Design a Mealy machine for a 20 cent candy dispensing machine which accepts nickel (n) , dime (d) and quarter(q) . It gives candy as well change in the form of nickels only. To help you, here are the some of the elements of the machine: States: 0, 5, 10 and 15 represent the amount of money already inserted in the machine. n, d and q representing coins inserted in the machine Output: c0 (candy and no change), c1 (one nickel as change), c2(two nickels as change ), c3 (three nickels as change), and c4 (four nickels as change)
Answers
Find the answer in the attachment
Related Questions
EP Electric has identified two new methods to treat its cooling water. Alternative D(for inflow) would treat the raw water with a conventional reverse osmosis system so that the cycles of concentration could be increased from 5 to 20, This will result in water cost savings of $360,000 per year and chemical cost savings of $56,000 per year. The initial cost of the equipment will be $2.3 million with an operating cost of $125,000 per year Alternative B (for blowdown) will treat the cooling tower blowdown water using a highpressure seawater reverse osmosis system to recover most of the water that is sent to an evaporation pond. This option will result in water sav- ings of $270,000 per year. The cost of the system will be $1.2 million with an operating cost of $105,000 per year. Assuming one of the two methods must be installed, determine which is preferred on the basis of the incremental ROR value using MARRof 5%per year, which is a typically low return expected of government projects. Use a 10-year study period with no salvage value for either system. (Note: See Problem 8.41 for more on this situation.)
Answers
Consider a torsionally elastic (GJ = 8000 lb-in2) wind tunnel model of a uniform wing, the ends of which are rigidly fastened (fixed) to the wind tunnel walls. The model is a symmetric airfoil with a span of 3 ft and a chord length of 6 in. The total lift-curve slope is 6 per rad. The aerodynamic center is located at the quarter-chord, and both the mass centroid and the elastic axis are at the mid-chord. (a) Calculate the divergence dynamic pressure at sea level. (b) Calculate the divergence airspeed at sea level.
Answers
Answer:
(a) 148.148 lb/ft^2
(b) 62.245 ft/s
Explanation:
In this question, we are asked to calculate the divergence dynamic pressure at sea level and the divergence airspeed at sea level of a torsionally elastic wind tunnel of model of uniform wing.
Please check attachment for complete step by step solution
Answer:
Explanation:
Find attached the solution
The y and z keys swapping position is messing with your touch typing. You decide to write out your email as if the keys were in the correct position and then use Python to swap all ys and zs. Your task here is to write a function called fix_yz. This function takes a single argument which will be a string. Your function needs to return this string with all of the ys and zs swapped, and all of the Ys and Zs swapped. Here are some example calls to your function:
s = fix_yz('What did zou saz?')print(s)What did you say?s = fix_yz('Zour tip about the yoo was a great one!')print(s)Your tip about the zoo was a great one!s = fix_yz('We onlz have one week left')print(s)We only have one week left :(HintThe auto-marker is expecting you to submit only your fix_yz function definition. You should not include any calls to your function.
Answers
Answer:
# the function fix_yz is defined
# it takes a string as parameter
def fix_yz(word):
# new_word is to hold the new corrected string
new_word = ""
# loop through the string
# and check for any instance of y or z.
# if any instance is found, it is replaced accordingly
for each_letter in word:
if each_letter == 'z':
new_word += 'y'
elif each_letter == 'Z':
new_word += 'Y'
elif each_letter == 'y':
new_word += 'z'
elif each_letter == 'Y':
new_word += 'Z'
else:
new_word += each_letter
# the value of new string is returned
return new_word
Explanation:
The function is written in Python 3 and it is well commented. An image is attached showing the output of the given example.
The function take a string as input. It then loop through the string and check for any instance of 'y' or 'z'; if any instance is found it is swapped accordingly and then append to the new_word.
The value of bew_word is returned after the loop.
Compute L, T, M, LC, and R and stations of the BC and EC for the circular curve with the given data of: I (delta) = 22°15′00" and E = 23.4604 and the P.I. station = 16+61.27. Express answers to .01′ (ft.).
Answers
Answer:
L = 475.718
T = 240.89 ft
M = 23.0195
LC = 472.728
R = 1225 ft
Explanation:
See the attached file for the calculation.
Write down one metal or alloy that is best suited for each of the following applications:
a. The block of an internal combustion engine
b. Condensing heat exchanger for hot steam
c. Drill bit
d. Container for strong acid
e. As a pyrotechnic (i.e., in flares and fireworks)
f. High-temperature furnace elements to be used in oxidizing atmospheres
Answers
Answer:
Explanation:
a. Cast iron or Aluminium alloy are typically used. Aluminium is much lighter in weight and it can transfer heat better to the coolant. While Cast Iron is typically stronger and is thus still used by the manufacturers.
b. Copper can be used as a condensing heat exchanger for hot steam due to its optimal thermal properties and its ability to resist corrosion.
c. high-speed steel are perfect for producing drill bits because of its hardness and resistance to heat to an extent. Drill bits tend to produce heat as a result of the friction between it and the material to be drilled.
d. lead can be used as a container for strong acids because of its anti-corrosive properties
e.zinc and copper can be used as fuel in pyrotechnics mainly due to the fact that burn with refreshing colours. Aluminium can also be used.
f. Platinum is the metal that best suits this purpose because of its high melting point and resistivity to oxidation.
Two forces P can be applied separately or at the same time to a plate that is welded to a solid circular bar of radius r. Determine the largest compressive stress in the circular bar, (a) when both forces are applied, (b) when only one of the forces is applied.
Answers
This answer discusses the calculation of the largest compressive stress in a circular bar when two forces are applied either together or individually, highlighting the difference in stress levels due to the application of forces.
Explanation:The question involves determining the largest compressive stress in a circular bar when two forces, P, are applied either simultaneously or separately. This problem is grounded in the principles of mechanics of materials, particularly in analyzing stress and strain in materials under tension or compression.
For part (a), when both forces are applied, the compressive stress would likely be higher, as the forces cumulatively contribute to the stress. The compressive stress (σ) in a circular bar can be calculated using the formula σ = P/A, where P is the total force applied, and A is the cross-sectional area of the bar. In the case of two forces applied simultaneously, P would be the summation of both forces.
For part (b), when only one of the forces is applied, the compressive stress in the bar would be lower, as it comes from a single source. The calculation still follows σ = P/A, but with P representing only one of the forces.
Compute the electrical resistivity of a cylindrical silicon specimen 7.0 mm (0.28 in.) diameter and 57 mm (2.25 in.) in length in which a current of 0.25 A passes in an axial direction. A voltage of 24 V is measured across two probes that are separated by 45 mm (1.75 in.).
Answers
The Atbash Cipher encrypts messages by reversing lowercase letters, so ‘a’ becomes ‘z’, ‘b’ becomes ‘y’, ‘c’ becomes ‘x’, etc... Also, any space or punctuation mark gets repeated. For example, hello human! encrypts to svool sfnzm!! Encrypt msg and save the answer to a variable called encrypted (you don’t have to display anything). Note: msg will only have lowercase letters, punctuation and spaces. msg = input('Enter secret message: ', 's');
Answers
Answer:
See Explanation Below
Explanation:
// Program is written in C++ Programming Language
// Comments are used for explanatory purpose
// Program starts here
#include<iostream>
#include <bits/stdc++.h>
using namespace std;
int main()
{
// Declare 2 string variables to store the secret message and to store the encrypted text
string message, result;
// Prompt user to enter a secret message
cout<<"Enter a secret message: ";
cin>message;
// Convert the input string to char array
int n = message.length();
char char_array[n + 1];
strcpy(char_array, message.c_str());
// Initialise result
result = "";
// Declare an array of all possible alphabets a-z
char possible[26] = { 'a','b','c','d','e','f','g,','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v',w','x','y','z'};
// Generate output string
// Start by getting string position
int count = 0;
while(count<n)
{
// If current character is blank or !
if(char_array[count] = '!' || char_array[count] = ' ')
{
result+=char_array[count];
}
else
{
for(int I = 0; I<26; I++)
{
if(char_array[count] = possible[I])
{
result+=possible[25-I];
}
}
}
count++;
}
// No output required; the program stops here
return 0;
}
// End of program
Open the"stateData3.c" program and try to understand how the tokenization works. If you open the input file "stateData.txt", you can clearly seethat a comma separates the state name and its population. The tokenizer portion of the programs separates each line into two tokens and stores them into two different arrays. You need to display the arrayelements such that each line has a state and itspopulation. Also calculate the total population of USA.Then,examine the portion of the code how it writes the array elements into a binary datafile.You need to write similar logic where it writes the array elements into a text file "stateDataOutput2.txt". Please check the syntax and usage of fprintf(); and use that here.
Answers
Answer:
Kindly see explaination
Explanation:
Code
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#define size 200
int main(void)
{
int const numStates = 50;
char tempBuffer[size];
char tmp[size];
char fileName[] = "stateData.txt"; // Name of the text file (input file) which contains states and its populations
char outFile[] = "stateDataOutput1.txt"; // Output file name
// Open the input file, quit if it fails...
FILE *instream = fopen(fileName, "r");
/* Output File variable */
FILE *opstream;
if(instream == NULL) {
fprintf(stderr, "Unable to open file: %s\n", fileName);
exit(1);
}
//TODO: Open the output file in write ("w") mode
/* Opening output file in write mode */
opstream = fopen(outFile, "w");
//TODO: Read the file, line by line and write each line into the output file
//Reading data from file
while(fgets(tmp, size, instream) != NULL)
{
//Writing data to file
fputs(tmp, opstream);
}
// Close the input file
fclose(instream);
//TODO: Close the output file
/* Closing output file */
fclose(opstream);
return 0;
}
You are an engineer at company XYZ, and you are dealing with the need to determine the maximum load you can apply to a set of bolted/clamped plates, such that you get infinite life. a) (1.5 pt) If you have an coarse thread M12 x 1.75 bolt of SAE Class 5.8, what would you recommend as the initial tightening force in the bolt? Is this initial force in the bolt the same as the clamped members?Hint: Use Fi = 0.9AtSp. b) (2 pt) If the clamped members have an effective stiffness twice that of the bolt and an external separating load varies between 0 and 5 kN, what are the alternating and mean forces on the bolt? What are the mean and alternating stresses in the bulk of the bolt? c) (1.5 pt) If the fatigue stress concentration factor is Kf=2.2 for the threads and we account for a yield strength of 400 MPa, what are the ef
Answers
Answer:
a)
[tex]F_i = 28.8 kN[/tex]
b)
Alternating and mean forces on the bolt are 0.85 kN and 29.65 kN respectively.
Mean and alternating stresses in the bulk of the bolt are 378 MPa and 10 MPa.
c)
Safety factor = 5.5
Explanation:
Basic Dimension of Isometric Screw thread" is considered for analysis.
At Nominal diameter, d = 12 mm , [tex]A_t = 84.3mm^{2}[/tex] , [tex]F_i = 0.9A_tS_p[/tex] , [tex]S_p = 380 MPa[/tex]
Rolled threads; [tex]K_f = 2.2[/tex]
Air at 27°C and a velocity of 5 m/s passes over the small region As (20 mm × 20 mm) on a large surface, which is maintained at Ts = 127°C. For these conditions, 0.5 W is removed from the surface As. To increase the heat removal rate, a stainless steel (AISI 304) pin fin of diameter 5 mm is affixed to As, which is assumed to remain at Ts = 127°C. (a) Determine the maximum possible heat removal rate through the fin. (Hint: look back to Chapter 3 to see which fin case produces the maximum heat transfer.) (b) What fin length would provide a close approximation to the heat rate found in part (a)? (Hint: refer to Example 3.9.) (c) Determine the fin effectiveness, εf. (d) What is the percentage increase in the heat rate from As due to installation of the fin? In other words, what is the percentage increase in q obtained by adding the fin as compared to not having the fin?
Answers
Answer:
a) The maximum possible heat removal rate = 2.20w
b) Fin length = 37.4 mm
c) Fin effectiveness = 89.6
d) Percentage increase = 435%
Explanation:
See the attached file for the explanation.
In this module you learned about searching, sorting and algorithms in C++ and how to implement these elements in your C++ programs. In this assignment you will implement your knowledge of searching, sorting and algorithms for charge account validation system. Write a program that lets the user enter a charge account number. Be sure to include comments throughout your code where appropriate. The program should determine if the number is valid by checking for it in the following list of valid account numbers:
814942053331743080098675596395269814449539938719751047262931356428263717502196086650316483824195904578589971890467499412545408Initialize a one-dimensional array with these values. Then use a simple linear search to locate the number entered by the user. If the user enters a number that is in the array, the program should display a message indicating that the account number is valid. If the user enters a number not in the array, the program should display a message indicating it is invalid.Next, using the same one-dimensional array, use a bubble-sort to put the array in order and perform a binary search to locate the number entered by the user. If the user enters a number that is in the array, the program should display a message indicating that the account number is valid. If the user enters a number not in the array, the program should display a message indicating it is invalid.
Answers
Question is not well presented:
The search strings in the question are not well formatted.
First thing, I'll arrange the account number in order of 7;
Meaning that the length of each account number is 7
8149420 5333174 3080098 6755963 9526981 4449539 9387197 5104726 2931356 4282637 1750219 6086650 3164838 2419590 4578589 9718904 674994 12545408
Answer:
// This program is written in C++ Programming Language
// Comments are used for explanatory purpose
// Program starts here
#include <iostream>
using namespace std;
// Initialise Function
int ArrayList(int[], int, int);
// Main Method starts at
int main()
{
// The list of account is 18; so, we Initialise an integer Variable to hold this value
const int acctlist = 18;
// Initialise an array to hold the account numbers
int List[acctlist] = { 8149420 5333174 3080098 6755963 9526981 4449539 9387197 5104726 2931356 4282637 1750219 6086650 3164838 2419590 4578589 9718904 674994 12545408
};
// Declare account number and search result Variables of type integer
int accountnumber, Result;
// Prompt user for account number
cout << "Enter a valid account number: ";
cin >> accountnumber;
// Determine if valid or not
Result = ArrayList(List, acctlist, accountnumber);
if(Result == -1){
cout << "Account number " << AcctNum << " is invalid.\n";}
else{
cout << "Account number " << List[Result] << " is valid\n";
}
return 0;
}
// Function starts here
int ArrayList(int list[], int size, int value)
{
int index = 0,
pos = -1;
bool seen = false;
// Check if Account number is seen in list
while (!seen && index < size)
{
if (list[index] == value)
{
seen = true;
}
pos = index;
}
index++;
}
return pos;
}
A steam power plant receives heat from a furnace at a rate of 280 GJ/h. (that is
‘giga’ joules.) Heat losses to the surrounding air from the steam as it passes through the pipes and other components are estimated to be about 8 GJ/h. If the waste heat is transferred to the cooling water at a rate of 145 GJ/h, determine (a) the net power output and (b) the thermal efficiency of this power plant.
Answers
Answer:
a) [tex]\dot W_{out} = 35.278\,MW[/tex], b) [tex]\eta_{th} = 45.357\,\%[/tex]
Explanation:
a) The net power output is:
[tex]\dot W_{out} = \dot Q_{in} - \dot Q_{out} - \dot Q_{loss}[/tex]
[tex]\dot W_{out} = \left(280\,\frac{GJ}{h} - 145\,\frac{GJ}{h} - 8\,\frac{GJ}{h}\right)\cdot \left(\frac{1000\,MJ}{1\,GJ}\right)\cdot \left(\frac{1\,h}{3600\,s} \right)[/tex]
[tex]\dot W_{out} = 35.278\,MW[/tex]
b) The thermal efficiency of the power plant is:
[tex]\eta_{th} = \frac{\dot Q_{in}-\dot Q_{out} - \dot Q_{loss}}{\dot Q_{in}}\times 100\,\%[/tex]
[tex]\eta_{th} = \frac{280\,\frac{GJ}{h}-145\,\frac{GJ}{h}-8\,\frac{GJ}{h} }{280\,\frac{GJ}{h} } \times 100\,\%[/tex]
[tex]\eta_{th} = 45.357\,\%[/tex]
Energy is released when hydrogen and oxygen react to produce water. This energy comes from the fact that the final hydrogen-oxygen bonds represent a lower total energy states compared to the original hydrogen-hydrogen and oxygen-oxygen bonds. Calculate how much energy (in kilojoules per mole of product) is released by the reaction:
1/2 O2 + H2 → H2O
at constant pressure and given the following standard bond enthalp enthalpies denote the enthalpy absorbed when bonds are broken at ture and pressure (298 K and 1 atm).
H-H : 432 kJ/mol
O=O : 494 kJ/mol
H-O : 460 kJ/mol
Answers
Answer:
The energy released is -241 kJ/mol
Explanation:
Please look at the solution in the attached Word file
- if `check_1` and `check_2` variables are both True, it should set the value of a variable `outcome` to the string 'BOTH' - elif `check_1` is True and `check_2` is False, it should set the value of a variable `outcome` to the string 'ONE' - elif `check_1` is False and `check_2` is True, it should set the value of a variable `outcome` to the string 'TWO' - else (meaning both must be False), it should set the value of a variable `outcome` to the string 'NEITHER'
Answers
Answer:
See Explaination
Explanation:
if(check1 and check2):
outcome = "BOTH"
elif(check1):
outcome = "ONE"
elif(check2):
outcome = "TWO"
else:
outcome = "NEITHER"
A stainless steel recycling facility wants to be completely off the grid and supply their energy entirely via renewables. Their choice of renewable energy was wind coupled with energy storage. Since you are the expert engineer hired to help them, they ask you if they will need more than one wind turbine to supply the power they need. What is your answer to them (provide all calculations to prove them)?
You gather some information to perform your calculations. The facility is planning to heat up the stainless steel from an initial temperature of 25 °C and melt 1000 metric tons of material per day 5 days a week. The model of wind turbine to be used has blades that are 60 m long, the facility is at sea level and at a prime location where the wind blows 16 hours per day at 10 m/s and is run for 7 days per week. Also, the wind turbine has an overall efficiency of 40%.
Answers
Answer:
3.25 turbines
Explanation:
Please kindly check attachment for the detailed answer.
y ????, 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 of the prototype?
Answers
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]
6.3 In a particular technology, a small BJT operating at conducts a collector current of 100 μA. What vBE = 28VT is the corresponding saturation current? For a transistor in the same technology but with an emitter junction that is 32 times larger, what is the saturation current? What current will this transistor conduct at base–emitter voltage of the latter transistor at iC AE1 =400 μm × 400 μm and AE2 vBE vBE = 28VT ? What is the =1mA? Assume active-mode operation in all cases.
Answers
Answer:
Saturation current
Isi = 6.91 * 10^-17 A
The saturation current Is2 of transistor Q2 is 2.2 * 10-15 A
Collector current Ic2 at Vbe2 is 3.2mA
Base emitter voltage Vbe2 is 26.84 VT
Explanation:
Complete question is as follows;
Problem 6.3: In a particular technology, a small BJT operating at vBE = 28VT conducts a collector current of 100
µA. What is the corresponding saturation current?
For a transistor in the same technology but with an emitter junction that is 32 times larger, what is the saturation
current?
What current will this (second) transistor conduct at vBE = 28VT ?
What is the base-emitter voltage of the latter transistor at iC = 1 mA? Assume active-mode operation in all cases.
solution;
Please check attachment for complete solution and step by step explanation
As the temperature of a thermal radiator is increased Group of answer choices the object appears redder. the object appears bluer. the object emits more power for the same area. the object emits less power for the same area. the obhect expands to keep the same power per units area.
Answers
Answer:
As the temperature of a thermal radiator is increased
Group of answer choices
the object appears redder. the object appears bluer. the object emits more power for the same area. the object emits less power for the same area. the object expands to keep the same power per units area.When the temperature of a thermal radiator increases ;
the object emits more power for the same arethe object becomes bluerExplanation:
Thermal radiation involves the transfer of heat between molecules of two substances without direct contact with each other. When a body is heated to a given temperature it begins to emit light which is transferred to nearby objects as thermal radiation. The medium through which the heat is transferred could be liquid, solid, or in a vacuum.
How temperature affects thermal radiation.Temperature determines the amount of heat that is been radiated from a body. An increase in temperature would increase the thermal radiation of the body. The increase in the heat radiation results to increase in the thermal energy of the body. Also when a body is heated it tends to be bluer than a cool object, this is caused by the rapid movement of the molecules.
Therefore When the temperature of a thermal radiator increases ;
the object emits more power for the same arethe object becomes bluerConsider a 25-mm-diameter and 15-m-long smooth tube that is used for heating fluids. The wall is heated electrically to provide a constant surface heat flux along the entire tube. Fluids enter the tube at 508C and exit at 1508C. If the mass flow rate is maintained at 0.01 kg/s, determine the convection heat transfer coefficients at the tube outlet for water, engine oil, and liquid mercury.
Answers
Answer:
Water: h = 35.53 W/m².k
Engine oil: h = 18.84 W/m².k
Mercury: h = 1.19 W/m².k
Explanation:
Assuming the steady state, one-dimensional heat flow, it is clear that the added to the fluid by tube heat will be equal to the heat transfer through convection outside the tube.
Therefore,
mCΔT = hAΔT
mC = hA
h = mC/A
where,
h = convection coefficient
m = mass flow rate = 0.01 kg/s
C = specific heat capacity of fluid
A = surface area of tube = 2πrL = 2π(0.0125 m)(15 m) = 1.178 m²
FOR WATER:
C = 4186 J/Kg.k
Therefore,
h = (0.01 kg/s)(4186 J/Kg.k)/(1.178 m²)
h = 35.53 W/m².k
FOR ENGINE OIL:
C = 2220 J/Kg.k
Therefore,
h = (0.01 kg/s)(2220 J/Kg.k)/(1.178 m²)
h = 18.84 W/m².k
FOR LIQUID MERCURY:
C = 140 J/Kg.k
Therefore,
h = (0.01 kg/s)(140 J/Kg.k)/(1.178 m²)
h = 1.19 W/m².k
Two concentric helical compression springs made of steel and having the same length when loaded and when unloaded are used to support a static load of 3 kN. The outer spring has D = 50 mm, d = 9 mm, and N = 5; the inner spring D = 30 mm, d = 5 mm, and N = 10. Determine the deflection and the maximum stress in each spring.
Answers
Answer:
see explaination for all the answers and full working.
Explanation:
deflection=8P*DN/Gd^4
G(for steel)=70Gpa=70*10^9N/m^2=70KN/mm^2
for outer spring,
deflection=8*3*50^3*5/(70*9^4)=32.66mm
for inner spring
deflection=8*3*30^3*10/(70*5^4)=148.11mm
max stress=k*8*P*C/(3.14*d^2)
for outer spring
c=50/9=5.55
k=(4c-1/4c-4)+.615/c=1.2768
max stress=1.2768*8*3*5.55/(3.14*9^2=.66KN.mm^2
for inner spring
c=6
k=1.2525
max stress=2.29KN/mm^2
Explanation:
Data
Load = 3kn = 3000N
Modulus of rigidity = 80Gpa= 80000mpa
Outer spring diameter = 50mm
d. = 9mm
N = 5
Inner spring diameter = 30mm
d = 5mm
N = 10
Fo = outer force
Fi = inner force
Ki = stiffness of inner spring
Ko = stiffness of outer spring
Ks = stress factor
A powerplant is emitting 80 g/s NO and has an effective stack height of 100 m. The windspeed is measured to be 4 m/s at a height of 10 m. It is a clear summer day and the sun is located directly overhead. Find: a. Ground-level NO concentration 2 km directly downwind of the powerplant. b. Maximum NO concentration at the ground-level. c. Ground-level NO concentration 2 km downwind and 0.1 km off the downwind axis.
Answers
Answer:
Explanation:
The step by step solution is in the attached file.
A horizontal 100-mm-diameter pipe passing hot oil is to be used in the design of an industrial water heater. Based on a typical water draw rate, the velocity of the water over the pipe is 0.5 m/s. the hot oil maintains the outer surface of the pipe at 85 °C and the water temperature is 37 °C. Investigate the effect of water flow direction on the heat rate (W/m) for (a) horizontal, (b) downward, and (c) upward flow.
Answers
Answer:
The answer and explanation is in the attached file
Explanation:
To a certain extent, there are few factors that affect the path a river is expected take. First of all, water runs downhill due to law of gravity. a flow is expected to go southward or northward, to the west or to the east, but always downhill.
Flow direction influences the particular direction water will flow in a given cell. Based on the steepest descent direction of the in each cell, we measure flow direction.
Kindly check the attached images below to get the step by step explanation to the question above.
A 10-m-long countercurrent-flow heat exchanger is being used to heat a liquid food from 20 to 808C. The heating medium is oil, which enters the heat exchanger at 1508C and exits at 608C. The specific heat of the liquid food is 3.9 kJ/(kg K). The overall heat-transfer coefficient based on the inside area is 1000 W/ (m2 K). The inner diameter of the inside pipe is 7 cm. a. Estimate the flow rate of the liquid food. b. Determine the flow rate of the liquid food if the heat exchanger is operated in a concurrent-flow mode for the same conditions of temperatures at the inlet and exit from the heat exchanger.
Answers
Answer:
mlf=0.5038kg/s
Explanation:
a. Please kindly check attachment for the step by step solution
b. B) in concurrent flow heat exchanger same exit temperatures for both fluids cannot be obtained. Since tho<tco
Case ii) only liquid food is considered tci=20 &tco=80
Cp=3.9kj/kgk
&If Q is same then mlf=0.5038kg/s this is same as case a
A vapor-compression refrigeration cycle operates at steady state with Refrigerant 134a as the working fluid. Saturated vapor enters the compressor at 2 bar, and saturated liquid exits the condenser at 8 bar. The isentropic compressor efficiency is 80%. The mass flow rate of refrigerant is 7 kg/min.
1-Write the assumptions no less than three
2-Draw clear Schematic
(3) the compressor power, in kW.
(4) the refrigeration capacity, in tons.
(5) the coefficient of performance.
Answers
Answer:
(3) the compressor power, in kW. = 4.17KW
(4) the refrigeration capacity, in tons.= 4.905tons
(5) the coefficient of performance. = 4.1336
check attached files for other answers.
Explanation:
To save steel-handling costs, an alternative design is proposed for the beam in Problem 1 using two No. 9 Grade 75 bars to provide approximately the same steel strength as the originally proposed four No. 7
Grade 60 bars. Check to determine if the redesigned beam is satisfactory with respect to cracking according to the ACI Code. What modification could you suggest that would minimize the number of bars to reduce cost, yet satisfy requirements of crack control?
Answers
Answer:
See the explanation for the answer
Explanation:
Check to determine the redesigned beam is satisfactory for cracking:
crack width is controlled by establishing a minimum spacing.
Steps followed to check for cracking in the beams
i) According to aci code 10.6.7 if depth of beam isgreater than 36 in then skin reinforcement has to provide.the skin reinforcement to be provided should be such that it should not be greater than the actual main tension reinforcement.
ii) In second step steel stress is determined:
steel stressfs=Ms/(As*(d-hf/2)) or fs= 0.60fy
where Ms=service load moment
As*(d-hf/2)=area of reinforcement *moment area
iii) S=540/fs-2.5Cc is less than or equal to 12*(36/fs)
here Cc = clear spacing
if S=center to center spacing is with in the limit as specified above then the cracking is with in the control if not then redesign has to done.
In the given problem the data given is :
grade of steel in desigened beam is 75 and in redesigned beam is 60 so the stress in steel is 75*0.6=45ksi and 0.6*60=36ksi respectively
now the spacing is calculated for the two design and redesigned beams
the center to center spacing is given by S=540/fs-2.5Cc
For designed beam
S=540/45-(2.5*2.25)=6.375in which is less than 12*36/fs=12*36/45=9.6in hence it is safe
For redesigned beam
S=540/36-(2.5*2.25)=9.375in and it is less than the maximum spacing which is given by 12*36/fs=12*36/36=12in
Hence, the beam is within the limits and the beam is safe against the cracking.
Modifications to reduce the number of reinforcing bars
The addition of steel does not prevent cracking due to restrained shrinkage but it limits the width of crack by causing the formation of the number of narrow cracks rather than single wide crack.
Larger size bars leads to fewer cracks but wider cracks while smaller size bars leads to number of narrow cracks hence it is advisable to provide number of smaller diameter bars of equal strength of designed bars rather than larger bars.
Air at a pressure of 1 atm and a temperature of 50 °C is in parallel flow over the top surface of a flat plate that is heated to a uniform temperature of 100 °C. The plate has a length of 0.20 m (in the flow direction) and a width of 0.10 m. The Reynolds number based on the plate length is 40,000.
a. What is the rate of heat transfer from the plate to the air?
b. If the free stream velocity of the air is doubled and the pressure is increased to 10 atm, what is the rate of heat transfer?
Answers
It is given that :
Let the mean bulk temperature [tex]$=\frac{50+100}{2}$[/tex]
[tex]$=75^\circ C$[/tex]
From the property table at 1 bar and [tex]$75^\circ C$[/tex],
[tex]$K=0.02917 \ W/\mu K, \ Pr = 0.71055 $[/tex]
Flow is laminar as Re = 4000 for laminar.
Flow Nusselt Number is given by :
[tex]$\overline{Nu} = 0.664 (Re)^{0.5} Pr^{1/3} = \frac{hd}{K}$[/tex]
[tex]$\theta = 4 \times 0.2 \times 0.1 \times (100-50)$[/tex]
[tex]$=17.32$[/tex]
At 10 bar and [tex]$75^\circ C$[/tex],
[tex]$\rho = 9.999 \ kg/m^3 , \ \mu =20.91 \times 10^{-6}$[/tex]
[tex]$K=30.05 \times 10^{-7} \ W/\mu K, \ Pr = 0.7092, \ C_p=1.019 \ kJ/kg K$[/tex]
[tex]$Re_2 = \frac{9.999 \times 2 \times V}{1 \times 20.9 \times 10^{-6}}$[/tex]
Initial, [tex]$Re_i = \frac{1 \times V}{1 \times 20.82 \times 10^{-6}}$[/tex]
[tex]$=40000$[/tex]
[tex]$V=40000 \times 0.2 \times 20.82 \times 10^{-6}$[/tex]
[tex]$Re_2 = \frac{9.999 \times 2 \times 40000}{1 \times 20.9 \times 10^{-6}}$[/tex]
[tex]$Re_2=796477.01$[/tex]
Flow is turbulent.
This Nusselt number is given by :
[tex]$Nu=(0.037)(Re)^{0.8}- 8\pi Pr^{1/3}=958.75$[/tex]
[tex]$h=\frac{958.75 \times k}{0.2}$[/tex]
[tex]$=144.05 \ W /\mu^2C$[/tex]
[tex]$\theta =144.05 \times 0.2 \times 0.1 \times (100.5)$[/tex]
[tex]$=144.05 \ \omega$[/tex]
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Water flows through a Xylan tube at 300 K temperature and 0.5 kg/s flow rate. The inner and outer radii of the Xylan tube is 20 and 30 mm, respectively. A thin electrical heating tape wrapped around the outer surface of the Xylan tube delivers a uniform surface heat flux of 1500 W/m², while a convection coefficient of 20 W/ m K is maintained on the outer surface of the tape by ambient air at 310 K. (a) What is the outer surface temperature of the Xylan tube? (b) What is the fraction of the power dissipated by the tape, which is transferred to the water? Please draw the thermal circuit. Assume that the thermal properties of water at 300 K are as follows: u = 1100x10kg/s.m; k = 0.555 W/m.K; Pr = 7.45. Thermal conductivity of Xylan tube is k = 0.25 W/ mK.
Answers
Answer:
Find the attachments for the complete solution
Note
In the above Question heat flux is balanced with ambient heat loss and tube heating of water.
In Tube, a hollow cylinder heat flow equation has been used.
Consider a single crystal of some hypothetical metal that has the BCC crystal structure and is oriented such that a tensile stress is applied along a [121] direction. If slip occurs on a (101) plane and in a direction, compute the stress at which the crystal yields if its critical resolved shear stress is 2.9 MPa.
Answers
Complete Question
Consider a single crystal of some hypothetical metal that has the BCC crystal structure and is oriented such that a tensile stress is applied along a [121] direction. If slip occurs on a (101) plane and in a [111] direction, compute the stress at which the crystal yields if its critical resolved shear stress is 2.9 MPa.
Answer:
The stress is [tex]\sigma = 10. 655 MPa[/tex]
Explanation:
From the question we are told that
The critical yield resolved shear stress is [tex]\sigma = 2.9Mpa[/tex]
First we obtain the angle [tex]\lambda[/tex] between the slip direction [121] and [111]
[tex]\lambda = cos^{-1} [\frac{(u_1 u_2 + v_1 v_2 + w_1 w_2}{\sqrt{u_1^2 + v_1 ^2+ w_1^2})\sqrt{( u_2^2 + v_2^2 + w_2 ^2)} } ][/tex]
Where [tex]u_1 ,u_2 ,v_1 , v_2 , w_1 , w_2[/tex] are the directional indices
[tex]\lambda = cos ^-[ \frac{(1) (-1) + (2) (1) + (1) (1)}{\sqrt{((1)^2 +(2)^2 + (1)^2)}\sqrt{((-1)^2 + (1)^2 + (1)^2 ) } } ][/tex]
[tex]= cos^{-1} [\frac{2}{\sqrt{6} \sqrt{3} } ][/tex]
[tex]= 61.87^0[/tex]
Next is to obtain the angle [tex]\O[/tex] between the direction [121] and [101]
[tex]\O = cos^{-1} [\frac{(u_1 u_3 + v_1 v_3 + w_1 w_3}{\sqrt{u_1^2 + v_1 ^2+ w_1^2})\sqrt{( u_3^2 + v_3^2 + w_3 ^2)} } ][/tex]
Substituting 1 for [tex]u_1[/tex] , 2 for [tex]v_1[/tex] , 1 for [tex]w_1[/tex] , 1 for [tex]u_2[/tex], 0 for [tex]v_2[/tex], and 1 for [tex]w_2[/tex]
[tex]\O = cos^{-1} [\frac{1* 1 + 2*0 + 1*1 }{\sqrt{1^2 + 2^2 + 1^2 } \sqrt{(1^2 + 0^2 + 1^2 )} } ][/tex]
[tex]\O = cos^{-1} [\frac{2}{\sqrt{6} \sqrt{2} } ][/tex]
[tex]= 54.74 ^o[/tex]
The stress is mathematically represented as
[tex]\sigma = \frac{\tau_c}{cos \O cos \lambda }[/tex]
[tex]= \frac{2.9}{cos 54.74^o cos 61.87^o}[/tex]
[tex]= \frac{2.9}{0.2722}[/tex]
[tex]\sigma = 10. 655 MPa[/tex]
An insulated rigid tank is initially evacuated. A valve is opened, and atmospheric air at 95 kPa and 17 ºC enters the tank until the pressure in the tank reaches 95 kPa, at which point the valve is closed. Determine the final temperature of the air in the tank. Assume constant specific heats.
Answers
The final temperature of the air in the tank is 290.15 K which is 17 °C
The details of the steps used to arrive at the above response are presented as follows;
The type of tank the air is evacuated and then enters = Insulated rigid tank
Initial pressure of air entering the tank, P₁ = 95 kPa
Initial temperature of the air entering the tank, T₁ = 17 °C
17 °C = 290.15 K
The final pressure of the air in the tank, P₂ = 95 kPa
The final temperature of the ait in the tank, T₂ is required
Gay-Lussac's law states that at constant volume, the pressure of a specified mass of gas is directly proportional to the absolute temperature
Mathematically, P₁·T₁ = P₂·T₂
Therefore, we get; 95 kPa × 290.15 K = 95 kPa × T₂
[tex]T_2 = 95 \, kPa \times\frac{290.15\, K}{95 \, kPa}[/tex]
[tex]95 \, kPa \times\frac{290.15\, K}{95 \, kPa}= 290.15 \, K[/tex]
T₂ = 290.15 K
290.15 K = 17 °C
T₂ = 17 °C (The temperature of the air is the same as the initial air temperature)
The final temperature, T₂ = 17 °C
The final temperature of the air in the tank remains 17°C, as pressure remains constant in the isobaric process.
To solve this problem, we can use the ideal gas law along with the assumption of constant specific heats. The ideal gas law states:
PV = nRT
Where:
- P ) = Pressure
- ( V ) = Volume
- n ) = Number of moles
- ( R ) = Gas constant
- ( T ) = Temperature
Given that the process is isobaric (constant pressure), we can use:
[tex]\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \][/tex]
Where:
[tex]- \( V_1 \) and \( T_1 \)[/tex] are the initial volume and temperature, respectively.
[tex]- \( V_2 \) and \( T_2 \)[/tex] are the final volume and temperature, respectively.
Since the tank is initially evacuated, [tex]\( V_1 = 0 \).[/tex]
Given:
- ( P = 95 ) kPa
- ( T_1 = 17 ) °C = ( 17 + 273.15 = 290.15 ) K (converting to Kelvin)
Since the pressure inside the tank reaches 95 kPa, the final pressure is also 95 kPa.
Let's denote [tex]\( V_2 \)[/tex] as the volume of air that entered the tank.
Using the ideal gas law for the initial and final states:
[tex]\[ P_1V_1 = nRT_1 \][/tex]
Since [tex]\( V_1 = 0 \):[/tex]
[tex]\[ P_1 \times 0 = nRT_1 \]\[ nR = \frac{P_1 \times 0}{T_1} \]\[ nR = 0 \][/tex]
For the final state:
[tex]\[ P_2V_2 = nRT_2 \]\[ P_2 \times V_2 = nRT_2 \][/tex]
We can rewrite this equation to find [tex]\( T_2 \):[/tex]
[tex]\[ T_2 = \frac{P_2 \times V_2}{nR} \][/tex]
But since [tex]\( nR = 0 \), \( T_2 \)[/tex] is undefined.
This means that we can't directly use the ideal gas law to find the final temperature because the initial volume is zero.
However, we can still infer the final temperature using the fact that the pressure remains constant throughout the process, so the final temperature will be the same as the initial temperature, [tex]\( T_2 = T_1 = 290.15 \) K.[/tex]
Therefore, the final temperature of the air in the tank is [tex]\( 290.15 \) K, or \( 17 \) °C.[/tex]