To determine the rate of heat supply by the fan-motor assembly, the electrical input power is calculated based on the 0.25-hp shaft output and 60% efficiency of the motor. The resulting heat supply to the room is the same as the electrical power input, which is 310.83 watts.
Explanation:The question asks to determine the rate of heat supply by a fan-motor assembly used to circulate chilled water through a heat exchanger for cooling a room. Given that the fan has a shaft output of 0.25 horsepower (hp) and that small electric motors driving such equipment typically have an efficiency of 60 percent, we can calculate the electrical power input needed to run the fan.
The electrical power input (Pinput) can be calculated as:
Pinput = Poutput / Efficiency
Where Poutput is the shaft output power (0.25 hp) and 'Efficiency' is the efficiency of the electric motor (60%, or 0.60 in decimal form).
Pinput = (0.25 hp) / 0.60
To convert horsepower to watts, we use the conversion factor 1 hp = 746 watts.
Pinput = (0.25 hp × 746 watts/hp) / 0.60
Pinput = 310.83 watts (rounded to two decimal places)
The rate of heat supply to the room will be equal to the electrical power input to the fan, which is 310.83 watts. This accounts for both the useful work done and all inefficiencies in the system that convert electrical energy into heat.
You are given a semiconductor resistor made from silicon with an impurity concentration of resistivity 1.00×10−3Ωm1.00×10−3Ωm. The resistor has a height of HH =0.5 mmmm, a length of LL = 2 mmmm, and a width of WW = 1.25 mmmm. The resistor can absorb (dissipate) up to PP = 7.81WW. What is the resistance of the resistor (RR), the maximum voltage (VV), and the maximum current (II)?
Answer:
The resistance (R) of the resistor is 2.4 ohm
The maximum voltage (V) is 4.33V
The maximum current (I) is 1.80A
Explanation:
Resistance (R) = resistivity×length/area
Resistivity = 0.003 ohm meter, length = 2mm = 2/1000 = 0.002m, width = 1.25mm = 1.25/1000 = 0.00125m, height = 0.5mm = 0.0005m, area = width × height = 0.00125m × 0.0005m = 6.25×10^-7m^2
R = 0.003×0.002/6.25×10^-7 = 3.2 ohm
Power (P) = V^2/R
V^2 = P × R = 7.81 × 3.2= 24.992
V = √24.992 = 4.99V
P = IV
I = P/V = 7.81/4.99 = 1.57A
A signal whose timing is completely influenced by the traffic volumes, when detected, on all of the approaches operates in the following mode: a. Pretimed b. Semi-actuated c. Fully-actuated
Answer: c. Fully actuated
Explanation: fully actuated signals are completely influenced by volumes if traffic and employs sensors at all approaches for its detection.
In studying of traffic flow at a highway toll booth over the course of 60 minutes, it is determined that the arrival and departure rates are deterministic, but not uniform. The arrival rate is found to vary according to the function A(t) = 1.8 + 0.25t - 0.0030t^2. The departure rate function is D(t) = 1.4 + 0.11t. In both of these functions, t is in minutes after the beginning of the observation and A(t) and D(t) are in vehicles per minute. At what time does the maximum queue length occur?
a. 2.7 min
b. 9.4 min
c. 49.4 min
d. 60.0 min
The time that the maximum queue length occurs would be c. 49.4 min
To find the time at which the maximum queue length occurs, we need to determine when the arrival rate equals the departure rate.
The queue length increases when the arrival rate exceeds the departure rate and decreases when the departure rate exceeds the arrival rate. The maximum queue length occurs when the arrival rate equals the departure rate.
Substituting the given functions into the equation, we get:
[tex]\[ 1.8 + 0.25t - 0.0030t^2 = 1.4 + 0.11t \][/tex]
Rearranging the terms, we get a quadratic equation:
[tex]\[ -0.0030t^2 + 0.25t - 0.11t + 1.8 - 1.4 = 0 \][/tex]
[tex]\[ -0.0030t^2 + 0.14t + 0.4 = 0 \][/tex]
Now, we can solve this quadratic equation to find the value(s) of t at which the maximum queue length occurs. We can use the quadratic formula:
[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Where:
a = -0.0030
b = 0.14
c = 0.4
Calculate the values of t using this formula. Then, we'll choose the appropriate value based on the physical meaning of the problem.
Using the quadratic formula:
[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
[tex]\[ t = \frac{-0.14 \pm \sqrt{(0.14)^2 - 4(-0.0030)(0.4)}}{2(-0.0030)} \][/tex]
[tex]\[ t = \frac{-0.14 \pm \sqrt{0.0196 + 0.0048}}{-0.0060} \][/tex]
[tex]\[ t = \frac{-0.14 \pm 0.156}{-0.0060} \][/tex]
Now, we have two possible values for t :
1. [tex]\( t_1 = \frac{-0.14 + 0.156}{-0.0060} \)[/tex]
2. [tex]\( t_2 = \frac{-0.14 - 0.156}{-0.0060} \)[/tex]
Calculate these values:
1. [tex]\( t_1 = \frac{0.016}{-0.0060} = -2.67 \)[/tex]
2. [tex]\( t_2 = \frac{-0.296}{-0.0060} = 49.4[/tex]
Since t represents time, it cannot be negative. The maximum queue length occurs at approximately t = 49.4 minutes after the beginning of the observation.
Primary U.S. interstate highways are numbered 1-99. Odd numbers (like the 5 or 95) go north/south, and evens (like the 10 or 90) go east/west. Auxiliary highways are numbered 100-999, and service the primary highway indicated by the rightmost two digits. Thus, the 405 services the 5, and the 290 services the 90 Given a highway number, indicate whether it is a primary or auxiliary highway. If auxiliary, indicate what primary highway it serves. Also indicate if the (primary) highway runs north/south or east/west. Ex: If the input is: 90 the output is: The 90 is primary, going east/west. Ex: If the input is: 290 the output is: The 290 is auxiliary, serving the 90, going east/west Ex: If the input is G here to search The 290 is auxiliary, serving the 90, going eas Ex: If the input is: 0 or any number not between 1 and 999, the output is: 0 is not a valid interstate highway number. See Wikipedia for more info on highway numbering LAB 4.16.1: LAB: Interstate highway numbers ACTIVITY LabProgr- 1 import java.util.Scanner; 2 3 public class LabProgram public static void main(Stringl] args) t Scanner scnr new Scanner(System.in); int highwayNumber; int primaryNumber; 4 5 6 7 highwayNumber scnr.nextInt(); 1e Type your code
Answer:
The Java code is given below with appropriate variable names and tags for better understanding
Explanation:
import java.util.Scanner;
public class LabProgram{
public static void main(String[] args) {
Scanner scnr = new Scanner(System.in);
int highwayNumber;
int primaryNumber;
highwayNumber = scnr.nextInt();
if(highwayNumber<1 || highwayNumber>999)
System.out.println(highwayNumber+" is not a valid interstate highway number.");
else{
if(highwayNumber>=1 && highwayNumber<=99){
System.out.print("The "+highwayNumber+" is primary, going ");
if(highwayNumber%2==1)
System.out.println("north/south.");
else
System.out.println("east/west.");
}
else{
primaryNumber = highwayNumber%100;
System.out.print("The "+highwayNumber+" is auxillary, serving the "+primaryNumber+", going ");
if(primaryNumber%2==1)
System.out.println("north/south.");
else
System.out.println("east/west.");
}
}
}
}
The program is an illustration of conditional statements.
Conditional statements are statements whose execution is dependent on its truth value.
The missing code segment in Java where comments are used to explain each line is as follows:
import java.util.Scanner;
public class Main {
public static void main(String [] args){
Scanner scnr = new Scanner(System.in);
int highwayNumber; int primaryNumber;
highwayNumber = scnr.nextInt();
//This checks if the highwayNumber is not between 1 and 999 (inclusive)
if (highwayNumber <1 || highwayNumber > 999){
//If yes, the highwayNumber is invalid
System.out.print(highwayNumber+" is not a valid interstate highwayNumber number.");
}
//If otherwise
else{
//This checks if highwayNumber is less than 100
if (highwayNumber< 100){
if (highwayNumber%2 == 0){
//Even highwayNumber are primary going east/west
System.out.print("I-"+highwayNumber+" is primary, going east/west.");
}
else{
//Odd highwayNumber are primary going north/south
System.out.print("I-"+highwayNumber+" is primary, going north/south.");
}
}
//Otherwise
else{
if ((highwayNumber%100) % 2 == 0){
//Even highwayNumber are auxiliary going east/west
System.out.print("I-"+highwayNumber+" is auxiliary, going east/west.");
}
else{
//Even highwayNumber are auxiliary going north/south
System.out.print("I-"+highwayNumber+" is auxiliary, going north/south.");
}
}
}
}
}
Read more about similar programs at:
https://brainly.com/question/22078310
A four-lane divided multilane highway (two lanes in each direction) in rolling terrain has five access points per mile and 11-ft lanes with a 4-ft shoulder on the right side and 2-ft shoulder on the left. The peak-hour factor is 0.84 and the traffic stream has a heavy vehicle adjustment factor of 0.847. If the analysis flow rate is 1250 pc/h/ln, what is the peak-hour volume? 976 veh/h 1345 veh/h 1389 veh/h 1779 veh/h
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
A scrubber on a coal-fired power plant is useful to remove
a. sulfur dioxide.
b. carbon dioxide.
c. particulate matter.
Answer:
sulfur dioxide
Explanation:
The scrubber is an apparatus installed in a coal-fired power plant to clean the passing gas through the smokestack. Due to the norm enacted through the clean air Act, almost all the scrubber used in the U.S is used to remove sulfur concentration from coal. it can remove approx 90-95% SO_2 from the smokestack.
Answer:
sulfur dioxide.
Explanation:
A certain battery has terminals labeled a and b. The battery voltage is vab = 12 V. a) To increase the chemical energy stored in the battery by 900J, how much charge must move through the battery? b) Should electrons move from a to b or from b to a?
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
Write a function called pyramid(height) that acceptsa parameter ""height"". It then prints a pyramid of that height
Answer:
I am writing a function in C++
Explanation:
C++ program#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
}
Liquid toluene is flowing through a pipe at a rate of 175 m 3 / h 175m3/h . What is the mass flow rate of this stream in kg/min? What is the molar flow rate in mol/s? In fact, the answer to Part (a) is only an approximation that is almost certain to be slightly in error. What did you have to assume to obtain the answer?
Answer:
(a) Mass flow rate is 2528.75 kg/min
(b) Molar flow rate is 458.1 mol/s
(c) Toulene is in-compressible at standard temperature and pressure
Explanation:
The volume flow rate of liquid toulene is given:
Volume Flow Rate = 175 m^3/h
Density of Liquid Toulene = 867 kg/m^3
Molar mass of Toulene = 92 g/mol = 0.092 kg/mol)
(a)
Thus, the mass flow rate of toulene will be:
Mass Flow Rate = (Volume Flow Rate)(Density)
Mass Flow Rate = (175 m^3/h)(867 kg/m^3)
Mass Flow Rate = (151725 kg/h)(1 h/60 min)
Mass Flow Rate = 2528.75 kg/min
(b)
Now, for molar flow rate we use formula:
Molar Flow Rate = (Mass Flow Rate)/(Molar Mass)
Molar Flow Rate = (2528.75 kg/min)/(0.092 kg/mol)
Molar Flow Rate = (27486.41 mol/min)(1 min/60sec)
Molar Flow Rate = 458.1 mol/s
(b)
The assumption is that Toulene is in-compressible at standard temperature and pressure. So, that its density can be taken constant.
why is the thermal conductivity of super insolation order of magnitude lower than the thermal conductivity of ordinary insulation?
Answer:
Super insulation are obtained by using layers of highly reflective sheets separated by glass fibers in an vacuumed space. Radiation heat transfer between any of the surfaces is inversely proportional to the number of sheets used and thus heat lost by radiation will be very low by using these highly reflective sheets which will an effective way of heat transfer.
Explanation:
Give two circumstances in which in situ treatment would be better, and two in which ex situ treatment would be better. Both using chemical oxidation as the remediation technique.
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;
a large volume of soil is to be treated at oncea permeable sandy soil (un-compacted) is to be treatedex situ treatment would be better would be better if;
a wider range of contaminants and soil types are to be treatedthere is more certainty about the uniformity of treatment because of the ability to homogenize, screen, and continuously mix the soil.Water flowing through both a small pipe and a large pipe can fill a water tank in 4 hours. Water flowing through the small pipe alone can fill the tank in 15 more hours. How many hours would it take to fill the tank using only the small pipe?
Answer with explanation:
As is the question the answer would be 19 hours, and the key to solving it is in the phrase in 15 more hours, basically what they are saying is that the small pipe takes 15 hours more than both the big and the small to fill the tank. Since both pipes working together can fill the tank in 4 hours we need to add 4 and 15 to solve the problem.
If the question is how many hours would it take to fill the tank using only the big pipe? Then we could solve t for the following equation:
[tex]\frac{1}{4+15} + \frac{1}{t} = \frac{1}{4}[/tex]
Getting as a result: 5.06
Note that the equation is the result of taking the rate of the small pipe (what we solved before), plus the unknown rate of the big one equals the rate of both.
Using phasors, the value of 37 sin 50t + 30 cos(50t – 45°) is _________ cos(50t+(_____°)). Please report your answer so the magnitude is positive and all angles are in the range of negative 180 degrees to positive 80 degrees
Answer: 62 cos(50t - 70°)
Explanation:
First we need to convert all sines into cosines because phasor forms are represented through cosine. For that we will use the fact that sin(wt + ∅) = cos(wt + ∅ - 90°)
Therefore, 37sin50t = 37cos(50t - 90°)
Now we have 37cos(50t - 90°)+ 30 cos(50t – 45°). We need to convert them into phasor form to add the terms. For that we will use the fact Acos(wt+∅)=A∠∅ which can be represented using real and imaginary parts as A [cos(∅)+jsin(∅)].
So,
37cos(50t - 90°)
= 37∠-90°
= 37[cos(-90°)+jsin(-90°)
=37[0+j(-1)]
= -j37
Similarly,
30 cos(50t – 45°)
=30∠-45
=30[cos(-45)+jsin(-45)
=30[0.707-j0.707]
=21.21 - j21.21
37cos(50t - 90)+ 30 cos(50t – 45°) = -j37+21.21 - j21.21 = 21.21 - j58.21
Now we need to convert the real and imaginary parts back to cosine form. We will first calculate the magnitude by the formula √a²+b² where a and b are the real and imaginary parts respectively.
Here a=21.21 and b=58.21
magnitude = √(21.21)²+(58.21)²=61.95≅62
For calculating phase ∅ the formula is ∅=inversetan (b/a) where a and b are the real and imaginary parts respectively.
∅=inversetan(-58.21/21.21)
= -69.9°≅-70°
So the final answer is 62cos(50t-70°)
The value of [tex]\(37 \sin 50t + 30 \cos(50t - 45^\circ)\)[/tex] is [tex]\(61.97 \cos(50t - 70^\circ)\)[/tex].
To solve [tex]\(37 \sin 50t + 30 \cos(50t - 45^\circ)\)[/tex] using phasors, we follow these steps:
Step-by-Step Calculation:
1. Express each term as a phasor:
- The term [tex]\(37 \sin 50t\)[/tex] :
- Convert to cosine form: [tex]\(37 \sin 50t = 37 \cos(50t - 90^\circ)\)[/tex]
- Phasor form: [tex]\(37 \angle -90^\circ\)[/tex]
- The term [tex]\(30 \cos(50t - 45^\circ)\)[/tex] :
- Phasor form: [tex]\(30 \angle -45^\circ\)[/tex]
2. Convert the phasors to rectangular form:
- [tex]\(37 \angle -90^\circ\)[/tex] :
[tex]\[ 37 \cos(-90^\circ) + j 37 \sin(-90^\circ) = 0 - j 37 = -j 37 \][/tex]
- [tex]\(30 \angle -45^\circ\)[/tex]:
[tex]\[ 30 \cos(-45^\circ) + j 30 \sin(-45^\circ) = 30 \left(\frac{\sqrt{2}}{2}\right) - j 30 \left(\frac{\sqrt{2}}{2}\right) = 21.2132 - j 21.2132 \][/tex]
3. Add the rectangular components:
[tex]\[ -j 37 + (21.2132 - j 21.2132) \][/tex][tex]\[ = 21.2132 - j (37 + 21.2132) \][/tex]
[tex]\[ = 21.2132 - j 58.2132 \][/tex]
4. Convert the result back to polar form:
- Magnitude:
[tex]\[ R = \sqrt{21.2132^2 + 58.2132^2} = \sqrt{449.54 + 3388.78} = \sqrt{3838.32} = 61.97 \][/tex]
- Angle:
[tex]\[ \theta = \tan^{-1}\left(\frac{-58.2132}{21.2132}\right) = \tan^{-1}(-2.743) \approx -70^\circ \][/tex]
5. Express the final result:
Using the positive magnitude, the expression in cosine form is:
[tex]\[ 37 \sin 50t + 30 \cos(50t - 45^\circ) = 61.97 \cos(50t - 70^\circ) \][/tex]
Recycled materials content is environmental information that is typically: a. self-declared by the manufacturer. b. independently certified by third-party entities. c. listed in the building code. d. part of an environmental label. e. None of the above
Answer:
a. Self declared by the manufacturer
Explanation:
Recycled content refers to the portion of materials used in a product that have been diverted from the solid waste stream. If those materials are diverted during the manufacturing process, they are be referred to as pre-consumer recycled content (sometimes referred to as post-industrial). If they are diverted after consumer use, they are
post-consumer .
What are nonexecuting statements that you can use to document or add notes to assist in the use of the program?
Answer:
Program comments
A certain lead-acid storage battery has a mass of 33 kg . Starting from a fully charged state, it can supply 6 A for 20 hours with a terminal voltage of 24 V before it is totally discharged.
a. If the energy stored in the fully charged battery is used to lift the battery with 100-percent efficiency, what height is attained? Assume that the acceleration due to gravity is 9.8 m/s2 and is constant with height.
b. If the stored energy is used to accelerate the battery with 100 percent efficiency, what velocity is attained?
c. Gasoline contains about rho1 = 4.5×107 J/kg. Compare this with the energy content per unit mass for the fully charged battery (rho2).
ans: rho2/ rho1=?
Answer: attached below. rate brainliest if correct please
The viscosity of a fluid is to be measured by an viscometer constructed of two 5 ft long concentric cylinders. The inner diameter of the outer cylinder is 6 in, and the gap between the two cylinders is 0.035 in. The outer cylinder is rotated at 250 rpm, and the torque is measured to be 1.2lbf *ft.
A) Determine the viscosity of the fluid.
Answer:
[tex] \mu= \frac{1.2 lb ft}{7.85ft^2 * \frac{6.545 ft/s}{0.00292ft} *0.25 ft}=2.728x10^{-4} \frac{lbf s}{ft^2}[/tex]
Explanation:
For this case we need to remember first thet the torque T is defined as:
[tex] T = FR[/tex]
Where T represent the torque, F the force acting in the inner cylinder and R the radius for the inner cylinder.
For the inner cylinder the force acting can be expressed as:
[tex] F = \mu A \frac{v}{l}[/tex]
Where [tex] \mu[/tex] represent the viscosity of the fluid, A the area of the inner cylinder, v represent the tangential velocity and l the thickness of fluid between the two cylinders.
And the tangential velocity for this case can be esxpressed as [tex] v = wR[/tex]
The info given is:
[tex] l = 0.035 in *\frac{1ft}{12in}=0.00292 ft[/tex]
[tex] R= \frac{D}{2} =\frac{6 in}{2}= 3 in*\frac{1ft}{12 in}=0.25 ft[/tex]
[tex] L = 5 ft[/tex] from the info given
N= 250 rpm represent the reveolutions per minute
[tex] T = 1.2 lbf ft[/tex] represent the torque given
We can find the surface area for the cylinder with this formula:
[tex] A= 2\pi R L[/tex]
And if we replace we got:
[tex] A= 2\pi 0.25 ft *5 ft= 7.85 ft^2[/tex]
Now we can find the tangential velocity like this:
[tex] v=wR= \frac2\pi *250 rpm* \frac{1min}{60s} * 0.25 ft=6.55\frac{ft}{s}[/tex]
Now we can set up the following equation for the torque:
[tex] T = FR[/tex]
[tex] T = \mu A \frac{v}{l} R[/tex]
And we can find the value for the viscosity [tex]\mu[/tex] like this:
[tex] \mu = \frac{T}{A \frac{v}{l} R}[/tex]
And if we replace we got:
[tex] \mu= \frac{1.2 lb ft}{7.85ft^2 * \frac{6.545 ft/s}{0.00292ft} *0.25 ft}=2.728x10^{-4} \frac{lbf s}{ft^2}[/tex]
A pipe in a district heating network is transporting over-pressurized hot water (10 atm) at a mass flow of 0.5 kg/s. The pipe is 5 m long, has an inner radius of 50 cm and pipe wall thickness of 50 mm. The pipe has a thermal conductivity of 20 W/m-K, and the inner pipe surface is at a uniform temperature of 110 ºC. The convection heat transfer coefficient of the air surrounding the pipe is 100W/m2 -K. The temperature of the water at inlet of pipe is 130 ºC and the constant pressure specific heat of hot water is 4000 J/kg-ºC. If the temperature of the air surrounding the pipe is 20 ºC, determine the exit temperature of the water at the end of the pipe.
The calculation of the exit temperature of water in the heated pipe involves using the energy balance equation, considering the heat lost through the pipe walls by convection, and then finding the change in thermal energy of the water via the heat transfer equation to solve for the exit temperature.
Explanation:Exit Temperature of Water in a Heated Pipe
To determine the exit temperature of water at the end of an over-pressurized heated pipe, we must consider the energy balance for the water flowing through the pipe. Based on the first law of thermodynamics, the change in thermal energy of the water will be equal to the heat lost through the pipe walls by convection:
Q = mc_p
(Exit Temperature - Inlet Temperature)
In this case, Q will be negative, since the water is losing heat to the surrounding air. The heat transfer from the pipe to the air is given by:
Q = hA(T_surface - T_air)
The area A for heat transfer is the external surface area of the pipe (
2
classes Math.PI
* radius * length of the pipe). Since we have the heat transfer coefficient h, the surface temperature of the pipe T_surface, and the air temperature T_air, we can calculate Q. Then we can use the mass flow rate m and the constant pressure specific heat c_p to find the exit temperature of the water.
To solve, we first calculate Q, then rearrange the first equation to solve for the Exit Temperature.
Create a conditional expression that evaluates to string "negative" if userVal is less than 0, and "non-negative" otherwise. Ex: If userVal is -9, output is: -9 is negative.
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".
A coal-fired power plant equipped with a SO2 scrubber is required to achieve an overall SO2 removal efficiency of 85%. The existing scrubber is 95% efficient. Rather than treating the entire gas stream to 95% removal, the plant proposes to treat part of the flue gas to 95% removal, and to bypass the remainder around the scrubber. Calculate the fraction of the flue gas stream that can be bypassed around the scrubber (i.e., Qbypass/Q) and still satisfy the regulatory requirement.
Answer:
bypassed fraction B will be B= 0.105 (10.5%)
Explanation:
doing a mass balance of SO₂ at the exit
total mass outflow of SO₂ = remaining SO₂ from the scrubber outflow + bypass stream of SO₂
F*(1-er) = Fs*(1-es) + Fb
where
er= required efficiency
es= scrubber efficiency
Fs and Fb = total mass inflow of SO₂ to the scrubber and to the bypass respectively
F= total mass inflow of SO₂
and from a mass balance at the inlet
F= Fs+ Fb
therefore the bypassed fraction B=Fb/F is
F*(1-er) = Fs*(1-es) + Fb
1-er= (1-B)*(1-es) +B
1-er = 1-es - (1-es)*B + B
(es-er) = es*B
B= (es-er)/es = 1- er/es
replacing values
B= 1- er/es=1-0.85/0.95 = 2/19 = 0.105 (10.5%)
A simple ideal Rankine cycle with water as the working fluid operates between the pressure limits of 4 MPa in the boiler and 20 kPa in the condenser and a turbine inlet temperature of 700°C. The boiler is sized to provide a steam flow of 50 kg/s. Determine the power produced by the turbine and consumed by the pump.
Answer:
a) 69,630KW
b) 203 KW
Explanation:
The data obtained from Tables A-4, A-5 and A-6 is as follows:
[tex]h_{1} = h_{f,@20KPa} = 251.42 KJ/kg\\v_{1} = v_{f,@20KPa} = 0.001017 KJ/kgK\\\\w_{p,in} = v_{1} * (P_{2} - P_{1})\\w_{p,in} = (0.001017)*(4000-20)\\\\w_{p,in} = 4.05 KJ/kg\\\\h_{2} = h_{1} - w_{p,in} \\h_{2} = 251.42 + 4.05\\\\h_{2} = 255.47KJ/kg\\\\P_{3} = 4000KPa\\T_{3} = 700 C\\s_{3} = 7.6214 KJ/kgK\\\\h_{3} = 3906.3 KJ/kg\\\\P_{4} = 20 KPa\\s_{3} = s_{4} = 7.6214KJ/kgK\\s_{f} = 0.8320 KJ/kgK\\s_{fg} = 7.0752 KJ/kgK\\\\[/tex]
[tex]x_{4} = \frac{s_{4} - s_{f} }{s_{fg} } \\\\x_{4} = \frac{7.6214-0.8320}{7.0752} = 0.9596\\\\h_{f} = 251.42KJ/kg \\h_{fg} = 2357.5KJ/kg \\\\h_{4} = h_{f} + x_{4}*h_{fg} = 251.42 + 0.9596*2357.5 = 2513.7KJ/kg\\\\[/tex]
The power produced and consumed by turbine and pump respectively are:
[tex]W_{T,out} = flow(m) *(h_{3} - h_{4}) \\W_{T,out} = 50 *(3906.3-2513.7)\\\\W_{T,out} = 69,630 KW\\\\W_{p,in} = flow(m) *w_{p,in} = 50*4.05 = 203 KW[/tex]
For an atmospheric pressure of 101 kPa (abs) determine the heights of the fluid columns in
barometers containing one of the following liquids a) mercury, b) water, c) ethyl alcohol.
Calculate the heights including the effect of vapour pressure and compare the results with those obtained
neglecting vapour pressure. Do these results support the widespread use of a particular fluid for
barometers? Explain the reason.
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.
Niobium has a BCC crystal structure, an atomic radius of 0.143 nm and an atomic weight of 92.91 g/mol. Calculate the theoretical density for Nb.
To calculate the theoretical density of Niobium with a BCC structure, we use its atomic radius and atomic weight, converting these into the cube's edge length using the BCC relation, then apply the density formula.
Explanation:To calculate the theoretical density of Niobium (Nb), which has a body-centered cubic (BCC) crystal structure, we first use the known values: atomic radius = 0.143 nm (or 0.143 × 10-9 m) and atomic weight = 92.91 g/mol. The formula for the density (ρ) in a BCC structure is ρ = (2 × M) / (a3 × NA), where M is the atomic mass, a is the edge length of the cube, and NA is Avogadro's number (6.022 × 1023 atoms/mol).
Since it's a BCC structure, the atomic radius relates to the cube's edge length (a) as a = 4r / √3. Substituting the given atomic radius, we find a = 4 * 0.143 × 10-9 m / √3. Then, to find the density, we substitute M (92.91 g/mol), a, and NA into the density formula. This calculation will give us the theoretical density of Niobium in g/cm3.
At Westtown High School, the mean score on the French final examination was with a standard deviation of , while the mean score on the Spanish final examination was with a standard deviation of . To earn a language honor at graduation, students must score in the th percentile on all their language final exams. Brynne scored on both the French exam and the Spanish exam. Is Brynne qualified for honors?
Answer:
The score for both exams 88 is above the 90 percentile, so then Brynne qualified for honors. See the explanation below.
Explanation:
Assuming the following question:"At Westtown High School, the mean score on the French final examination was 81 with a standard deviation of 5, while the mean score on the Spanish final examination was 72 with a standard deviation of 12.
To earn a language honor at graduation, students must score in the 90th percentile on all their language final exams. Brynne scored 88 on both the French exam and the Spanish exam. Is Brynne qualified for honors?"
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
French case
Let X the random variable that represent the scores for the French exam of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(81,5)[/tex]
Where [tex]\mu=81[/tex] and [tex]\sigma=5[/tex]
For this part we want to find a value a, such that we satisfy this condition:
[tex]P(X>a)=0.1[/tex] (a)
[tex]P(X<a)=0.90[/tex] (b)
Both conditions are equivalent on this case. We can use the z score again in order to find the value a.
As we can see on the figure attached the z value that satisfy the condition with 0.90 of the area on the left and 0.1 of the area on the right it's z=1.28. On this case P(Z<1.28)=0.90 and P(z>1.28)=0.1
If we use condition (b) from previous we have this:
[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.90[/tex]
[tex]P(z<\frac{a-\mu}{\sigma})=0.90[/tex]
But we know which value of z satisfy the previous equation so then we can do this:
[tex]z=1.28<\frac{a-81}{5}[/tex]
And if we solve for a we got
[tex]a=81 +1.28*5=87.4[/tex]
So the value for the scores that separates the bottom 90% of data from the top 10% is 87.4 (90th percentile).
And since the score of Brynne is 88 is above the 90 percentile
Spanish case
Let X the random variable that represent the scores for the Spanish exam of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(72,12)[/tex]
Where [tex]\mu=72[/tex] and [tex]\sigma=12[/tex]
For this part we want to find a value a, such that we satisfy this condition:
[tex]P(X>a)=0.1[/tex] (a)
[tex]P(X<a)=0.90[/tex] (b)
Both conditions are equivalent on this case. We can use the z score again in order to find the value a.
As we can see on the figure attached the z value that satisfy the condition with 0.90 of the area on the left and 0.1 of the area on the right it's z=1.28. On this case P(Z<1.28)=0.90 and P(z>1.28)=0.1
If we use condition (b) from previous we have this:
[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.90[/tex]
[tex]P(z<\frac{a-\mu}{\sigma})=0.90[/tex]
But we know which value of z satisfy the previous equation so then we can do this:
[tex]z=1.28<\frac{a-72}{12}[/tex]
And if we solve for a we got
[tex]a=72 +1.28*12=87.36[/tex]
So the value for the scores that separates the bottom 90% of data from the top 10% is 87.36 (90th percentile).
And since the score of Brynne is 88 is above the 90 percentile
Determine the tension developed in cord DE required for equilibrium of the lamp. Express your answer to three significant figures and include the appropriate units.
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)
dentify a semiconducting material and provide the value of its band gap) that could be used in: (a) (1 point) red LED (b) (1 point) UV LED c) (1 point) infrared detector (d) (1 point) blue LED (e) (1 point) green LED
Answer:
(a) Aluminum Indium Gallium Phosphide (AlInGaP). Band gap = 1.81eV ≈ 2eV
(b) Gallium Nitride (GaN). Band Gap = 3.4eV
(c) Aluminium Gallium Arsenide (AlGaA). Band Gap = 1.42eV ≈ 2.16eV
(d) Zinc Selenide (ZnSe). Band Gap = 2.82eV
(e) Gallium Phosphide (GaP). Band Gap = 2.24eV
Explanation:
LED's are semi-conducting materials that convert electrical energy to light energy. The light color emitted from the LED depends on the semi-conducting material and other compositions.
The band gap of the semi-conductor determines its wavelength. High band gap semi-conductors emit lower wavelengths which means greater power(UV semi-conducting macterials fall under this category).
If E = 94.2 mJ of energy is transferred when Q = 1.66 C of charge flows through a circuit element, what is the voltage across the circuit element?
Answer:
V = 56.8 mV
Explanation:
When a current I flows across a circuit element, if we assume that the dimensions of the circuit are much less than the wavelength of the power source creating this current, there exists a fixed relationship between the power dissipated in the circuit element, the current I and the voltage V across it, as follows:
P = V*I
By definition, power is the rate of change of energy, and current, the rate of change of the charge Q, so we can replace P and I, as follows:
E/t = V*q/t ⇒ E = V*Q
Solving for V:
V = E/Q = 94.2 mJ /1.66 C = 56.8 mV
assume a strain gage is bonded to the cylinder wall surface in the direction of the axial strain. The strain gage has nominal resistance R0 and a Gage Factor GF . It is connected in a Wheatstone bridge configuration where all resistors have the same nominal resistance; the bridge has an input voltage Vin. (The strain gage is bonded and the Wheatstone bridge balanced with the vessel already pressurized.) Develop an expression for the voltage change ?V across the bridge if the cylinder pressure changes by ?P.
Explanation:
Note: For equations refer the attached document!
The net upward pressure force per unit height p*D must be balanced by the downward tensile force per unit height 2T, a force that can also be expressed as a stress, σhoop, times area 2t. Equating and solving for σh gives:
Eq 1
Similarly, the axial stress σaxial can be calculated by dividing the total force on the end of the can, pA=pπ(D/2)2 by the cross sectional area of the wall, πDt, giving:
Eq 2
For a flat sheet in biaxial tension, the strain in a given direction such as the ‘hoop’ tangential direction is given by the following constitutive relation - with Young’s modulus E and Poisson’s ratio ν:
Eq 3
Finally, solving for unknown pressure as a function of hoop strain:
Eq 4
Resistance of a conductor of length L, cross-sectional area A, and resistivity ρ is
Eq 5
Consequently, a small differential change in ΔR/R can be expressed as
Eq 6
Where ΔL/L is longitudinal strain ε, and ΔA/A is –2νε where ν is the Poisson’s ratio of the resistive material. Substitution and factoring out ε from the right hand side leaves
Eq 7
Where Δρ/ρε can be considered nearly constant, and thus the parenthetical term effectively becomes a single constant, the gage factor, GF
Eq 8
For Wheat stone bridge:
Eq 9
Given that R1=R3=R4=Ro, and R2 (the strain gage) = Ro + ΔR, substituting into equation above:
Eq9
Substituting e with respective stress-strain relation
Eq 10
The article "Display of Health Risk Behaviors on MySpace by Adolescents"† described a study in which researchers looked at a random sample of 500 publicly accessible MySpace web profiles posted by 18-year-olds. The content of each profile was analyzed. One of the conclusions reported was that displaying sport or hobby involvement was associated with decreased references to risky behavior (sexual references or references to substance abuse or violence).
a. It's not entirely reasonable to generalize the conclusion to all 18-year-olds with publicly accessible MySpace profiles due to the study's limited sample size and potential biases.
b. b. No, it's not reasonable to generalize the conclusion to all 18-year-old MySpace users because not all users have publicly accessible profiles.
c. While somewhat reasonable, generalizing to MySpace users with publicly accessible profiles requires caution, considering potential individual differences within this subgroup.
The study only looked at a random sample of 500 profiles, which might not be fully representative of all 18-year-olds on MySpace. Generalizing to all 18-year-old MySpace users is not reasonable as not all users have publicly accessible profiles, which could skew the findings.
Additionally, factors such as cultural differences, regional variations, and individual preferences could influence whether adolescents choose to display sport or hobby involvement on their profiles.
See text
The article "Display of Health Risk Behaviors on MySpace by Adolescents" (Archives of Pediatrics and Adolescent Medicine [2009]:
) described a study in which researchers looked at a random sample of 500 publicly accessible MySpace web profiles posted by 18-year-olds. The content of each profile was analyzed. One of the conclusions reported was that displaying sport or hobby involvement was associated with decreased references to risky behavior (sexual references or references to substance abuse or violence).
a. Is it reasonable to generalize the stated conclusion to all 18-year-olds with a publicly accessible MySpace web profile? What aspect of the study supports your answer?
b. Not all MySpace users have a publicly accessible profile. Is it reasonable to generalize the stated conclusion to all 18-year-old MySpace users? Explain.
c. Is it reasonable to generalize the stated conclusion to all MySpace users with a publicly accessible profile? Explain.
Chlorate (ClO3 − ) is a toxic anion in polluted water from the use of high energy materials, herbicide, and drinking water disinfection with chlorine chemicals. In a reaction for the catalytic reduction of ClO3 − into the non-toxic chloride (Cl− ), the reaction rate constant was measured as 9.98 hr−1 , then how many minutes does the reaction take to reduce 90% of the chlorate in the water?
Answer:
Time taken = 5.41mins
Explanation:
A step by step calculations with detailed explanation has been attached below.