Oil with a kinematic viscosity of 4 10 6 m2 /s fl ows through a smooth pipe 12 cm in diameter at 2.3 m/s. What velocity should water?

Answer:

Updated question

The big ben clock tower in London has clocks on all four sides. If each clock has a minute hand that is 11.5 feet in length, how far does the tip of each hand travel in 52 minutes?

The distance traveled by the tip of the minute hand of the clock would be 62.59 ft

Explanation:

Let us assume the shape of the clock is circular.

the minute hand is equal to the radius = 11.5 ft

Diameter = radius x 2

Diameter = 11.5 x 2 = 23 ft

The distance traveled by the tip of the minute hand can be calculated thus;

the fraction of the circumference traveled by the minute hand would be;

52/60 = 0.8667

Circumference of the clock would be;

C = pi x d

where C is the circumference

pi is a constant

d is the diameter

C = 3.14 x 23

C = 72.22 ft

Therefore the fraction of the circumference covered by the minute hand would be;

72.22 ft x 0.8667 = 62.59 ft

Therefore the distance traveled by the tip of the minute hand of the clock would be 62.59 ft

Answer:

D. Both "Materials" and "Packaging"

Explanation:

Product liability may refer to the manufacturer or the seller being held responsible or liable for providing any defective product into the hands of the consumer or the customer. Responsibility or liability for a defective product which causes injuries lies with all the sellers of the product from the manufacturer to the distributor to the seller.

There are majorly three product defects. They are :

1. Manufacturing defect

2. Design defect

3. Marketing defect

Answer:

This graph shows linear

y = f(x) and y = g(x).

Find the solution to the equation f(x) - g(x) = 0

Answer:

K = 96 and the design speed of the curve = 50mph

Explanation:

109+00 = 10900

Elevation = 950ft

110+77 = 11077

Elevation = 947.34ft

Station of low point = 110+50 = 11050

To get grade of curve

Gi = 947.34-950/11077-10900

= -2.66/177

= -0.015x100

= -1.5%

Locate of low point (XL)

= 11050-10900

= 150

To get the value of K

XL = |GL| x K

When we substitute values

150 = 1.5 x K

150 = 1.5K

K = 150/1.5

K = 100

The suitable and most nearest value is K = 96

Then we use the standard chart to get the design speed for K = 96

On this chart, the design speed for the curve = 50mph

Therefore K = 96 and speed = 50mph

This question is incomplete, the complete question is;

A thin, flat pate that is 0.2m by 0.2m on a side is oriented parallel to an atmospheric airstream having a velocity of 40m/s. The air is at a temperature of T∞ = 20 °C, while the plate is maintained at Ts = 120°C. The air flows over the top and bottom surfaces of the plate, and measurement of the drag force reveals a value of 0.075 N.

What is the rate of heat transfer from both sides of the plate to the air?

Answer:

the rate of heat transfer from both sides of the plate to the air is 236.54 W

Explanation:

Given the data in the question,

first we calculate the  Reynold's number for the flow

Re = pu∞d / Ц

Re = (1.12 × 40 × 0.2) / 1.983 × 10⁻⁵

Re = 451840

Now the Local skin friction coefficient is given as;

Cfx =  T / ( 1/2pu∞²)

Cfx = (Fd/A) / ( 1/2pu∞²)

Cfx = (0.075/(2×0.2×0.2)) / ( 1/2 × 1.12 × 40²)

= 0.9375 / 896

= 0.0010463

Cfx = 1.0463 × 10⁻³

Apply Reynold's- cOLBURN analogy

Cfx/2 = StₓPr^2/3

so

1.0463 × 10⁻³ / 2 = (h/pu∞Cp) × ( 0.711)^2/3

5.2315 × 10⁻⁴ × 1.12 × 40 × 1.005 × 1000 = h(0.711)^2/3

h = 23.554 / 0.7966

h = 29.56 W/m².K

so

The heat transfer rate from both the sides of the plate will be;

Q = 2 × 29.56 × 0.2 × 0.2 × ( 120 - 20 )

Q = 236.54 W

Therefore the rate of heat transfer from both sides of the plate to the air is 236.54 W

Answer:

Hello the needed data given is not properly arranged attached below is the properly arranged data

Answer:

b) 62.5 * 10^3 MPa

c) ≈ 285 MPa

d)  370Mpa

e)  16%

Explanation:

Given Data:

cylindrical aluminum diameter = 12.8 mm

Gauge length = 50.8 mm

A) plot of engineering stress vs engineering strain

attached below

B ) calculate Modulus of elasticity

Modulus of elasticity = Δб / Δ ε

                                   = ( 200 - 0 ) / (0.0032 - 0 ) = 62.5 * 10^3 MPa

C) Determine the yield strength

at strain offset = 0.002

hence yield strength ≈ 285 MPa

D) Determine tensile strength of the alloy

The tensile strength can be approximated at 370Mpa because that is where it corresponds to the maximum stress on the stress  vs strain ( complete plot )

E) Determine approximate ductility in percent elongation

ductility in percent elongation = plastic strain at fracture * 100

total strain = 0.165 , plastic strain = 0.16

therefore Ductility in percent elongation = 0.16 * 100 = 16%

Answer:

robotic technology    

Explanation:

Innovation is nothing but the use of various things such as ideas, products, people to build up a solution for the benefit of the human. It can be any product or any solution which is new and can solve people's problems.

Innovation solution makes use of technology to provide and dispatch new solutions or services which is a combination of both technology and ideas.

One such example of an innovative solution we can see is the use of "Robots" in medical science or in any military operations or rescue operation.

Sometimes it is difficult for humans to do everything or go to everywhere. Thus scientist and engineers have developed many advance robots or machines using new ideas and technology to find solutions to these problems.

Using innovations and technologies, one can find solutions to many problems which is difficult for the peoples. Robots can be used in any surveillance operation or in places of radioactive surrounding where there is a danger of humans to get exposed to such threats. They are also used in medical sciences to operate and support the patient.  

Answer:

B. Larger than that of the larger-diameter cylinder

Explanation:

By looking at the equation for the Nusselt number for a cylinder in cross flow, Nu = hd/k, and assuming both cylinders are made of the same material, you can see that the heat transfer coefficient h will have to be much larger for the smaller cylinder. You can verify this by using random numbers for each variable (shown below).

Verification:

ds = 1

db = 10

k = 12

Nu = 10

h small = ?

h big = ?

Nu = hd/k

10 = h small * ds / 12

120 = 1 * h small

h small = 120

Nu = hd/k

10 = h big*db / 12

120 = 10 * h big

h big = 12

This shows that the heat transfer coefficient for the smaller diameter, ds, must be bigger than the heat transfer coefficient for the larger diameter, db.

For two different air velocities, the Nusselt number for two different diameter cylinders in cross flow is the same. The average heat transfer coefficient for the smaller-diameter cylinder is " Smaller than that of the larger-diameter cylinder." (Option C)

The given information states that for two different air velocities, the Nusselt number for two different diameter cylinders in cross flow is the same.

The Nusselt number is a dimensionless number that relates the convective heat transfer rate to the conductive heat transfer rate.

Since the Nusselt number is the same for both cylinders, and the heat transfer coefficient is directly related to the Nusselt number, the smaller-diameter cylinder will have a smaller average heat transfer coefficient compared to the larger-diameter cylinder.

Thus option C is the right answer.

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Answer:

0.75 kg

Explanation:

c = Damping coefficient = 3 kg/s

x = Displacement of spring = 0.5 m

F = Force = 1.5 N

From Hooke's law we get

[tex]F=kx\\\Rightarrow k=\dfrac{F}{x}\\\Rightarrow k=\dfrac{1.5}{0.5}\\\Rightarrow k=3\ \text{N/m}[/tex]

In the case of critical damping we have the relation

[tex]c^2-4mk=0\\\Rightarrow m=\dfrac{c^2}{4k}\\\Rightarrow m=\dfrac{3^2}{4\times 3}\\\Rightarrow m=0.75\ \text{kg}[/tex]

Mass that would produce critical damping is 0.75 kg.

0.75 kg is the mass that would produce critical damping. As spring with an m-kg mass and a damping constant 3 (kg/s) can be held stretched 0.5 meters beyond its natural length by a force of 1.5 newtons.

A change in time and position is referred to as an object's velocity. When there is no movement of the object, the velocity of the object is said to be 0.

For any body in planar motion, the velocity is always instantaneously 0 at some point in the plane of motion (if it were rigidly connected to the body). This place is known as the instantaneous center of zero velocity, or IC.

Example: The gravitational pull of the earth pushes the ball away from the thrower when a ball is thrown upwards on Earth at a constant speed. The speed of the ball increases until it reaches its maximum, at which point it starts to plummet.

Thus, it is 0.75 kg.

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The image of the load applied to the polystyrene is missing, so i have attached it.

Answer:

a_new = 2.00302 in

b_new = 2.00552

Explanation:

From the image attached, we can see that the load of 500 lb/in is applied in the x-direction while the load of 350 lb/in acts in the y-direction.

Now, formula for stress is;

Stress(σ) = Force/Area

We are not given force and area but the load and plate thickness.

Thus, stress = load/thickness

We are given;

Load in x - direction = 500 lb/in.

Load in y - direction = 350 lb/in.

Thickness; t = 0.25 in

Thus;

σ_x = 500/0.25

σ_x = 2000 ksi

σ_y = 350/0.25

σ_y = 1400 ksi

From Hooke's law for 2 dimensions, strain is given by the formula;

ε_x = (1/E)(σ_x - vσ_y)

ε_y = (1/E)(σ_y - vσ_x)

We are given v_p = 0.25 and Ep = 597 × 10³ psi

Thus;

ε_x = (1/(597 × 10^(3)))(2000 - (0.25 × 1400)

ε_x = 0.00276

ε_y = (1/(597 × 10^(3)))(1400 - (0.25 × 2000)

ε_y = 0.00151

From elongation formula, we know that;

Startin is: ε = ΔL/L

Thus; ΔL = Lε

We are given a = 2 and b = 2

Thus;

ΔL_x = 2 × 0.00276

ΔL_x = 0.00552

ΔL_y = 2 × 0.00151

ΔL_y = 0.00302

New dimensions are;

a_new = 2 + 0.00302

a_new = 2.00302 in

b_new = 2 + 0.00552

b_new = 2.00552

Calculate the normal stress along the x-direction.

[tex]\to \sigma_x =\frac{500}{t}=\frac{500}{0.25}= 2000 \ \frac{lb}{in^2}\\\\[/tex]

Calculate the normal stress along the y-direction.

[tex]\to \sigma_y =\frac{350}{t}=\frac{350}{0.25}= 1400 \ \frac{lb}{in^2}[/tex]

Calculate the strain along the x-direction.

[tex]\to \varepsilon_y=\frac{1}{E}[\sigma_x-v(\sigma_y-\sigma_z)]\\\\[/tex]

        [tex]=\frac{1}{597\times 10^3} [2000-0.25(1400+0)] \\\\= 2.764\times 10^{-3}\\\\[/tex]

Calculate the strain along the y-direction.

[tex]\to \varepsilon_y=\frac{1}{E}[\sigma_x-v(\sigma_x+\sigma_z)][/tex]

        [tex]=\frac{1}{597\times 10^3}[1400 -0.25 (2000+0)]\\\\=1508\times 10^{-3}[/tex]

Calculate the strain along the z-direction.

[tex]\to \varepsilon_x= \frac{1}{E}[\sigma_z-v(\sigma_x +\sigma_y)][/tex]

        [tex]=\frac{1}{597\times 10^3} [0-0.25(2000+1400)] \\\\=-1.424\times 10^{-3}[/tex]

Calculate the new dimensions.

[tex]\to b' =b+ \varepsilon_xb\\\\[/tex]

       [tex]= 2+2.764\times 10^{-3} \times 2\\\\= 2.005528 \ in\\\\[/tex]

[tex]\to a' = a + \varepsilon_y a\\\\[/tex]

        [tex]= 2+1.508\times 10^{-3} \times 2\\\\= 2.003016\ in\\\\[/tex]

[tex]\to c'= c + \varepsilon_x c\\\\[/tex]

        [tex]= 0.25 +(-1.424\times 10^{-3}) \times 0.25\\\\= 0.249644\ in\\\\[/tex]

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Answer:

± 0.003 ft

Explanation:

Since our distance is 10,000 ft and we need to use a full tape measure of 100 ft. We find that 10,000 = 100 × 100.

Let L' = our distance and L = our tape measure

So, L' = 100L

Now by error determination ΔL' = 100ΔL

Now ΔL' = ± 0.30 ft

ΔL = ΔL'/100

= ± 0.30 ft/100

= ± 0.003 ft

So, the maxim error per tape is ± 0.003 ft

Answer:

The Answer is C!

Explanation:

Answer:

mountain ranges may be

Answer:

design

Explanation:

Design is a process used to solve problems systematically.

Human beings have specific needs and desires, which require a design process to interpret those needs and make them real from a product or service.

Design uses specific methods and techniques integrating ideals, creativity, technology and innovation to satisfy users' needs and solve problems.

Answer:

The total mass in the tank = 0.45524  kg

The amount of heat transferred = 3426.33 kJ

Explanation:

Given that:

The volume of the tank V = 0.06 m³

The pressure of the liquid and the vapor of H2O (p) = 15 bar

The initial quality of the mixture [tex]\mathbf{x_{initial} - 0.20}[/tex]

By applying the energy rate balance equation;

[tex]\dfrac{dU}{dt} = Q_{CV} - m_eh_e[/tex]

where;

[tex]m_e =- \dfrac{dm_{CV}}{dt}[/tex]

Thus, [tex]\dfrac{dU}{dt} =Q_{CV} + \dfrac{dm_{CV}}{dt}h_e[/tex]

If we integrate both sides; we have:

[tex]\Delta u_{CV} = Q_{CV} + h _e \int \limits ^2_1 \ dm_{CV}[/tex]

[tex]m_2u_2 - m_1 u_1 = Q_{CV} + h_e (M_2-m_1) \ \ \ --- (1)[/tex]

We obtain the following data from the saturated water pressure tables, at p = 15 bar.

Since:

[tex]h_e =h_g[/tex]

Then: [tex]h_g = h_e = 2792.2 \ kJ/kg[/tex]

[tex]v_f = 1.1539 \times 10^{-3} \ m^3 /kg[/tex]

[tex]v_g = 0.1318 \ m^3/kg[/tex]

Hence;

[tex]v_1 = v_f + x_{initial} ( v_g-v_f)[/tex]

[tex]v_1 = 1.1529 \times 10^{-3} + 0.2 ( 0.1318-1.159\times 10^{-3} )[/tex]

[tex]v_1 = 0.02728 \ m^3/kg[/tex]

Similarly; we obtained the data for [tex]u_f \ \& \ u_g[/tex] from water pressure tables at p = 15 bar

[tex]u_f = 843.16 \ kJ/kg\\\\ u_g = 2594.5 \ kJ/kg[/tex]

Hence;

[tex]u_1 = u_f + x_{initial } (u_g -u_f)[/tex]

[tex]u_1 =843.16 + 0.2 (2594.5 -843.16)[/tex]

[tex]u_1 = 1193.428[/tex]

However; the initial mass [tex]m_1[/tex] can be calculated by using the formula:

[tex]m_1 = \dfrac{V}{v_1}[/tex]

[tex]m_1 = \dfrac{0.06}{0.02728}[/tex]

[tex]m_1 = 2.1994 \ kg[/tex]

From the question, given that the final quality; [tex]x_2 = 1[/tex]

[tex]v_2 = v_f + x_{final } (v_g - v_f)[/tex]

[tex]v_2 = 1.1539 \times 10^{-3} + 1(0.1318 -1.1539 \times 10^{-3})[/tex]

[tex]v_2 = 0.1318 \ m^3/kg[/tex]

Also;

[tex]u_2 = u_f + x_{final} (u_g - u_f)[/tex]

[tex]u_2 = 843.16 + 1 (2594.5 - 843.16)[/tex]

[tex]u_2 = 2594.5 \ kJ/kg[/tex]

Then the final mass can be calculated by using the formula:

[tex]m_2 = \dfrac{V}{v_2}[/tex]

[tex]m_2 = \dfrac{0.06}{0.1318}[/tex]

[tex]m_2 = 0.45524 \ kg[/tex]

Thus; the total mass in the tank = 0.45524  kg

FInally; from the previous equation (1) above:

[tex]m_2u_2 - m_1 u_1 = Q_{CV} + h_e (M_2-m_1) \ \ \ --- (1)[/tex]

[tex]Q = (m_2u_2-m_1u_1) - h_e(m_2-m_1)[/tex]

Q = [(0.45524)(2594.5) -(2.1994)(1193.428)-(2792.2)(0.45524-2.1994)]

Q = [ 1181.12018 - 2624.825543 - (2792.2)(-1.74416 )]

Q = 3426.33 kJ

Thus, the amount of heat transferred = 3426.33 kJ

This question is incomplete, the complete question is;

All of the questions for this quiz involve a drone with a mass of 1.2 kg.

The drone hovers for 29 seconds at a height of 53 meters above the ground while taking a picture of a crowd. What is the vertical impulse by the lift force (combined force of the four rotors) during this time (in Newton-seconds)?

Answer: the vertical impulse is 341.388 Ns

Explanation:

Given that;

mass m = 1.2 kg

t = 29 sec

h = 53 m

vertical impulse = ?

Vertical impulse by lift force is given as;

Vertical Impulse = F × t

= mg × t

so we substitute

Vertical impulse = 1.2 ×9.81 × 29

= 341.388 Ns

therefore the vertical impulse is 341.388 Ns

Answer:

C.

Explanation:

Let's have

Q = heat transfer surface

∆T = average temperature

F = area of the heat surface

Then the heat transfer coefficient = Q/∆T*F

In a flow over flat plate, the average friction and the heat transfer coefficient are determined by the integration of local friction and also great transfer coefficients over the plate entirely and then dividing by the plates length.

Therefore answer option C is the answer to this question.

Answer:

Time constant t = 0.0083 (Approx)

Explanation:

Given:

L = 0.025

Missing resistance r = 3Ω

Find:

Time constant t

Computation:

Time constant t = L / r

Time constant t = 0.025 / 3

Time constant t = 0.0083 (Approx)

Using the inductance - resistance relation to calculate the time constant , the time constant would be 0.08333

Given the Parameters :

The time taken for current to reach the wire can be calculated thus :

Time constant = 0.025 ÷ 3 = 0.083333

Therefore, the time constant for current to reach is 0.08333

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Answer:

1 pulse rotate = 9 degree

Explanation:

given data

incremental encoder rotating = 15 rpm

wheel holes = 40

solution

we get here first 1 revolution time

as 15 revolution take = 60 second

so 1 revolution take = [tex]\frac{60}{15}[/tex]

1 revolution take = 4 seconds

and

40 pulse are there for 1 revolution

40 pulse for 360 degree

so 1 pulse rotate is = [tex]\frac{360}{40}[/tex]

1 pulse rotate = 9 degree

Answer:

banker’s algorithm,

(2) detect deadlock and kill thread, releasing all resources,

(3) reserve all resources in advance,

(4) restart thread and release all resources if thread needs to wait,

(5) resource ordering, and

(6) detect deadlock and roll back thread’s actions.

b) Another criterion is efficiency; in other words, which requires the least processor overhead. Rank order the approaches from 1 to 6, with 1 being the most efficient, assuming that deadlock is a very rare event. Comment on your ordering

Explanation:

Solution :

The nuclear reaction for boron is given as :

[tex]$^{10}\textrm{B}_5 + ^{1}\textrm{n}_0 \rightarrow ^{7}\textrm{Li}_3 + ^{4}\textrm{He}_2$[/tex]

And the reaction for Cadmium is :

[tex]$^{113}\textrm{Cd}_48 + ^{1}\textrm{n}_0 \rightarrow ^{114}\textrm{Cd}_48 + \gamma [5 \ \textrm{MeV}]$[/tex]

We know that it is easier that to shield or stop an alpha particle (i.e. He nucli) as they can be stopped or obstructed by only a few centimetres of the material. However, the gamma rays ( γ ) can penetrate through the material to a greater distance. Therefore, we can choose the first one.

Answer:

155fts

Explanation:

We apply the bernoulli's equation to get the depth of water.

We have the following information

P1 = pressure at top water surface = 0

V1 = velocity at too water surface = 0

X1 = height of water surface = h

Hf = friction loss = 0

P2 = pressure at exit = 0

V2 = velocity at exit if penstock = 100ft/s

X2 = height of penstock = 0

g = acceleration due to gravity = 32.2ft/s²

Applying these values to the equation

0 + 0 + h = 0 + v2²/2g +0 + 0

= h = 100²/2x32.2

= 10000/64.4

= 155.28ft

= 155

Answer:

common fate

Explanation:

The gestalt effect may be defined as the ability of our brain to generate the whole forms from the groupings of lines, points, curves and shapes. Gestalt theory lays emphasis on the fact that whole of anything is much greater than the parts.

Some of the principles of Gestalt theory are proximity, similarity, closure, symmetry & order, figure or ground and common fate.

Common fate : According to this principle, people will tend to group things together which are pointed towards or moving in a same direction. It is the perception of the people that objects moving together belongs together.

Answer:

The friction factor is 0.303.

Explanation:

The flow velocity ([tex]v[/tex]), measured in meters per second, is determined by the following expression:

[tex]v = \frac{4\cdot \dot V}{\pi \cdot D^{2}}[/tex] (1)

Where:

[tex]\dot V[/tex] - Flow rate, measured in cubic meters per second.

[tex]D[/tex] - Diameter, measured in meters.

If we know that [tex]\dot V = 0.01\,\frac{m^{3}}{s}[/tex] and [tex]D = 0.05\,m[/tex], then the flow velocity is:

[tex]v = \frac{4\cdot \left(0.01\,\frac{m^{3}}{s} \right)}{\pi\cdot (0.05\,m)^{2}}[/tex]

[tex]v \approx 5.093\,\frac{m}{s}[/tex]

The density and dinamic viscosity of the glycerin at 20 ºC are [tex]\rho = 1260\,\frac{kg}{m^{3}}[/tex] and [tex]\mu = 1.5\,\frac{kg}{m\cdot s}[/tex], then the Reynolds number ([tex]Re[/tex]), dimensionless, which is used to define the flow regime of the fluid, is used:

[tex]Re = \frac{\rho\cdot v \cdot D}{\mu}[/tex] (2)

If we know that [tex]\rho = 1260\,\frac{kg}{m^{3}}[/tex], [tex]\mu = 1.519\,\frac{kg}{m\cdot s}[/tex], [tex]v \approx 5.093\,\frac{m}{s}[/tex] and [tex]D = 0.05\,m[/tex], then the Reynolds number is:

[tex]Re = \frac{\left(1260\,\frac{kg}{m^{3}} \right)\cdot \left(5.093\,\frac{m}{s} \right)\cdot (0.05\,m)}{1.519 \frac{kg}{m\cdot s} }[/tex]

[tex]Re = 211.230[/tex]

A pipeline is in turbulent flow when [tex]Re > 4000[/tex], otherwise it is in laminar flow. Given that flow has a laminar regime, the friction factor ([tex]f[/tex]), dimensionless, is determined by the following expression:

[tex]f = \frac{64}{Re}[/tex]

If we get that  [tex]Re = 211.230[/tex], then the friction factor is:

[tex]f = \frac{64}{211.230}[/tex]

[tex]f = 0.303[/tex]

The friction factor is 0.303.

Answer:

10uF

Explanation:

A higher value of capacitance is the best option when we are trying to filter power supply outputs in other to reduce hum.

The greater the capacitance or the voltage of a circuit is, the more energy it can the particular circuit can store. When capacitors are being connected in series, the total value of the capacitance reduces but contrarily, the voltage of the same system increases anyway. Connecting circuits in parallel helps to keep the voltage rating the same but on the other hand, it increases the total capacitance.

A 10 μF capacitor is better.

This is because, to filter out high frequency noise, our capacitor is connected in parallel with the voltage supply. This parallel connection causes the capacitance of the circuit to increase but the voltage stays constant.

Since there is an increase in capacitance, this causes the circuit to filter out high frequency noise.

So, a high value capacitance connected in parallel with the voltage source is a better filter for high frequency noise.

So, the 10 μF capacitor is better.

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Answer:

commercial freezing

Explanation:

smaller ice crystals are formed this causes less damage to cell membranes so the quality is less effected

Answer:

a) the  rate of cooling provided is 18604.8 Btu/h

b) the rate of exergy destruction in the evaporator is 0.46 Btu/Ibm

c) the second-law efficiency of the evaporator is 37.39%

Explanation:

Given that;

Temperature of sink TL = 50°F = 510 R

Temperature at evaporator inlet TI = 10°F = 470 R

mass flow rate m" = 0.08 lbm/s

quality of refrigerant at evaporator inlet x1 = 0.3

quality of refrigerant at evaporator exit x2 = 1.0

T₀ = 77°F = 537 R

h1 = 107.5 Btu/lbm

s1 = 0.2851 Btu/lbm·R,

h2 = 172.1 Btu/ lbm,

s2 = 0.4225 Btu/lbm·R.

a) rate of cooling provided, in Btu/h.

QL = m"( h2 - h1)

we substitute

QL = 0.08( 172.1 - 107.5

= 0.08 × 64.6

= 5.168 Btu/s

we convert to Btu/h

5.168 × 60 × 60

QL = 18604.8 Btu/h

Therefore the  rate of cooling provided is 18604.8 Btu/h

b) The rate of exergy destruction in the evaporator

Entropy generation can be expressed as;

S_gen = m"(s2 - s1) - QL/TL

so we substitute

S_gen = 0.08( 0.4225 -  0.2851  ) - 5.168 / 510

= 0.010992 - 0.01013

S_gen = 0.00086 Btu/ibm.R

now the energy destroyed expressed as;

X_dest = T₀ × S_gen

so

X_dest =  537 × 0.00086

X_dest = 0.46 Btu/Ibm

Therefore the rate of exergy destruction in the evaporator is 0.46 Btu/Ibm

c)  The second-law efficiency of the evaporator.

Energy expended is expressed as;

X_exp = m"(h1 - h2) - m"T₀(s1 - s2)

we substitute

= 0.08( 107.5 - 172.1 ) - [0.08 × 537 ( 0.2851 - 0.4225 )

= -5.168 - [ - 5.9027)

= -5.168 + 5.9027

= 0.7347 Btu/s

Now second law efficiency is expressed as;

nH = 1 - (X_dest / X_esp)

= 1 - ( 0.46 / 0.7347 )

= 1 - 0.6261

= 0.3739

nH = 37.39%

Therefore the second-law efficiency of the evaporator is 37.39%

Answer:

8 kmol/s

Explanation:

From the given information:

The combustion reaction equation for Octane in a stoichiometric condition can be expressed as:

[tex]C_{8}H_{18} +12.5(O_2 + \dfrac{79}{21} N_2) \to 9H_2O +8CO_2 + 12.5(\dfrac{79}{21}N_2)[/tex]

[tex]C_{8}H_{18} +12.5(O_2 + 3.76N_2) \to 9H_2O +8CO_2 + 12.5(3.76 \ N_2)[/tex]

In the combustor, it is said that 60% of excess air and 1 mole of Octane is present.

Thus;

the air supplied = 1.6  × 12.5 = 20

The equation can now be re-written as:

[tex]C_{8}H_{18} +20(O_2 + 3.76N_2) \to 9H_2O +8CO_2 + 7.5 \ O_2+ 75.2 \ N_2[/tex] because for 1 mole of Octane, 8 moles of CO2 can be found in the combustion product.

Thus, for 1 kmol/s of Octane also produce 8 kmol/s of CO2.

∴

The mole flow rate in Kmol/s of CO2 in the product stream = 8 kmol/s