Is it possible to maintain a pressure of 10 kpa in a condenser that is being cooled by river water entering at 20 C?

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.

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

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

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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:

a) 6.4  mg/l

b) 5.6 mg/l

Explanation:

Given data:

effluent Discharge Q_w = 1.0 m^3.s

Ultimate BOD L_w = 40 mg/l

Discharge of stream Q_r = 10 m^3.s

Stream ultimate BOD L_r = 3  mg/l

a) Ultimate BOD of mixture[tex] = \frac{Q_w l_w + Q_r L_r}{Q_w + Q_r}[/tex]

                                         [tex] = \frac{1*40 + 10*3}{10 +1} = 6.4 mg/l[/tex]

b) utlimate BOD at 10,000 m downstream

[tex]t =\frac{distance}{speed} = \frac{10,000}{\frac{Q_r +Q+w}{55}} \times \frac{hr}{3600} \times  \frac{day}{24 hr}[/tex]

putting [tex]Q_r + Q_w = 1+ 10 = 11 m^3/s[/tex]

t = 0.578  days

we know

[tex]L_t = L_o e^{-kt}[/tex]

[tex]L_t = 6.4 \times e^{-0.22 \times 0.578}[/tex]

[tex]L_t = 5.6 mg/l[/tex]

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:

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.

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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Answer:

in situ treatment would be better if; (1) a large volume of soil is to be treated at once, and (2) a permeable sandy soil (un-compacted) is to be treated.

ex situ treatment would be better would be better if; (1) a wider range of contaminants and soil types are to be treated, and (2) there is more certainty about the uniformity of treatment because of the ability to homogenize, screen, and continuously mix the soil.

Explanation:

in situ treatment would be better if;

ex situ treatment would be better would be better if;

Answer:

Integrity: involves maintaining and assuring the accuracy of data over its life-cycle

Explanation:

Confidentiality: This is a CIA triad designed to prevent sensitive information from reaching the wrong people, while making sure that the right people have access to it.

Integrity: This is a CIA triad that involves maintaining the consistency, accuracy, and trustworthiness of data over its entire life cycle.

Availability: This is a CIA triad that involves hardware repairs and maintaining a correctly functioning operating system environment that is free of software conflicts.

Authentication:This is a security control that is used to protect the system with regard to the CIA properties.

Answer:

Part A:

In W-h:

Energy Stored=1440 W-h

In Joules:

[tex]Energy Stored=5.184*10^6Joules\\Energy Stored=5.184 MJ[/tex]

Part B:

In W-h:

Energy left=240 W-h

In Joules:

Energy left= 8.64*10^5 J

Explanation:

Part A:

We are given rating 120A-h and voltage 12 V

Energy Stored= Rating*Voltage              (Gives us units W-h)

Energy Stored=120A-h*12V

Energy Stored=1440 W-h

Converting it into joules (watt=joules/sec)

Energy Stored=[tex]1440 Joules * \frac{3600sec}{h}h[/tex]

Energy Stored=5184000 Joules

[tex]Energy Stored=5.184*10^6Joules\\Energy Stored=5.184 MJ[/tex]

Part B:

Energy used by lights for 8h=150*8

Energy used by lights for 8h=1200W-h

Energy left= Energy Stored(Calculated above)- Energy used by lights for 8h

Energy left=1440-1200

Energy left=240 W-h

Energy left=[tex]240 Joules * \frac{3600sec}{h}h[/tex]

Energy left=864000 Joules

Energy left= 8.64*10^5 J

The total energy stored in the battery is 1440 Wh. If the headlights are left on overnight (8h), the energy remaining in the battery in the morning is 1320 Wh.

(a) The total energy stored in the battery can be calculated by multiplying the battery's capacity (120 A-h) by the voltage (12 V). So, the total energy stored in the battery is 1440 Wh.

(b) To calculate the energy consumed by the headlights in 8 hours, we need to find the power (P) using the formula P = V × I, where V is the voltage (12 V) and I is the current (the total power rating divided by the voltage). Then, we can find the energy consumed by multiplying the power by the time (8 hours). We can subtract this energy consumption from the total energy stored in the battery to find the energy remaining in the morning. So, the energy remaining in the battery is 1440 Wh - 12 kWh = 1320 Wh.

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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:

(userVal < 0) ? "negative" : "non-negative" ;

Explanation:

The above code has been written using Java's ternary operator (? : ).

Java uses this operator to write conditional expressions that are similar to the regular if ... else statements.

This expression has three parts,

i. the conditional statement: this is the expression before the ? mark. In this case, (userVal < 0). This expression contains the condition to be tested for and it returns true or false.

ii. the second part is the statement after the ? mark. In this case, "negative". This expression will be executed if the conditional statement returns true.

iii. the third part is the statement after the : mark. In this case, "non-negative". This expression will be executed if the conditional statement returns false.

So in this case, if userVal < 0, the string "negative" will be evaluated. Otherwise, "non-negative".

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:

Fde = 1080 N (3 sig fig)

Explanation:

Taking point E

Sum of forces at E in x direction:

Fde cos (30) - Fcd = 0  ..... Eq 1

Sum of forces at E in y direction:

Fde sin (30) - Wf = 0   ..... Eq 2

Wf = m*g = 55 * 9.81 = 539.55 N

Using Eq 2 to evaluate Fde

Fde = Wf / sin (30) = 539.55 / sin (30) = 1079.1 N

Answer: Fde = 1080 N (3 sig fig)

Answer:

v = 1779 veh/h

Explanation:

We will calculate the peak hour volume by using the following formula:-

Vp = v / ( PHF)(N)(Fg)(Fhv)

here,

1. v is the peak hour volume

2. vp is the analysis flow rate

3. PHF is the peak hour factor

4. N is the number of lanes

5. Fg is the grade adjustment factor which is 0.99 for rolling terrain > 1200 pc/h/ln

6. Fhv is the heavy vehicle adjustment factor

vp = v / (PHF) (N) (Fg) (Fhv)

1250 = v / (0.84)(2)(1)(0.847)

v = 1779 veh/h

Answer:

(a) The charges of 75 C must flow through battery.

(b) The electrons must flow from a to b.

Explanation:

(a)

The relationship between the energy and voltage is as follows:

Energy = (Voltage)(Current)(Time)

but, (current)(time) = charge

therefore,

Energy = (Voltage)(Charge)

Charge = (Energy)/(Voltage)

Charge = (900 J)/(12 V)

Charge = 75 C

(b)

Since, Vab = Va - Vb = 12 V

Hence, the equation suggest that Va is at higher potential then Vb.

As, the current flows from higher to lower potential.

Thus, electrons must flow from a to b

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

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.

The temperature at which the resistance is [tex]1 k\Omega[/tex] is [tex]89.4^{\circ}C[/tex]

Explanation:

For the thermistor in this problem, the relationship between temperature and resistance is linear.

We have:

[tex]R_1 = 5000 \Omega[/tex] when the temperature is [tex]T=25^{\circ}C[/tex]

[tex]R_2=340 \Omega[/tex] when the temperature is [tex]T=100^{\circ}C[/tex]

Assuming a straight-line relationship, we can find the slope of the line:

[tex]m=\frac{R_2-R_1}{T_2-T_1}=\frac{340-5000}{100-25}=-62.1 \Omega/^{\circ}C[/tex]

Now that we know the slope, we can extrapolate the temperature when the resistance is

[tex]R_3 = 1 k\Omega = 1000 \Omega[/tex]

In fact, we can write:

[tex]m=\frac{R_3-R_2}{T_3-T_2}[/tex]

And solving for [tex]T_3[/tex],

[tex]m(T_3-T_2)=R_3-R_2\\T_3 = T_2 +\frac{R_3-R_2}{m}=100 + \frac{1000-340}{-62.1}=89.4^{\circ}C[/tex]

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The temperature at which the thermistor's resistance equals 1 kΩ is approximately 11.88 °C.

The relationship between the resistance of a thermistor and temperature can be modeled by a straight line equation, which is given by:

R = Ro(1 + αΔT)

where R is the resistance at a given temperature, Ro is the resistance at a reference temperature, α is the temperature coefficient of resistance, and ΔT is the change in temperature from the reference temperature.

To find the temperature at which the thermistor's resistance equals 1 kΩ, we can rearrange the equation:

ΔT = (R/Ro - 1) / α

Plugging in the given values:

Ro = 5 kΩ, R = 1 kΩ, α = (340 Ω - 5 kΩ) / (100 °C - 25 °C)

ΔT = (1 kΩ / 5 kΩ - 1) / [(340 Ω - 5 kΩ) / (100 °C - 25 °C)]

Simplifying the equation, we find that ΔT ≈ -14.12 °C.

Therefore, the temperature at which the thermistor's resistance equals 1 kΩ is approximately 11.88 °C.

Answer: 30joules

Explanation

voltage of battery =12volts

2.5coulombs is the same as 2.5ampere-seconds

Energy required =12V*2.5A-S

=30watts-seconds

Hence the required energy is equal to 30joules.

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);  

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:

I am writing a function in C++                              

Explanation:

#include <iostream>

using namespace std;

int pyramid(int height) // function pyramid with parameter height

{int distance; //variable for spaces

   for(int i = 1, j = 0; i <= height; ++i, j= 0)  //handle the rows

   {

for(distance= 1; distance<= height-i; ++distance)

// handles the spaces & columns

    { cout <<"  ";        } //prints spaces between stars

       while(j!= 2*i-1)  //handles shape and spaces

       {   cout << "* "; // printing stars

           ++j;        }

       cout << '\n';    }   } // for the next line

int main()

{

int height; //declare height variable

cout <<"Enter height of pyramid "; //asks user to enter height of pyramid

cin>>height; //stores value of height

pyramid(height); //calls pyramid function

}

Answer:

a) 0.759 m , 0.759 m

b) 10.1 m , 10.3 m

c) 12.3 m , 13 m

Explanation:

Vapour Pressure included:

[tex]P_{atm,vp} = y*h + P_{vp} \\\\h = \frac{P_{atm,vp}-P_{vp}}{y} ... Eq1[/tex]

mercury

[tex]h_{Hg,vp} = \frac{101*10^3-1.6*10^(-1)}{133*10^3} \\\\h_{Hg,vp} = 0.759m[/tex]

[tex]h_{Hg} = \frac{101*10^3}{133*10^3} \\\\h_{Hg} = 0.759m[/tex]

water

[tex]h_{H2O,vp} = \frac{101*10^3-1.77*10^3}{9.8*10^3} \\\\h_{H2O,vp} = 10.1m[/tex]

[tex]h_{H2O} = \frac{101*10^3}{9.8*10^3} \\\\h_{H20} = 10.3m[/tex]

Ethyl Alcohol

[tex]h_{EA,vp} = \frac{101*10^3-5.9*10^3}{7.74*10^3} \\\\h_{EA,vp} = 12.3m[/tex]

[tex]h_{EA} = \frac{101*10^3-5}{7.74*10^3} \\\\h_{EA} = 13m[/tex]

For mercury barometers the effects of vapour pressure are negligible and required height for mercury barometer is reasonable as compared to that of water and ethyl alcohol.

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;

 }

}

Answer:

True

Explanation:

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: attached below. rate brainliest if correct please

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]