A water treatment plant processes 30,000 cubic meters of water each day. A square rapid-mix tank with vertical baffles and flat impeller blades will be used. The design detention time and velocity gradient are 30 seconds and 900 s-1 Determine the power input, if the temperature of the water is 20°C. µ = 1 x 10-3 kg/m•s. 1kW = 1000 J/s, 1J = 1N•m = 1 kg•m2/s2. Note: you can use the equation in its current version.

Answer:

4.8 years

Explanation:

The rate of corrosion (CPR) is defined as the rate at which a metal corrodes in a specific environment. It depends on the environmental condition and the type of metal. It is expressed in inches per year or melts per year. CPR is given by:

[tex]CPR=\frac{KW}{\rho At}[/tex]

Where K is a constant = 534, W is the weight corroded = 2.1 kg = 2.1 × 10⁶mg, A is the area = 17 in², ρ is the density = 7.9 g/cm³

From the CPR equation:

[tex]t=\frac{KW}{\rho A(CPR)}=\frac{534*2.1*10^6}{7.9*17*200}= 4.17*10^4 hrs=4.8years[/tex]

Answer:

See explaination

Explanation:

We can describe Aspiration Effect as a phenomenon of providing an allowance for the release of air from the mold cavity during the metal pouring.

See the attached file for detailed solution of the given problem.

Answer:

The diameter at the bottom of the sprue is =0.725 in, and the sprue will not occur when 0.021<0.447.

Explanation:

Solution

The first step to take is to define the Bernoulli's eqaution

h effective = v²top/2g + ptop /ρg = hbottom + v² bottom/2g + p bottom/ ρg

h effective  + 0 +0= 0 +v² bottom/2g + 0

Thus,

v bottom = √ 2gh total

=√ 2 (32. 6 ft/ s²) + (12/12 ft)

Which is = 8.074 ft/s

We now, express the relation for flow rate.

Q =π/4 D² bottom v bottom

= 40 in 3/s = π/4 D²₃ ( 8.074 ft/s) (12  in/ ft)

so,

D bottom = 0.725 in.

Then,

We express the relation to avoid aspiration

A₃/A₂ < √ h top /h total

= π/4 D²₃/ π/4 D²₂ < √3/15

= 0.725²/5² < √3/15

=0.021<0.447

Therefore, the aspiration will not happen or occur

Answer:

See explaination

Explanation:

package sample;

import javafx.application.Application;

import javafx.fxml.FXMLLoader;

import javafx.geometry.*;

import javafx.scene.Parent;

import javafx.scene.Scene;

import javafx.scene.control.Label;

import javafx.geometry.Insets;

import javafx.geometry.Pos;

import javafx.scene.Scene;

import javafx.scene.control.Button;

import javafx.scene.control.Label;

import javafx.scene.control.TextField;

import javafx.scene.layout.GridPane;

import javafx.stage.Stage;

public class Main extends Application {

atOverride // Replace the at with at symbol

public void start(Stage primaryStage) throws Exception{

primaryStage.setTitle("Calculator");

GridPane rootNode = new GridPane();

rootNode.setPadding(new Insets(15));

rootNode.setHgap(5);

rootNode.setVgap(5);

rootNode.setAlignment(Pos.CENTER);

Scene myScene = new Scene(rootNode, 300, 200);

rootNode.add(new Label("Amount:"), 0, 0);

TextField firstValue = new TextField();

rootNode.add(firstValue, 1, 0);

rootNode.add(new Label("Toatal is:"), 0, 5);

Button aButton = new Button("Calculate");

rootNode.add(aButton, 1, 2);

GridPane.setHalignment(aButton, HPos.LEFT);

TextField result = new TextField();

result.setEditable(false);

rootNode.add(result, 1, 5);

TextField tax = new TextField();

rootNode.add(new Label("Tax:"), 0, 3);

tax.setEditable(false);

rootNode.add(tax,1,3);

TextField tip = new TextField();

rootNode.add(new Label("Tip:"), 0, 4);

tip.setEditable(false);

rootNode.add(tip,1,4);

aButton.setOnAction(e -> {

Float value1 = Float.valueOf(firstValue.getText());

Float value2 =(value1*18)/100;

Float value3 = (value1*7)/100;

Float r = value1+value2+value3 ;

tax.setText(value3.toString());

tip.setText(value2.toString());

result.setText(r.toString());

});

primaryStage.setScene(myScene);

primaryStage.show();

}

public static void main(String[] args) {

launch(args);

}

}

The question is about creating a JavaFX application that calculates and displays the cost of a restaurant meal, including an 18% tip and 7% sales tax. To create the application, an interface is needed with a text field for user input, a button for the calculation, and labels to display the results. The calculation logic is added to the button event handler.

The subject of this question is in the area of Computers and Technology, specifically application development using JavaFX. To create the desired application, we need to first create an interface, typically using FXML, with a text field for user input (food charge), a button for the calculation, and three labels to display the tip, sales tax, and total respectively.

On the button event handler, we would include the logic to calculate the 18% tip, 7% sales tax, and add all three (food charge, tip, sales tax) to get the total amount. This is done using the following formulas: tip = food charge * 0.18, sales tax = food charge * 0.07, and total = food charge + tip + sales tax. Finally, we would display the calculated amounts on the respective labels.

The code is written in JavaFX, a software platform used to create and deliver desktop applications, along with rich internet applications that can run across a wide variety of devices.

https://brainly.com/question/31593283

#SPJ3

Answer:

24.78 mg/L

Explanation:

The step-to-step explanation is written legibly with clear explanation in the diagram attached below.

Answer:

Answer: Input rate = 2.288 x 10⁶mg/s

             Output rate =78 x 10³Cmg/s

            Decay rate = 28.94 x 10 ⁵C mg/s

Explanation:

Assuming that complete and instantaneous mixing occurs in the lake, this implies that the concentration in the lake C is the same as the concentration of the mix leaving the lake Cm

   Input rate = Output rate  + KCV

Input rate = Q₁C₁ + QwCw

                = (75.0m³/s x 25.5mg/L + 3.0m³/s x 125.0mg/L) x 10³L/m³

                = 2.288 x 10⁶mg/s

Output rate = QmCm = (Q₁ +Qw)C

                   =(75 + 3.0)m³/s x Cmg/L x 10³L/m³ = 78 x 10³Cmg/s

Decay rate = KCV = 5/d x Cmg/L x 50 x 10⁶ x 10³L/m³  

                                           24 hr/d x 3600s/hr

                     = 28.94 x 10 ⁵C mg/s

So,

                      =2.288 x 10⁶ = 78 x 10³C + 28.94 x 10 ⁵C  = 29.72 X 10⁵C

                      = 2.288 x 10⁶          

                        29.72 X 10⁵         = 0.77 mg/l

                   C =

Answer:

(a) 0.0968 (b) the probability that a randomly selected LU graduate will have a salary of  exactly $30,400 is 0.0000 (c) 29.12% (d) 3000 students graduates this year from this university

Explanation:

Solution

For the problem given,

The average salary starting for this year's graduates at a large university (LU) is = 20,000

So,

The mean μ = $ 20,000

Standard deviation is б = $ 8000

Note: kindly find the complete steps taken to get the solution to this questions attached below.

Answer:

HW=1.71m

Explanation:

Please kindly check attachment for the step by step solution of the given problem.

Answer:

The shortest distance d to the edge of the plate is 66.67 mm

Concepts and reason

Moment of a force:

Moment of a force refers to the propensity of the force to cause rotation on the body it acts upon. The magnitude of the moment can be determined from the product of force’s magnitude and the perpendicular distance to the force.

Moment(M) = Force(F)×distance(d)

Moment of inertia ( I )

It is the product of area and the square of the moment arm for a section about a reference. It is also called as second moment of inertia.

First prepare the free body diagram of sectioned plate and apply moment equilibrium condition and also obtain area and moment of inertia of rectangular cross section. Finally, calculate the shortest distance using the formula of compressive stress (σ) in combination of axial and bending stress

Solution and Explanation:

[Find the given attachments]

The shortest distance (d) to the edge of the rectangular-plate is equal to 66.65 mm.

Given the followin data:

Thickness (length) of plate = 10 mm to m = 0.001 m.

Width of plate = 200 mm to m = 0.2 m.

c = [tex]\frac{0.2}{2}[/tex] = 0.1 m.

First of all, we would determine the area of the rectangular-plate and its moment of inertia.

For area:

[tex]A=LW\\\\A=0.001 \times 0.2\\\\A= 0.002 \;m^2[/tex]

For moment of inertia:

Mathematically, the moment of inertia of a rectangular-plate is given by this formula:

[tex]I=\frac{b^3d}{12} \\\\I=\frac{0.2^3 \times 0.01}{12} \\\\I=\frac{0.008 \times 0.01}{12}\\\\I=\frac{0.0008 }{12}\\\\I=6.67 \times 10^{-6}\;m^4[/tex]

The compressive stress of a rectangular-plate with respect to axial and bending stress is given by this formula:

[tex]\sigma = \frac{P}{A} -\frac{Mc}{I} \\\\\sigma = \frac{P}{0.002} -\frac{P(0.1-d)\times 0.1}{6.67 \times 10^{-6}} \\\\\sigma = \frac{P}{0.002} -\frac{0.01P-0.1Pd}{6.67 \times 10^{-6}} \\\\\frac{P}{0.002}=\frac{0.01P-0.1Pd}{6.67 \times 10^{-6}}\\\\500P=\frac{0.01P-0.1Pd}{6.67 \times 10^{-6}}\\\\3.335 \times 10^{-3}P=0.01P-0.1Pd\\\\[/tex]

[tex]0=0.01P-0.1Pd-3.335 \times 10^{-3}P\\\\0=(0.01-0.1d-3.335 \times 10^{-3})P\\\\0.01-0.1d-3.335 \times 10^{-3}=0\\\\0.1d=0.01-3.335 \times 10^{-3}\\\\d=\frac{6.665\times 10^{-3}}{0.1} \\\\d=6.665\times 10^{-2}\;m\\\\[/tex]

d = 66.65 mm.

Read more on moment of inertia here: https://brainly.com/question/3406242

Answer:

[tex]L = 0.319\,m[/tex]

Explanation:

Let suppose that temperature of air is 15°C. The heating process of the solar concentrator is modelled after the First Law of Thermodynamics:

[tex]\dot Q = h\cdot \pi\cdot D\cdot L\cdot (\bar T-T_{\infty})[/tex]

The required length is:

[tex]L = \frac{\dot Q}{h\cdot \pi\cdot D\cdot (\bar T-T_{\infty})}[/tex]

But,

[tex]\dot Q = \dot m \cdot c_{w}\cdot (T_{o}-T_{i})[/tex]

[tex]\dot Q = \left(0.015\,\frac{kg}{s}\right)\cdot \left(4180\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (85^{\circ}C-15^{\circ}C)[/tex]

[tex]\dot Q = 4389\,W[/tex]

[tex]\bar T = \frac{15^{\circ}C + 85^{\circ}C}{2}[/tex]

[tex]\bar T = 50^{\circ}C[/tex]

[tex]L = \frac{4389\,W}{\left(2500\,\frac{W}{m^{2}} \right)\cdot \pi \cdot (0.05\,m)\cdot (50^{\circ}C-15^{\circ}C)}[/tex]

[tex]L = 0.319\,m[/tex]

Answer:

2.83 years

Explanation:

Please kindly see the attached file at thw attachment area for a detailed and step by step solution to the given problem.

Answer:

See explaination

Explanation:

Please kindly check attachment for the step by step solution of the given problem

Answer:

a) [tex]w_{out} = 281.55\,\frac{kJ}{kg}[/tex], b) [tex]s_{gen} = 0.477\,\frac{kJ}{kg\cdot K}[/tex]

Explanation:

a) The process within the turbine is modelled after the First Law of Thermodynamics:

[tex]-q_{out} - w_{out} + h_{in}-h_{out} = 0[/tex]

[tex]w_{out} = h_{in} - h_{out}-q_{out}[/tex]

[tex]w_{out} = c_{p}\cdot (T_{in}-T_{out})-q_{out}[/tex]

[tex]w_{out} = \left(1.005\,\frac{kJ}{kg\cdot K}\right)\cdot (980\,K-670\,K)-30\,\frac{kJ}{kg}[/tex]

[tex]w_{out} = 281.55\,\frac{kJ}{kg}[/tex]

b) The entropy production is determined after the Second Law of Thermodynamics:

[tex]-\frac{q_{out}}{T_{surr}} + s_{in}-s_{out} + s_{gen} = 0[/tex]

[tex]s_{gen} = \frac{q_{out}}{T_{surr}}+s_{out}-s_{in}[/tex]

[tex]s_{gen} = \frac{q_{out}}{T_{surr}}+c_{p}\cdot \ln\left(\frac{T_{out}}{T_{in}} \right)[/tex]

[tex]s_{gen} = \frac{30\,\frac{kJ}{kg} }{315\,K} + \left(1.005\,\frac{kJ}{kg\cdot K} \right)\cdot \ln\left(\frac{980\,K}{670\,K} \right)[/tex]

[tex]s_{gen} = 0.477\,\frac{kJ}{kg\cdot K}[/tex]

Answer:

Explanation:

Part c:

(Q)out = m(u1 - u2) = 5(2506-193.67) = 11562KJ

Total Entropy change = Δs + (Qout/Tsurrounding) = -33.36 + (11562/393) = 5.44KJ/K

Answer:

The energy, that is dissipated in the resistor during this time interval is 153.6 mJ

Explanation:

Given;

number of turns, N = 179

radius of the circular coil, r = 3.95 cm = 0.0395 m

resistance, R = 10.1 Ω

time, t = 0.163 s

magnetic field strength, B = 0.573 T

Induced emf is given as;

[tex]emf= N\frac{d \phi}{dt}[/tex]

where;

ΔФ is change in magnetic flux

ΔФ  = BA = B x πr²

ΔФ  = 0.573 x π(0.0395)² = 0.002809 T.m²

[tex]emf = N\frac{d \phi}{dt} = 179(\frac{0.002809}{0.163} ) = 3.0848 \ V[/tex]

According to ohm's law;

V = IR

I = V / R

I = 3.0848 / 10.1

I = 0.3054 A

Energy = I²Rt

Energy = (0.3054)² x 10.1 x 0.163

Energy = 0.1536 J

Energy = 153.6 mJ

Therefore, the energy, that is dissipated in the resistor during this time interval is 153.6 mJ

Answer:

164.4 W

Explanation:

We can define Heat loss as a measure of the total transfer of heat through the fabric of a building from inside to the outside, either from conduction, convection, radiation, or any combination of the these.

Please kindly check attachment for the solution of the heat loss as asked in the question.

The attached file have the solved problem.

Answer:

Explanation:

The solutions to this question can be seen in the screenshot taken from the solution manual.

The question involves applying principles of static equilibrium and mechanical advantage in physics to calculate forces exerted by a hydraulic cylinder and at a support point in a construction telescoping arm scenario.

The question relates to the application of static equilibrium and mechanical advantage concepts in physics to solve for forces in a telescoping arm used in construction. Specifically, it involves calculating force exerted by a hydraulic cylinder and the force at a support point when a mass is placed at a specific location on the arm.

In order to solve for the force exerted at point B by the hydraulic cylinder (part a) and the force exerted on the supporting carriage at A (part b) when the arm makes an angle  heta = 25 degrees with the horizontal, one would typically use the principles of torques and static equilibrium. Summation of torques around a point, often the pivot, and summation of forces in horizontal and vertical directions are standard procedures. This requires knowledge of the geometry of the construction, the location of the center of gravity, and the mass of the workers and platform.

The information given about the forces in a wheelbarrow and cranes are analogous examples that illustrate similar principles of physics applied to different scenarios. These examples demonstrate how to calculate the mechanical advantage and forces based on lever arms and mass distribution.

The force exerted at B by the hydraulic cylinder BD and the force exerted on the supporting carriage at A can be calculated using principles of moments and equilibrium of forces, respectively.

For part (a): The force exerted at B by the hydraulic cylinder BD can be calculated using the principle of moments. By summing the moments about point A, you can find the force at B.

For part (b): The force exerted on the supporting carriage at A can be determined by considering.

Find solution in attachments below

Answer:

Find the attachments sequence wise for complete solution.

Answer:

Explanation:

Solution:-

- To categorize the flow conditions of any fluid we utilize a dimensionless number, called Reynold's number ( Re ) to study the behaviour of the fluid.

- Reynold's number is proportional to the ratio of inertial forces ( forces that resist any change in motion of a unit mass ) and viscous forces ( forces that resist any iner-plane deformations between layers of fluid ).

- Considering 2-Dimensional viscous flow over a flat plate, the Reynold number (Re) is mathematically expressed as a function of distance "x" denoted from leading edge and along the length of the plate:

                             [tex]Re_x = \frac{U*x}{v}[/tex]

Where,

               U: The free stream velocity of the fluid

               ν: The kinematic viscosity of the fluid

- The distance "x" along the length of the plate is substituted in the above formula and the corresponding Reynold number is evaluated. This gives a highly localized value about "x".

- The purpose of the Reynold number is the substitution of dynamically similar fluids i.e Fluid with the same Reynold's number when testing models to see how they would behave in a specific environment.

- The Reynolds number has a set of ranges above and below the critical range defined by the critical length "xc" along the plate.  The range below the critical has laminar flow characteristics, whereas the range above the critical has turbulent flow. The laminar region has flow along smooth streamlines, while turbulent region is characterized by 3 - dimensional random eddy.

- The critical length " xc " is determined from the critical Reynold number i.e ( 5 x 10^5 ) which is a small region that has mixed characteristics of laminar and turbulent conditions.

- The flow at the boundaries has zero velocity, there is a steep velocity gradient from the boundary into the flow. This velocity gradient in a real fluid sets up shear forces near the boundary that reduce the flow speed to that of the boundary. That fluid layer which has had its velocity affected by the boundary shear is called the "boundary layer"

-  For smooth upstream boundaries, " x << xc " or " Re_x < 5 x 10^5 " , the boundary layer starts out as a laminar boundary layer in which the fluid particles move in smooth layer.

- As the laminar boundary layer increases in thickness, it becomes unstable as changes in motion become more predominant ( inertial ) than viscous effect of fluid layers. This leads to a transformation of laminar boundary layer into turbulent boundary layer in which fluid particles move in haphazard paths.  " x > xc " or " Re_x > 5 x 10^5 "

- The boundary layer thickness/height d increases as x increases. The relationships for laminar and turbulent regions of boundary layer are given as follows:

                  [tex]\frac{d}{x} = \left \{ {{\frac{5}{\sqrt{Re_x} } , 10^3 < Re_x < 10^6} \\\\\atop {\frac{0.16}{\frac{1}{7} \sqrt{Re_x} } , 10^6 < Re_x}} \right.[/tex]

- To construct a function of boundary layer thickness " d " and length from leading edge of the plate " x ". We use the Re_x relation and substitute, we get the following proportionalities for our sketch:

                  d ∝ √x  .... Laminar region

                  d ∝ [tex]x^\frac{6}{7}[/tex]  .... Turbulent region

- Use the above relation to develop sketch for the boundary layer along the length "x" from leading edge.

- The sketch is given as an attachment.

Answer:

D

Explanation:

the way vertices are connected may be different so having same number of edges do not mean that total degree will also be same.

Answer:

Zx = 176In³

Explanation:

See attached image file

Answer:

Step 1

Given

Diameter of circular grill,   D = 0.3m

Distance between the coal bricks and the steaks,  L = 0.2m

Temperatures of the hot coal bricks,  T₁ = 950k

Temperatures of the steaks, T₂ = 5°c

Explanation:

See attached images for steps 2, 3, 4 and 5

Answer:

volume flow rate Q  = 53.23 in³/s

Explanation:

given data

diameter = 4.22 inches

length = 75 inches

screw rotates = 65 revolutions per minute

depth = 0.23 in

flight angle = 21.4 degrees

head pressure = 705 lb/in²

viscosity = 145 x [tex]10^{-4}[/tex] lb·sec/in²

solution

we get here volume flow rate of platstic in barrel that is express as

volume flow rate Q = volume flow rate of die - volume flow of extruder barrel   ................1

here

volume flow rate extruder barrel is

flow rate   = [tex]\frac{\pi \times 705 \times 4.22 \times 0.23\times sin21.4 }{12\times 145 \times 10^{-4}\times 75 }[/tex]    

flow rate = 60.10  

and

volume flow rate of die is express as

volume flow rate = 0.5 × π² × D² × Ndc ×  sinA × cosA   .............2

put here value and w eget

volume flow rate = 0.5 × π² × 4.22² × 0.23 × 1 × sin21.4 × cos21.4

volume flow rate = 6.866

so put value in equation 1 we get

volume flow rate Q  = 60.10 - 6.866  

volume flow rate Q  = 53.23

Answer:

[tex] z = \frac{3-4.18}{1.19}=-0.992[/tex]

[tex] z = \frac{5-4.18}{1.19}=0.689[/tex]

And we can find this probability with this difference:

[tex] P(-0.992<z<0.689) = P(z<0.689) -P(z<-0.992) =0.752 -0.161=0.591[/tex]

And then we can conclude that the probability that someone watches between 3 and 5 hours a day is approximately 0.591 using a normal distribution

Explanation:

For this case we can define the random variable X as "hours that a person watches television". For this case we don't have the distribution for X but we have the following parameters:

[tex]\mu = 4.18,\sigma =1.19[/tex]

We can assume that the distribution for X is normal

[tex] X \sim N(\mu = 4.18 , \sigma =1.19)[/tex]

And we want to find this probability:

[tex] P(3 <X<5)[/tex]

And we can use the z score formula given by:

[tex] z=\frac[X- \mu}{\sigma}[/tex]

And we can find the z score for each limit and we got:

[tex] z = \frac{3-4.18}{1.19}=-0.992[/tex]

[tex] z = \frac{5-4.18}{1.19}=0.689[/tex]

And we can find this probability with this difference:

[tex] P(-0.992<z<0.689) = P(z<0.689) -P(z<-0.992) =0.752 -0.161=0.591[/tex]

And then we can conclude that the probability that someone watches between 3 and 5 hours a day is approximately 0.591 using a normal distribution

Answer:

(a) 561.12 W/ m² (b) 196.39 MW

Explanation:

Solution

(a) Determine the energy and power of the wave per unit area

The energy per unit are of the wave is defined as:

E = 1 /16ρgH²

= 1/16 * 1025 kg/ m3* 9.81 m/s² * (2.5 m )²

=3927. 83 J/m²

Thus,

The power of the wave per unit area is,

P = E/ t

= 3927. 83 J/m² / 7 s = 561.12 W/ m²

(b) The average and work power output of a wave power plant

W = E * л * A

= 3927. 83 J/m² * 0.35 * 1 *10^6 m²

= 1374.74 MJ

Then,

The power produced by the wave for one km²

P = P * л * A

= 5612.12 W/m² * 0.35 * 1* 10^6 m²

=196.39 MW

Answer:

Find the complete solution in the given attachments

Answer: The feed rate is

17,020kg/he and the rate is 13,520kg/h

Check the attachment for step by step explanation

Answer:

Given:

P₁ = 500 kPa

T₁ = 860 K

P₂= 100 kPa

T₂ = 460 K

Let's take entropy properties of T1 and T2 from ideal properties of air,

at T = 860K, s(T₁) = 2.79783 kJ/kg.K

at T = 460K, s(T₂) = 2.13407 kJ/kg.K

using entropy balance equation:

[tex] \frac{\sigma _cv}{m} = s(T_2)- s(T_1) - R In [\frac{P_2}{P_1}] [/tex]

[tex] \frac{\sigma _cv}{m} = 2.79783 - 2.13407 - 0.287 In [\frac{100}{500}] [/tex]

= - 0.2018 kJ/kg. K

In this case the entropy is negative, which means the value of exit temperature is not correct, beacause entropy should always be positive(>0).

Answer:

Check the attached images below.

Explanation:

It Is required to develop an approximate relation between . Reynolds and Grashor numbers such that the heat-transfer coefficients for pure forced convection and pure .e convection are equal, assuming laminar flow, by comparing the heat-transfer coefficients for forced or free convection over vertical hat plates.  

write the equation or heat transfer coefficient Mr (arced comedian.

Kindly check the attached images below.

Answer:

See attachment

Explanation:

Gain= Vo/Vin

If we set Vout=9.62V corresponding to Vin=2.6V, then gain will be 3.7

Using above value of gain, let's design non-inverting op-amp configuration

Gain= 1+Rf/Rin

3.7= 1= Rf/Rin

2.7= Rf/Rin

If Rin=100Ω then Rf= 270Ω