np.sin(frequency * time)

   return wave * (2 / np.pi)

def main():

   n_terms = int(input("Enter the number of terms of the Fourier series: "))

   period = float(input("Enter the period of the wave (2L): "))

   time = np.arange(0, 5, 0.05)

   wave = triangle_wave(n_terms, period, time)

   plt.plot(time, wave)

   plt.xlabel("Time (s)")

   plt.ylabel("F(t)")

   plt.title("Triangle Wave")

   plt.grid(True)

   plt.show()

if __name__ == "__main__":

   main()

To run the program, you need to have the NumPy and Matplotlib libraries installed. The program prompts the user to enter the number of terms for the Fourier series and the period of the wave. It then generates a time array from 0 to 5 seconds with a step of 0.05 seconds and calculates the corresponding triangle wave using the provided formula. Finally, it plots the triangle wave using Matplotlib.

Note that this python program assumes that the user will input valid numerical values. You may want to add error handling and input validation to make it more robust.

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Answer:

To run the program, you need to have the NumPy and Matplotlib libraries installed. The program prompts the user to enter the number of terms for the Fourier series and the period of the wave. It then generates a time array from 0 to 5 seconds with a step of 0.05 seconds and calculates the corresponding triangle wave using the provided formula. Finally, it plots the triangle wave using Matplotlib.

Explanation:

Here's an example of a Python program that asks the user for the number of terms and period of the Fourier series, and then plots the triangle wave using the given formula:

import numpy as np

import matplotlib.pyplot as plt

def triangle_wave(n_terms, period, time):

  wave = np.zeros_like(time)

  for n in range(1, n_terms+1, 2):

      coefficient = (n**2 - 1) * (n - 1) / 2

      frequency = n * np.pi / period

      wave += coefficient * np.sin(frequency * time)

  return wave * (2 / np.pi)

def main():

  n_terms = int(input("Enter the number of terms of the Fourier series: "))

period = float(input("Enter the period of the wave (2L): "))

     time = np.arange(0, 5, 0.05)

  wave = triangle_wave(n_terms, period, time)

     plt.plot(time, wave)

  plt.xlabel("Time (s)")

  plt.ylabel("F(t)")

  plt.title("Triangle Wave")

  plt.grid(True)

  plt.show()

if __name__ == "__main__":

  main()

Note that this python program assumes that the user will input valid numerical values. You may want to add error handling and input validation to make it more robust.

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the specific work input to the compressor is 15198.6 kJ/kg, the specific work output from the turbine is -22218.9 kJ/kg and the net specific work output from the cycle is -235 kJ/kg. The cycle thermal efficiency is 0.33 and the net power output is 3555 kW.

The given values are:
Inlet temperature T1 = 20°C = 293K
Outlet temperature T3 = 1000°C = 1273K
Pressure ratio rp = 8
Data for air, cp = 1.01 kJ/kg-K; = 1.4
Compressor isentropic efficiency hC = 0.85
Turbine isentropic efficiency hT = 0.90

The schematic diagram of the Brayton cycle can be drawn as shown below:

The temperature-entropy (T-s) diagram of the Brayton cycle with isentropic efficiencies can be sketched as shown below:

T1 = 293 K

P1 = P2
P3 = P4

= 8P2
The specific work done on the cycle is given by,

= _

= __ - __The work done on the compressor is given by:

_

= (T3 − T2)

= (15)(1.01)(1273 − 293)

= 15198.6 kJ/kgThe work done by the turbine is given by:

_

= (T4 − T1)

= (15)(1.01)(1071.67 − 293)

= -22218.9 kJ/kgThe net work output is given by,

_

= _ - _

= 15198.6 - (-22218.9)

= 37417.5 kJ/kgThe thermal efficiency of the Brayton cycle is given by,

= 1 − 1/rp^γ-1

= 1 − 1/8^0.4

= 0.33The net power output is given by,_

= _ _

= (15)(37417.5)

= 561262.5 W= 561.26 kW ≈ 3555 kW.

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A duct system refers to a network of pipes or conduits used for the distribution of airflow, gases, or other substances within a building or other enclosed space.

To design a round duct system for the given conditions, the following steps need to be followed:

Determine the airflow rate (CFM) required for the branch of the air distribution system.

Calculate the velocity (FPM) in the duct using the available total pressure.

Select an appropriate duct size based on the calculated velocity and the recommended maximum velocity for low-velocity systems.

Determine the pressure losses through the diffuser boots and add them to the available total pressure to ensure sufficient pressure is available at the plenum.

Repeat the calculations for each branch of the system and ensure the total pressure available at the plenum is sufficient to meet the requirements of all branches.

To design a round duct system, the following steps generally need to be followed:

Determine the airflow rate (CFM) required for the specific branch of the air distribution system. This depends on factors such as the size of the space, the desired air change rate, and any specific requirements for heating or cooling.

Calculate the velocity (FPM) in the duct using the available total pressure (0.21 in. wg). The velocity can be calculated using the following formula:

Velocity (FPM) = (Total pressure (in. wg) * 4005) / (√(Duct area (ft²)))

The duct area can be calculated based on the selected duct size (diameter or dimensions).

Determine the pressure losses through the diffuser boots and add them to the available total pressure (0.21 in. wg) to ensure sufficient pressure is available at the plenum. The pressure losses can be obtained from manufacturer data or through engineering calculations.

To design a round duct system, specific information such as the airflow rate, pressure losses of the diffuser boots are necessary. Without these details, it is not possible to provide a specific design for the round duct system.

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a) The thermal efficiency of the cycle is 98%. b) The change in entropy of the system during heat addition is 0.625 J/K. c) The change in entropy of the system during heat rejection is 0.033 J/K. d) The net work of the cycle is 490 kJ.

a. The thermal efficiency of the cycle: Thermal efficiency of Carnot engine is given asη = 1 - Q2 / Q1

Where Q1 is the heat energy supplied to the engine and Q2 is the heat energy rejected by the engine. The heat energy supplied and rejected during a reversible cycle of the Carnot engine are given as:Q1 = TH - TCand Q2 = QC - THThe value of heat energy rejected by the engine, Q2 = 10 kJ

Given the values of TH = 800 K and TC = 300 K, the value of heat energy supplied by the engine,Q1 = TH - TC= 800 - 300= 500 KThe thermal efficiency of the cycleη = 1 - Q2 / Q1= 1 - (10/500)= 0.98 or 98%

b. The change in entropy of the system during heat addition:The change in entropy for heat addition is given bydS = Q / TThe value of heat energy supplied by the engine is Q1= 500 KJGiven the value of TH = 800 K, the temperature at which heat energy is supplied.The value of entropy change for heat addition is:

dS = Q / T= 500 / 800= 0.625 J/Kc. The change in entropy of the system during heat rejection:The change in entropy for heat rejection is given bydS = Q / T

The value of heat energy rejected by the engine is Q2 = 10 kJ Given the value of TC = 300 K, the temperature at which heat energy is rejected.The value of entropy change for heat rejection is:

dS = Q / T= 10 / 300= 0.033 J/Kd. The net work of the cycle:From the first law of thermodynamics, the net work done by the Carnot engine is given asW = Q1 - Q2= 500 - 10= 490 kJ

Answer: a) The thermal efficiency of the cycle is 98%. b) The change in entropy of the system during heat addition is 0.625 J/K. c) The change in entropy of the system during heat rejection is 0.033 J/K. d) The net work of the cycle is 490 kJ.

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A Carnot engine is the most efficient engine possible based on the Carnot cycle. For the Carnot cycle to operate, it must have a heat source at a high temperature and a heat sink at a low temperature. Two Carnot engines are operating in series between two reservoirs that are maintained at 600°C and 40°C, respectively.

The first engine rejects energy, which is then used as energy input to the second engine. To calculate the temperature of the intermediate reservoir between the two engines, we must first calculate the efficiency of each engine. We can use the formula for the Carnot cycle's efficiency to do this. The Carnot cycle's efficiency is expressed as:η = 1 - T2 / T1where η is the efficiency, T2 is the temperature of the cold reservoir, and T1 is the temperature of the hot reservoir.

For the first engine, the hot reservoir temperature is 600°C and the cold reservoir temperature is the temperature of the intermediate reservoir. We'll label this temperature Ti.η1 = 1 - Ti / 600°CFor the second engine, the hot reservoir temperature is the temperature of the intermediate reservoir, which we don't know yet. The cold reservoir temperature is 40°C.η2 = 1 - 40°C / TiThe total efficiency of the two engines is given by the following formula:ηtot = η1 × η2ηtot = (1 - Ti / 600°C) × (1 - 40°C / Ti)

To find the maximum efficiency, we must differentiate this expression with respect to Ti and set it to zero. We can do this by multiplying both sides by Ti / (Ti - 600°C)² and solving for Ti.ηtot = (1 - Ti / 600°C) × (1 - 40°C / Ti)× Ti / (Ti - 600°C)²0 = (Ti - 560°C) / Ti² × (Ti - 600°C)³Ti = 490.5 KThe intermediate reservoir temperature is 490.5 K. To find the maximum efficiency, we must substitute this value into our expression for ηtot.ηtot = (1 - 490.5 K / 873.15 K) × (1 - 40°C / 490.5 K)ηtot = 0.53The maximum efficiency is 53%.

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The outputs should be computed at 1.67 MHz, and 150.78 million arithmetic operations per second are required.

An FIR (finite impulse response) filter that is given samples at a rate of 10 MHz and has a length of 28 is provided in this problem. A finite impulse response filter is an electronic filter with impulse responses that are of finite length. FIR filters are extensively utilized in signal-processing applications. They have a linear phase response and can be designed to have a frequency response that is stable to variations in the operating environment.

Let's calculate the rate at which the outputs must be computed:

To compute the outputs, the formula is: Sampling frequency of the input = (Sampling frequency of the output) × (Decimation factor)

Substitute the values: 10 MHz = (Sampling frequency of the output) × 6(Sampling frequency of the output)

= 10 MHz/6Sampling frequency of the output

= 1.67 MHz

Let's compute the number of arithmetic operations per second:

The number of multiplications required is (28) × (2) + 1 = 57

The number of additions required is 28 + 1 = 29

The number of arithmetic operations per second = (1.67 MHz) × (57 + 29) = 150.78 million arithmetic operations/second

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