Answer:
Diameter will be 27394.76 m
Explanation:
Power P = 0.5 kW = 500 W
Time t required for grinding = 1 hr = 3600 sec
Energy required E = P x t
E = 500 x 3600 = 1800000 J
Flow rate of water Q = 400 ltr/min
We convert to m3/sec
400 ltr/min = 400/(1000 x 60) m3/ses
Q = 0.0067 m3/sec
Energy provided by flow will be
E = pgQd
Where p = density of water = 1000 kg/m3
g = acceleration due to gravity 9.81 m/s2
d = diameter of wheel.
Equating both energy, we have,
1800000 = 1000 x 9.81 x 0.0067 x d
1800000 = 65.73d
d = 1800000/65.73
d = 27394.76 m
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
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]
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.
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.
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');
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
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.
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.
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
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.
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.
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;
}
Specifically, the following methods must be implemented in the LinkedList class: (You should utilize listIterator() method already defined in the LinkedList class to obtain its LinkedListIterator object, and use the methods in the LinkedListIterator class to traverse from the first element to the last element of the linked list to define the following methods.)
Answer:
Attached to this solution is a Seventeen pages of code. Cheers!
Explanation:
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.
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.
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?
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.
A conical enlargement in a vertical pipeline is 5 ft long and enlarges the pipe diameter from 12 in. to 24 in. diameter. Calculate the magnitude and direction of the vertical force on this enlargement when 10 f t3/s of water flow upward through the line and the pressure at the smaller end of the enlargement is 30 psi.
Answer:
F_y = 151319.01N = 15.132 KN
Explanation:
From the linear momentum equation theory, since flow is steady, the y components would be;
-V1•ρ1•V1•A1 + V2•ρ2•V2•A2 = P1•A1 - P2•A2 - F_y
We are given;
Length; L = 5ft = 1.52.
Initial diameter;d1 = 12in = 0.3m
Exit diameter; d2 = 24 in = 0.6m
Volume flow rate of water; Q2 = 10 ft³/s = 0.28 m³/s
Initial pressure;p1 = 30 psi = 206843 pa
Thus,
initial Area;A1 = π•d1²/4 = π•0.3²/4 = 0.07 m²
Exit area;A2 = π•d2²/4 = π•0.6²/4 = 0.28m²
Now, we know that volume flow rate of water is given by; Q = A•V
Thus,
At exit, Q2 = A2•V2
So, 0.28 = 0.28•V2
So,V2 = 1 m/s
When flow is incompressible, we often say that ;
Initial mass flow rate = exit mass flow rate.
Thus,
ρ1 = ρ2 = 1000 kg/m³
Density of water is 1000 kg/m³
And A1•V1 = A2•V2
So, V1 = A2•V2/A1
So, V1 = 0.28 x 1/0.07
V1 = 4 m/s
So, from initial equation of y components;
-V1•ρ1•V1•A1 + V2•ρ2•V2•A2 = P1•A1 - P2•A2 - F_y
Where F_y is vertical force of enlargement pressure and P2 = 0
Thus, making F_y the subject;
F_y = P1•A1 + V1•ρ1•V1•A1 - V2•ρ2•V2•A2
Plugging in the relevant values to get;
F_y = (206843 x 0.07) + (1² x 1000 x 0.07) - (4² x 1000 x 0.28)
F_y = 151319.01N = 15.132 KN
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.
Answer:
Explanation:
The step by step solution is in the attached file.
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.
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.
- 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'
Answer:
See Explaination
Explanation:
if(check1 and check2):
outcome = "BOTH"
elif(check1):
outcome = "ONE"
elif(check2):
outcome = "TWO"
else:
outcome = "NEITHER"
(Using Python)Part 2aNumerology is the "study of the purported mystical or special relationship between a number and observed or perceived events." It has been used throughout human history as a way to attach meaning to a name, object or event using mathematics. It is considered a "pseudoscience" by modern scientists since it has no basis in observable phenomena. With that said, it makes a great programming challenge so we're going to go with it! :)What you want to do for this project is to ask the user to type in their name. Next, you will need to use a technique called "theosophical reduction" to convert their name into a number. With this technique we assign each letter of the alphabet its own number. For example, the letter "a" is equal to the number 1. "b" = 2, "c" = 3, "z" = 26, etc. You should ignore non-alphabetic characters (i.e. numbers, spaces and special characters)Once you've gotten all of the letters converted into numbers you can add them up into one single number. This is the "numerology number" for the name that the user entered.So for the name "craig" the numerology number would be:c = 3r = 18a = 1i = 9g = 73 + 18 + 1 + 9 + 7 = 38Here's are a few sample runnings of this program:Name: craigYour 'cleaned up' name is: craigReduction: 38Name: craig kappYour 'cleaned up' name is: craigkappReduction: 82Name: rumple stil skinYour 'cleaned up' name is: rumplestilskinReduction: 198Name: !rumple!stil!skinYour 'cleaned up' name is: rumplestilskinReduction: 198Name: pikachu!pikapika!Your 'cleaned up' name is: pikachupikapikaReduction: 143Name: PIKACHUpikapikaYour 'cleaned up' name is: pikachupikapikaReduction: 143Some hints:Convert the user's name to all uppercase or all lowercase before you do anything elseRemove any spaces, numbers or special characters from the name to ensure that you are only working with the letters A-ZThe ord() function may be userful to convert each character into an ASCII index
Answer:
See explaination for python programming code
Explanation:
Python programming code below
import re
s = "abc" # enter string here
#s = "hello world! HELLOW INDIA how are you? 01234"
# Short version
print filter(lambda c: c.isalpha(), s)
# Faster version for long ASCII strings:
id_tab = "".join(map(chr, xrange(256)))
tostrip = "".join(c for c in id_tab if c.isalpha())
print s.translate(id_tab, tostrip)
# Using regular expressions
s1 = re.sub("[^A-Za-z]", "", s)
s2 = s1.lower()
print s2
import string
values = dict()
for index, letter in enumerate(string.ascii_lowercase):
values[letter] = index + 1
sum = 0
for ch2 in s2:
for ch1 in values:
if(ch2 == ch1):
sum = sum + values[ch1]
print sum
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.
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
Neglecting the presence of friction, air drag, and other inefficiencies, how much gasoline is consumed when a 1300 kg automobile accelerates from rest to 80 km/h. Assume the density and enthalpy of gasoline are 680 kg/m3 and 45 MJ/kg respectively. Express your answer in the units of mL. Show all work and use SI units for full credit. Box your final answer(s)
Answer:
Explanation:
Given that, .
Mass of car is
M = 1300kg
Velocity of car
V = 80km/h = 80 × 1000/3600
V = 22.22m/s
Calculate the kinetic energy of the vehicle as follows:
K.E = ½ MV²
K.E = ½ × 1300 × 22.22²
K.E = 320,987.65 J
Given that,
Enthalpy is 45MJ / kg
h = 45MJ / kg
Then, enthalpy is given as.
Enthalpy = Energy / mass
h = E / m
45 × 10^6 = 320,987.65 / m
m = 320,987.65 / 45 × 10^6
m = 7.133 × 10^-3 kg
m = 7.133 mg
Also, given that, density is 680kg/m³
Density is given as
Density = mass / Volume
ρ = m / v
Then, v = m / ρ
v = 7.133 × 10^-3 / 680
v = 1.049 × 10^-5 m³
We know that
1mL = 10^-6 m³
Therefore,
v = 1.049 × 10^-5 m³ × 1mL / 10^-6m³
v = 10.49 mL
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.
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]
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?
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]
Determine the angle φ at which the applied force P should act on the pipe so that the magnitude of P is as small as possible for pulling the pipe up the incline. What is the corresponding value of P? The pipe weighs W and the slope α is known. Express the answer in terms of the angle of static friction, θ = tan-1 μs.
Answer:
∅=Ф
P = W sin([tex]\alpha[/tex] + Ф)
Explanation:
First, we'll isolate and draw the free-body diagram of the pipe
Note that since the pipe is moving, the friction force is equal to the product of normal reaction force and the kinetic coefficient of friction
F = F_max = u_kN
Also note that the weight makes with the y-axis angle a because the x-axis makes the same angle with the horizontal
The expression for angle of friction is:
B = tan-1 (u_k)
From here we can express the coefficient of friction as:
u_k = tan(Ф)
Replace u_s by tan(Ф) in the expression for the friction force
F = N tan(Ф)
diagram is attached
By equating sum of forces in y-direction to zero, we can write the expression for the normal reaction force
ΣF_y = 0
N — W cos[tex]\alpha[/tex]- P sin Ф= 0
From here we can express N as:
N = W cos[tex]\alpha[/tex] -— P sin Ф
Replace N by the expression above in the expression for friction force F(written in step 1)
F = (W cos[tex]\alpha[/tex] — P sin Ф) tan( Ф) (1)
Now, we'll equate sum of forces in x-direction to zero
ΣF_x = 0
-F - W cos[tex]\alpha[/tex] + P sin Ф =0
Replace F by expression (1)
— (W cos[tex]\alpha[/tex] — P sin Ф) tan(Ф) — W sin[tex]\alpha[/tex]+pcosФ=0
-W cos [tex]\alpha[/tex] tan(Ф) + P sin Ф tan(Ф) — W sin[tex]\alpha[/tex] +pcosФ=0
P(sin Ф tan(Ф) + cosФ) — W(cos [tex]\alpha[/tex] tan(Ф) + sin [tex]\alpha[/tex])
From here we can express the force P needed to pull the pipe as:
P = W(cos[tex]\alpha[/tex] tan(Ф) + sin[tex]\alpha[/tex])/sinФ*tansФ+cosФ (2)
All we have to do now is to simplify the expression (2). We'll start by sin replacing tan(Ф) with sinФ/cosФ
P = W(cos *sinФ/cosФ + sin)/sinФ*sinФ/cosФ+cosФ *cosФ/cosФ
We can multiply the right side of equation by cosФ/cosФ
P = W(cos[tex]\alpha[/tex] *sinФ + sin[tex]\alpha[/tex]cosФ)/sin∅*sinФ+cos∅cosФ *cosФ/cosФ
Finally, we'll replace (cos[tex]\alpha[/tex] *sinФ + sin[tex]\alpha[/tex]cosФ) by sin([tex]\alpha[/tex] + Ф) and (sin∅ sinФ + cos∅ cos Ф) by cos( ∅— Ф)
P wsin([tex]\alpha[/tex] + Ф) /cos(∅ — Ф) (3)
Since the first derivative of the function is actually tangens of the angle which tangent makes with the x-axis, we'll find it by equating the first derivative by zero(this means that the tangent of the function is horizontal, i.e. that the function is at its maximum or minimum)
Note that the variable in the expression (3) is 0, since both B and a are known
dP/d∅ =d/d∅ [sin(Ф+)/cos(∅-Ф) ]
Note that sin(Ф+[tex]\alpha[/tex]) is constant since both Ф and a are known
dP/d∅ = sin(Ф+[tex]\alpha[/tex]) d/dФ [1/cos(∅-Ф) ]
Next, we'll apply the reciprocal rule
= -dP/d∅[cos(∅-Ф)]/cos^2(∅-Ф)*sin(Ф+[tex]\alpha[/tex])
Next, we'll apply the differentiation rule
=(-sin(∅-Ф))*d/d∅[∅-Ф]*sin(Ф+[tex]\alpha[/tex])/cos^2(∅-Ф)
=(d/d∅[∅]+d/d∅[-∅])*sin(Ф+[tex]\alpha[/tex])sin(∅-Ф)/cos^2(∅-Ф)
dP/d∅ =sin(Ф+[tex]\alpha[/tex])*sin(∅-Ф)/cos^2(∅-Ф) (4)
Next step will be to equate the expression (4) to zero, to determine the value of # when the function is minimum
sin(Ф+[tex]\alpha[/tex])*sin(∅-Ф)/cos^2(∅-Ф) =0
Note that sin(Ф+[tex]\alpha[/tex]) is constant, so in order for the equation above to be correct, sin(∅-Ф) needs to be equal to zero
sin(∅-Ф) = 0
Since sin 0° = sin 180° = 0, two possible solutions for ∅ are:
∅-Ф=0 Ф=∅
or
∅-Ф = 180° ∅ = 180° + Ф
Since the function for P is only good over the range 0 < ∅ < 90°, since when > 90° the friction force will change its direction, we can conclude that the minimum force P is required to move the pipe at angle:
∅=Ф
Finally, replace # by 8 in expression (3) to determine the minimum force P required to move the pipe
P = W sin([tex]\alpha[/tex] + Ф ) / cos ∅ — ∅)
P = W sin([tex]\alpha[/tex] + Ф)
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%.
Answer:
3.25 turbines
Explanation:
Please kindly check attachment for the detailed answer.
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?
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]
Learn More :
https://brainly.in/question/43296102
https://brainly.in/question/48196879
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.
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]
The air velocity in the duct of a heating system is to be measured by a Pitot-static probe inserted into the duct parallel to the flow. The differential height between the water columns connected to the two outlets of the probe is 0.126 m.Take the density of water to be 1000 kg/m3. The gas constant of air is R = 0.287 kPa-m3/kg-K.The air temperature and pressure in the duct are 352 K and 98 kPa, respectively.
Answer:
Flow velocity
50.48m/s
Pressure change at probe tip
1236.06Pa
Explanation:
Question is incomplete
The air velocity in the duct of a heating system is to be measured by a Pitot-static probe inserted into the duct parallel to the flow. If the differential height between the water columns connected to the two outlets of the probe is 0.126m, determine (a) the flow velocity and (b) the pressure rise at the tip of the probe. The air temperature and pressure in the duct are 352k and 98 kPa, respectively
solution
In this question, we are asked to calculate the flow velocity and the pressure rise at the tip of probe
please check attachment for complete solution and step by step explanation
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.
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
A circular bar will be subjected to an axial force (P) of 2000 lbf. The bar will be made of material that has a strength (S) of 24 kpsi. After some calculation, the designer has selected a standard 1/2 in diameter (d) bar. The factor of safety (n) is ____. Round the answers to three significant digits.
Answer:
[tex]n = 2.36[/tex]
Explanation:
The stress experimented by the circular bar is:
[tex]\sigma = \left[\frac{2000\, lbf}{\frac{\pi}{4}\cdot (0.5\,in)^{2}}\right]\cdot \left(\frac{1\,kpsi}{1000\,psi} \right)[/tex]
[tex]\sigma = 10.186\,kpsi[/tex]
The safety factor is:
[tex]n = \frac{24\,kpsi}{10.186\,kpsi}[/tex]
[tex]n = 2.36[/tex]
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?
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.
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.
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
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
Answer:
The energy released is -241 kJ/mol
Explanation:
Please look at the solution in the attached Word file
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.)
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.
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]