Consider steady one-dimensional heat conduction through a plane wall, a cylindrical shell, and a spherical shell of uniform thickness with constant thermophysical properties and no thermal energy generation. The geometry in which the variation of temperature in the direction of heat transfer will be linear is (a) Cylindrical wall (b) Plane shell (c) Spherical shell (d) All of them (e) None of them

Answer:

331809.5gallon/hr or 92.16gallon/s

Explanation:

What is the peak runoff discharge for a 1.3 in/hr storm event from a 9.4-acre concrete-paved parking lot (C

convert 9.4 acre to inches we have=5.896*10^7

How to calculate Peak runoff discharge

1. take the dimension of the roof

2. multiply the dimension by the n umber of inches of rainfall

3. Divide by 231 to get gallon equivalence (because 1 gallon = 231 cubic inches)

5.896*10^7*1.3

7.66*10^7 cubic inches/hr

1 gallon=231 cubic inches

7.66*10^7 cubic inches=331809.5gallon/hr or 92.16gallon/s

this is gotten by converting 1 hr to seconds

331809.5gallon/hr /3600s=92.16gallon/s

Answer:

symbolic machine code.

Explanation:

The instructions in the language are closely linked to the machine's architecture.

Answer:

Work done = 12,000J = 12KJ

Explanation:

This process is an Isobaric one i.e a process at constant pressure

From the formula for Workdone = Integral (PdV)

Work done = P ( Vf - Vi)

Plugging the values in the equation,

Work done = 200,000 ( 0.1 - 0.04)

Work done = 12,000J = 12KJ

Answer:

c.attenuation

Explanation:

In telecommunication, it is called attenuation of a signal, be it acoustic, electrical or optical, to the loss of power suffered by it when passing through any transmission medium.

Attenuation is usually not expressed as a difference in powers but in logarithmic units such as decibels, which are more comfortable to use when calculating.

The attenuation is expressed in decibels (db) and the power is measured with this formula:

α = 10 * log (P1 / P2)

where P is the power both initial and final.

Answer:

1.1451 x 104 (11451.13)mg/m3

Explanation:

1 ppmv is defined as one volume of a contaminant or solid(CO)(mL) in 1 x 106 volume of solvent/water.

1ppmv = 1mL/m3

Concentration in mg/m3 = volume in ppm x molecular weight x pressure(kPa)/( gas constant x temperature(K)

Molecular weight of CO = 12 + 16

= 28g/mol

Temperature = 273.15 + 25

= 298.15K

Pressure = 1 x 101.325kPa

= 101.325kPa

Ppmv = 1 x 10-4ppmv

Gas constant, R = 8.3144 L.kPa/mol.K

Concentration in mg/m3 = (1 x 104 * 28 * 101.325)/(8.3144 * 298)

= 1.1451 x 104mg/m3

= 11451.13 mg/m3

Answer:

The three phase full load secondary amperage is 2775.7 A

Explanation:

Following data is given,

S = Apparent Power = 1000 kVA

No. of phases = 3

Secondary Voltage: 208 V/120 V (Here 208 V is three phase voltage and 120 V is single phase voltage)

Since,

[tex]V_{1ph} =\frac{ V_{3ph}}{\sqrt{3} }\\V_{1ph) = \frac{208}{\sqrt{3} }\\[/tex]

[tex]V_{1ph} = 120 V[/tex]

The formula for apparent power in three phase system is given as:

[tex]S = \sqrt{3} VI[/tex]

Where:

S = Apparent Power

V = Line Voltage

I = Line Current

In order to calculate the Current on Secondary Side, substituting values in above formula,

[tex]1000 kVA = \sqrt{3} * (208) * (I)\\1000 * 1000 = \sqrt{3} * (208) * (I)\\I = \frac{1000 * 1000}{\sqrt{3} * (208) }\\ I = 2775.7 A[/tex]

The displacement of a particle when its velocity is zero can be found by differentiating the displacement function to get the velocity function, then determining when this velocity is zero. By substituting the corresponding time back into the displacement equation, we calculate the displacement at that time. Additionally, for acceleration, we can integrate to find velocity and displacement with given initial conditions.

The question involves determining the displacement of a particle when its velocity is zero. Since velocity is the derivative of displacement with respect to time, we first need to find the velocity function by differentiating the displacement function x = t3
- 6t2 + 9t + 3. Once we have the velocity function, we can locate the time t at which the velocity is zero. The displacement at this time will be our required value. Similarly, if given an acceleration function, a(t) = t - 1, and initial conditions for velocity and position, we integrate the acceleration to find the velocity, and again to find the displacement.

To address the specific question of the student regarding the particle's position, acceleration, and total distance traveled at t = 5 seconds, we would substitute t into our established functions for position and acceleration accordingly.

If the initial acceleration of a particle is to be calculated from the provided differential equation dv(t)/dt = 6.0 - 3v(t), with the initial condition of the particle being at rest, we would substitute t = 0 into this equation to find the initial acceleration.

Answer:

appPane.getChildren().add(nameField);

Explanation:

The question is related to JavaFx which is a set of packages used for making interactive graphical user interface that contains graphical components for better user experience.

Pane is a JavaFx component which is used for adjusting the position and size of a graphical component for a given scene.

We can create a pane object by appPane = new Pane(); command.

TextField is a JavaFx object that is used for displaying a line of text. We have various functions available to set properties of TextField object.

We can create a TextField object by  nameField = new TextField(); command.

A pane can have multiple set of graphical components for various parts of an application. These are called children.

Hence we can write the following statement to add a TextField object nameField to a Pane object appPane.

appPane.getChildren().add(nameField);

Answer:

It serves as a guarantee that the contractor who wins the bid will honor the terms of the bid after the contract is signed.

Explanation:

A bid bond is a type of construction bond that protects the obligee in a  construction bidding process.

A bid bond typically involves three parties:

The obligee; the owner or developer of the construction project under bid. The principal; the bidder or proposed contractor.

The surety; the agency that issues the bid bond to the principal example insurance company or bank.

A bid bond generally serves as a guarantee that the contractor who wins the bid will honor the terms of the bid after the contract is signed.

Contractors are required to submit a bid bond to ensure they are financially able and willing to complete the project they bid on, preserving the integrity of the bidding process by preventing cost overruns, favoritism, and corruption.

A contractor is normally required to submit a bid bond when making a proposal to an owner on a competitively bid contract to ensure that the contractor is serious and financially able to carry out the project. The bid bond is a form of guarantee that the contractor will enter into the contract at the bid price if awarded the project and provides a financial penalty if the contractor fails to do so. Municipalities and other entities require this bond to prevent situations where a contractor might win a bid and then refuse to complete the work or ask for additional funds to finish the project (“cost overruns”). This is significant in a competitive bidding environment where fairness and the avoidance of favoritism or corruption are essential, as it stops anyone from being favored over others and ensures the project will be completed at the bid price.

Tenders, which are bid documents that outline the project plan and cost, play a crucial role in the construction industry. A well-prepared tender demonstrates that the projected cost reflects a balance between safety, environmental friendliness, and profitability. In contrast, a sealed-bid auction often used in these processes, requires strategic bidding and is susceptible to collusion or bribery to uncover competitor's bid details. Hence, a bid bond safeguards the process by providing a financial deterrent against such unethical practices.

Answer:

The peak amplitude velocity is found to be 0.028 m/s.

Explanation:

Given that the displacement of an oscillating mass is:

Displacement = x = 0.004 Sin(7t)

Now, to find out the velocity of this particle, me have to take derivative of 'x' with respect to 't'.

Velocity = V = dx/dt = (7)(0.004)Cos (7t)

V = 0.028 Cos (7t)

Now, for the maximum displacement, the value of the Cos function must be maximum. And we know that the maximum value of Cos function is 1. Thus, to get maximum displacement, we set the value of Cos (7t) to be equal to 1.

Vmax = 0.028(1)

Vmax = 0.028 m/s (Peak Amplitude of Velocity)

Answer:

Part A:

[tex]1500\ KPa= 31328.145 \frac{lb}{ft^2}[/tex]

Part B:

[tex]1500 KPa=217.55656 \frac{lb}{in^2}[/tex](Psi)

Explanation:

Part A:

Line Pressure is 1500 KPa

We need a conversion factor which converts KPa to lb/ft^2.

[tex]20.88543 \frac{lb}{ft^2}= 1\ KPa[/tex]

In order to convert 1500 KPa to lb/ft^2, we proceed as:

[tex]1\ KPa=20.88543 \frac{lb}{ft^2} \\1500\ KPa= 1500 KPa*20.88543 \frac{lb}{ft^2.KPa}\\1500\ KPa= 31328.145 \frac{lb}{ft^2}[/tex]

1500 KPa is 31328.145 lb/ft^2

Part B:

We will use the same procedure we did in Part A:

1 ft= 12 in

[tex](1\ ft)^2=(12\ in)^2\\1 ft^2=144 in^2[/tex]

Converting [tex]1500 KPa\ into\ \frac{lb}{in^2}[/tex]

[tex]1500\ KPa= 1500 KPa*20.88543 \frac{lb}{ft^2.KPa} * \frac{ft^2}{144\ in^2}[/tex]

[tex]1500 KPa=217.55656 \frac{lb}{in^2}[/tex](Psi)

1500 KPa is 217.55656 lb/in^2 (psi)

Answer:

This hydroelectric power plant has the capability of producing around 78.4 MW of electric energy.

Explanation:

First, we calculate the pressure the water exerts or gains when it travels 200 m below the intake. W have the data:

Density of water = ρ = 1000 kg/m^3

Acceleration due to gravity = g = 9.8 m/s²

Height lost = h = 200 m

Volume flow rate = 40 m^3/s

The pressure difference, is now given as:

ΔP = ρgh

ΔP = (1000 kg/m^3)(9.8 m/s²)(200 m)

ΔP = 1960000 Pa = 1.96 MPa

Now, we calculate the power capacity of this flow:

Power = ΔP(Volume flow rate)

Power = (1960000 N/m²)(40 m^3/s)

Power = 78400000 Watt

Power = 78.4 MW

Thus, this plant can produce at max 78.4 MW. The actual power produced will be slightly less than this due of the losses and efficiency of turbine system used.

Answer:

Answer code is given below along with comments to explain each step

Explanation:

Method 1:

// size of the array numbers to store 1 to 5 integers

int Size = 5;  

// here we have initialized our array numbers, the type of array is integers and size is 5, right now it is empty.

ArrayList<Integer> numbers = new ArrayList<Integer>(Size);

// lets store values into the array numbers one by one

numbers.add(1)

numbers.add(2)

numbers.add(3)

numbers.add(4)

numbers.add(5)

// here we are finding the first element in the array numbers by get() function, the first element is at the 0th position

int firstElement = numbers.get(0);

// to print the first element in the array numbers

System.out.println(firstElement);  

Method 2:

int Size = 5;  

ArrayList<Integer> numbers = new ArrayList<Integer>(Size);

// here we are using a for loop instead of adding manually one by one to store the values from 1 to 5

for (int i = 0; i < Size; i++)  

{

// adding the values 1 to 5 into array numbers

  numbers.add(i);  

}

//here we are finding the first element in the array numbers

int firstElement = numbers.get(0);  

// to print the first element in the array

System.out.println(firstElement);  

Answer:

The pressure exerted by this man on ground

(a) if he stands on both feet is 8.17 KPa

(b) if he stands on one foot is 16.33 KPa

Explanation:

(a)

When the man stand on both feet, the weight of his body is uniformly distributed around the foot imprint of both feet. Thus, total area in this case will be:

Area = A = 2 x 480 cm²

A = 960 cm²

A = 0.096 m²

The force exerted by man on his area will be equal to his weight.

Force = F = Weight

F = mg

F = (80 kg)(9.8 m/s²)

F = 784 N

Now, the pressure exerted by man on ground will be:

Pressure = P = F/A

P = 784 N/0.096 m²

P = 8166.67 Pa = 8.17 KPa

(b)

When the man stand on one foot, the weight of his body is uniformly distributed around the foot imprint of that foot only. Thus, total area in this case will be:

Area = A = 480 cm²

A = 0.048 m²

The force exerted by man on his area will be equal to his weight, in this case, as well.

Force = F = Weight

F = mg

F = (80 kg)(9.8 m/s²)

F = 784 N

Now, the pressure exerted by man on ground will be:

Pressure = P = F/A

P = 784 N/0.048 m²

P = 16333.33 Pa = 16.33 KPa

Answer:

0.867

Explanation:

The driver population factor ([tex]f_{p}[/tex])can be estimated using the equation below:

[tex]f_{p} = \frac{V}{PHF*N*f_{HV}*v_{p}}[/tex]

The value of the heavy vehicle factor ([tex]f_{HV}[/tex]) is determined below:

The values of the [tex]E_{T}[/tex] = 2 and [tex]E_{R}[/tex] = 3 are gotten from the tables for the RVs, trucks and buses upgrades for passenger-car equivalents. Therefore:

[tex]f_{HV}[/tex] = 1/[1+0.08(2-1)+0.06(3-1)] = 1/[1+0.08+0.12] = 1/1.2 = 0.833

Furthermore, the vp is taken as 2250 pc/(h*In) from the table of LOS criteria for lane freeway using the 15 minutes flow rate. Therefore:

[tex]f_{p}[/tex] = 3900/[0.8*3*0.833*2250] = 3900/4498.2 = 0.867

Answer:

The potential voltage range across a 2.2 kΩ 10% tolerance resistor when current of  4 sin 44t mA is flowing through the element is between a range of 7.92sin44t and 9.68sin44t volts.

Explanation:

Given that there is 10% tolerance for the 2.2 kΩ resistor, this implies that the resistance would range between 2,200 — 10% of 2,200 and 2,200 + 10% of 2,200, which is:

(i) 2,200 — 10% of 2,200 = 2,200 — 220 = 1,980 Ω, and

(ii) 2,200 + 10% of 2,200 = 2,200 + 220 = 2,420 Ω

Therefore, we will calculate the potential voltage for 1,980 Ω and 2,420 Ω if the current flowing through the element is 4sin44t mA:

(a) The potential voltage for a resistance of 1,980 Ω: we will use the formula: potential voltage v = i × R

Where i = 4sin44t mA = 0.004sin44t A, and R = 1,980 Ω

The potential voltage = v = 1,980 × 0.004sin44t = 7.92sin44t (in volts)

(b) The potential voltage for a resistance of 2,420 Ω: we will use the formula: potential voltage v = i × R

Where i = 4sin44t mA = 0.004sin44t A, and R = 2,420 Ω

The potential voltage = v = 2,420 × 0.004sin44t = 9.68sin44t (in volts)

Answer:

a) Rankine

Net work output = 719.1 KJ/kg

Thermal Eff = 0.294

a) Carnot

Net work output = 563.2 KJ/kg

Thermal Eff = 0.294

For T-s diagrams see attachments

Explanation:

Part a    Rankine Cycle

The obtained data from water property tables:

[tex]P_{L,sat liquid} = 50 KPa \\v_{1} = 0.00103m^3/kg\\\\ h_{1} = 340.54KJ/kg\\\\P_{H} = 5000KPa\\h_{2} = h_{1} + v_{1} *(P_{H} - P_{L} )\\h_{2} =350.54 + (0.00103)*(5000 - 50)\\\\h_{2} = 345.64KJ/kg\\\\P_{H,satsteam} = 5000KPa\\s_{3} = 5.9737KJ/kgK\\\\h_{3} = 2794.2KJ/kg\\\\s_{3} = s_{4} = 5.9739KJ/kgK\\P_{L}= 50KPa\\\\h_{4}= 2070KJ/kg\\\\[/tex]

Heat transferred from boiler

[tex]q_{b} = h_{3}-h_{2}\\q_{b}=2794.2-345.64\\\\q_{b} =2448.56KJ/kg\\\\[/tex]

Heat transferred from condenser

[tex]q_{c} = h_{4}-h_{1}\\q_{b}=2070-340.54\\\\q_{b} =1729.46KJ/kg\\\\[/tex]

Thermal Efficiency

[tex]u_{R} = 1- \frac{q_{c}}{q_{b}}\\\\u_{R} = 1 - \frac{1729.46}{2448.56}\\\\u_{R} =0.294[/tex]

Net work output

[tex]w_{R} = q_{b}-q_{c}\\w_{R} = 2448.56-1729.46\\\\w_{R}=719.1KJ/kg[/tex]

Part b    Carnot Cycle

The obtained data from water property tables:

[tex]P_{H,sat-steam} = 5000KPa\\T_{3} = 263.94 C\\s_{3} = 5.9737KJ/kgK\\\\h_{3} = 2794.2KJ/kg\\\\T_{2,sat-liquid} = T_{3} = 263.94C\\s_{2} = 2.920KJ/kgK\\\\h_{2} = 1150KJ/kg\\\\P_{L} = 50KPa\\s_{1}=s_{2} = 2.920KJ/kgK\\\\h_{1} = 989KJ/kg\\\\s_{3} = s_{4} = 5.9737KJ/kgK\\P_{L} = 50KPa\\\\h_{4} = 2070KJ/kg[/tex]

Heat transferred from boiler

[tex]q_{b} = h_{3}-h_{2}\\q_{b}=2794.2-1150\\\\q_{b} =1644.2KJ/kg\\\\[/tex]

Heat transferred from condenser

[tex]q_{c} = h_{4}-h_{1}\\q_{b}=2070-989\\\\q_{b} =1081KJ/kg\\\\[/tex]

Thermal Efficiency

[tex]u_{C} = 1- \frac{q_{c}}{q_{b}}\\\\u_{C} = 1 - \frac{1081}{1644.2}\\\\u_{C} =0.343[/tex]

Net work output

[tex]w_{C} = q_{b}-q_{c}\\w_{C} = 1644.2-1081\\\\w_{C}=563.2KJ/kg[/tex]

Answer:

The final pressure in the vessel is 2.270atm

Explanation:

Mass of PCl3 = 24.5g, MW of PCl3 = 137.5g/mole, number of moles of PCl3 = 24.5/137.5 = 0.178mole

Mass of O2 = 3g, MW of O2 = 32g/mole, number of moles of O2 = 3/32 = 0.094moles

Total moles of rectant (n) = 0.178mole + 0.094mole = 0.272mole

From ideal gas equation

PV = nRT

P (pressure of reactants) = nRT/V

n = 0.272mole, R = 82.057cm^3.atm/gmol.K, T = 15°C = 15+273K = 288K, V = 4.6L = 4.6×1000cm^3 = 4600cm^3

P = 0.272×82.057×288/4600 = 1.397atm

From pressure law

P1/T1 = P2/T2

P2 (final pressure in the vessel) = P1T2/T1

P1 = 1.397atm, T1 = 288K, T2 = 195°C = 195+273K = 468K

P2 = 1.397×468/288 = 2.270atm

Answer:

[tex]Ma=\frac{260}{330} \\Ma=0.7878[/tex]

So, Ma < 1  Flow is Subsonic

Explanation:

Mach Number:

Mach Number is the ratio of speed of the object to the speed of the sound. It is used to categorize the speed of the object on the basis of mach number as sonic, supersonic and hyper sonic. (It is a unit less quantity)

Mach < 1       Subsonic

Mach > 1       Supersonic

Ma= Speed of the object/Speed of the sound

[tex]Ma=\frac{260}{330} \\Ma=0.7878[/tex]

So, Ma < 1 Flow is Subsonic

To provide 2 kW of electrical power, the fuel cell requires an active area of approximately 1333.3 cm2, calculated by dividing the total power required by the fuel cell's power density.

The question involves calculating the area of a fuel cell required to deliver a specific power output, given its operational parameters. This problem is grounded in the principles of electrochemical engineering and requires an understanding of how power density influences the design and performance of fuel cells. Given that the fuel cell operates at a current density (j) of 3 A/cm2 and a power density (P) of 1.5 W/cm2, we are tasked with determining the active area needed to supply 2 kW of electrical power.

The formula to calculate the area is straightforward:

Therefore, to deliver 2 kW of electrical power, a fuel cell with an active area of approximately 1333.3 cm2 is required.

Explanation:

The obtained data from water properties tables are:

Point 1 (condenser exit) @ 8 KPa, saturated fluid

[tex]h_{f} = 173.358 \\h_{fg} = 2402.522[/tex]

Point 2 (Pump exit) @ 18 MPa, saturated fluid & @ 4 MPa, saturated fluid

[tex]h_{2a} = 489.752\\h_{2b} = 313.2[/tex]

Point 3 (Boiler exit) @ 18 MPa, saturated steam & @ 4 MPa, saturated steam

[tex]h_{3a} = 2701.26 \\s_{3a} = 7.1656\\h_{3b} = 2634.14\\s_{3b} = 7.6876[/tex]

Point 4 (Turbine exit) @ 8 KPa, mixed fluid

[tex]x_{a} = 0.8608\\h_{4a} = 2241.448938\\x_{b} = 0.9291\\h_{4b} = 2405.54119[/tex]

Calculate mass flow rates

Part a) @ 18 MPa

mass flow

[tex]\frac{100*10^6 }{w_{T} - w_{P}} = \frac{100*10^3 }{(h_{3a} - h_{4a}) - (h_{2a} - h_{f})}\\\\= \frac{100*10^ 3}{(2701.26 - 2241.448938 ) - (489.752 - 173.358)}\\\\= 697.2671076 \frac{kg}{s} = 2510161.587 \frac{kg}{hr}[/tex]

Heat transfer rate through boiler

[tex]Q_{in} = mass flow * (h_{3a} - h_{2a})\\Q_{in} = (697.2671076)*(2701.26-489.752)\\\\Q_{in} = 1542011.787 W[/tex]

Heat transfer rate through condenser

[tex]Q_{out} = mass flow * (h_{4a} - h_{f})\\Q_{out} = (697.2671076)*(2241.448938-173.358)\\\\Q_{out} = 1442011.787 W[/tex]

Thermal Efficiency

[tex]n = \frac{W_{net} }{Q_{in} } = \frac{100*10^3}{1542011.787} \\\\n = 0.06485[/tex]

Part b) @ 4 MPa

mass flow

[tex]\frac{100*10^6 }{w_{T} - w_{P}} = \frac{100*10^3 }{(h_{3b} - h_{4b}) - (h_{2b} - h_{f})}\\\\= \frac{100*10^ 3}{(2634.14 - 2405.54119 ) - (313.12 - 173.358)}\\\\= 1125 \frac{kg}{s} = 4052374.235 \frac{kg}{hr}[/tex]

Heat transfer rate through boiler

[tex]Q_{in} = mass flow * (h_{3b} - h_{2b})\\Q_{in} = (1125.65951)*(2634.14-313.12)\\\\Q_{in} = 2612678.236 W[/tex]

Heat transfer rate through condenser

[tex]Q_{out} = mass flow * (h_{4b} - h_{f})\\Q_{out} = (1125)*(2405.54119-173.358)\\\\Q_{out} = 2511206.089 W[/tex]

Thermal Efficiency

[tex]n = \frac{W_{net} }{Q_{in} } = \frac{100*10^3}{1542011.787} \\\\n = 0.038275[/tex]

For (a) 18 MPa:

- The mass flow rate of steam is approximately 238,050 kg/h.

- The heat transfer rates are approximately 674,970 kW (boiler) and 481,360 kW (condenser).

- The thermal efficiency is approximately 14.81%.

For (b) 4 MPa:

- The mass flow rate of steam is approximately 187,320 kg/h.

- The heat transfer rates are approximately 480,040 kW (boiler) and 380,450 kW (condenser).

- The thermal efficiency is approximately 20.83%.

Explanation and Calculation

To analyze the ideal Rankine cycle, we need to follow these steps for each case:

Case (a): 18 MPa

1. Determine Specific Enthalpies

- Condenser exit: Saturated liquid at 8 kPa.

 From steam tables: [tex]\( h_1 \approx 191.81 \, \text{kJ/kg} \)[/tex]

-Pump exit: Compressed liquid at 18 MPa.

 Using [tex]\( h_2 = h_1 + v_1 \Delta P \)[/tex]

 [tex]\[ v_1 \approx 0.001 \, \text{m}^3/\text{kg} \][/tex]

 [tex]\[ h_2 \approx 191.81 + 0.001 \times (18000 - 8) \approx 209.81 \, \text{kJ/kg} \][/tex]

- Boiler exit: Saturated vapor at 18 MPa.

 From steam tables: [tex]\( h_3 \approx 2821.6 \, \text{kJ/kg} \)[/tex]

-Turbine exit: Expanding to 8 kPa.

 From steam tables: [tex]\( h_4 \approx 2201.4 \, \text{kJ/kg} \)[/tex]

2. Mass Flow Rate of Steam

Using the power output:

[tex]\[ \dot{W}_{\text{net}} = \dot{m} (h_3 - h_4 - (h_2 - h_1)) \][/tex]

[tex]\[ 100 \times 10^6 = \dot{m} (2821.6 - 2201.4 - (209.81 - 191.81)) \][/tex]

[tex]\[ 100 \times 10^6 = \dot{m} \times 420.2 \][/tex]

[tex]\[ \dot{m} \approx 238050 \, \text{kg/h} \][/tex]

3. Heat Transfer Rates

- Boiler:

 [tex]\[ \dot{Q}_{\text{in}} = \dot{m} (h_3 - h_2) \][/tex]

 [tex]\[ \dot{Q}_{\text{in}} = 238050 \times (2821.6 - 209.81) \approx 674.97 \times 10^3 \, \text{kW} \][/tex]

- Condenser:

 [tex]\[ \dot{Q}_{\text{out}} = \dot{m} (h_4 - h_1) \][/tex]

 [tex]\[ \dot{Q}_{\text{out}} = 238050 \times (2201.4 - 191.81) \approx 481.36 \times 10^3 \, \text{kW} \][/tex]

4. Thermal Efficiency

[tex]\[ \eta = \frac{\dot{W}_{\text{net}}}{\dot{Q}_{\text{in}}} \][/tex]

[tex]\[ \eta = \frac{100 \times 10^3}{674.97 \times 10^3} \approx 14.81\% \][/tex]

Case (b): 4 MPa

1. Determine Specific Enthalpies

- Condenser exit: Saturated liquid at 8 kPa.

 From steam tables: [tex]\( h_1 \approx 191.81 \, \text{kJ/kg} \)[/tex]

- Pump exit: Compressed liquid at 4 MPa.

 Using [tex]\( h_2 = h_1 + v_1 \Delta P \)[/tex]

 [tex]\[ v_1 \approx 0.001 \, \text{m}^3/\text{kg} \][/tex]

 [tex]\[ h_2 \approx 191.81 + 0.001 \times (4000 - 8) \approx 195.81 \, \text{kJ/kg} \][/tex]

- Boiler exit: Saturated vapor at 4 MPa.

 From steam tables: [tex]\( h_3 \approx 2749.7 \, \text{kJ/kg} \)[/tex]

- Turbine exit: Expanding to 8 kPa.

 From steam tables: [tex]\( h_4 \approx 2222.0 \, \text{kJ/kg} \)[/tex]

2. Mass Flow Rate of Steam

Using the power output:

[tex]\[ \dot{W}_{\text{net}} = \dot{m} (h_3 - h_4 - (h_2 - h_1)) \][/tex]

[tex]\[ 100 \times 10^6 = \dot{m} (2749.7 - 2222.0 - (195.81 - 191.81)) \][/tex]

[tex]\[ 100 \times 10^6 = \dot{m} \times 531.9 \][/tex]

[tex]\[ \dot{m} \approx 187320 \, \text{kg/h} \][/tex]

3. Heat Transfer Rates

- Boiler:

 [tex]\[ \dot{Q}_{\text{in}} = \dot{m} (h_3 - h_2) \][/tex]

 [tex]\[ \dot{Q}_{\text{in}} = 187320 \times (2749.7 - 195.81) \approx 480.04 \times 10^3 \, \text{kW} \][/tex]

- Condenser:

[tex]\[ \dot{Q}_{\text{out}} = \dot{m} (h_4 - h_1) \][/tex]

 [tex]\[ \dot{Q}_{\text{out}} = 187320 \times (2222.0 - 191.81) \approx 380.45 \times 10^3 \, \text{kW} \][/tex]

4. Thermal Efficiency

[tex]\[ \eta = \frac{\dot{W}_{\text{net}}}{\dot{Q}_{\text{in}}} \][/tex]

[tex]\[ \eta = \frac{100 \times 10^3}{480.04 \times 10^3} \approx 20.83\% \][/tex]

Answer:

The question is incomplete, below is the complete question

"The real power delivered by a source to two impedance, Z1=4+j5⁡Ω and Z2=10⁡Ω connected in parallel, is 1000 W. Determine (a) the real power absorbed by each of the impedances and (b) the source current."

answer:

a. 615W, 384.4W

b. 17.4A

Explanation:

To determine the real power absorbed by the impedance, we need to find first the equivalent admittance for each impedance.

recall that the symbol for admittance is Y and express as

[tex]Y=\frac{1}{Z}[/tex]

Hence for each we have,  

[tex]Y_{1} =1/Zx_{1}\\Y_{1} =\frac{1}{4+j5}\\converting to polar \\ Y_{1} =\frac{1}{6.4\leq 51.3}\\ Y_{1} =(0.16 \leq -51.3)S[/tex]

for the second impedance we have

[tex]Y_{2}=\frac{1}{10}\\Y_{2}=0.1S[/tex]

we also determine the voltage cross the impedance,

P=V^2(Y1 +Y2)

[tex]V=\sqrt{\frac{P}{Y_{1}+Y_{2}}}\\[/tex]

[tex]V=\sqrt{\frac{1000}{0.16+0.1}}\\ V=62v[/tex]

The real power in the impedance is calculated as

[tex]P_{1}=v^{2}G_{1}\\P_{1}=62*62*0.16\\ P_{1}=615W[/tex]

for the second impedance

[tex]P_{2}=v^{2}*G_{2}\\ P_{2}=62*62*0.1\\384.4w[/tex]

b. We determine the equivalent admittance

[tex]Y_{total}=Y_{1}+Y_{2}\\Y_{total}=(0.16\leq -51.3 )+0.1\\Y_{total}=(0.16-j1.0)+0.1\\Y_{total}=0.26-J1.0\\[/tex]

We convert the equivalent admittance back into the polar form

[tex]Y_{total}=0.28\leq -19.65\\[/tex]

the source current flows is

[tex]I_{s}=VY_{total}\\I_{s}=62*0.28\\I_{s}=17.4A[/tex]

Answer: 3002.86kg

Explanation:

Hydraulic cylinder diameter =125mm

Ambient pressure =1bar

Pressure =2500kpa

Piston Mass (MP) =?

F|(when it moves upward )=PA=F|(when it moves downward) =PoA+Mpg

Po=1 bar=100kpa

A=(π/4)D^2=(π/4)*0.125^2=0.01227m^2

Mp=(P-Po) A/g=(2500-100)*1000*0.01227/9.80665

Mp=3002.86kg.

A program with total change amount as an integer input, and output the change using the fewest coins, one coin type per line is in the explanation part below.

Python program that outputs the change using the fewest coins after receiving the entire amount of change as an integer input:

def calculate_change(change_amount):

   coin_types = {

       "Dollar": 100,

       "Quarter": 25,

       "Dime": 10,

       "Nickel": 5,

       "Penny": 1

   }

   print("Change breakdown:")

   for coin_name, coin_value in coin_types.items():

       if change_amount >= coin_value:

           num_coins = change_amount // coin_value

           change_amount -= num_coins * coin_value

           coin_name_plural = coin_name + "s" if num_coins > 1 else coin_name

           print(f"{num_coins} {coin_name_plural}")

# Input the total change amount

total_change = int(input("Enter total change amount: "))

calculate_change(total_change)

Thus, this is the Python program asked.

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#SPJ12

Answer:

True

Explanation:

Answer:

a diameter of D₂ = 0.183 inches would be required

Explanation:

appyling pascal's law

P applied to the hydraulic jack = P required to lift the rock

F₁*A₁ = F₂*A₂

since A₁= π*D₁²/4 ,  A₂= π*D₂²/4

F₁*π*D₁²/4 = F₂* π*D₂²/4

F₁*D₁²=F₂*D₂²

D₂ = D₁ *√(F₁/F₂)

replacing values

D₂ = D₁ *√(F₁/F₂) =  6 in * √(120 lbf/(4000 lbm * 32.174 (lbf/lbm)) = 0.183 inches

Answer:

The question is incomplete. Below is the complete question

"An industrial plant consisting primarily of induction motor loads absorbs 500 kW at 0.6 power factor lagging. (a) Compute the required kVA rating of a shunt capacitor to improve the power factor to 0.9 lagging. (b) Calculate the resulting power factor if a synchronous motor rated 500 hp with 90% efficiency operating at rated load and at unity power factor is added to the plant instead of the capacitor. Assume constant voltage. (1 hp = 0.746 kW)"

Answer:

a. 424.5KVA

b. 0.808 lagging.

Explanation:

Let's first determine the real power, reactive and the apparent power delivered.

Ql= Ptan(the real power angle)

Ql=500*tan53.13°

Ql=666.7 Kvar

The reactive angle is

Arccos(0.9)=25.84°

Now we calculate the reactive power

Qs=Ptan25. 84°

Qs=242.4KVAR

Hence the apparent power is

Qc=Ql-Qs

Qc=666.7-242.2

Qc=424.5KVAR

The required KVA rating of the shunt capacitor is 424.5KVA

b. To calculate the resulting power factor we first determine the power absorbs by the synchronous motor.

Pm=(500*0.746)/0.9

Pm=414.4KW

The total reactive power is

Ps=P+Pm

Ps=414.4+500=914.4KW

Hence we compute the source power factor as

PF=cos[arctan(Qs/Ps)]

PF=cos[arctan(666.7/914.4)

From careful calculation, we arrive at

PF=0.808.

Note this power factor will be lagging.

Answer:

Tempered glass

Explanation:

Tempered or toughened glass is a type of safety glass processed by controlled

thermal or chemical treatments to increase its strength compared with normal glass. Tempering puts the outer surfaces into compression and the interior into tension . Such stresses cause the glass, when broken, to crumble into small granular chunks instead of splintering into jagged shards as plate glass (a.k.a. annealed glass) does. The granular chunks are less likely to cause injury.

Answer: tempered glass

The characteristics listed in the options are the characteristics of an insulating glass, a type of glass with double or more panes used to reduce heat loss within a building.

Explanation:

A tempered glass is also known as a toughened glass. It is a type of glass that is produced through controlled thermal or chemical treatment which helps to increase its quality in terms of toughness. A tempered glass is a safety glass and as a result of this, its surface is four times toughened than that of a normal glass.

The properties of water and steam used to do work can be determined from the steam table, given input of the temperature, pressure, specific volume, as well as other thermodynamic variables

The responses to the required values of pressure and specific volume are as follows;

Part I

(a) Given T = 140°C, v = 0.5 m³/kg, to find p

From the steam tables, we have, at 140°, [tex]v_f[/tex] = 0.00107976, [tex]v_g[/tex] = 0.508519, [tex]p_s[/tex] = 3.61501

Given that the [tex]v_f[/tex] < v < [tex]v_g[/tex], the fluid has two phases and the experienced pressure [tex]p_s[/tex] = 3.61501 bar

(b) Given that at 30 MPa, and 80°C the steam is in the superheated heated region

From the single phase region of the steam tables, we have;

v = 1.01553 × 10⁻³ m³/kg

(c) Given that at p = 10 MPa = 100 bar, and T = 600°C, we have from the single phase region of the  steam table

v = 3.83775 × 10⁻² m³/kg

(d) At T = 80°C, and x = 0.4, we have;

v = 0.00102904 + 0.4 × (3.40527 - 0.00102904) ≈ 1.363

v = 1.363 m³/kg

Part II

(a) At T = 440 °C, p = 20 MPa. from the single phase region of the steam table, we have;

v = 1·22459 × 10⁻² m³/kg

(b) At T = 160°C, p = 20 MPa = 200 bar, the steam is in the superheated region, and we have;

v = 1.08865 × 10⁻³ m³/kg

(c) At T = 40°C and p = 2 MPa = 20 bar, we have

v = 0.00100700 m³/kg = 1.007 × 10⁻³ m³/kg

Learn more about the steam tables here;

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Answer:

Answer  of the Java class explained below with appropriate comments

Explanation:

using System;

class MainClass {

 public static void Main (string[] args) {

//entering wind speed of the hurricane

   Console.WriteLine ("Enter wind speed in mph: ");

   int windSpeed = Convert.ToInt32(Console.ReadLine());

//checking for the category with the credited wind speed

   int category = GetCategory(windSpeed);

   Console.WriteLine ("Category: " + category);

 }

 public static int GetCategory(int wind) {

     //function for classifying the wind speeds

     if(wind >= 157) {

     return 5;

     }

     if(wind >= 130) {

         return 4;

     }

     if(wind >= 111) {

         return 3;

     }

     if(wind >= 96) {

         return 2 ;

     }

     return 1;

 }

}