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RoseWind [281]
3 years ago
7

An ideal vapor-compression refrigeration cycle that uses refrigerant-134a as its working fluid maintains a condenser at 800 kPa

and an evaporator at -20°C. Determine this system’s COP and the amount of power required to service a 150 kW cooling load. (Take the required values from saturated refrigerant-134a tables.)

Engineering
1 answer:
mote1985 [20]3 years ago
4 0

Answer:

COP = 3.828

W' = 39.18 Kw

Explanation:

From the table A-11 i attached, we can find the entropy for the state 1 at -20°C.

h1 = 238.43 KJ/Kg

s1 = 0.94575 KJ/Kg.K

From table A-12 attached we can do the same for states 3 and 4 but just enthalpy at 800 KPa.

h3 = h4 = hf = 95.47 KJ/Kg

For state 2, we can calculate the enthalpy from table A-13 attached using interpolation at 800 KPa and the condition s2 = s1. We have;

h2 = 275.75 KJ/Kg

The power would be determined from the energy balance in state 1-2 where the mass flow rate will be expressed through the energy balance in state 4-1.

W' = m'(h2 - h1)

W' = Q'_L((h2 - h1)/(h1 - h4))

Where Q'_L = 150 kW

Plugging in the relevant values, we have;

W' = 150((275.75 - 238.43)/(238.43 - 95.47))

W' = 39.18 Kw

Formula foe COP is;

COP = Q'_L/W'

COP = 150/39.18

COP = 3.828

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A well-established way of power generation involves the utilization of geothermal energy-the energy of hot water that exists nat
jeka94

Answer:

the maximum thermal efficiency is 29%

Explanation:

the maximum efficiency for a thermal engine that works between a cold source and a hot source is the one of a Carnot engine. Its efficiency is given by

Maximum efficiency= 1 - T2/T1

where

T2= absolute temperature of the cold sink (environment)= 20°C + 273 = 293

T2= absolute temperature of the hot source (hot water supply) = 140°C + 273 = 413

therefore

Maximum efficiency= 1 - T2/T1 = 1 - 293/413 = 0,29 =29%

3 0
2 years ago
In this assignment, you will write a user interface for your calculator using JavaFX. Your graphical user interface (GUI) should
Zolol [24]

Answer:

Kindly note that, you're to replace "at" with shift 2 as the brainly text editor can't take the symbol

Explanation:

import javafx.application.Application;

import javafx.stage.Stage;

import javafx.scene.Group;

import javafx.scene.Scene;

import javafx.scene.layout.VBox;

import javafx.scene.layout.HBox;

import javafx.scene.control.TextField;

import javafx.scene.control.Button;

public class Calculator extends Application {

public static void main(String[] args) {

// TODO Auto-generated method stub

launch(args);

}

"at"Override

public void start(Stage primaryStage) throws Exception {

// TODO Auto-generated method stub

Group root = new Group();

VBox mainBox = new VBox();

HBox inpBox = new HBox();

TextField txtInput = new TextField ();

txtInput.setEditable(false);

txtInput.setStyle("-fx-font: 20 mono-spaced;");

txtInput.setText("0.0");

txtInput.setMinHeight(20);

txtInput.setMinWidth(200);

inpBox.getChildren().add(txtInput);

Scene scene = new Scene(root, 200, 294);

mainBox.getChildren().add(inpBox);

HBox rowOne = new HBox();

Button btn7 = new Button("7");

btn7.setMinWidth(50);

btn7.setMinHeight(50);

Button btn8 = new Button("8");

btn8.setMinWidth(50);

btn8.setMinHeight(50);

Button btn9 = new Button("9");

btn9.setMinWidth(50);

btn9.setMinHeight(50);

Button btnDiv = new Button("/");

btnDiv.setMinWidth(50);

btnDiv.setMinHeight(50);

rowOne.getChildren().addAll(btn7,btn8,btn9,btnDiv);

mainBox.getChildren().add(rowOne);

HBox rowTwo = new HBox();

Button btn4 = new Button("4");

btn4.setMinWidth(50);

btn4.setMinHeight(50);

Button btn5 = new Button("5");

btn5.setMinWidth(50);

btn5.setMinHeight(50);

Button btn6 = new Button("6");

btn6.setMinWidth(50);

btn6.setMinHeight(50);

Button btnMul = new Button("*");

btnMul.setMinWidth(50);

btnMul.setMinHeight(50);

rowTwo.getChildren().addAll(btn4,btn5,btn6,btnMul);

mainBox.getChildren().add(rowTwo);

HBox rowThree = new HBox();

Button btn1 = new Button("1");

btn1.setMinWidth(50);

btn1.setMinHeight(50);

Button btn2 = new Button("2");

btn2.setMinWidth(50);

btn2.setMinHeight(50);

Button btn3 = new Button("3");

btn3.setMinWidth(50);

btn3.setMinHeight(50);

Button btnSub = new Button("-");

btnSub.setMinWidth(50);

btnSub.setMinHeight(50);

rowThree.getChildren().addAll(btn1,btn2,btn3,btnSub);

mainBox.getChildren().add(rowThree);

HBox rowFour = new HBox();

Button btnC = new Button("C");

btnC.setMinWidth(50);

btnC.setMinHeight(50);

Button btn0 = new Button("0");

btn0.setMinWidth(50);

btn0.setMinHeight(50);

Button btnDot = new Button(".");

btnDot.setMinWidth(50);

btnDot.setMinHeight(50);

Button btnAdd = new Button("+");

btnAdd.setMinWidth(50);

btnAdd.setMinHeight(50);

rowFour.getChildren().addAll(btnC,btn0,btnDot,btnAdd);

mainBox.getChildren().add(rowFour);

HBox rowFive = new HBox();

Button btnEq = new Button("=");

btnEq.setMinWidth(200);

btnEq.setMinHeight(50);

rowFive.getChildren().add(btnEq);

mainBox.getChildren().add(rowFive);

root.getChildren().add(mainBox);

primaryStage.setScene(scene);

primaryStage.setTitle("GUI Calculator");

primaryStage.show();

}

}

4 0
3 years ago
Conduct online research and write a short report on the origin and evolution of the meter as a measurement standard. Discuss how
valina [46]

Answer:

People have come up with all sorts of inventive ways of measuring length. The most intuitive are right at our fingertips. That is, they are based upon the human body: the foot, the hand, the fingers or the length of an arm or a stride.

In ancient Mesopotamia and Egypt, one of the first standard measures of length used was the cubit. In Egypt, the royal cubit, which was used to build the most important structures, was based on the length of the pharaoh’s arm from elbow to the end of the middle finger plus the span of his hand. Because of its great importance, the royal cubit was standardized using rods made from granite. These granite cubits were further subdivided into shorter lengths reminiscent of centimeters and millimeters.

piece of black rock with white Egyptian markings

Fragment of a Cubit Measuring Rod

Credit: Gift of Dr. and Mrs. Thomas H. Foulds, 1925

Later length measurements used by the Romans (who had taken them from the Greeks, who had taken them from the Babylonians and Egyptians) and passed on into Europe generally were based on the length of the human foot or walking and multiples and subdivisions of that. For example, the pace—one left step plus one right step—is approximately a meter or yard. (On the other hand, the yard did not derive from a pace but from, among other things, the length of King Henry I of England’s outstretched arm.) Mille passus in Latin, or 1,000 paces, is where the English word “mile” comes from.

And thus, the meter has and likely will remain so elegantly defined in these terms for the foreseeable future.

Explanation:

is this short enough

5 0
2 years ago
Air exits a compressor operating at steady-state, steady-flow conditions at 150 oC, 825 kPa, with a velocity of 10 m/s through a
ioda

Answer:

a) Qe = 0.01963 m^3 / s , mass flow rate m^ = 0.1334 kg/s

b) Inlet cross sectional area = Ai = 0.11217 m^2 , Qi = 0.11217 m^3 / s    

Explanation:

Given:-

- The compressor exit conditions are given as follows:

                  Pressure ( Pe ) = 825 KPa

                  Temperature ( Te ) = 150°C

                  Velocity ( Ve ) = 10 m/s

                  Diameter ( de ) = 5.0 cm

Solution:-

- Define inlet parameters:

                  Pressure = Pi = 100 KPa

                  Temperature = Ti = 20.0

                  Velocity = Vi = 1.0 m/s

                  Area = Ai

- From definition the volumetric flow rate at outlet ( Qe ) is determined by the following equation:

                   Qe = Ae*Ve

Where,

           Ae: The exit cross sectional area

                   Ae = π*de^2 / 4

Therefore,

                  Qe = Ve*π*de^2 / 4

                  Qe = 10*π*0.05^2 / 4

                  Qe = 0.01963 m^3 / s

 

- To determine the mass flow rate ( m^ ) through the compressor we need to determine the density of air at exit using exit conditions.

- We will assume air to be an ideal gas. Thus using the ideal gas state equation we have:

                   Pe / ρe = R*Te  

Where,

           Te: The absolute temperature at exit

           ρe: The density of air at exit

           R: the specific gas constant for air = 0.287 KJ /kg.K

             

                ρe = Pe / (R*Te)

                ρe = 825 / (0.287*( 273 + 150 ) )

                ρe = 6.79566 kg/m^3

- The mass flow rate ( m^ ) is given:

               m^ = ρe*Qe

                     = ( 6.79566 )*( 0.01963 )

                     = 0.1334 kg/s

- We will use the "continuity equation " for steady state flow inside the compressor i.e mass flow rate remains constant:

              m^ = ρe*Ae*Ve = ρi*Ai*Vi

- Density of air at inlet using inlet conditions. Again, using the ideal gas state equation:

               Pi / ρi = R*Ti  

Where,

           Ti: The absolute temperature at inlet

           ρi: The density of air at inlet

           R: the specific gas constant for air = 0.287 KJ /kg.K

             

                ρi = Pi / (R*Ti)

                ρi = 100 / (0.287*( 273 + 20 ) )

                ρi = 1.18918 kg/m^3

Using continuity expression:

               Ai = m^ / ρi*Vi

               Ai = 0.1334 / 1.18918*1

               Ai = 0.11217 m^2          

- From definition the volumetric flow rate at inlet ( Qi ) is determined by the following equation:

                   Qi = Ai*Vi

Where,

           Ai: The inlet cross sectional area

                  Qi = 0.11217*1

                  Qi = 0.11217 m^3 / s    

- The equations that will help us with required plots are:

Inlet cross section area ( Ai )

                Ai = m^ / ρi*Vi  

                Ai = 0.1334 / 1.18918*Vi

                Ai ( V ) = 0.11217 / Vi   .... Eq 1

Inlet flow rate ( Qi ):

                Qi = 0.11217 m^3 / s ... constant  Eq 2

               

6 0
3 years ago
Problem 4.079 SI A rigid tank whose volume is 3 m3, initially containing air at 1 bar, 295 K, is connected by a valve to a large
salantis [7]

Answer:

Q_{cv} = -1007.86kJ

Explanation:

Our values are,

State 1

V=3m^3\\P_1=1bar\\T_1 = 295K

We know moreover for the tables A-15 that

u_1 = 210.49kJ/kg\\h_i = 295.17kJkg

State 2

P_2 =6bar\\T_2 = 296K\\T_f = 320K

For tables we know at T=320K

u_2 = 228.42kJ/kg

We need to use the ideal gas equation to estimate the mass, so

m_1 = \frac{p_1V}{RT_1}

m_1 = \frac{1bar*100kPa/1bar(3m^3)}{0.287kJ/kg.K(295k)}

m_1 = 3.54kg

Using now for the final mass:

m_2 = \frac{p_2V}{RT_2}

m_2 = \frac{1bar*100kPa/6bar(3m^3)}{0.287kJ/kg.K(320k)}

m_2 = 19.59kg

We only need to apply a energy balance equation:

Q_{cv}+m_ih_i = m_2u_2-m_1u_1

Q_{cv}=m_2u_2-m1_u_1-(m_2-m_1)h_i

Q_{cv} = (19.59)(228.42)-(3.54)(210.49)-(19.59-3.54)(295.17)

Q_{cv} = -1007.86kJ

The negative value indidicates heat ransfer from the system

7 0
3 years ago
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