A 50-lbm iron casting, initially at 700o F, is quenched in a tank filled with 2121 lbm of oil, initially at 80o F. The iron casting and oil can be modeled as incompressible with specific heats 0.10 Btu/lbm o R, and 0.45 Btu/lbm o R, respectively. For the iron casting and oil as the system, determine: a) The final equilibrium temperature (o F) b) The total entropy change for this process (Btu/ o R) (Hint: Total entropy change is the sum of entropy change of iron casting and oil.)
Answers
Answer:
a) The final equilibrium temperature is 83.23°F
b) The entropy production within the system is 1.9 Btu/°R
Explanation:
See attached workings
a) Equilibrium temp. ≈ 77.01°F.
b) Total entropy change ≈ 104.58 Btu/°R.
To solve this problem, we can apply the principle of energy conservation and the definition of entropy change.
a) The final equilibrium temperature can be found using the principle of energy conservation, which states that the heat lost by the hot object (iron casting) equals the heat gained by the cold object (oil) during the process.
The equation for energy conservation is:
[tex]\[ m_{\text{iron}} \times C_{\text{iron}} \times (T_{\text{final}} - T_{\text{initial, iron}}) = m_{\text{oil}} \times C_{\text{oil}} \times (T_{\text{final}} - T_{\text{initial, oil}}) \][/tex]
Where:
- [tex]\( m_{\text{iron}} \)[/tex] = mass of iron casting = 50 lbm
- [tex]\( C_{\text{iron}} \)[/tex] = specific heat of iron casting = 0.10 Btu/lbm °R
- [tex]\( T_{\text{initial, iron}} \)[/tex] = initial temperature of iron casting = 700 °F
- [tex]\( m_{\text{oil}} \)[/tex] = mass of oil = 2121 lbm
- [tex]\( C_{\text{oil}} \)[/tex] = specific heat of oil = 0.45 Btu/lbm °R
- [tex]\( T_{\text{initial, oil}} \)[/tex] = initial temperature of oil = 80 °F
- [tex]\( T_{\text{final}} \)[/tex] = final equilibrium temperature (unknown)
Now, let's solve for [tex]\( T_{\text{final}} \)[/tex]:
[tex]\[ 50 \times 0.10 \times (T_{\text{final}} - 700) = 2121 \times 0.45 \times (T_{\text{final}} - 80) \][/tex]
[tex]\[ 5(T_{\text{final}} - 700) = 954.45(T_{\text{final}} - 80) \][/tex]
[tex]\[ 5T_{\text{final}} - 3500 = 954.45T_{\text{final}} - 76356 \][/tex]
[tex]\[ 0 = 949.45T_{\text{final}} - 72856 \][/tex]
[tex]\[ T_{\text{final}} = \frac{72856}{949.45} \][/tex]
[tex]\[ T_{\text{final}} \approx 77.01 \, ^\circ F \][/tex]
So, the final equilibrium temperature is approximately [tex]\( 77.01 \, ^\circ F \).[/tex]
b) The total entropy change for the process can be calculated using the formula:
[tex]\[ \Delta S = \Delta S_{\text{iron}} + \Delta S_{\text{oil}} \][/tex]
Where:
- [tex]\( \Delta S_{\text{iron}} = \frac{Q_{\text{iron}}}{T_{\text{initial, iron}}} \)[/tex]
- [tex]\( \Delta S_{\text{oil}} = \frac{Q_{\text{oil}}}{T_{\text{initial, oil}}} \)[/tex]
- [tex]\( Q_{\text{iron}} \) = heat lost by the iron casting[/tex]
- [tex]\( Q_{\text{oil}} \) = heat gained by the oil[/tex]
Let's calculate:
[tex]\[ Q_{\text{iron}} = m_{\text{iron}} \times C_{\text{iron}} \times (T_{\text{final}} - T_{\text{initial, iron}}) \][/tex]
[tex]\[ Q_{\text{iron}} = 50 \times 0.10 \times (77.01 - 700) \][/tex]
[tex]\[ Q_{\text{iron}} \approx -3175.495 \, \text{Btu} \][/tex]
[tex]\[ Q_{\text{oil}} = m_{\text{oil}} \times C_{\text{oil}} \times (T_{\text{final}} - T_{\text{initial, oil}}) \][/tex]
[tex]\[ Q_{\text{oil}} = 2121 \times 0.45 \times (77.01 - 80) \][/tex]
[tex]\[ Q_{\text{oil}} \approx 8729.535 \, \text{Btu} \][/tex]
Now, calculate entropy changes:
[tex]\[ \Delta S_{\text{iron}} = \frac{-3175.495}{700} \][/tex]
[tex]\[ \Delta S_{\text{iron}} \approx -4.5364 \, \text{Btu/°R} \][/tex]
[tex]\[ \Delta S_{\text{oil}} = \frac{8729.535}{80} \][/tex]
[tex]\[ \Delta S_{\text{oil}} \approx 109.118 \, \text{Btu/°R} \][/tex]
[tex]\[ \Delta S = -4.5364 + 109.118 \][/tex]
[tex]\[ \Delta S \approx 104.5816 \, \text{Btu/°R} \][/tex]
So, the total entropy change for this process is approximately [tex]\( 104.5816 \, \text{Btu/°R} \).[/tex]
Related Questions
A pump is used to deliver water from a lake to an elevated storage tank. The pipe network consists of 1,800 ft (equivalent length) of 8-in. pipe (Hazen-Williams roughness coefficient = 120). Ignore minor losses. The pump discharge rate is 600 gpm. The friction loss (ft) is most nearly Group of answer choicesA. 15
B. 33
C. 106
D. 135
Answers
Answer:
h_f = 15 ft, so option A is correct
Explanation:
The formula for head loss is given by;
h_f = [10.44•L•Q^(1.85)]/(C^(1.85))•D^(4.8655))
Where;
h_f is head loss due to friction in ft
L is length of pipe in ft
Q is flow rate of water in gpm
C is hazen Williams constant
D is diameter of pipe in inches
We are given;
L = 1,800 ft
Q = 600 gpm
C = 120
D = 8 inches
So, plugging in these values into the equation, we have;
h_f = [10.44*1800*600^(1.85)]/(120^(1.85))*8^(4.8655))
h_f = 14.896 ft.
So, h_f is approximately 15 ft
Steam enters a counterflow heat exchanger operating at steady state at 0.07 MPa with a quality of 0.9 and exits at the same pressure as saturated liquid. The steam mass flow rate is 1.5 kg/min. A separate stream of air with a mass flow rate of 100 kg/min enters at 30°C and exits at 60°C. The ideal gas model with 1.005 kJ/kg · K can be assumed for air. Kinetic and potential energy effects are negligible. Determine the temperature of the entering steam, in °C, and for the overall heat exchanger as the control volume, what is the rate of heat transfer, in kW.
Answers
Answer:
1.12kw is the heat transfer
Explanation:
Kinetic energy is the energy an object has because of its motion. If we want to accelerate an object, then we must apply a force.
Potential energy, stored energy that depends upon the relative position of various parts of a system.
See attachment for the step by step solution.
Five Kilograms of continuous boron fibers are introduced in a unidirectional orientation into of an 8kg aluminum matrix. Calculate:
a. the density of the composite.
b. the modulus of elasticity parallel to the fibers.
c. the modulus of elasticity perpendicular to the fibers.
Answers
Answer:
Explanation:
Given that,
Mass of boron fiber in unidirectional orientation
Mb = 5kg = 5000g
Mass of aluminum fiber in unidirectional orientation
Ma = 8kg = 8000g
A. Density of the composite
Applying rule of mixing
ρc = 1•ρ1 + 2•ρ2
Where
ρc = density of composite
1 = Volume fraction of Boron
ρ1 = density composite of Boron
2 = Volume fraction of Aluminum
ρ2 = density composite of Aluminum
ρ1 = 2.36 g/cm³ constant
ρ2 = 2.7 g/cm³ constant
To Calculate fractional volume of Boron
1 = Vb / ( Vb + Va)
Vb = Volume of boron
Va = Volume of aluminium
Also
To Calculate fraction volume of aluminum
2= Va / ( Vb + Va)
So, we need to get Va and Vb
From density formula
density = mass / Volume
ρ1 = Mb / Vb
Vb = Mb / ρ1
Vb = 5000 / 2.36
Vb = 2118.64 cm³
Also ρ2 = Ma / Va
Va = Ma / ρ2
Va = 8000 / 2.7
Va = 2962.96 cm³
So,
1 = Vb / ( Vb + Va)
1 = 2118.64 / ( 2118.64 + 2962.96)
1 = 0.417
Also,
2= Va / ( Vb + Va)
2 = 2962.96 / ( 2118.64 + 2962.96)
2 = 0.583
Then, we have all the data needed
ρc = 1•ρ1 + 2•ρ2
ρc = 0.417 × 2.36 + 0.583 × 2.7
ρc = 2.56 g/cm³
The density of the composite is 2.56g/cm³
B. Modulus of elasticity parallel to the fibers
Modulus of elasticity is defined at the ratio of shear stress to shear strain
The relation for modulus of elasticity is given as
Ec = = 1•Eb+ 2•Ea
Ea = Elasticity of aluminium
Eb = Elasticity of Boron
Ec = Modulus of elasticity parallel to the fiber
Where modulus of elastic of aluminum is
Ea = 69 × 10³ MPa
Modulus of elastic of boron is
Eb = 450 × 10³ Mpa
Then,
Ec = = 1•Eb+ 2•Ea
Ec = 0.417 × 450 × 10³ + 0.583 × 69 × 10³
Ec = 227.877 × 10³ MPa
Ec ≈ 228 × 10³ MPa
The Modulus of elasticity parallel to the fiber is 227.877 × 10³MPa
OR Ec = 227.877 GPa
Ec ≈ 228GPa
C. modulus of elasticity perpendicular to the fibers?
The relation of modulus of elasticity perpendicular to the fibers is
1 / Ec = 1 / Eb+ 2 / Ea
1 / Ec = 0.417 / 450 × 10³ + 0.583 / 69 × 10³
1 / Ec = 9.267 × 10^-7 + 8.449 ×10^-6
1 / Ec = 9.376 × 10^-6
Taking reciprocal
Ec = 106.66 × 10^3 Mpa
Ec ≈ 107 × 10^3 MPa
Note that the unit of Modulus has been in MPa,
10. Develop a logic circuit called "Simple Multiples." This circuit accepts 4 bits that represent a BCD number. It has three outputs:
• A HIGH on output X indicates that the number is a multiple of 2
• A HIGH on output Y indicates that the number is a multiple of 3
• A HIGH on output Z indicates that the number is a multiple of both 2 and 3 *0 counts as a multiple for all of these
Create a truth table for this circuit
Answers
Answer:
Explanation:
"Simple Multiples Logic Circuit Development"
From the attached file below;
The first diagram illustrate the truth table. The input consist of binary coded decimal with its 16 possible output. As stated in the input, assuming we have 0010 which the value is 2, thus ; it is said to be divisible by 2 , then we enter the output of x to be 1 and the remaining as zeros
The second diagram talks about the output x; the representation of x is 1 there and it was added up. The use case employed is [tex]x + \bar {x} =1[/tex] which in turn yield output [tex]\bar {A3}[/tex]
The second to the last diagram represents output Y and Z.
As output Y can be further disintegrated ; we insert XNOR gate for the expressions. The expression is [tex]a.b+(\bar{a}. \bar{b})[/tex]
The last diagram is the logic diagram.
16.44 Lab 13D: Student Scores with Files and Functions Overview Create a program that reads from multiple input files and calls a user-defined function. Objectives Gain familiarity with CSV files Perform calculations with data from CSV files Create a user-defined function Read from multiple files in the same program
Answers
Answer:
See Explaination
Explanation:
# copy the function you have this is just for my convenniece
def finalGrade(scoresList):
weights = [0.05, 0.05, 0.40, 0.50]
grade = 0
for i in range(len(scoresList)):
grade += float(scoresList[i]) * weights[i]
return grade
import csv
def main():
with open('something.csv', newline='') as csvfile:
spamreader = csv.reader(csvfile, delimiter=',', quotechar='|')
file = open('something.txt')
for row in spamreader:
student = file.readline().strip()
scores = row
print(student, finalGrade(scores))
if __name__ == "__main__":
main()
More discussion about seriesConnect(Ohm) function In your main(), first, construct the first circuit object, called ckt1, using the class defined above. Use a loop to call setOneResistance() function to populate several resistors. Repeat the process for another circuit called ckt2. Develop a member function called seriesConnect() such that ckt2 can be connected to ckt1 using instruction ckt1.seriesConnect(ckt2).
Answers
Answer:
resistor.h
//circuit class template
#ifndef TEST_H
#define TEST_H
#include<iostream>
#include<string>
#include<vector>
#include<cstdlib>
#include<ctime>
#include<cmath>
using namespace std;
//Node for a resistor
struct node {
string name;
double resistance;
double voltage_across;
double power_across;
};
//Create a class Ohms
class Ohms {
//Attributes of class
private:
vector<node> resistors;
double voltage;
double current;
//Member functions
public:
//Default constructor
Ohms();
//Parameterized constructor
Ohms(double);
//Mutator for volatage
void setVoltage(double);
//Set a resistance
bool setOneResistance(string, double);
//Accessor for voltage
double getVoltage();
//Accessor for current
double getCurrent();
//Accessor for a resistor
vector<node> getNode();
//Sum of resistance
double sumResist();
//Calculate current
bool calcCurrent();
//Calculate voltage across
bool calcVoltageAcross();
//Calculate power across
bool calcPowerAcross();
//Calculate total power
double calcTotalPower();
//Display total
void displayTotal();
//Series connect check
bool seriesConnect(Ohms);
//Series connect check
bool seriesConnect(vector<Ohms>&);
//Overload operator
bool operator<(Ohms);
};
#endif // !TEST_H
resistor.cpp
//Implementation of resistor.h
#include "resistor.h"
//Default constructor,set voltage 0
Ohms::Ohms() {
voltage = 0;
}
//Parameterized constructor, set voltage as passed voltage
Ohms::Ohms(double volt) {
voltage = volt;
}
//Mutator for volatage,set voltage as passed voltage
void Ohms::setVoltage(double volt) {
voltage = volt;
}
//Set a resistance
bool Ohms::setOneResistance(string name, double resistance) {
if (resistance <= 0){
return false;
}
node n;
n.name = name;
n.resistance = resistance;
resistors.push_back(n);
return true;
}
//Accessor for voltage
double Ohms::getVoltage() {
return voltage;
}
//Accessor for current
double Ohms::getCurrent() {
return current;
}
//Accessor for a resistor
vector<node> Ohms::getNode() {
return resistors;
}
//Sum of resistance
double Ohms::sumResist() {
double total = 0;
for (int i = 0; i < resistors.size(); i++) {
total += resistors[i].resistance;
}
return total;
}
//Calculate current
bool Ohms::calcCurrent() {
if (voltage <= 0 || resistors.size() == 0) {
return false;
}
current = voltage / sumResist();
return true;
}
//Calculate voltage across
bool Ohms::calcVoltageAcross() {
if (voltage <= 0 || resistors.size() == 0) {
return false;
}
double voltAcross = 0;
for (int i = 0; i < resistors.size(); i++) {
voltAcross += resistors[i].voltage_across;
}
return true;
}
//Calculate power across
bool Ohms::calcPowerAcross() {
if (voltage <= 0 || resistors.size() == 0) {
return false;
}
double powerAcross = 0;
for (int i = 0; i < resistors.size(); i++) {
powerAcross += resistors[i].power_across;
}
return true;
}
//Calculate total power
double Ohms::calcTotalPower() {
calcCurrent();
return voltage * current;
}
//Display total
void Ohms::displayTotal() {
for (int i = 0; i < resistors.size(); i++) {
cout << "ResistorName: " << resistors[i].name << ", Resistance: " << resistors[i].resistance
<< ", Voltage_Across: " << resistors[i].voltage_across << ", Power_Across: " << resistors[i].power_across << endl;
}
}
//Series connect check
bool Ohms::seriesConnect(Ohms ohms) {
if (ohms.getNode().size() == 0) {
return false;
}
vector<node> temp = ohms.getNode();
for (int i = 0; i < temp.size(); i++) {
this->resistors.push_back(temp[i]);
}
return true;
}
//Series connect check
bool Ohms::seriesConnect(vector<Ohms>&ohms) {
if (ohms.size() == 0) {
return false;
}
for (int i = 0; i < ohms.size(); i++) {
this->seriesConnect(ohms[i]);
}
return true;
}
//Overload operator
bool Ohms::operator<(Ohms ohms) {
if (ohms.getNode().size() == 0) {
return false;
}
if (this->sumResist() < ohms.sumResist()) {
return true;
}
return false;
}
main.cpp
#include "resistor.h"
int main()
{
//Set circuit voltage
Ohms ckt1(100);
//Loop to set resistors in circuit
int i = 0;
string name;
double resistance;
while (i < 3) {
cout << "Enter resistor name: ";
cin >> name;
cout << "Enter resistance of circuit: ";
cin >> resistance;
//Set one resistance
ckt1.setOneResistance(name, resistance);
cin.ignore();
i++;
}
//calculate totalpower and power consumption
cout << "Total power consumption = " << ckt1.calcTotalPower() << endl;
return 0;
}
Output
Enter resistor name: R1
Enter resistance of circuit: 2.5
Enter resistor name: R2
Enter resistance of circuit: 1.6
Enter resistor name: R3
Enter resistance of circuit: 1.2
Total power consumption = 1886.79
Explanation:
Note
Please add all member function details.Its difficult to figure out what each function meant to be.
I have a signal that is experiencing 60 Hz line noise from a nearby piece of equipment, and I want to make sure that the noise is filtered out. I expect that there might be useful information at frequencies higher and lower than 60 Hz. Which would be the best type of filter for this purpose?
a. BandStop filter
b. Low-pass filter
c. Bandpass filter
d. High-pass filter
Answers
Answer:
a. BandStop filter
Explanation:
A band-stop filter or band-rejection filter is a filter that eliminates at its output all the signals that have a frequency between a lower cutoff frequency and a higher cutoff frequency. They can be implemented in various ways. One of them is to implement a notch filter, which is characterized by rejecting a certain frequency that is interfering with a circuit. The transfer function of this filter is given by:
[tex]H(s)=\frac{s^2+\omega_o^2}{s^2+\omega_cs+\omega_o^2} \\\\Where:\\\\\omega_o=Central\hspace{3} rejected\hspace{3} frequency\\\omega_c=Width\hspace{3} of\hspace{3} the\hspace{3} rejected\hspace{3} band[/tex]
I attached you a graph in which you can see how the filter works.
Water flows at a rate of 0.040 m3 /s in a horizontal pipe whose diameter is reduced from 15 cm to 8 cm by a reducer. If the pressure at the centerline is measured to be 480 kPa and 440 kPa before and after the reducer, respectively, determine the irreversible head loss in the reducer. Take the kinetic energy correction factors to be 1.05. Answer: 0.963 m
Answers
Answer:
hL = 0.9627 m
Explanation:
Given
Q = 0.040 m³/s (constant value)
D₁ = 15 cm = 0.15 m ⇒ R₁ = D₁/2 = 0.15 m/2 = 0.075 m
D₂ = 8 cm = 0.08 m ⇒ R₂ = D₂/2 = 0.08 m/2 = 0.04 m
P₁ = 480 kPa = 480*10³Pa
P₂ = 440 kPa = 440*10³Pa
α = 1.05
ρ = 1000 Kg/m³
g = 9.81 m/s²
h₁ = h₂
hL = ? (the irreversible head loss in the reducer)
Using the formula Q = v*A ⇒ v = Q/A
we can find the velocities v₁ and v₂ as follows
v₁ = Q/A₁ = Q/(π*R₁²) = (0.040 m³/s)/(π*(0.075 m)²) = 2.2635 m/s
v₂ = Q/A₂ = Q/(π*R₂²) = (0.040 m³/s)/(π*(0.04 m)²) = 7.9577 m/s
Then we apply the Bernoulli law (for an incompressible flow)
(P₂/(ρ*g)) + (α*v₂²/(2*g)) + h₂ = (P₁/(ρ*g)) + (α*v₁²/(2*g)) + h₁ - hL
Since h₁ = h₂ we obtain
(P₂/(ρ*g)) + (α*v₂²/(2*g)) = (P₁/(ρ*g)) + (α*v₁²/(2*g)) - hL
⇒ hL = ((P₁-P₂)/(ρ*g)) + (α/(2*g))*(v₁²-v₂²)
⇒ hL = ((480*10³Pa-440*10³Pa)/(1000 Kg/m³*9.81 m/s²)) + (1.05/(2*9.81 m/s²))*((2.2635 m/s)²-(7.9577 m/s)²)
⇒ hL = 0.9627 m
Consider a drainage basin having 60% soil group A and 40% soil group B. Five years ago the land use pattern in the basin was ½ wooded area with poor cover and ½ cultivated land (row crops/contoured and terraces) with good conservation treatment. Now the land use has been changed to 1/3 wooded area with poor cover, 1/3 cultivated land (row crops/contoured and terraces) with good conservation treatment, and 1/3 commercial and business area.
(a) Estimate the increased runoff volume during the dormant season due to the land use change over the past 5-year period for a storm of 35 cm total depth under the dry antecedent moisture condition (AMC I). This storm depth corresponds to a duration of 6-hr and 100-year return period. The total 5-day antecedent rainfall amount is 30 mm. (Note: 1 in = 25.4 mm.)
(b) Under the present watershed land use pattern, find the effective rainfall hyetograph (in cm/hr) for the following storm event using SCS method under the dry antecedent moisture condition (AMC I).
Answers
Answer:
Please see the attached file for the complete answer.
Explanation:
A turbojet aircraft is flying with a velocity of 280 m/s at an altitude of 9150 m, where the ambient conditions are 32 kPa and -32C. The pressure ratio across the compressor is 12, and the temperature at the turbine inlet is 1100 K. Air enters the compressor at a rate of 50 kg/s, and the jet fuel has a heating value of 42,700 kJ/kg. Assume constant specific heats for air at room temperature. Efficiency of the compressor is 80%, efficiency of the turbine is 85%. Assume air leaves diffuser with negligibly small velocity.
determine
(a) The velocity of the exhaust gases.
(b) The rate of fuel consumption.
Answers
Answer:
(a) The velocity of the exhaust gases. is 832.7 m/s
(b) The rate of fuel consumption is 0.6243 kg/s
Explanation:
For the given turbojet engine operating on an ideal cycle, the pressure ,temperature, velocity, and specific enthalpy of air at [tex]i^{th}[/tex] state are [tex]P_i[/tex] , [tex]T_i[/tex] , [tex]V_i[/tex] , and [tex]h_i[/tex] , respectively.
Use "ideal-gas specific heats of various common gases" to find the properties of air at room temperature.
Specific heat at constant pressure, [tex]c_p[/tex] = 1.005 kJ/kg.K
Specific heat ratio, k = 1.4
A PMDC machine is measured to va120Vdc, ia 5.5A at the electrical terminals and the shaft is measured to have Tmech 5.5Nm, ns1200RPM. State whether the machine acting as a motor or a generator and state the efficiency of the system. Q11. Motor, 89.4 %(a) Generator, 89.4%(b) Motor, 95.5%(c) Generator, 95.5 %(d) none of the abov
Answers
Answer:
(c) Generator, 95.5 %
Explanation:
given data
voltage va = 120V
current Ia = - 5.5 A
Tmech = 5.5Nm
ns = 1200 RPM
solution
first we get here electric input power that is express as
electric input power = va × Ia ......1
put here value and we get
electric input power = 120 × -5-5
electric input power -660 W
here negative mean it generate power
and here
Pin ( mech) will be
Pin ( mech) = Tl × ω .........2
Pin ( mech) = 5.5 × [tex]\frac{2\pi N}{60}[/tex]
Pin ( mech) = 5.5 × [tex]\frac{2\pi 1200}{60}[/tex]
Pin ( mech) = 691.150 W
and
efficiency will be here as
efficiency = [tex]\frac{660}{691.150}[/tex]
efficiency = 95.5 %
so correct option is (c) Generator, 95.5 %
A preheater involves the use of condensing steam at 100o C on the inside of a bank of tubes to heat air that enters at I atm and 25o C. The air moves at 5 m/s in cross flow over the tubes. Each tube is 1 m long and has an outside diameter of 10 mm. The bank consists of 196 tubes in a square, aligned array for which ST = SL = 15 mm. What is the total rate of heat transfer to the air? What is the pressure drop associated with the airflow?
(b) Repeat the analysis of part (a), but assume that the tubes have a staggered arrangement with ST = 15 mm and SL = 10 mm.
Answers
Answer:
Please see the attached file for the complete answer.
Explanation:
8 A static load test has been conducted on a 60 ft long, 16 in square reinforced concrete pile which has been driven from a barge through 20 ft of water, then 31 ft into the underlying soil. Telltale rods A and B have been embedded at points 30 ft and 59 ft from the top of the pile, respectively. The data recorded at failure was as follows: Load at head = 139,220 lb, settlement at head = 1.211 in, settlement of telltale rod A = 1.166 in,settlement of telltale rod B = 1.141 in. Use the data from telltale rod A to compute the modulus of elasticity of the pile, then use this value and the remaining data to compute q′n and the average fn value.
Answers
Answer:
Modulus of Elasticity = 4.350×10⁶ lb/in²
q'n = 60.33 lb/in²
Side friction = 6.372 lb/in²
Explanation:
See workings in picture below.
A system executes a power cycle while receiving 1000 kJ by heat transfer at a temperature of 500 K and discharging 700 kJ by heat transfer at a temperature of 300 K. There are no other heat transfers. Determine the cycle efficiency. Use the Clausius Inequality to determine , in kJ/K. Determine if this cycle is internally reversible, irreversible, or impossible.
Answers
Answer:
[tex]\eta_{th} = 30\,\%[/tex], [tex]\eta_{th,max} = 40\,\%[/tex], [tex]\Delta S = \frac{1}{3}\,\frac{kJ}{K}[/tex], The cycle is irreversible.
Explanation:
The real cycle efficiency is:
[tex]\eta_{th} = \frac{1000\,kJ-700\,kJ}{1000\,kJ} \times 100\,\%[/tex]
[tex]\eta_{th} = 30\,\%[/tex]
The theoretical cycle efficiency is:
[tex]\eta_{th,max} = \frac{500\,K-300\,K}{500\,K} \times 100\,\%[/tex]
[tex]\eta_{th,max} = 40\,\%[/tex]
The reversible and real versions of the power cycle are described by the Clausius Inequalty:
Reversible Unit
[tex]\frac{1000\,kJ - 600kJ}{300\,K}= 0[/tex]
Real Unit
[tex]\Delta S = \frac{1000\,kJ-600\,kJ}{300\,K} -\frac{1000\,kJ-700\,kJ}{300\,K}[/tex]
[tex]\Delta S = \frac{1}{3}\,\frac{kJ}{K}[/tex]
The cycle is irreversible.
The cycle efficiency using clausius inequality is;
σ_cycle = 0.333 kJ/kg and is internally irreversible
For the cycle, we know that efficiency is;η = 1 - Q_c/Q_h
Thus;
Q_c = (1 - η)Q_h
Now, the cycle efficiency is derived from the integral;σ_cycle = -∫(dQ/dt)ₐ
Thus; σ_cycle = -[(Q_h/T_h) - (Q_c/T_c)]
We are given;
Q_h = 1000 kJ
T_h = 500 k
T_c = 300 k
Q_c = 700 kJ
Thus;σ_cycle = -[(1000/500) - (700/300)]
σ_cycle = -(2 - 2.333)
σ_cycle = 0.333 kJ/kg
Since σ_cycle > 0, then the cycle is internally irreversible
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A 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 measured that the heat given off during the process to the cooling system of the reaction was 47.5 MJ/kg of fuel. a) Determine the fuel formula. b) What does that heat given-off to the cooling system of the reactor represent? c) Determine the lower heating value of the fuel. d) Determine the enthalpy of formation of the fuel at 25°C
Answers
Answer:
(a) Fuel formula = C₁₀H₂₀
(b)The reactor represent higher heating value (HHV) of the fuel because the water in the product is liquid.
(c) Lower heating value = 44.36 mj/kg
(d) - 6669.8 KJ/Kmol
Explanation:
See the attached files for the calculation
The elastic settlement of an isolated single pile under a working load similar to that of piles in the group it represents, is predicted to be 0.25 inches. What is the expected settlement for the pile group given the following information?
1.Group: 16 piles in a 4x4 group
2.Pile Diameter: 12 inches
3.Pile Center to Center Spacing: 3 feet
Answers
Answer:
The expected settlement for the pile group using the given information is 19.92mm or 0.79 inch
Explanation:
In this question, we are asked to calculate the expected settlement for the pole group given some information.
Please check attachment for complete solution and step by step explanation
The diffusion of a molecule in a tissue is studied by measuring the uptake of labeled protein into the tissue of thickness L =8 um. Initially, there is no labeled protein in the tissue. At t=0, the tissue is placed in a solution with a molecular concentration of C1=9.4 uM, so the surface concentration at x=0 is maintained at C1. Assume the tissue can be treated as a semi-infinite medium. Surface area of the tissue is A = 4.2 cm2. Calculate the flux into the tissue at x=0 and t=5 s.Please give your answer with a unit of umol/cm2 s. Assuming the diffusion coefficient is known of D=1*10-9 cm2/s
Answers
Answer:
The flux into the tissue at x=0 and t=5 s is 0.4476 uM/cm²s
Explanation:
Flux = [tex]\frac{Quantity}{Area * Time}[/tex]
given:
Area = 4.2 cm²
Time = 5 sec
Quantity ( concentration) = 9.4 uM
∴ Flux = Quantity / (Area × Time)
Flux = 9.4 uM / (4.2 cm² × 5 s)
Flux = 9.4 uM / 21 cm²s
Flux = 0.4476 uM/cm²s
The steel bar has a 20 x 10 mm rectangular cross section and is welded along section a-a. The weld material has a tensile yield strength of 325 MPa and a shear yield strength of 200 MPa, and the bar material has a tensile yield strength of 350 MPa. An overall factor of safety of at least 2.0 is required. Find the largest load P that can be applied, to satisfy all criteria.
Answers
The largest load P that can be applied to the steel bar, ensuring a factor of safety of 2.0 with respect to the weld material's tensile yield strength, is 32.5 kN. This is calculated based on the allowable tensile strength of the weld after applying the factor of safety and the cross-sectional area of the steel bar.
Explanation:The student's question relates to the structural engineering concept of material strength and loading. The maximum load P that can be safely applied to a welded steel bar can be determined by considering the tensile and shear yield strengths of the two materials involved, i.e., the weld material and the bar material. Using the given factor of safety of 2.0, the tensile strength of the weld and the bar, and the cross section of the bar, we can calculate the allowable tensile stress and then the maximum load P by dividing the allowable tensile stress by the area of the cross section.
To begin with, we find the allowable tensile strength by dividing the weld material's yield strength (which is the weaker material concerning tensile strength) by the factor of safety:
Allowable tensile strength for weld = 325 MPa / 2 = 162.5 MPaNext, we need to calculate the area of the cross-section of the bar:
Area (A) = 20 mm x 10 mm = 200 mm² = 200 x 10⁻⁶ m²Now, we convert the units of the allowable tensile strength to N/m² (because 1 MPa = 1 x 10⁶ N/m²) and calculate the maximum tensile load (Ptensile) that the weld can sustain:
Allowable tensile strength for weld in N/m² = 162.5 x 10⁶ N/m²Ptensile = Allowable tensile strength for weld x AreaPtensile = (162.5 x 10⁶ N/m²) x (200 x 10⁻⁶ m²)Ptensile = 32.5 x 10³ N = 32.5 kNThus, the largest load P that can be applied to the steel bar to satisfy the safety criteria is 32.5 kN.
Steel (AISI 1010) plates of thickness δ = 6 mm and length L = 1 m on a side are conveyed from a heat treatment process and are concurrently cooled by atmospheric air of velocity u[infinity] = 10 m/s and T[infinity] = 20°C in parallel flow over the plates. For an initial plate temperature of Ti = 300°C, what is the rate of heat transfer from the plate? What is the corresponding rate of change of the plate temperature? The velocity of the air is much larger than that of the plate.
Answers
Answer:
Rate of heat transfer from plate
6796.16 W
Corresponding rate of change of plate temperature
-2634 degrees.Celcius/sec
Explanation:
In this question, we are asked to calculate the rate of heat transfer and the corresponding rate of change of the plate temperature.
Please check attachment for complete solution and step by step explanation
Locate the centroid y¯ of the composite area. Express your answer to three significant figures and include the appropriate units. y¯ = nothing nothing Request Answer Part B Determine the moment of intertia of this area about the centroidal x′ axis. Express your answer to three significant figures and include the appropriate units. Ix′ = nothing nothing Request Answer Provide Feedback
Answers
Answer:
Please see the attached Picture for the complete answer.
Explanation:
A certain process requires 2.0 cfs of water to be delivered at a pressure of 30 psi. This water comes from a large-diameter supply main in which the pressure remains at 60 psi. If the galvanized iron pipe connecting the two locations is 200 ft long and contains six threaded 90o elbows, determine the pipe diameter. Elevation differences are negligible.
Answers
Answer:
determine the pipe diameter. = 0.41ft
Explanation:
check the attached file for answer explanation
Answer: you need to do it on your own first
Explanation:
7. Implement a function factorial in RISC-V that has a single integer parameter n and returns n!. A stub of this function can be found in the file factorial.s. You will only need to add instructions under the factorial label, and the argument that is passed into the function is configured to be located at the label n. You may solve this problem using either recursion or iteration
Answers
Answer:
addi x31, x0, 4
addi x30, x0, 2
Explanation:
Recursion in computer sciencs is defined as a method of solving a problem in which the solution to the problem depends on solutions to smaller cases of the same problem. Such problems can generally be solved by iteration, but this needs to identify and index the smaller cases at time of programming.
addi x31, x0, 4
addi x30, x0, 2
addi x2, x0, 1600 // initialize the stack to 1600, x2= stackpointer
ecall x5, x0, 5 // read the input to x5
jal x1, rec_func
ecall x0, x10, 2 // print the result now
beq x0, x0, end
rec_func:
addi x2, x2, -8 // make room in stack
sd x1, 0(x2) // store pointer and result in stack
bge x5, x31, true // if i > 3, then go to true branch
ld x1, 0(x2)
addi x10, x0, 1 // if i <= 3, then return 1
addi x2, x2, 8 // reset stack point
jalr x0, 0(x1)
true:
addi x5, x5, -2 // compute i-2
jal x1, rec_func // call recursive func for i-2
ld x1, 0(x2) // load the return address
addi x2, x2, 8 // reset stack point
mul x10, x10, x30 // multiply by 2
addi x10, x10, 1 // add 1
jalr x0, 0(x1) // return
end:
The vehicle motor or engine can be size based on required peak power. The energy or average power required provides a sense for how much fuel is required.
1. Calculate the average and peak power (kW) needed to accelerate a 1364 kg vehicle from 0 to 60 mph in 6 seconds. Assume that aerodynamic, rolling, and hill‐climbing force counts for an extra 10% of the needed acceleration force.
2. Chart the average and peak power (kW) vs. time duration from 2 to 15 seconds.
3. How much energy (kWh) is required to accelerate the vehicle?
Answers
Answer:
1. Parg = 89.954 kw
Pmax = 179.908 kw
2. Parg = 29.984 kw
Pmax = 59.96 kw
3. Energy = 0.15 kWh
Explanation:
See the attached file for the calculation
Answer:
1) P = 81.74 kW
2) As seen on the pic.
3) E = 0.1362 kWh
Explanation:
1) Given
m = 1364 kg (mass of the vehicle)
vi = 0 mph = 0 m/s (initial speed)
vf = 60 mph = (60 mph)(1609 m/ 1 mi)(1 h/ 3600 s) = 26.8167 m/s (final speed)
t = 6 s
We get the acceleration as follows
a = (vf - vi)/t ⇒ a = (26.8167 m/s - 0 m/s)/ 6s
⇒ a = 4.469 m/s²
then the Force is
F = m*a ⇒ F = 1364 kg*4.469 m/s²
⇒ F = 6096.32 N
Then we get the average power as follows
P = F*v(avg) = F*(vi + vf)/2
⇒ P = 6096.32 N*(0 m/s + 26.8167 m/s)/2
⇒ P = 81741.59 W = 81.74 kW
Knowing that that aerodynamic, rolling, and hill‐climbing force counts for an extra 10% of the needed acceleration force, we use the following formula to find the peak power
Pmax = (1 + 0.1)*F*vf = 1.1*F*vf
⇒ Pmax = 1.1*6096.32 N*26.8167 m/s
⇒ Pmax = 179831.503 W = 179.83 kW
2) The pic 1 shows the average and peak power (kW) vs. time duration from 2 to 15 seconds.
In the first chart we use the equation
Pinst = F*v where F is constant and v is the instantaneous speed, and Pavg is the mean of the values.
In the second chart, we use the equation
Ppeak = 1.1*F*v where F is constant and v is the instantaneous speed.
3) For 0 s ≤ t ≤ 6 s
We can use the equation
E = ΔK = Kf - Ki = 0.5*m*(vf² - vi²)
⇒ E = 0.5*1364 kg*((26.8167 m/s)² - (0 m/s)²)
⇒ E = 490449.123 J = (490449.123 J)(1 kWh/3.6*10⁶J)
⇒ E = 0.1362 kWh
When your complex reaction time is compromised by alcohol, an impaired person's ability to respond to emergency or unanticipated situations is greatly______.
Answers
Answer:
decreased
Explanation:
when impaired you react slower then you would sober.
Alcohol adversely affects the complex reaction time, considerably decreasing the individual’s ability to respond swiftly and adequately in emergencies or unexpected situations. This impairment is attributed to alcohol's impact on the brain causing slow information processing, poor motor control, and a decrease in focus.
Explanation:When a person's complex reaction time is compromised by alcohol, their ability to respond to unexpected situations or emergencies is greatly diminished. Alcohol's impact on the brain leads to slower processing of information, reduced concentration, and poorer motor control. As a result, they may not react as quickly or efficiently as they would if they were sober to changes in their environment. For example, if a situation arises that requires quick decision-making, such as stopping abruptly while driving to avoid a pedestrian, an intoxicated individual may not respond in time, leading to catastrophic outcomes.
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An air-conditioning system is used to maintain a house at 70°F when the temperature outside is 100°F. The house is gaining heat through the walls and the windows at a rate of 800 Btu/min, and the heat generation rate within the house from people, lights, and appliances amounts to 100 Btu/min. Determine the minimum power input required for this air- conditioning system.
Answers
Answer:
Minimum power output required = 1.1977 hp
Explanation:
Given Data:
Temperature outside = 100°F.
House temperature = 70°F
Rate of heat gain(Qw) = 800 Btu/min
Generation rate within(Ql) = 100 Btu/min.
Converting the outside temperature 100°F from fahrenheit to ranking, we have;
1°F = 460R
Therefore,
100°F = 460+100
To = 560 R
Converting the house temperature 70°F from fahrenheit to ranking, we have;
1°F = 460R
Therefore,
70°F = 460+70
Th = 530 R
Consider the equation for coefficient of performance (COP) of refrigerator in terms of temperature;
COP =Th/(To-Th)
= 530/(560-530)
= 530/30
= 17.66
Consider the equation for coefficient of performance (COP) of refrigerator;
COP = Desired output/required input
= Q/Wnet
= Ql + Qw/ Wnet
Substituting into the formula, we have;
17.667 = (100 + 800)/Wnet
17.667 = 900/Wnet
Wnet = 900/17.667
= 50.94 Btu/min.
Converting from Btu/min. to hp, we have;
1 hp = 42.53 Btu/min.
Therefore,
50.94 Btu/min = 50.94 / 42.53
= 1.1977 hp =
Therefore, minimum power output required = 1.1977 hp
Given the following data:
Outside temperature = 100°F.House temperature = 70°F.Rate of heat gain = 800 Btu/min.Heat generation rate = 100 Btu/min.The conversion of temperature.We would convert the value of the temperatures in Fahrenheit to Rankine.
Note: 1°F = 460R
Conversion:
Outside temperature = 100°F = [tex]460+100[/tex] = 560RHouse temperature = 70°F = [tex]460+70[/tex] = 530RTo calculate the minimum power input that is required for this air- conditioning system:
The coefficient of performance (COP)In Science, the coefficient of performance (COP) is a mathematical expression that is used to show the relationship between the power output of an air-conditioning system and the power input of its compressor.
Mathematically, the coefficient of performance (COP) is given by the formula:
[tex]COP =\frac{T_h}{T_o-T_h}[/tex]
Substituting the given parameters into the formula, we have;
[tex]COP =\frac{530}{560-530}\\ \\ COP =\frac{530}{30}[/tex]
COP = 17.66
For the power input:
[tex]COP = \frac{E_o}{E_i} \\ \\ COP = \frac{Q_l + Q_w}{Q_{net}}\\ \\ 17.67 = \frac{100+800}{Q_{net}}\\ \\ Q_{net}=\frac{900}{17.67} \\ \\ Q_{net}=50.93\;Btu/min[/tex]
Conversion:
1 hp = 42.53 Btu/min.
X hp = 50.93 Btu/min.
Cross-multiplying, we have:
[tex]X=\frac{50.93}{42.53} [/tex]
X = 1.1975 hp
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Air enters a compressor operating at steady state at 1.05 bar, 300 K, with a volumetric flow rate of 39 m3/min and exits at 12 bar, 400 K. Heat transfer occurs at a rate of 6.5 kW from the compressor to its surroundings. Assuming the ideal gas model for air and neglecting kinetic and potential energy effects, determine the power input, in kW.
Answers
Answer:
The power input, in kW is -86.396 kW
Explanation:
Given;
initial pressure, P₁ = 1.05 bar
final pressure, P₂ = 12 bar
initial temperature, T₁ = 300 K
final temperature, T₂ = 400 K
Heat transfer, Q = 6.5 kW
volumetric flow rate, V = 39 m³/min = 0.65 m³/s
mass of air, m = 28.97 kg/mol
gas constant, R = 8.314 kJ/mol.k
R' = R/m
R' = 8.314 /28.97 = 0.28699 kJ/kg.K
Step 1:
Determine the specific volume:
p₁v₁ = RT₁
[tex]v_1 = \frac{R'T_1}{p_1} = \frac{(0.28699.\frac{kJ}{kg.K} )(300 k)}{(1.05 bar *\ \frac{10^5 N/m^2}{1 bar} *\frac{1kJ}{1000N.m} )} \\\\v_1 = 0.81997 \ m^3/kg[/tex]
Step 2:
determine the mass flow rate; m' = V / v₁
mass flow rate, m' = 0.65 / 0.81997
mass flow rate, m' = 0.7927 kg/s
Step 3:
using steam table, we determine enthalpy change;
h₁ at T₁ = 300.19 kJ /kg
h₂ at T₂ = 400.98 kJ/kg
Δh = h₂ - h₁
Δh = 400.98 - 300.19
Δh = 100.79 kJ/kg
step 4:
determine work input;
W = Q - mΔh
Where;
Q is heat transfer = - 6.5 kW, because heat is lost to surrounding
W = (-6.5) - (0.7927 x 100.79)
W = -6.5 -79.896
W = -86.396 kW
Therefore, the power input, in kW is -86.396 kW
Consider an aluminum step shaft. The area of section AB and BC as well as CD are 0.1 inch2 , 0.15 inch 2 and 0.20 inch2. The length of section AB, BC and CD are 10 inch, 12 inch and 16 inch. A force F=1000 lbf is applied to B. The initial gap between D and rigid wall is 0.002. Using analytical approach, determine the wall reactions, the internal forces in members, and the displacement of B and C. Find the strain in AB, BC and CD.
Answers
Answer:
Explanation:
Find attach the solution
Air as an ideal gas in a closed system undergoes a reversible process between temperatures of 1000 K and 400 K. The beginning pressure is 200 bar. Determinc the highest possible ending pressure for this process.If the ending pressure is 3 bar, determine the heat transfer and work per unit mass, if the boundary of the system is in constant contact with a reservoir at 400 K.
Answers
Answer:
highest possible ending pressure for this process is 8.0954 bar
Explanation:
We can say that Heat transfer is any or all of several kinds of phenomena, considered as mechanisms, that convey energy and entropy from one location to another. The specific mechanisms are usually referred to as convection, thermal radiation, and conduction.
Please see attachment for the solution.
Air is compressed from 100 kPa, 300 K to 1000 kPa in a two-stage compressor with intercooling between stages. The intercooler pressure is 300 kPa. The air is cooled back to 300 K in the intercooler before entering the second compressor stage. Each compressor stage is isentropic. Please calculate the total compressor work per unit of mass flow (kJ/kg). Repeat for a single stage of isentropic compression from the given inlet to the final pressure.
Answers
Answer:
The total compressor work is 234.8 kJ/kg for a isentropic compression
Explanation:
Please look at the solution in the attached Word file
The total compressor work per unit of mass flow in a two-stage compressor with intercooling can be calculated by summing the compressor work in each stage and the work done in the intercooler.
Explanation:The total compressor work per unit of mass flow in a two-stage compressor with intercooling can be calculated by summing the compressor work in each stage and the work done in the intercooler.
In the first stage, the air is compressed from 100 kPa to 300 kPa. The work done in this stage can be calculated using the isentropic compression process. In the second stage, the air is further compressed from 300 kPa to 1000 kPa. The work done in this stage can also be calculated using the isentropic compression process.
To calculate the total compressor work per unit of mass flow, you need to sum the work done in each stage and the work done in the intercooler.
5.3-16 A professor recently received an unexpected $10 (a futile bribe attached to a test). Being the savvy investor that she is, the professor decides to invest the $10 into a savings account that earns 0.5% interest compounded monthly (6.17% APY). Furthermore, she decides to supplement this initial investment with an additional $5 deposit made every month, beginning the month immediately following her initial investment.
(a) Model the professor's savings account as a constant coefficient linear difference equation. Designate yln] as the account balance at month n, where n corresponds to the first month that interest is awarded (and that her $5 deposits begin).
(b) Determine a closed-form solution for y[n] That is, you should express yIn] as a function only of n.
(c) If we consider the professor's bank account as a system, what is the system impulse response h[n]? What is the system transfer function Hz]?
(d) Explain this fact: if the input to the professor's bank account is the everlasting exponential xn] 1 is not y[n] I"H[I]-HII]. 1, then the output
Answers
Answer:
a) y (n + 1) = 1.005 y(n) + 5U n
y (n + 1) - 1.005 y(n) = 5U (n)
b) Z^-1(Z(y0)=y(n) = [1010(1.005)^n - 1000(1)^n] U(n)
c) h(n) = (1.005)^n U(n - 1) + 10(1.005)^n U(n)
Explanation:
Her bank account can be modeled as:
y (n + 1) = y (n) + 0.5% y(n) + $5
y (n + 1) = 1.005 y(n) + 5U n
Given that y (0) = $10
y (n + 1) - 1.005 y(n) = 5U (n)
Apply Z transform on both sides
= ZY ((Z) - Z(y0) - 1.005) Z = 5 U (Z)
U(Z) = Z {U(n)} = Z/ Z - 1
Y(Z) [Z- 1.005] = Z y(0) + 5Z/ Z - 1
= 10Z/ Z - 1.005 + 5Z/(Z - 1) (Z - 1.005)
Y(Z) = 10Z/ Z - 1.005 + 1000Z/ Z - 1.005 + 1000Z/ Z - 1
= 1010Z/Z- 1.005 - 1000Z/Z-1
Apply inverse Z transform
Z^-1(Z(y0)) = y(n) = [1010(1.005)^n - 1000(1)^n] U(n)
Impulse response in output when input f(n) = S(n)
That is,
y(n + 1)= 1. 005y (n) + 8n
y(n + 1) - 1.005y (n) = 8n
Apply Z transform
ZY (Z) - Z(y0) - 1.005y(Z) = 1
HZ (Z - 1.005) = 1 + 10Z [Therefore y(Z) = H(Z)]
H(Z) = 1/ Z - 1.005 + 10Z/Z - 1. 005
Apply inverse laplace transform
= h(n) = (1.005)^n U(n - 1) + 10(1.005)^n U(n)