Answer:
a) [tex]\frac{P}{\rho v^{3}D^{2} } = f(\frac{\Omega D}{v}, n)[/tex]
b) P = 2184.57 kW
c) [tex]\Omega = 122.5 rpm[/tex]
Explanation:
Pressure = P
Density = [tex]\rho[/tex]
Diameter = D
Velocity = v
Rate of rotation = Ω
number of blades = n
a) The relationship for the model is given by:
[tex]\frac{P}{\rho v^{3}D^{2} } = f(\frac{\Omega D}{v}, n)[/tex]
b) The Power developed geometrically and dynamically similar prototype:
The data corresponding to the model:
[tex]P_{mod} = 3.8 kW\\v_{mod} = 40 m/s\\D_{mod} = 50 cm = 0.5 m\\\rho_{mod} = 1.2255 kg/m^{3}\\ \Omega = 4200 rpm[/tex]
The data corresponding to the prototype
[tex]P_{prot} = ? \\v_{prot} = 35 m/s\\D_{prot} = 15 m\\\rho_{prot} = 1.1685 kg/m^{3}[/tex]
[tex](\frac{P}{\rho v^{3}D^{2} } )_{mod} = (\frac{P}{\rho v^{3}D^{2} } )_{prot}[/tex]
[tex]\frac{3.8}{1.2255 * 40^{3} * 0.5^{2} }= \frac{P}{1.1685 * 35^{3} * 15^{2} }\\P = \frac{3.8 * 1.1685 * 35^{3} * 15^{2}}{1.2255 * 40^{3} * 0.5^{2} }[/tex]
P = 2184.57 kW
c) To calculate the appropriate rotation rate for the prototype
[tex](\frac{\Omega D}{v})_{model} = (\frac{\Omega D}{v})_{prot}[/tex]
[tex]\frac{4200*0.5}{40}= \frac{\Omega * 15}{35}\\\Omega = \frac{4200*0.5*35}{40*15}\\\Omega = 122.5 rpm[/tex]
Ray L. Zapp is thinking about testing strategies for his new HashTable class, which uses Rainforest's cloud storage service to make hash tables persistent. For the next 4 questions, write the letter of the one BEST testing technique (from the list below) to achieve the listed testing goal. (Some testing techniques may be used more than once or not at all.) A. Black-box testing B. Fuzz testing C. Stubbing& mocking D. Mutation testing E. White-box/glass-box testing 4 13.(1 pts) Making sure the hash table is robust when presented with nonsensical input 15. (1 pts) Making sure his service behaves correctly when Rainforest's cloud storage service is down 16. (1 pts) Improving the thoroughness of coverage of his test suite by uncovering test cases he missed 一( 1 pts) Making sure hash collisions are handled properly
Answer:
As Ray L zapp is thinking about new strategies to test his new hash Table,hence therefore the best testing technique is Stubbing& mocking .
Change the open-loop transfer function into G(s) = 1 s(s+2) , Design a lag compensation so that the dominant poles of the closed-loop system are located at s = –1 ± j and the steady-state error to a unit-ramp input is less than 0.2. Verify your design by providing the compensated closed-loop system root locus and calculating the steady-state error to a unit-ramp input.
Answer:
[tex]D_c(s) = 1.8(\frac{s+1.11}{s+0.2})[/tex]
Explanation:
Code attached with picture.
Evaporation in Double-Effect Reverse-Feed Evaporators. A feed containing 2 wt % dissolved organic solids in water is fed to a double-effect evaporator with reverse feed. The feed enters at 100°F and is concentrated to 25% solids. The boiling-point rise can be considered negligible as well as the heat of solution. Each evaporator has a 1000-ft2 surface area and the heat-transfer 700 btu/h -ft2-°F. The feed enters evaporator number 2 and steam at 100 psia is fed to number 1. The pressure iin the vapor space of evaporator number 2 is 0.98 psia. Assume that the heat capacity of all liquid solutions is that of liquid water. Calculate the feed rate F and the product rate Li of a solution containing 25% solids. (Hint: Assume a feed rate of, say, F 1000 lbm/h. Calculate the area. Then calculate the actual coefficients are U1 500 and U2- feed rate by multiplying 1000 by 1000/calculated area.) Ans. F- 133 800 lbm/h (60 691 kg/h), L,- 10700lb»/h (4853 kg/h)
Answer:
472,826 lb/hr
Explanation:
As per the given data:
Solids in feed= 2%
Solids in concentrate= 25%
HTA1 = HTA2 = 1000 ft^2
U1 = 500 Btu/ h ft^2 F & U2 = 700 Btu/ h ft^2 F
Overall material balance: Feed= Distillate + concentrate ----> Eq-1
Component balance: Feed * 0.02 = Distillate * 0 + concentrate * 0.25
Feed = 12.5 * concentrate ---> Eq-2
Boiling point rise = negligible, so solution & solvent vapor temperature will be same.
Assumed that the 1st effect is operating under atmospheric pressure (Boiling point - 212F).
As per the data:
Latent heat 212F = 300 Btu/lb
Latent heat 100F = 320 Btu/lb
As per material balance:
Vapor flowrate * latent heat = Overall HT coefficient * HTA * DT
1st effect: M-1= (500 * 1000 * (326-212)) / 300 = 190,000 lb/hr
2nd effect: M-2= (700*1000 * (212-100)) / 320 = 245,000 lb/hr
Distillate = M-1 + M-2 = 190,000+245,000 = 435,000 lb/hr
Substituting the above in Eq-1
Feed = 435,000 + concentrate
Substitute Eq-2 in the above
12.5 * concentrate = 435,000 + concentrate
Concentrate, L1 = 435,000/11.5 = 37826 lb /hr
Feed, F = 435,000 + 37826 = 472,826 lb/hr
1st effect operating pressure is not given, That may be the reason we are not getting the given answer. But procedure is right.
[ Find the figure in the attachment ]
Traditional password entry schemes are susceptible to "shoulder surfing" in which an attacker watches an unsuspecting user enter their password or PIN number and uses it later to gain access to the account. One way to combat this problem is with a randomized challenge-response system. In these systems, the user enters different information every time based on a secret in response to a randomly generated challenge. Consider the fol- lowing scheme in which the password consists of a five-digit PIN number (00000 to 99999). Each digit is assigned a random number that is 1, 2, or 3. The user enters the random numbers that correspond to their PIN instead of their actual PIN numbers.For example, consider an actual PIN number of 12345. To authenticate the user would be presented with a screen such as:PIN: 0 1 2 3 4 5 6 7 8 9 NUM: 3 2 3 1 1 3 2 2 1 3The user would enter 23113 instead of 12345. This doesn’t divulge the password even if an attacker intercepts the entry because 23113 could correspond to other PIN numbers, such as 69440 or 70439. The next time the user logs in, a different sequence of random numbers would be generated, such as: PIN: 0 1 2 3 4 5 6 7 8 9 NUM: 1 1 2 3 1 2 2 3 3 3Your program should simulate the authentication process. Store an actual PIN number in your program. The program should use an array to assign random numbers to the digits from 0 to 9. Output the random digits to the screen, input the response from the user, and output whether or not the user’s response correctly matches the PIN number.I have this code so far, but would like to input cstrings and vectors to fulfill the requirements, I need help with that. This is for c++ for beginners#include #include #include #include using namespace std;void generateRandomNumbers(int *random){ // Use current time as seed for random generatorsrand(time(0));for(int i=0;i<10;i++){random[i] = 1 + rand() % 3;}}bool isMatch(string pin,string randomPin,int *random){int index;for(int i=0;i<(int)pin.length();i++){ //converting pin number to int so that we can check the random number at that indexindex = pin[i]-'0';if((randomPin[i]-'0') != random[index-1])return false;}return true;}int main(){string pin = "12345";string randomPin;int random[10];generateRandomNumbers(random);cout << "Randomly Generated numbers " << endl;for(int i=0;i<10;i++){cout << random[i] << " ";}cout << endl;cout << "Now Enter your pin interms of random numbers: ";cin >> randomPin;if(isMatch(pin,randomPin,random)){cout << "Both matches" << endl;}else{cout << "Sorry you entered wrong pin.." << endl;}}
The following code or the program will be used:
Explanation:
import java.util.Scanner;
public class Authenticate
{
public static void main(String[] args)
{
// Actual password is 99508
int[] actual_password = {9, 9, 5, 0, 8};
// Array to hold randomly generated digits
int[] random_nums = new int[10];
// Array to hold the digits entered by the user to authenticate
int[] entered_digits = new int[actual_password.length];
// Randomly generate numbers from 1-3 for
// for each digit
for (int i=0; i < 10; i++)
{
random_nums[i] = (int) (Math.random() * 3) + 1;
}
// Output the challenge
System.out.println("Welcome! To log in, enter the random digits from 1-3 that");
System.out.println("correspond to your PIN number.");
System.out.println();
System.out.println("PIN digit: 0 1 2 3 4 5 6 7 8 9");
System.out.print("Random #: ");
for (int i=0; i<10; i++)
{
System.out.print(random_nums[i] + " ");
}
System.out.println();
System.out.println();
// Input the user's entry
Scanner keyboard = new Scanner(System.in);
System.out.println("Enter code.");
String s = keyboard.next();
String s = keyboard.next();
// Extract the digits from the code and store in the entered_digits array
for (int i=0; i<s.length(); i++)
{
entered_digits[i] = s.charAt(i) - '0'; // Convert char to corresponding digit
}
// At this point, if the user typed 12443 then
// entered_digits[0] = 1, entered_digits[1] = 2, entered_digits[2] = 4,
// entered_digits[3] = 4, and entered_digits[4] = 3
/****
TO DO: fill in the parenthesis for the if statement
so the isValid method is invoked, sending in the arrays
actual_password, entered_digits, and random_nums as
parameters
***/
if (isValid (actual_password, entered_digits, random_nums)) // FILL IN HERE
{
System.out.println("Correct! You may now proceed.");
}
else
{
System.out.println("Error, invalid password entered.");
}
/***
TO DO: Fill in the body of this method so it returns true
if a valid password response is entered, and false otherwise.
For example, if:
actual = {9,9,5,0,8}
randnums = {1,2,3,1,2,3,1,2,3,1}
then this should return true if:
entered[0] == 1 (actual[0] = 9 -> randnums[9] -> 1)
entered[1] == 1 (actual[1] = 9 -> randnums[9] -> 1)
entered[2] == 3 (actual[2] = 5 -> randnums[5] -> 3)
entered[3] == 1 (actual[3] = 0 -> randnums[0] -> 1)
entered[4] == 3 (actual[4] = 8 -> randnums[8] -> 3)
or in other words, the method should return false if any of
the above are not equal.
****/
public static boolean isValid(int[] actual, int[] entered, int[] randnums)
{
int Index = 0;
boolean Valid = true;
while (Valid && (Index < actual.length))
{
int Code = actual[Index];
if (entered [Index] != randnums [Code])
{
Valid = false;
}
Index++;
}
return Valid;
}
Which of the following are TRUE concerning rectifier circuits? Select all that apply.
a. The output of an ideal full wave rectifier is zero volts only when the input is zero volts.
b. Half-wave rectifier circuits need a minimum of 4 diodes to operate.
c. In an ideal full wave bridge rectifier, half of the diodes are in the ON state and half of the diodes are in the OFF state at any given time the input voltage is not zero.
d. The output of an ideal half wave rectifier is zero volts only when the input is zero volts.
e. A turn-on voltage of a diode (y,) greater than zero can cause the output of a full wave rectifier to be zero volts even when the input is not zero volts.
f. An advantage of the half wave rectifier is that is can use a smoothing capacitor, while a full wave rectifier cannot.
Explanation:
a. The output of an ideal full wave rectifier is zero volts only when the input is zero volts.
True
The output of an ideal full wave rectifier is zero volts only when the input is zero since 1st diode is forward biased during one half cycle of the input and 2nd diode is forward biased during the other half cycle of the input therefore, it fully utilizes both the input cycles so the output voltage is only zero when the input is zero.
b. Half-wave rectifier circuits need a minimum of 4 diodes to operate.
False
A Half-wave rectifier circuits need a minimum of 1 diode to operate, whereas a full-wave bridge rectifier need minimum of 4 diodes to operate.
c. In an ideal full wave bridge rectifier, half of the diodes are in the ON state and half of the diodes are in the OFF state at any given time the input voltage is not zero.
True
A full-wave bridge rectifier consists of 4 diodes, where 2 diodes are functional in half of the cycle(so the other 2 are off) and other 2 diodes are functional in the other half cycle( so the other 2 are off).
d. The output of an ideal half wave rectifier is zero volts only when the input is zero volts.
False
The output of an ideal half wave rectifier is zero during half of the cycle when the diode is reversed biased and doesn't conduct even though input voltage is not zero volts at this point.
e. A turn-on voltage of a diode (y,) greater than zero can cause the output of a full wave rectifier to be zero volts even when the input is not zero volts.
False
A turn-on voltage of a diode (y,) greater than zero cannot cause the output of a full wave rectifier to be zero rather there will be a little voltage drop across the output of full wave rectifier due to this turn-on voltage of diode which is usually 0.7 volts for silicon based diodes.
f. An advantage of the half wave rectifier is that is can use a smoothing capacitor, while a full wave rectifier cannot.
False
Smoothing capacitor can be used in both half wave rectifier as well as full wave rectifier.
An average person produce 0.25 kg of moisture while taking a shower and 0.05 kg while bathing in a tub. Consider a family of four who each shower once a day in a bathroom that is not ventilated. Taking the heat of vaporization of water to be 2450 kJ/kg, determine the contribution of showers to the latent heat load of the air conditioner per day in summer?
Answer:
Q = 2450 KJ/day
Explanation:
We are given;
The amount of moisture produced by a person while taking shower; m = 0.25 kg/person
Total number of persons in the family; N = 4
Latent heat of vaporization of water:
h_fg = 2450 KJ/kg
Since amount of moisture per person is 0.25kg per day,thus for 4 people, total mass; m_total = 4 x 0.25 = 1 kg per day
The formula for the amount of latent heat added to the space due to shower per day, is given as;
Q = m_total x h_fg
Plugging in the relevant values;
Q = 1 x 2450 = 2450 KJ/day
An offset h must be introduced into a metal tube of 0.75-in. outer diameter and 0.08-in. wall thickness. Knowing that the maximum stress after the offset is introduced must not exceed 4 times the stress in the tube when it is straight, determine the largest offset that can be used
Complete Question
The diagram for this question is shown on the first uploaded image
Answer:
The largest offset that can be used is [tex]h = 0.455 \ in[/tex]
Explanation:
From the question we are told that
The diameter of the metal tube is [tex]d_m = 0.75 \ in[/tex]
The thickness of the wall is [tex]D = 0.08 \ in[/tex]
Generally the inner diameter is mathematically evaluated as
[tex]d_i = d_m -2D[/tex]
[tex]= 0.75 - 2(0.08)[/tex]
[tex]= 0.59 \ in[/tex]
Generally the tube's cross-sectional area can be evaluated as
[tex]a = \frac{\pi}{4} (d_m^2 - d_i^2)[/tex]
[tex]= \frac{\pi}{4} (0.75^2 - 0.59^2)[/tex]
[tex]= 0.1684 \ in^2[/tex]
Generally the maximum stress of the metal is mathematically evaluated as
[tex]\sigma = \frac{P}{A}[/tex]
[tex]\sigma = \frac{P}{ 0.1684}[/tex]
The diagram showing when the stress is been applied is shown on the second uploaded image
Since the internal forces in the cross section are the same with the force P and the bending couple M then
[tex]M = P * h[/tex]
Where h is the offset
The maximum stress becomes
[tex]\sigma_n = \frac{P}{A} + \frac{M r_m }{I}[/tex]
Where [tex]r_m[/tex] is the radius of the outer diameter which is evaluated as
[tex]r_m = \frac{0.75}{2}[/tex]
[tex]r_m = 0.375 \ in[/tex]
and I is the moment of inertia which is evaluated as
[tex]I = \frac{\pi}{64} (d_m^4 - d_i^4 )[/tex]
[tex]= \frac{\pi}{64}(0.75^4 - 0.59^4)[/tex]
[tex]= 0.009583 \ in^4[/tex]
So the maximum stress becomes
[tex]\sigma' = \frac{P}{0.1684} + \frac{Phr}{0.009583}[/tex]
Now the question made us to understand that the maximum stress when the offset was introduced must not exceed the 4 times the original stress
So
[tex]\sigma ' = 4 \sigma[/tex]
=> [tex]\frac{P}{0.1684} + \frac{Phr_m }{0.009583} = 4 [\frac{P}{0.1684} ][/tex]
The P would cancel out
[tex]\frac{1}{0.1684} + \frac{h(0.375)}{0.009583} = \frac{4}{0.1684}[/tex]
[tex]5.94 + 39.13h = 23.753[/tex]
[tex]39.13h = 17. 813[/tex]
[tex]h = 0.455 \ in[/tex]
Ronny has a hydraulic jack. The input force is 250 N, while the output force is 7,500 N. If the area of the pipe below the input is 0.2 m2, what is the area of the pipe below the load?
Answer:
6 m²
Explanation:
application of fluid pressure according to Pascal's principle for the two pistons is given as:
[tex]P_1=P_2[/tex]
Where P₁ is the pressure at the input and P₂ is the pressure at the output.
But P₁ = F₁ / A₁ and P₂ = F₂ / A₂
Where F₁ and F₂ are the forces applied at the input and output respectively and A₁ and A₂ are the area of the input pipe and output pipe respectively
Since, [tex]P_1=P_2[/tex]
[tex]\frac{F_1}{A_1} =\frac{F_2}{A_2}\\[/tex]
But A₁ = 0.2 m², F₁ = 250 N, F₂ = 7500 N. Substituting values to get:
[tex]\frac{F_1}{A_1} =\frac{F_2}{A_2}\\\frac{250}{0.2}=\frac{7500}{A_2}\\ A_2=\frac{7500*0.2}{250} = 6m^2[/tex]
Therefore, the area of the pipe below the load is 6 m²
Go online and search for information about companies that have been harmed or bankrupted by a disaster. Choose one such company and create a brief case study about it. Successful narratives will focus on the manner in which the organization was impacted, including financial losses, losses of sales, or the need for layoffs.
Answer:
See explaination
Explanation:
In recent times and in the past, so many companies have been dealt with or bankrupted by disaster.
This disasters are in the form of floods, earthquake and tsunami. This has led to many companies across world to have faced problem. As Japan is particularly vulnerable to natural disaster because of its climate and topography let me take example of japan .The tsunami that struck Japan in March this year which has lead to a debt of worth nearly $8 billion due to failed businesses. A total of 341 firms in japan with a combined 6,376 employees has got affected by this disaster.
One of company that hit hard by these disaster is Toyota. Toyota is a Japanese multinational automotive manufacturer headquartered in Toyota, Aichi, Japan. Actually due to flood in thailand toyota has faced a big loss. As flood hit thailand the parts of vehicle which is produced in thailand has stopped . It leads to interruption in the supply chain of some Thailand made components.
This lead to suspension of production as thai made component was unavailable.It has been reported until the situation of thailand will be good toyota plants in Indiana, Kentucky and Ontario, Canada, will be shut down . And production rate at plant in North America will get slow down .This flood leads to a loss of 37500 vehicles .Due to this reason toyota was forced to open its plants in Southeast asian country just to enhance the production rate.Sales of vehicle has also got reduced in thailand . Hence toyota faced a total loss of 1.6 billion dollar.
In Nigeria, due to the insurgency of the Boko Haram, many companies have entirely stopped operations in some regions, this harsh realities leads to massive loss of revenue for such companies.
The piping system that connects one reservoir to a second reservoir consists of 150-ft of 3-in. cast iron pipe that has four flanged elbows, a well-rounded entrance, and sharp-edged exit, and a fully open gate valve. For 75 gal/min of water at 50 °F, determine the elevation difference between the two reservoirs (in ft).
The elevation difference between the two reservoirs is 10.28 ft
Given information:
- Length of the 3-in. cast iron pipe = 150 ft
- Number of flanged elbows = 4
- Entrance condition: well-rounded
- Exit condition: sharp-edged
- Fully open gate valve
- Flow rate = 75 gal/min
- Water temperature = 50 °F
Step 1: Convert the flow rate from gal/min to ft³/s.
Flow rate in ft³/s = (75 gal/min) × (1 ft³/7.48 gal) × (1 min/60 s) = 0.167 ft³/s
Step 2: Calculate the velocity of flow.
Velocity of flow, V = Flow rate / Cross-sectional area of the pipe
Cross-sectional area of the 3-in. cast iron pipe = (π/4) × (3/12)² ft² = 0.0491 ft²
Velocity of flow, V = 0.167 ft³/s / 0.0491 ft² = 3.4 ft/s
Step 3: Calculate the Reynolds number to determine the flow regime.
Reynolds number, Re = (ρVD) / μ
Where,
ρ = density of water at 50 °F = 1.94 slugs/ft³ (from tables)
V = velocity of flow = 3.4 ft/s
D = diameter of the pipe = 3/12 ft
μ = dynamic viscosity of water at 50 °F = 2.73 × 10⁻⁵ lb.s/ft² (from tables)
Re = (1.94 slugs/ft³) × (3.4 ft/s) × (3/12 ft) / (2.73 × 10⁻⁵ lb.s/ft²) = 1.9 × 10⁵
Since Re > 4000, the flow is turbulent.
Step 4: Calculate the head loss due to friction in the straight pipe.
Head loss due to friction, hf = f × (L/D) × (V²/2g)
Where,
f = friction factor (depends on Reynolds number and pipe roughness)
L = length of the pipe = 150 ft
D = diameter of the pipe = 3/12 ft
V = velocity of flow = 3.4 ft/s
g = acceleration due to gravity = 32.2 ft/s²
From the Moody diagram or friction factor tables for cast iron pipe, assuming a relative roughness of 0.00085, and Re = 1.9 × 10⁵, the friction factor, f = 0.021.
Substituting the values, hf = 0.021 × (150 × 12/3) × [(3.4)²/(2 × 32.2)] = 8.4 ft
Step 5: Calculate the head loss due to fittings (elbows, entrance, exit, and valve).
Head loss due to fittings, hf_fittings = Σ(K × V²/2g)
Where K is the loss coefficient for each fitting.
For 4 flanged elbows, K = 0.3 (for each elbow)
For a well-rounded entrance, K = 0.03
For sharp-edged exit, K = 1.0
For a fully open gate valve, K = 0.2
Substituting the values, hf_fittings = [4 × 0.3 + 0.03 + 1.0 + 0.2] × [(3.4)²/(2 × 32.2)] = 1.7 ft
Step 6: Calculate the velocity head.
Velocity head, hv = V²/2g = (3.4)²/(2 × 32.2) = 0.18 ft
Step 7: Calculate the total head loss.
Total head loss, H = hf + hf_fittings + hv
H = 8.4 ft + 1.7 ft + 0.18 ft = 10.28 ft
Therefore, the elevation difference between the two reservoirs is 10.28 ft for the given piping system and a flow rate of 75 gal/min of water at 50 °F.
A square hole is to be cut using ECM through a plate of low alloy steel that is 12 mm thick. The hole is 35 mm on each side, but the electrode used to cut the hole is slightly less than 35 mm on its sides to allow for overcut, and its shape includes a hole in its center to permit the flow of electrolyte and to reduce the area of the cut. This will produce a center core that can be removed after the tool breaks through. This tool design results in a frontal area of 245 mm2 . The applied current = 1200 amps.
Using an efficiency of 95%, how long will it take to cut the hole?
It will take approximately 0.0238 s to cut the hole.
To calculate the time required to cut the hole using Electrochemical Machining (ECM), we can use the following formula:
[tex]\text { Time }=\frac{\text { Volume of Material to be Removed }}{\text { Material Removal Rate (MRR) }}[/tex]
First, let's calculate the volume of material to be removed:
Volume of Material=Frontal Area × Thickness of Plate
Given:
[tex]Frontal Area =245 \mathrm{~mm}^2\\Thickness of Plate $=12 \mathrm{~mm}$[/tex]
Volume of Material = [tex]245 \mathrm{~mm}^2 \times 12 \mathrm{~mm}=2940 \mathrm{~mm}^3[/tex]
Next, we need to calculate the Material Removal Rate (MRR). MRR is usually given in units of volume removed per unit time per unit current density. Let's assume it is given as [tex]X \mathrm{~mm}^3 / \mathrm{min} / \mathrm{A} / \mathrm{mm}^2 .[/tex] Given the current density [tex]\text { Current Density }=\frac{\text { Applied Current }}{\text { Frontal Area }}[/tex], we can calculate the MRR as:
MRR=X×Current Density
Given:
Applied Current = 1200 A
Efficiency = 95%
Frontal Area = 245 [tex]mm^2[/tex]
[tex]\begin{aligned}& \text { Current Density }=\frac{\text { Applied Current }}{\text { Frontal Area }}=\frac{1200 \mathrm{~A}}{245 \mathrm{~mm}^2} \\& \text { Current Density } \approx 4.897 \mathrm{~A} / \mathrm{mm}^2\end{aligned}[/tex]
Let's assume X=0.2[tex]mm^3[/tex] /s/A/[tex]mm^2[/tex](hypothetical value).
[tex]\mathrm{MRR}=0.2 \times 4.897 \approx 0.979 \mathrm{~mm}^3 / \mathrm{s}[/tex]
To convert mm³/s to g/s, we need to know the density of the material. Let's assume the density of low alloy steel is approximately [tex]7.85 \mathrm{~g} / \mathrm{cm}^3[/tex], which is [tex]7.85 \times 10^{-6} \mathrm{~g} / \mathrm{mm}^3 .[/tex]
Now, we can calculate the MRR in grams per second:
[tex]MRR in $\mathrm{g} / \mathrm{s}=\mathrm{MRR} \times$ Density of material\\MRR in $\mathrm{g} / \mathrm{s}=0.979 \mathrm{~mm}^3 / \mathrm{s} \times 7.85 \times 10^{-6} \mathrm{~g} / \mathrm{mm}^3$\\MRR in $\mathrm{g} / \mathrm{s} \approx 0.00000768515 \mathrm{~g} / \mathrm{s}$[/tex]
Now, we can use these values to calculate the time required:
[tex]\text { Time }=\frac{2940 }{0.00000768515} \\& \text { Time } \approx 0.022594341[/tex]
However, we need to consider the efficiency. Since the efficiency is 95%, the actual time required will be:
[tex]\begin{aligned}& \text { Actual Time }=\frac{\text { Time }}{\text { Efficiency }} \\& \text { Actual Time }=\frac{0.022594341}{0.95}\end{aligned}[/tex]
≈ 0.0238 s
A solenoid with a length of 20 cm and a radius of 5 cm consists of 400 turns and carries a current of 12 A. If z = 0 represents the midpoint of the solenoid, generate a plot for |H(z)| along the axis of the solenoid for a range −20 cm ≤ z ≤ 20 cm. Please use MATLAB to generate the solution numerically and indicate how you did it. Include a copy of the computer code that you create to solve the problem
Answer:
Copy MATLAB code to plot the magnitude of magnetic field strength with respect to z on the axis of solenoid:
z=-20:0.01:20;
H=120.*(((20-(2.*z))./sqrt((20-(2.*z)).^2+100))+((20+(2.*z))./sqrt((20+(2.*z)).^2+100)));
plot(z,H)
title('plot of |H| vs z along the axis of solenoid')
ylabel('Magnitude of magnetic field 'H")
xlabel('position on axis of solenoid 'z")
Explanation:
full explanation is attached as picture and the resultant plot also.
A 2-lane highway is to be constructed across a 6-ft diameter metal culvert which is oriented perpendicular to the highway center line. The culvert is located at station 33+00, and the invert (bottom) of the culvert has an elevation of 150 ft. The elevation of station 21+00 on the tangent alignment is 140 ft. The alignment gradient is +4.0% from station 21+00 to station 29+00, and the gradient from station 29+00 to 37+00 is -3%.
a. Find the longest full station vertical curve that will give approximately 3 ft (3 plusminus 0.5 ft) of cover over the top of the culvert.
b. Compute the station and elevation of the PVC and PVT and the elevation of each full station on the vertical curve.
Answer:
Dude just like forget the highway and drive on the interstate bruh
Explanation:
The intercept of the CML is the origin while the intercept of the SML is RF CML consists of efficient portfolios, while the SML is concerned with all portfolios or securities CML could be downward sloping while that is impossible for the SML CML and the SML are essentially the same except in terms of the securities represented
Answer:
Here, The CML is used for efficient portfolios whereas the SML applies to all portfolios or securities
Considering the seperation theorem, The separation theorem states that the investment decision is sepaarte from the financing decision.
This implies that The SML can be used to analyze the relationship between risk and requried return for all assets.
Under the separation theorem investors should Hold the same portfolio of risky assets and the same expected return but at different levels of risk
Explanation:
See answer
Air is cooled and dehumidified as it flows over the coils of refrigeration system at 100 kPa from 30 ºC and relative humidityof 84%to 15 ºC and relative humidity of 100%. The mass flow rate of dry air is 0.4 kg/s.
Using the formulaand not the Psychrometric chart, determine:
a) the mass flow rate of water,
b)the heat removal from the air.
Answer:
(a) The mass glow rate of water is 0.0032kg/s
(b) The heat removal from air is 14.17KJ/sec
Explanation:
In this question, we are asked to use the formula to calculate the mass flow rate and the heat removal from water based on the data in the question.
To answer this question, we take values of absolute humidity and enthalpy values from table from corresponding temperature
Please check attachment for complete solution and step by step explanation
The attached program (studentsGpa.cpp) uses dynamic allocation to create an array of strings. It asks the user to enter a number and based on the entered number it allocates the array size. Then based on that number it asks the user that many times to enter student’s names. What you need to do:Add another array of doubles to store the gpa of each student as you enter themYou need to display both the student’s name and the gpa.NOTE: must use pointer notation not array subscript. Any submission that uses array subscript won’t be graded
Question: The program was not attached to your question. Find attached of the program and the answer.
Answer:
See the explanation for the answer.
Explanation:
#include <iostream>
using namespace std;
int main()
{
cout << "How many students will you enter? ";
int n;
cin>>n;
string *name = new string[n];
double *gpa = new double[n];
for(int i=0;i<n;i++){
cout<<"Enter student"<<(i+1)<<"'s name: ";
cin>>name[i];
cout<<"Enter student"<<(i+1)<<"'s gpa: ";
cin>>gpa[i];
}
cout<<"The list students"<<endl;
cout<<"Name GPA"<<endl;
cout<<"----------------------"<<endl;
for(int i=0;i<n;i++){
cout<<name[i]<<" "<<gpa[i]<<endl;
}
return 0;
}
OUTPUT : See the attached file.
Q2: Determine the number of 8-inch-high by 8-inch-wide by 16-inch-long concrete blocks required to complete the wall in Figures below. The overhead doors are 10 feet wide by 12 feet high. If lintel blocks are required wherever the #4 horizontal bars are located and above the doors, how many plain blocks and how many lintel blocks are needed for the wall?
The image of the elevation and wall section is missing, so i have attached them.
Answer:
Number of concrete blocks = 1020 blocks
Number of lintel blocks = 240
Number of plain blocks = 780
Explanation:
First of all, we'll find the net area of the wall as follows:
The height of the wall is 17' - 4", so we need to convert it to ft, thus, h = 17' + (4/12)' = 17.33'
So, Gross wall area = 80′ × 17.333' = 1,387 ft²
From the image, there are 4 doors, thus, Area of doors = 4 × 10′ × 12′ = 480 ft²
Thus, Net area = Gross wall area - area of doors
Net area = 1,387 – 480 = 907 ft²
We are told that the block is 8 inches length by 16inches width. Thus, converting to ft; (8/12)ft by (16/12)ft.
So area of one block = (8/12) x (16/12) = 128/144 ft² = 0.88889 ft²
So, number of blocks per ft² = 1/0.88889 = 1.125 blocks
So, there are 1.125 blocks per ft²
Thus, Number of concrete blocks = 907 ft² × 1.125 blocks per ft² = 1,020 blocks
From the image attached, there are five rows of lintel blocks, located at 4′, 8′, 12′, 16′, and 17′-4″. The bottom three rows pass through the doors and are only a total of (80′ – (4 × 10′) ) = 40′ long .
The top two rows are 80′ long. Thus, there is 40 feet (4 × 10′) of lintel block above the doors.
Lintel blocks = (3 × 40′) + (2 × 80′) + 40′ = 320 ft
Number of Lintel blocks = 320′ × 12 in per ft / 16″ = 240 blocks
Number of Plain blocks = 1,020 blocks – 240 blocks = 780 blocks
Without specific wall dimensions provided, it is not possible to calculate the exact number of plain and lintel concrete blocks required for the construction of the wall. However, the general procedure involves calculating the wall's total area, subtracting the area of doors or windows, and then dividing by the size of one block, while also considering the requirements for lintel blocks above openings and where supports are needed.
Explanation:Determining the Quantity of Concrete Blocks for a Wall
To determine the number of 8-inch-high by 8-inch-wide by 16-inch-long concrete blocks required to complete a wall, one must calculate the total volume of the wall and then divide by the volume of a single block. Unfortunately, the dimensions of the wall are not provided in the question, so we cannot calculate the exact number of blocks needed. As for the lintel blocks, which are used above the doors and wherever the #4 horizontal bars are located, the number would depend on the linear footage that needs to be covered by the lintels and the length of each lintel block.
Without specific wall dimensions, we must refer to the typical process where one would calculate the wall's total surface area, subtract the area occupied by doors or windows, and divide by the area covered by one block. Then, identify the areas requiring lintel blocks and count the number of lintels needed based on their standard lengths.
The plain blocks and lintel blocks calculation requires detailed measurements of the wall and openings. Since the details provided are related to the overhead doors with known dimensions, one would need to consider the door's dimensions when calculating the plane wall areas and the lintel requirements above them.
One method of reducing an aircarft’s landing distance is through the use of thrust reversers. Consider the turbofan engine in Fig. P2.5 with thrust reverser of the bypass airstream. It is given that 1500 lbm / s of air at 60 8 F and 14.7 psia enters the engine at a velocity of 450 ft / s and that 1250 lbm / s of bypass air leaves the engine at 60 deg to the horizontal, velocity of 890ft / s, and pressure of 14.7 psia. The remaining 250 lbm / s leaves the engine core at a velocity of 1200 fps and a pressure of 14.7 psia.
Determine the force on the strut [Fx]. Assume the outside of the engine sees a pressure of 14.7 psia.
Answer:
The force on the strut is 909000 lbm ft/s²
Explanation:
please look at the solution in the attached Word file
For each of the three products that are observed in the NMR spectrum, there is a diagnostic proton that can be used to accurately calculate the product ratios. Calculate the percentage of the mixture that is o-bromoaniline. To answer this question correctly, you will need to use the BEST diagnostic signals for each isomer.
1. Proton assignments for each isomer: p-bromoaniline: Hd signal around 7.5 ppm (doublet, 2H), Hb signal around 6.7 ppm (doublet, 2H) o-bromoaniline: Ha and Hc signals overlapping around 7.6-7.65 ppm (multiplets, 4H total) o,p-dibromoaniline: Hb signal around 6.6 ppm (doublet, 2H), Hd signal around 7.55 ppm (doublet, 1H) o,o,p-tribromoaniline: No clear signals observed, likely not formed.
2. The isomer not observed to be formed is o,o,p-tribromoaniline. A plausible experimental explanation is that the highly brominated product is less favored due to steric hindrance and deactivation of the ring towards further electrophilic substitution.
3. Diagnostic protons for product ratio calculations: p-bromoaniline: Hb around 6.7 ppm o-bromoaniline: Ha/Hc around 7.6-7.65 ppm o,p-dibromoaniline: Hd around 7.55 ppm o,o,p-tribromoaniline: N/A (not formed)
4. Calculating product ratios using diagnostic proton integrations: p-bromoaniline: Integration of Hb signal ≈ 2.0, relative ratio = 2.0/2.0 = 100% o-bromoaniline: Integration of Ha/Hc signals ≈ 2.8, relative ratio = 2.8/4.8 ≈ 58% o,p-dibromoaniline: Integration of Hd signal ≈ 1.0, relative ratio = 1.0/3.0 ≈ 33% o,o,p-tribromoaniline: 0% (not formed)
5. Theoretical reasons for relative product abundance: i. p-bromoaniline: Dominant product due to the directing effect of the amino group favoring para substitution. ii. o-bromoaniline: Formed in moderate amounts due to some ortho substitution, but less favored than para. iii. o,p-dibromoaniline: Formed in smaller amounts due to the decreased reactivity of the ring after the first substitution. iv. o,o,p-tribromoaniline: Not formed, likely due to high steric hindrance and deactivation of the ring towards further substitution
The complete question:
The chemist’s crude product mixture was submitted for NMR analysis to give Spectrum C. Using the expected splitting patterns and chemical shift changes you predicted in part a, assign each proton for each of the isomers above (see structures on Spectrum C) by labeling each signal on the spectrum with the appropriate letter. Keep the integration ratios in mind when assigning each proton.
2. One of the four potential products above was NOT observed to be formed. Which one? Propose a very brief experimental explanation for the absence of this product.
3. For each of the three products that are observed in Spectrum C, there is a diagnostic proton that can be used to accurately calculate the product ratios. (Remember that a good diagnostic proton has a “clean” signal that has minimal overlap with other signals- see the “Using Integrations to Calculate Compound Ratios” page in the Assignment Information folder on Top Hat.) What diagnostic proton can be used to calculate product ratios for each isomer? If the isomer was not formed, write “N/A.”
4. Using the diagnostic protons you listed in part c, calculate the percent of each isomer. Show calculations for each isomer. If an isomer was not formed, write “0%.”
5. Provide a theoretical reason for the relative abundance of the following products
i. p-bromoaniline
ii. o-bromoaniline
iii. o,p-dibromoaniline
iv. o,o,p-tribromoaniline
Spectrum C bromination of aniline, crude solvent: CDCl Notes 1. Spectrum C is the zoomed-in aromatic region (6.3-7.9 ppm); amine protons are not shown 2. There are 2 overlapping signals from relevant protons at 6.60-6.65 ppm. You should assign both NH2 NH2 NH2 NH2 Br Br Chloroform H Sovent Но Hb He Br Hd ignore (possible oxidation product) 7.8 7.7 7.6 7.5 7.4 7.3 7.2 7.1 7.0 6.9 6.8 6.7 6.6 6.5 6.4 ppm p-bromoaniline o-bromoaniline o,p- dibromoaniline o,o,p- tribromoaniline %p-bromoaniline %o-bromoaniline % o,p- dibromoaniline % o,o,p- tribromoaniline. show all working. solve in hum, and form
The rotor is driven at ????m = 120π rad/sec. If the stator winding carries a current of 5 A (rms) at 60 Hz, determine the instantaneous voltage and rms voltage induced in the rotor coil. Determine the frequency of the rotor induced voltage. (b) Suppose the stator and rotor coils are connected in series and a current of 5
Answer:
(a) Frequency induced voltage of motor is 120 Hz
(b) The speed at which the machine will produce an average torque is 240π rad/s. Maximum Torque is 1.0 Nm
Explanation:
Determine the enthalpy of foA novel gaseous hydrocarbon fuel CxHy is proposed for use in spark-ignition engines. An analysis of a sample of this fuel revealed that its molecular weight is 140 and its molar H/C ratio is 2.0. In order to evaluate some of its properties, the fuel was burned with stoichiometric standard air in a constant pressure, steady-flow reactor. The fuel and air entered the reactor at 25°C, and the products of complete combustion were cooled to 25°C. At the exit condition, water in the products was a liquid. It was measuredrmation of the fuel at 25°C
Answer:
a. The fuel formula is C10H20 (dec-1-ene)
b. The heat given-off to the cooling system of the reactor represents the Higher Heating Value (HHV) of the gaseous hydrocarbon fuel
c. LHV = 44.36 KJ/kg
d. ΔH (formation) = -6669.6 KJ/mol
Explanation:
Molecular weight = 140, Molar H/C ratio = 2.0
a) To determine the fuel formula, we represent the information we have as equations
Molecular weight of C = 12, Molecular weight of H = 1
CxHy = 140g
12 · x + 1 · y = 140 ·········Eqn 1
y = 2x ·········Eqn 2
Substitute y = 2x into Eqn 1, we have:
12 (x) + 1 (2x) = 140
12x + 2x = 140
14x = 140
x = 10
Substitute x into Eqns 2, we have:
y = 2 x 10 = 20
∴ the fuel formula is C10H20 (dec-1-ene)
b) The heat given-off to the cooling system of the reactor represents the Higher Heating Value (HHV) of the gaseous hydrocarbon fuel . HHV refers to the amount of heat given-off in the combustion of a specified amount of fuel which was initially at 25°C and cooled back to a temperature of 25°C.
c) We will use LHV formula given as:
LHV = HHV – [2.441 (0.09H + M) ÷ 1000]
where:
LHV = lower heating value of fuel in MJ/kg
HHV = higher heating value of fuel in MJ/kg
M = Weight % of moisture in fuel
H = Weight % of hydrogen in fuel
HHV = 47.5 MJ/kg, M = 0, H = mass of hydrogen ÷ total mass of fuel
⇒ H = 20 ÷ 140 = 0.143 * 100 = 14.3 kg per 100 kg, Latent heat of steam (at 25°C) = 2441 KJ/kg
LHV = 47.5 - [2441 * (0.09 * 14.3 + 0) ÷ 1000]
LHV = 47.5 - 3.14 = 44.36
LHV = 44.36 KJ/kg
d) To get the enthalpy of the combustion of the fuel, we have:
fuel + air - - - - > carbon dioxide + water
C10H20 + 15 O2 - - - - > 10 CO2 + 10 H20
Enthalpy (formation) = Enthalpy (product) - Enthalpy (reactant)
ΔH (CO2) = -393.5 kJ/mol, ΔH (H2O) = -285.8 kJ/mol, ΔH (C10h20) = - 123.4KJ/mol
ΔH (formation) = 10 * (-393.5) + 10 * (-285.8) - (-123.4)
ΔH (formation) = -3935 - 2858 + 123.4
ΔH (formation) = -6669.6 KJ/mol
Write a program to accept a one-line string (maximum of 100 characters) from the keyboard. Edit the string entered in Part 1a (with code that you write) to remove all the white space,digits, punctuation, and other special characters, leaving only the letters. Print out the resulting compressed string to the screen.
Answer:
// This program is written in C++ programming language
// Comments are used for explanatory purpose
/* The aim of this program is to to remove all the white space,digits, punctuation, and other special characters, leaving only the letters. */
// Program starts here
#include <stdio.h>
#include<iostream>
using namespace std;
int main()
{
// Declare Variable of 100 characters
char word[100];
// Prompt user for input
cout<<"Your input goes here (max, 100 characters)";
cin>>word;
// Iterate through string to check for non alphabetic characters
for (int i = 0; word[i] != '\0'; ++i) {
// Check for uppercase and lowercase letters
while (!((word[i] >= 'a' && word[i] <= 'z') || (word[i] >= 'A' && word[i] <= 'Z') || word[i] == '\0')) {
for (int j = i; word[j] != '\0'; ++j) {
word[j] = word[j + 1];
}
word[j] = '\0';
}
}
cout<<"The resulting compressed string: "<<word;
return 0;
}
Answer:
w = str(input("input your values: "))
values = ' '.join(filter(str.isalpha, w))
while len(w) < 100:
print(values)
break
Explanation:
The code is written in python
w = str(input("input your values: "))
This code ask the user to input any string values with characters, numbers, line spaces , letters etc.
values = ' '.join(filter(str.isalpha, w))
This code filters the inputted value to bring only letters. All the letter are then joined together
while len(w) < 100:
The code check if the inputted value is less than 100 characters. While it is less than 100 characters. If it is less than 100 character the next code will function.
print(values)
This code prints the joined letters after checking with a while loop to confirm the length of character is less than 100
break
The break function breaks the code whether it print the values or not.
Generally, the letters will only be printed if the character inputted is less than 100 and later break the while loop or will not print any letter if the character is greater than 100 and later break.
Lorraine Jean Hopping's viewpoint in "Bone Detective" is that Diane France's work has helped many people. Hopping's purpose in writing this text is to
Answer:
inform the readers about Diane and her unusual job.
Explanation:
The purpose of Lorraine Jean Hopping's viewpoint in "Bone Detective" is just to inform the readers about Diane as a person and how what she does has really helped people even though her job was really unusual. It was mainly for the purpose of passing information across to the readers and nothing else.
Answer:
Explanation:she is trying to inform you hope it helps
Under certain conditions, wind blowing past a rectangular speed limit Sign can cause the sing to oscillate with a frequency omega. Assume that omega is a function of the Sign width, b, Sign height, h, wind velocity, V, air density p, and an elastic constant, K, for the supporting pole. Hint: The constant, k, has dimensions of [force x length]. (i) How many (non-dimensional) Pi-groups are there? (ii) Find these non-dimensional groups. (iii) Can one of the pi-groups be considered a Reynolds number?
Answer:
Explanation:
Given that,
Omega is a function of the following
ω = f(b, h, v, ρ, k)
Where, all unit have a dimension of
ω = T^-1
b = L
h = L
V = LT^-1
ρ = FL^-4T²
k = FL
Then,
From the pie theorem
The required pi term is 6—3 = 3 terms,
So, we use V, p and b as a repeating term.
For first pi
π1 = ω•b^a•v^b•ρ^c.
Since
ω= T^-1, b = L, v = LT^-1 and
ρ= FL^-4T²
Since π is dimensionless then,
π = F^0•L^0•T^0
(T^-1)•(L^a)•(LT^-1)^b•(FL^-4T²)^c = F^0•L^0•T^0
Rearranging
F^0•L^0•T^0 = F^c•T^(2c-b-1)• L^(a+b-4c)
Comparing coefficient
c = 0
2c - b - 1 = 0
b = 2c - 1 = 0 - 1 = -1
a + b - 4c = 0
a = 4c - b = 0 - -1 = 0+1
a = 1
Then, a = 1, b = -1 and c = 0
So, π1 = ω•b^a•v^b•ρ^c.
π1 = ω•b^1•v^-1•ρ^0
π1 = ω•b / v
Check dimensions
ωb/v = (T^-1)L / LT^-1 = L^0•T^0 = 1
Then, π1 is dimensionless
For second pi
π2 = h•b^a•v^b•ρ^c.
Since
h = L, b = L, v = LT^-1 and ρ= FL^-4T²
Since π is dimensionless then,
π = F^0•L^0•T^0
(L)•(L^a)•(LT^-1)^b•(FL^-4T²)^c = F^0•L^0•T^0
Rearranging
F^0•L^0•T^0 = F^c•T^(2c-b)• L^(1+a+b-4c)
Comparing coefficient
c = 0
2c - b = 0
b = 2c = 0
1 + a + b - 4c = 0
a = 4c - b - 1 = 0 -0 - 1 = -1
a = -1
Then, a = -1, b = 0 and c = 0
So, π2 = h•b^a•v^b•ρ^c.
π2 = h•b^-1•v^0•ρ^0
π2 = h / b
Check dimensions
h / b = L / L = 1
Then, π2 is dimensionless
For third pi
π3 = k•b^a•v^b•ρ^c.
Since
k= FL, b = L, v = LT^-1 and ρ=FL^-4T²
Since π is dimensionless then,
π = F^0•L^0•T^0
(FL)•(L^a)•(LT^-1)^b•(FL^-4T²)^c = F^0•L^0•T^0
Rearranging
F^0•L^0•T^0 = F^(c+1)•T^(2c-b)• L^(1+a+b-4c)
Comparing coefficient
c + 1= 0
Then, c = -1
2c - b = 0
b = 2c = -2
1 + a + b - 4c = 0
a = 4c - b - 1 = -4 +2 - 1 = -3
a = -3
Then, a = -3, b = -2 and c = -1
So, π3 = k•b^a•v^b•ρ^c.
π3 = k•b^-3•v^-2•ρ^-1
Therefore,
π3 = k / b³•v²•ρ
Let check for dimension
π3 = FL / (L³• L²T^-2 • FL^-4T²)
π3 = FL / (L^(3+2-4) • T^(-2+2) •F)
π3 = FL / (L• T^(0) •F)
π3 = FL / LF = 1
π3 is also dimensions less
So.
I. There are three none dimensional pi
II. The none dimensional group are
π1 = ω•b / v
π2 = h / b
π3 = k / b³•v²•ρ
III. Reynolds Number. The Reynolds number is the ratio of inertial forces to viscous forces and it is dimensionless
So, the π3 can be considered as a Reynolds number
The power P generated by a certain windmill design depends upon its diameter D, the air density ????, the air viscosity µ, the wind velocity V, the rotation rate Ωand the number of blades n. (a) Write this relationship in dimensionless form. A model windmill, of diameter 50 cm, develops 3.8 kW at sea level when V 40 m/s and when rotating at 4200 rpm. (b) What power will be developed by a geometrically and dynamically similar prototype, of diameter 15 m, in winds of 35 m/s at 500 m standard altitude from sea level? (c) What is the appropriate rotation rate
Answer:
a) check the attached files below
b) What power will be developed by a geometrically and dynamically similar prototype, of diameter 15 m, in winds of 35 m/s at 500 m standard altitude from sea level = 2184.56KW
c) What is the appropriate rotation rate = 122.5 rpm
Explanation:
please kindly check the attachment below.
A and B connect the gear box to the wheel assemblies of a tractor, and shaft C connects it to the engine. Shafts A and Blie in the vertical yz plane, while shaft C is directed along the x axis. Replace the couples applied to the shafts with a single equivalent couple, specifying its magnitude and the direction of its axis.
The couplings are not all on the same axis or plane, but if the A and B connector were to be described, it would travel with a magnitude of roughly 15 by the yz axis diagonal to the x axis. The axis would be oriented so that it was pointing upward into the second quadrant.
What is magnitude?In terms of physics, magnitude is just "distance or quantity." It displays an object's size, direction, or motion in absolute or relative terms. It is employed to indicate the size or scope of something.
In physics, the term "magnitude" often refers to a size or amount. Magnitude is defined as "how much of a quantity." The magnitude can be used, for instance, to compare the speeds of a car and a bicycle.
It can also be used to indicate how far something has come or how much something weighs relative to its size.
Thus, The couplings are not all on the same axis or plane.
For more details about magnitude, click here:
https://brainly.com/question/14452091
#SPJ2
Refrigerant 134a is the working fluid in a vaporcompression heat pump that provides 35 kW to heat a dwelling on a day when the outside temperature is below freezing. Saturated vapor enters the compressor at 1.6 bar, and saturated liquid exits the condenser, which operates at 8 bar. Determine for isentropic compression a. the refrigerant mass flow rate, in kg/s. b. the compressor power, in kW. c. the coefficient of performance. Recalculate the quantities in parts (b) and (c) for an isentropic compressor efficiency of 75%.
Thermodynamic calculations for mass flow rate, compressor power, and coefficient of performance of a heat pump require refrigerant property data, which is not provided in the question. An energy balance and compressor efficiency adjustment are also essential for accurate calculations.
Explanation:To calculate the properties of the heat pump using refrigerant 134a and determine the refrigerant mass flow rate, compressor power, and coefficient of performance, we need to make use of thermodynamic equations and refrigerant property data under the given conditions (such as pressures and phase states).Normally, these calculations would involve using a thermodynamic properties table or software to find enthalpies for the refrigerant at the specified conditions. An energy balance would be applied around the compressor and condenser. The work done by the compressor can be calculated using the enthalpy values before and after the compressor and assuming an isentropic process initially, and then adjusting for the non-ideal compressor efficiency of 75% in the revised calculation.For the coefficient of performance (COP) of the heat pump, it is calculated as the ratio of heat output to work input for the heating mode. The provided heat output is given here as 35 kW. For ideal conditions, the COP can be calculated without the compressor efficiency factor; for the revised COP with a 75% efficient compressor, this factor is taken into account.Unfortunately, without the enthalpy values or additional data, it is not possible to provide the numerical answers to the student's question.
A three-phase source delivers 4.8 kVA to a wye-connected load with a phase voltage of 214 V and a power factor of 0.9 lagging. Calculate the source line current and the source line voltage.
Answer:
a) 7.5A
b) 370V
Explanation:
Note that the Line current is the current through any one line between a three-phase source and load.
From the given question, The source line current =7.5A(approximately)
While the Source line voltage = 370V.
Kindly go through the attached file for a step by step detail of how I arrived at these answers.
The source line current in the three-phase system is 2.08 A and the source line voltage is approximately 371 V.
Explanation:To calculate the source line current and the source line voltage in a three-phase system, we can use the formula:
I = P / (sqrt(3) * V * PF)
Where I is the source line current, P is the power delivered to the load, V is the phase voltage, and PF is the power factor.
Given that the power delivered is 4.8 kVA, the phase voltage is 214 V, and the power factor is 0.9 lagging, we can substitute these values into the formula and calculate:
I = 4.8 kVA / (sqrt(3) * 214 V * 0.9)
Calculating this gives us a source line current of 2.08 A.
To find the source line voltage, we can use the formula:
Vline = sqrt(3) * Vphase
Substituting the phase voltage of 214 V into the formula gives us a source line voltage of approximately 371 V.
Question 1: Consider a vapour compression refrigeration cycle that uses R-134a as refrigerant. The R-134a enters the compressor as a saturated vapour at 200 kPa, and exits the condenser as a saturated liquid at 900 kPa. The rate of refrigeration of the cycle is to be 6.0 tons of refrigeration (1 ton of refrigeration = 3.517 kW). The compressor isentropic efficiency is 80%. Determine: a) The temperature of evaporation and condensation of the refrigerant; b) Mass flow of the refrigerant R-134a, in kg/min; c) Coefficient of Performance (COP) of the refrigeration cycle. Typical vapour compression refrigeration systems use a throttling valve (or an orifice tube, or a capillary tube) to reduce the liquid refrigerant’s pressure. This is a highly irreversible process. If the cycle efficiency is to be improved by replacing the throttle with a rotary expander with 70% isentropic efficiency, d) What is the new COP? The expander delivers its work developed to the compressor.
Answer:
a) The temperature of evaporation is -10.09°C
The temperature of condensation is 35.53°C
b) The mass flow of the refrigerant is 0.149 kg/s
c) The COP is 3.65
d) The new COP is 4.05
Explanation:
please look at the solution in the attached Word file
Consider a very long, slender rod. One end of the rod is attached to a base surface maintained at Tb, while the surface of the rod is exposed to an air temperature of 400°C. Thermocouples imbedded in the rod at locations 25 mm and 120 mm from the base surface register temperatures of 335°C and 375°C, respectively.
(a) Calculate the rod base temperature (°C).
(b) Determine the rod length (mm) for the case where the ratio of the heat transfer from a finite length fin to the heat transfer from a very long fin under the same conditions is 99 percent.
Answer:
(a) Calculate the rod base temperature (°C). = 299.86°C
(b) Determine the rod length (mm) for the case where the ratio of the heat transfer from a finite length fin to the heat transfer from a very long fin under the same conditions is 99 percent. = 0.4325m
Explanation:
see attached file below