A balanced die is rolled. Find the probability of getting: a value of at least 3.

Answer: A sample of the representative outcomes or sequences of events for all possible consequences of a model is called a scenario.

The answer is A. Sensors

The maximum stress in the beam when the radius of the fillets is 4 mm is 72.4 MPa.

The maximum stress in the beam when the radius of the fillets is 8 mm is 20.1 MPa.

To find the maximum stress, the formula used is:

σm = (M/W)×y, where σm = maximum stress, M = bending moment, W = section modulus, y = distance from neutral axis to extreme fiber at section

1. Consider the case where the radius of the fillets is 4 mm:

The centroidal axis is obtained by finding the moment of areas about the neutral axis. (centroidal axis passing through point 1)

With respect to the neutral axis and centroidal axis, the top surface is in compression while the bottom surface is in tension. Maximum stress is obtained at the extreme fiber, which is at the top surface at a distance of 72 mm from the neutral axis, i.e., y = 72 mm

W = (bh/6) + 2A (use any method), where b = width = 40 mm, h = depth = 80 mm, A = area of fillet + area of semi-circle

Area of fillet = πr²/4 = π(4)²/4 = 4π

Area of semi-circle = (πr²/2) - (r²/2) = ((π×4²)/2) - (4²/2) = 8π

W = (bh/6) + 2A = (40×80/6) + 2(4π + 8π) = 1600/3 + 24π

Therefore, W = 241.14 × 10³ mm³

Maximum stress is obtained at the extreme fiber, which is at the top surface at a distance of 72 mm from the neutral axis, i.e., y = 72 mm

σm = (M/W)×y

σm = (242×10⁶/(241.14×10³))×72

= 72.42 MPa

Therefore, the maximum stress in the beam when the radius of the fillets is 4 mm is 72.4 MPa.

2. Similarly, now consider the case where the radius of the fillets is 8 mm:

The centroidal axis is obtained by finding the moment of areas about the neutral axis. (centroidal axis passing through point 2). With respect to the neutral axis and centroidal axis, the top surface is in compression while the bottom surface is in tension. Maximum stress is obtained at the extreme fiber, which is at the top surface at a distance of 64 mm from the neutral axis, i.e., y = 64 mm

W = (bh/6) + 2A (use any method), where b = width = 40 mm, h = depth = 80 mm, A = area of fillet + area of semi-circle

Area of fillet = πr²/4 = π(8)²/4 = 16π

Area of semi-circle = (πr²/2) - (r²/2) = ((π×8²)/2) - (8²/2) = 60π

W = (bh/6) + 2A = (40×80/6) + 2(16π + 60π)

= 1600/3 + 152π

Therefore, W = 2412.62 × 10³ mm³

Maximum stress is obtained at the extreme fiber, which is at the top surface at a distance of 64 mm from the neutral axis, i.e., y = 64 mm

σm = (M/W)×y

σm = (242×10⁶/(2412.62×10³))×64 = 20.14 MPa

Therefore, the maximum stress in the beam when the radius of the fillets is 8 mm is 20.1 MPa.

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Answer:

Explanation:

Consider the outer surface of the roof to be \(T_{out}\)

Since, the heat conducted is equal to the sum of the heat transferred through convection and the rest by radiation

\(Q_{cond}=Q_{rad}+Q_{conv}\)

Rewrite the equation as follows

\(kA_s\frac{\Delta T}{L} =\epsilon \sigma A_s(T_s^4-T_{\infty}^4)+hA_s \Delta T\\\\k\frac{\Delta T}{L} =\epsilon \sigma (T_s^4-T_{\infty}^4)+h \Delta T\)

\(k\frac{T_{in}-T_{out}}{L} =\epsilon \sigma (T_{out}^4-T_{rad}^4)+h(T_{out}-T_{\infty})\)

subsititute

k = 2 W/m

\(15^o C=T_{in}\)

0.22 for L

\(0.9 = \epsilon\)

\(5.67\times 10^{-8}W/m^2.K^4=\sigma\)

\(255=T_{rad}\)

\(15W/m^2.K = h\)

\(10^oC=T_{\infty}\)

\(2\times \frac{15 -T_{out}}{0.22} =[0.9\times(5.67\times10^-^8)\times((T_{out}+273)^4-255^4)]+[15\times(T_{out}-10)]\\\\T_{out}=7.7^oC\)

Hence, the temperature of outer surface of the roof is \(T_{out}=7.7^oC\)

Calculate the surface area of the roof

\(A_s=b\times l\)

Here, b is the width , l is the length

substitute 15 for b , 20 for l

\(A_s=15 \times 20\\\\=300m^2\)

Write the equation for conduction

\(Q_{cond}=kA_s\frac{\Delta T}{L}\)

substitute 2W/m.K for k

\(300m^2 \ \ for \ A_s\\\\(15-7.7)^oC \ \ for \ \Delta T\\\\0.22m \ for \ L\)

\(Q_{cond}=2\times 300 \times \frac{15-7.7}{0.22}\\\\=19,000\)

Therefore, the total heat transferred through conduction is \(Q_{cond}=19,9009W\)

Consider the amount of  natural gas required be R and the cost incurred in running  the furnace through the night be M

\(R=\frac{Q_{cond}}{0.85} \times T\)

Duration of time T = 14 x 3600s

\(R=\frac{19909\times(14\times3600)}{0.85}kJ\\\\= \frac{19909\times(14\times3600)}{0.85\times105500}therm\\\\=11.2 \ therm\)

And the required money for the gas M = 11.2 x $1.2

= $13.44

Therefore, the money lost through the roof due to the heat transfer M=$13.44

Answer:

= s[j k] for all 1 <= k <= m.

In other words, a repeated subsequence is a sequence of characters within the original string that appears in the same order at least twice. The indices i1, i2, ..., im refer to the positions of the characters in the original string that make up the repeated subsequence, and the indices j1, j2, ..., jm refer to the positions of the same characters in the second occurrence of the repeated subsequence.

Answer:

There are a few key benefits to a pre-construction and design specialist taking an interdisciplinary approach:

To maximize employability options. Having exposure to different fields and skills makes a specialist a more versatile and attractive job candidate. They can work across multiple areas of design, planning, architecture and construction.

To draw on a varied creative tool set. Exposure to fields like architecture, engineering, economics, sustainability and more provides many perspectives and solutions to tapping into. This leads to more innovative and well-rounded design thinking.

To integrate well with other team members. Understanding the methods, priorities, constraints and languages of fields like architecture, engineering, planning, and construction allows a specialist to collaborate more effectively. They can translate across disciplines and facilitate joined-up thinking.

To understand constraints and optimally balance priorities. No design or pre-construction problem has limitless options. An interdisciplinary approach helps gain insights into the constraints each field operates under and how to optimize solutions across ecological, economic, functional and other priorities.

To consider impacts holistically. Many impacts, like environmental sustainability, cost efficiency and user experience, span multiple disciplines. An interdisciplinary mindset is needed to evaluate options based on overall impact, not just impacts within one area of focus.

Of these options, I believe "To integrate well with other team members" is the strongest summary of why an interdisciplinary approach is so valuable for this type of specialist role. Understanding other fields in depth helps in collaborating, communicating clearly, addressing concerns proactively and finding optimal compromise - all of which are key to successfully interweaving different areas of expertise on a team.

Explanation:

When designing a set of simple test programs to determine the type compatibility rules of a C compiler to which you have access, it is important to consider the different data types that are used in C programming. An example of a set of test programs that can be used to determine the type compatibility rules of a C compiler:

Integer Test the compatibility of the C compiler with integer data types. It declares two variables of type int, initializes them with values, and then adds them together. The result is printed to the screen. If the program compiles and runs without any errors, then the C compiler is compatible with integer data types.

Floating-Point Test  the compatibility of the C compiler with floating-point data types. It declares two variables of type float, initializes them with values, and then adds them together. The result is printed to the screen. If the program compiles and runs without any errors, then the C compiler is compatible with floating-point data types.

By running the set of simple test programs described above, you can determine the type compatibility rules of a C compiler to which you have access. If any of the programs do not compile or run without errors, then you can determine which data types are not compatible with the C compiler and adjust your code accordingly.

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Answer:

fracture will occur since ( 31.8 Mpa√m ) is greater than the \(K_{Ic\) of the material ( 30 Mpa√m )

Explanation:

Given the data in the question;

To determine whether the aircraft component will fracture, given a fracture toughness of 30 Mpa√m, stress level of 355 and maximum internal crack length of 1.39 mm.

On a similar component, it has been said that fracture results at a stress of 237 MPa when the maximum (or critical) internal crack length is 2.78 mm.

so we first of all solve for the parameter Y in the condition where fracture occurred.

\(K_{Ic\) = 30 Mpa√m

σ = 237 MPa

2α = 2.78 mm = 2.78 × 10⁻³ m  

so

Y = \(K_{Ic\) / σ√πα

we substitute

Y = (30 Mpa√m) / (237 MPa)√(π(2.78 × 10⁻³ m / 2 ) )

Y =  (30 Mpa) / (237)( 0.06608187 )

Y = 30 / 15.6614

Y = 1.9155

Next we solve for Yσ√πα for the second case;

σ = 355 Mpa, 2α = 1.39 mm = 1.39 × 10⁻³ m

so

Yσ√πα = 1.9155 × 355 Mpa × √( π × (1.39 × 10⁻³ m / 2) )

= 1.9155 × 355 × 0.0467269

= 31.8 Mpa√m

so

( 31.8 Mpa√m ) > \(K_{Ic\) ( 30 Mpa√m )

Therefore, fracture will occur since ( 31.8 Mpa√m ) is greater than the \(K_{Ic\) of the material ( 30 Mpa√m )

To determine the highest value of Vr for which the transistor still operates in the active mode, we need to consider the transistor's voltage characteristics.

In the active mode, the transistor operates as an amplifier, and Vr should be within certain limits. One important parameter to consider is the base-emitter voltage (Vbe). When Vbe exceeds the threshold voltage, the transistor starts to enter saturation.

In the given circuit of Fig. P6.51, with a very high β (amplification factor), the base current (Ib) can be considered negligible. Therefore, the collector current (Ic) flows through the resistor Rc.

To find the highest value of Vr for active mode operation, we need to ensure that the voltage drop across Rc (Vrc) is sufficiently high to keep the transistor in active mode. This is typically achieved by ensuring that Vrc is at least a few times Vbe.

To find the value of Vr for which the transistor operates in saturation with a forced β of 3, we need to consider the transistor's saturation characteristics. In saturation mode, the transistor acts as a switch, and Vrc should be reduced to a low voltage level.

To provide a more accurate analysis and determine the specific values of Vr for active mode and saturation, the circuit diagram (Fig. P6.51) and transistor parameters (such as Vbe, β, etc.) need to be provided.

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Yes construction of the stator in the wound-rotor motor is identical to that of the squirrel cage motor .

Given,

Construction of stator in induction motors.

Here,

An induction motor is an AC electric motor in which the electric current in the rotor needed to produce torque is obtained by electromagnetic induction from the magnetic field of the stator winding.

Regarding the construction of wound rotor motor and squirrel cage motor, the main difference is in the rotor.

In a wound rotor motor, the rotor is made up of a set of wire windings that are connected to external resistance, while in a squirrel cage motor, the rotor is made up of a set of aluminum or copper bars that are short-circuited by end rings. Because of this, wound rotor motors have the ability to adjust the rotor resistance, which can be used to control the speed of the motor and improve the starting torque. On the other hand, squirrel cage motors are simpler, more reliable, and less expensive to manufacture and maintain.

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Answer:

take

play

are

sits

crouches

direct

indirect

direct

direct

indirect

Explanation:

i had just taken the test

When installing conduit in reinforced concrete, it is best to use a masonry bit.

Conduit is a tube or other non-flexible material in which electrical cables are housed. The use of conduit is necessary for protection against electric shock, mechanical injury, and chemical damage in the wiring system.

Conduit is utilized to run the wiring through various places to get them from one point to another. It provides extra support and protection to wires, reducing the risk of damage from external factors, and it is crucial in shielding wires in underground installations.

A masonry bit is a type of drill bit used to drill holes in concrete, brick, and other masonry materials. It is designed to drill through tough and hard materials that are difficult to drill through using other drill bits.

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When installing conduit in reinforced concrete, it is best to use a concrete saw to cut channels and then attach conduit clamps. Conduit is a type of protective casing used to house and protect electrical wiring. It is available in various materials such as PVC, metal, and fiberglass.

It is available in various materials such as PVC, metal, and fiberglass. Conduits are usually installed in various locations and may involve a range of techniques depending on the material used.The conduit can be installed in reinforced concrete, which is a type of concrete that has steel bars or fibers embedded to provide it with greater strength and durability. When installing conduit in reinforced concrete, it is best to use a concrete saw to cut channels and then attach conduit clamps. This process is also referred to as the "chasing method."The chasing method is a very common installation method. It involves cutting a channel into the concrete, laying the conduit in it, and then filling the channel with a patching compound.It is best to use a metal conduit that is resistant to damage, corrosion, and bending.The conduit should be cut to length and deburred to remove any sharp edges.

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Answer:

Q ≅ 1.53 m³/s

Explanation:

From the given information:

The flow rate of the orifice is:

\(v = c_v \sqrt{2gh}\)

\(v = 0.90 \times \sqrt{2*9.81 * 10}\)

where;

\(Q = c_d \times \sqrt{2gh} \times A\); &

\(c_d = c_c \times c_v\)

∴

\(Q = c_c \times c_v \sqrt{2gh} \times \dfrac{\pi}{4}\times d^2\)

\(Q = 0.90 \times 0.62 \sqrt{2*9.81*10} \times \dfrac{\pi}{4}\times 0.5^2\)

\(Q = 0.558 \times 14.00714104 \times 0.1963495408\)

Q ≅ 1.53 m³/s

At 1.2mpa pressure and 50c

What is pressure?
By pressing a knife against some fruit, one can see a straightforward illustration of pressure. The surface won't be cut if you press the flat part of the knife against the fruit. The force is dispersed over a wide area (low pressure).

a)Mass flow rate of the refrigerant
Therefore h1= condenser inlet enthalpy =278.28KJ/Kg
saturation temperature at 1.2mpa is 46.29C
Therefore the temperature of the condenser

T2 = 46.29C - 5
T2 = 41.29C

Now,
d)power consumed by compressor W = 3.3KW
Q4 = QL + w = Q4
QL = mR(h1-h2)-W
= 0.0498 x (278.26 - 110.19)-3.3
=5.074KW
Hence refrigerator load is 5.74Kg

(COP)r = 238/53
(Cop) = 4.490

Therefore the above values are the (a) mass flow rate of the refrigerant, b) the refrigerant load, c) the cop, and d) the minimum power input to the compressor for the same refrigeration load.

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The solution obtained from OPL is, X = 1, Y = 4, and the optimal value of Z is 10. She estimates that she is going to earn grade points depending on the number of periods spent on each course.

Tyler Galpin, a friend of Wednesday Addams, arrives in town and calls Wednesday Addams for a date. Assessing her situation, Wednesday decides that all she really needs is a total of 16 grade points gained from any of the courses to graduate.

She wants to allocate her time so that she spends the fewest number of study periods necessary to guarantee her receiving at least 16 grade points. We need to formulate this decision problem as an integer programming model and solve using OPL.

So she gets 4 grade points if she spends one period studying an easy course and 2 grade points if she spends one period studying a difficult course. Number of periods studied Grade points from Easy course Difficult course04                 0             0        14                 4             2        24                 4             2        37                 7             4        48                 8             6        58                 8             9        

 So, the given problem can be formulated as follows: Minimize \(Z = x11 + x12 + x13 + x14\)

Subject to\(4 x11 + 4 x12 + 7 x13 + 8 x14 ≥ 16\)

(Easy courses)\(2 x11 + 2 x12 + 4 x13 + 6 x14 ≥ 16\)

(Difficult courses\()Y ≥ 1x11, x12, x13, x14,\)

Y are integers\(xij ≥ 0 (i = 1,2,3,4; j = 1, 2, …, 10)\)

Below is the OPL code:

int easy\([1..4]=[4,4,7,8];\)

int hard\([1..4]=[2,2,4,6];\)

dvar int \(x[1..4][1..10] in 0..10\);

dvar int y in 1..4;

minimize sum\((i in 1..4, j in 1..10)\)

The solution obtained from OPL is, X = 1, Y = 4, and the optimal value of Z is 10.

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The closest equivalent present worth of John's deposits is less than $8,000.

To determine the closest equivalent present worth of John's deposits, we need to calculate the present value of the cash flows he will make.

The deposit of $1000 at the end of the next year can be considered a future value (FV). We need to calculate its present value (PV) using the formula:

PV = FV / (1 + r)^n

Where:

FV = $1000

r = interest rate = 10% = 0.10

n = number of years = 1

PV = $1000 / (1 + 0.10)^1 = $909.09

Next, we calculate the present value of the increasing deposits of $100 per year for 10 years. These cash flows form an arithmetic progression with a common difference of $100.

Using the formula for the sum of an arithmetic progression, we can find the present value of these cash flows:

PV = (n/2) * (2a + (n-1)d)

Where:

n = number of terms = 10

a = first term = $100

d = common difference = $100

PV = (10/2) * (2*100 + (10-1)*100) = 5 * (200 + 9 * 100) = $5,500

Now, we can sum up the present values of both cash flows:

PV = $909.09 + $5,500 = $6,409.09

The closest equivalent present worth is between $8,000 - $8,300. Since the calculated present value is lower than $8,000, the closest equivalent present worth is less than $8,000.

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False. when you call a string object's split method, the method extracts tokens from the string and returns them as integers.

When you call a string object's split method, the method extracts tokens from the string and returns them as strings, not integers. The split method divides a string into substrings based on a specified delimiter and returns those substrings as an array of strings. It does not perform any conversion of the extracted tokens to integers. If you want to convert the extracted tokens to integers, you would need to explicitly perform the conversion after splitting the string.

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The engineering method of separating costs is d. All of these choices are correct.

The engineering method of separating costs involves a step-by-step analysis of various elements involved in the production process to estimate the cost of activities and new products.

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Drum Brakes are mostly to be found on the rear of a modern front wheel drive vehicle.

What Is Drum Brake?

Drum brakes have been around since 1900 ,patented until 1902 by Louis Renault.

Types of drum brakes

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The density of the element can be calculated using the formula: density = (atomic mass * Avogadro's number) / (volume of the unit cell).

In this scenario, we start by using the scattering angle obtained from x-ray diffraction and the wavelength of the x-ray to determine the interplanar spacing (d) using the formula d = λ / (2 * sin(θ)), where λ is the wavelength and θ is the scattering angle. Once we have the interplanar spacing, we can determine the side length of the unit cell (a) using the relationship between the interplanar spacing and the Miller indices for the BCC structure.

In the BCC structure, the {110} planes have Miller indices of [1, 1, 0], which corresponds to a side length of a = d / sqrt(2). With the side length of the unit cell known, we can calculate the volume of the unit cell using V = (4 * side length^3) / 3. Finally, we can substitute the atomic mass (molar mass) of the element, Avogadro's number, and the volume of the unit cell into the formula for density to determine the density of the element.

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Answer:

(1). False, (2). True, (3). False, (4). False, (5). True.

Explanation:

The term ''contouring'' in this question does not have to do with makeup but it has to deal with the measurement of all surfaces in planes. It is a measurement in which the rough and the contours are being measured. So, let us check each questions again.

(1). In contouring, it is necessary to measure position and not velocity for feedback.

ANSWER : b =>False. IT IS NECESSARY TO MEASURE BOTH FOR FEEDBACK.

(2). In contouring during 2-axis NC machining, the two axes are moved at the same speed to achieve the desired contour.

ANSWER: a=> True.

(3). Job shop is another term for process layout.

ANSWER: b => False

JOB SHOP IS A FLEXIBLE PROCESS THAT IS BEING USED during manufacturing process and are meant for job Production. PROCESS LAYOUT is used in increasing Efficiency.

(4). Airplanes are normally produced using group technology or cellular layout.

ANSWER: b => False.

(5). In manufacturing, value-creating time is greater than takt time.

ANSWER: a => True.

Answer:

B

Explanation:

Vortex generators on the upper side of the wing surface will decrease the intensity of shock wave induced air separation and increase the critical Mach Number.

So, the correct answer is C.

Vortex generators are small devices that are placed on the upper side of an aircraft's wing surface. They are used to improve the aircraft's performance at high speeds.

The primary function of vortex generators is to control the airflow over the wing surface, which helps to reduce drag and increase lift.

When placed on the upper side of the wing surface, vortex generators can increase the critical Mach number, which is the speed at which air flowing over the wing surface reaches supersonic speed. This helps to prevent the formation of shock waves, which can lead to air separation and reduced lift.

Vortex generators also help to reduce the intensity of shock wave-induced air separation and decrease span wise flow at high Mach numbers, which improves the aircraft's stability and handling.

Hence the correct answer is C.

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Answer:

what?

Explanation:

The provided recursive method "containsVowel" determines if a string contains a vowel by checking the first character and recursively checking the remaining substring. The "isVowel" helper method identifies if a character is a vowel.

Here is the recursive, boolean-valued method that accepts a string and returns true if the string contains a vowel. A string contains a vowel if: The first character of the string is a vowel, or The rest of the string (beyond the first character) contains a vowel.

```
public static boolean containsVowel(String str) {
   if (str.length() == 0) {  // empty string doesn't have vowel
       return false;
   } else if (isVowel(str.charAt(0))) {  // check if first character is vowel
       return true;
   } else {  // first character is not vowel, check remaining string
       return containsVowel(str.substring(1));
   }
}

public static boolean isVowel(char c) {
   return "AEIOUaeiou".indexOf(c) != -1;
}
```

The contains Vowel method is a recursive method that returns a boolean value. It takes a string as input. Here is how it works:

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Answer:

a) \(\dot m = 16.168\,\frac{kg}{s}\), b) \(v_{out} = 680.590\,\frac{m}{s}\), c) \(\dot W_{out} = 18276.307\,kW\)

Explanation:

A turbine is a steady-state devices which transforms fluid energy into mechanical energy and is modelled after the Principle of Mass Conservation and First Law of Thermodynamics, whose expressions are described hereafter:

Mass Balance

\(\frac{v_{in}\cdot A_{in}}{\nu_{in}} - \frac{v_{out}\cdot A_{out}}{\nu_{out}} = 0\)

Energy Balance

\(-q_{loss} - w_{out} + h_{in} - h_{out} = 0\)

Specific volumes and enthalpies are obtained from property tables for steam:

Inlet (Superheated Steam)

\(\nu_{in} = 0.055665\,\frac{m^{3}}{kg}\)

\(h_{in} = 3650.6\,\frac{kJ}{kg}\)

Outlet (Liquid-Vapor Mix)

\(\nu_{out} = 5.89328\,\frac{m^{3}}{kg}\)

\(h_{out} = 2500.2\,\frac{kJ}{kg}\)

a) The mass flow rate of the steam is:

\(\dot m = \frac{v_{in}\cdot A_{in}}{\nu_{in}}\)

\(\dot m = \frac{\left(60\,\frac{m}{s} \right)\cdot (0.015\,m^{2})}{0.055665\,\frac{m^{3}}{kg} }\)

\(\dot m = 16.168\,\frac{kg}{s}\)

b) The exit velocity of steam is:

\(\dot m = \frac{v_{out}\cdot A_{out}}{\nu_{out}}\)

\(v_{out} = \frac{\dot m \cdot \nu_{out}}{A_{out}}\)

\(v_{out} = \frac{\left(16.168\,\frac{kg}{s} \right)\cdot \left(5.89328\,\frac{m^{3}}{kg} \right)}{0.14\,m^{2}}\)

\(v_{out} = 680.590\,\frac{m}{s}\)

c) The power output of the steam turbine is:

\(\dot W_{out} = \dot m \cdot (-q_{loss} + h_{in}-h_{out})\)

\(\dot W_{out} = \left(16.168\,\frac{kg}{s} \right)\cdot \left(-20\,\frac{kJ}{kg} + 3650.6\,\frac{kJ}{kg} - 2500.2\,\frac{kJ}{kg}\right)\)

\(\dot W_{out} = 18276.307\,kW\)

Answer:

The mass flow rate = 16.17kg/sec

The exit velocity = 680.84 m/sec

\(W_o_u_t=14562.37kw\)

Explanation:

Attached is a copy of the diagram to this question

Let's write out the given data

T= 600⁰C

\(v_1=0.055665m^3/kg\\h_1=3650.60kJ/kg\\at p = 25kpa\\v_f=0.00102m^3/kg\\v_g=6.2034m^3/kg\\hfg=2345.5kJ/kg\\hf=271.96kJ/kg\)

The enthalpy at exit is

\(h_2=271.96+\frac{95}{100} *2345.50=2500.185 kJ/kg\\\)

The specific volume is

\(v_2=v_f+x_2vfg\\\\v_2=0.001020+\frac{95}{100}(6.2034-0.00102)=5.893m^3/kg\)

a) To calculate the mass flow rate

\(m=\frac{V_1m}{v_1}=\frac{60*0.015}{0.055665}\\m=16.17kg/s\)

The mass flow rate is 16.17kg/s

b) Exit velocity

\(v_2=\frac{mV_2}{A_2}=\frac{16.17*5.893}{0.14}=680.64m/sec\)

The exit velocity is 680.64m/s

c) The power output

\(W_o_u_t=-mQ_o_u_t+m(h_1-h_2+\frac{v_1^2-v_2^2}{2})\\ W_o_u_t=-(16.17*20)+16.17(3650.60-2500.185+\frac{60^2-680.64^2}{2}*\frac{1}{1000} \\W_o_u_t=14562.37kw\)

The power output of the machine is 14562.37kw

Answer:

a. 6.48 kW b. 1678.34 kg c. 777.92 W d. 201.42 kg

Explanation:

(a) the rate of heat transfer to the iced water in the tank

The rate of heat transfer to the outer surface of the spherical tank is P = P₁ + P₂ where P₁ = rate of heat transfer to the outer surface by radiation and P₂ = rate of heat transfer to the outer surface by convection through air

P₁ = εσAT⁴ where ε = emissivity of outer surface ε= 1, σ = Stefan-Boltzmann constant = 5.67 × 10⁻⁸ W/m-K⁴, A = area of outer surface of spherical tank = 4πR² where R = outer radius of spherical tank = inner radius + thickness = inner diameter/2 + 5 cm = 2 m/2 + 0.05 m = 1 m + 0.05 m = 1.05 m and T = temperature of surroundings = 20 °C = 273 + 20 = 293 K.

P₁ = εσAT⁴

P₁ = 1 × 5.67 × 10⁻⁸ W/m-K⁴ × 4π(1.05 m)² × (293 K)⁴

P₁ = 1 × 5.67 × 10⁻⁸ W/m-K⁴ × 4π(1.1025 m²) × 7370050801 K⁴

P₁ = 184285909263.7647π × 10⁻⁸ W

P₁ = 578951258703.16 × 10⁻⁸ W

P₁ = 5789.51 W

P₂ = hA(T - T₁) where h = coefficient of thermal convection of air = 2.5 W/m²-K, A = outer surface area of spherical tank = 4πR², T = temperature of surroundings = 20 °C = 273 + 20 = 293 K and T₁ = temperature of outer surface of spherical tank = 0 °C = 273 + 0 = 273 K.  

P₂ = hA(T - T₁)

P₂ = 2.5 W/m²-K × 4π(1.05 m)² × (293 K - 273 K)

P₂ = 2.5 W/m²-K × 4π(1.1025 m²) × 20 K

P₂ = 220.5π W

P₂ = 692.72 W

So, P = P₁ + P₂ = 5789.51 W + 692.72 W = 6482.23 W

Since we are neglecting the thermal resistance of the spherical tank, the rate of heat absorption of the outer surface equals the rate of heat absorption in the inner surface. The rate of heat absorption at the inner surface equals the rate of heat transfer to the iced water.

So, rate of heat transfer to the iced water = P = 6482.23 W = 6.48223 kW 6.48 kW

(b) the amount of ice at 0°C that melts during a 24-h period. The heat of fusion of water is 333.7 kJ/kg.

Since the amount of heat, Q = Pt where P = heat transfer rate to iced water = 6482.23 W and t = time = 24 h = 24 h × 60 min/h × 60 s/min = 86400 s.

Also, Q = the latent heat required to melt the ice at 0 °C = mL where m = mass of ice melted and L = latent heat of fusion of ice = 333.7 kJ/kg

So, Pt = mL

m = Pt/L

= 6482.23 W × 86400 s/333.7 × 10³ J/kg

= 560064672/333.7 × 10³

= 1678.34 kg

(c) the rate of heat transfer to the iced water in the tank for this double-walled tank configuration

Since P is the rate of heat transfer to the outer surface, this is also the rate of heat transfer to the outer 0.5 cm thick wall = P₃ = 6482.23 W

P₃ = kA(T - T₃)/d where k = thermal conductivity of outer wall = 15 W/m²-K

A = surface area of outer wall = 4πR'² where R' = radius of outer wall = radius of inner wall + thickness of inner wall + thickness of vacuum + thickness of outer wall = 2.0 m/2 + 0.5 cm + 1.5 cm + 0.5 cm = 1 m + 2.5 cm = 1 m + 0.025 m = 1.025 m, T = temperature of surroundings = 20 °C = 273 + 20 = 293 K, T₃ = temperature of inner surface of outer wall of spherical tank and d = thickness of outer surface of tank = 0.5 cm = 0.05 m

P₃ = kA(T - T₃)/d

making T₃ subject of the formula, we have

P₃d = kA(T - T₃)

P₃d/kA = (T - T₃)

T₃ = T - P₃d/kA

substituting the values of the variables into the equation, we have

T₃ = 293 K - 6482.23 W × 0.05 m/[15 W/m-K × 4π(1.025 m)²]

T₃ = 293 K - 324.1115 Wm/[15 W/m-K × 4π(1.050625 m²)]

T₃ = 293 K - 324.1115 Wm/[63.0375π W/m-K)]

T₃ = 293 K - 324.1115 Wm/[198.0381 W/m-K)]

T₃ = 293 K - 1.64 K

T₃ = 291.36 K

Since the 1.5 cm thick air space is evacuated, all the heat gets to the inner 0.5 cm thick wall by radiation.

So P = εσAT₃⁴

P₄ = εσAT₃⁴ where ε = emissivity of outer surface ε = 0.15, σ = Stefan-Boltzmann constant = 5.67 × 10⁻⁸ W/m-K⁴, A = area of inner surface of outer wall of spherical tank = 4πR"² where R" = outer radius of inner thick wall of spherical tank = inner radius + thickness of inner wall = inner diameter/2 + 0.5 cm = 2 m/2 + 0.005 m = 1 m + 0.005 m = 1.005 m and T = temperature of outer wall = 291.36 K.

P₄ = 0.15 × 5.67 × 10⁻⁸ W/m-K⁴ × 4π(1.005 m)² × (291.36 K)⁴

P₄ = 0.15 × 5.67 × 10⁻⁸ W/m-K⁴ × 4π(1.010025 m²) × 7206422389.51 K⁴

P₄ = 24762024365.028π × 10⁻⁸ W

P₄ = 77792193833.18 × 10⁻⁸ W

P₄ = 777.92 W

Now P₄ is the heat transfer rate to the inner surface which is at temperature T₄

Since T₄ = 0 °C, P₄ is the rate of heat transfer to the iced water

So, rate of heat transfer to the iced water P₄ = 777.92 W

(d) the amount of ice at 0°C that melts during a 24-h period for this double-walled tank configuration

Since the amount of heat, Q = P₄t where P₄ = heat transfer rate to iced water = 777.92 W and t = time = 24 h = 24 h × 60 min/h × 60 s/min = 86400 s.

Also, Q = the latent heat required to melt the ice at 0 °C = mL where m = mass of ice melted and L = latent heat of fusion of ice = 333.7 kJ/kg

So, P₄t = mL

m = P₄t/L

= 777.92 W × 86400 s/333.7 × 10³ J/kg

= 67212288/333.7 × 10³

= 201.42 kg