An eddy current separator is to separate aluminum product from an input streamshredded MSW. The feed rate to the separator is 2,500 kg/hr. The feed is known to contain174 kg of aluminum and 2,326 kg of reject. After operating for 1 hour, a total of 256 kg ofmaterials is collected in the product stream. On close inspection, it is found that 140 kg ofproduct is aluminum. Estimate the % recovery of aluminum product and the % purity of thealuminum produc
Answer:
the % recovery of aluminum product is 80.5%
the % purity of the aluminum product is 54.7%
Explanation:
feed rate to separator = 2500 kg/hr
in one hour, there will be 2500 kg/hr x 1 hr = 2500 kg of material is fed into the machine
of this 2500 kg, the feed is known to contain 174 kg of aluminium and 2326 kg of rejects.
After the separation, 256 kg is collected in the product stream.
of this 256 kg, 140 kg is aluminium.
% recovery of aluminium will be = mass of aluminium in material collected in the product stream ÷ mass of aluminium contained in the feed material
% recovery of aluminium = 140kg/174kg x 100% = 80.5%
% purity of the aluminium product = mass of aluminium in final product ÷ total mass of product collected in product stream
% purity of the aluminium product = 140kg/256kg
x 100% = 54.7%
The guy wires AB and AC are attached to the top of the transmission tower. The tension in cable AB is 8.7 kN. Determine the required tension T in cable AC such that the net effect of the two cables is a downward force at point A. Determine the magnitude R of this downward force.
Answer:
[tex] T_A_C = 6.296 kN [/tex]
[tex] R = 10.06 kN [/tex]
Explanation:
Given:
[tex] T_A_B = 8.7 kN[/tex]
Required:
Find the tension TAC and magnitude R of this downward force.
First calculate [tex] \alpha, \beta, \gamma [/tex]
[tex] \alpha = tan^-^1 =\frac{40}{50} = 38. 36 [/tex]
[tex] \beta = tan^-^1 =\frac{50}{30} = 59.04 [/tex]
[tex] \gamma = 180 - 38.36 - 59.04 = 82.6 [/tex]
To Find tension in AC and magnitude R, use sine rule.
[tex] \frac{sin a}{T_A_C} = \frac{sin b}{T_A_B} = \frac{sin c}{R} [/tex]
Substitute values:
[tex]\frac{sin 38.36}{T_A_C} = \frac{sin 59.04}{8.7} = \frac{82.6}{R}[/tex]
Solve for T_A_C:
[tex] T_A_C = 8.7 * \frac{sin 38.36}{sin 59.04} = [/tex]
[tex] T_A_C = 8.7 * 0.724 = 6.296 kN [/tex]
Solve for R.
[tex] R = 8.7 * \frac{sin 82.6}{sin 59.04} = [/tex]
[tex] R = 8.7 * 1.156 [/tex]
R = 10.06 kN
Tension AC = 6.296kN
Magnitude,R = 10.06 kN
Choose two consumer services careers and research online to determine what kinds of professional organizations exist for these professions. Write a paragraph describing the purpose of the organization, the requirements for joining, and the benefits of membership.
Bank loan facilitator, and hospital emergency care specialist are the two consumer or customer services careers.
Bank loan facilitator is a consumer service facilitator who ask and provide people loan in emergency, for the purpose of education, treatment, family events, and for other reasons. For bank loan facilitator the professional organizations should be banking and finance sector. The purpose of these organizations is to help people in financial matter seeking benefit by getting interest from customers. The requirements for joining of the employee must include strong convincing power for the employee, time management, strong and tactful communication skills. Benefits of membership of the customers can help them to seek loans on need basis on lower interest. Hospital emergency care specialist provides help to the staff and the customers in medical emergency. These professionals are necessary for the hospital, clinics, and rehabilitation centers. Purpose of the organization is to provide medical care to the patients. The requirements for joining of the employee includes ability to give information to patients and staff during emergency conditions, facilitating ambulance to rescue patients from their homes, and from other areas, providing medicine, medical equipment, and other facilities to the patients and other medical staff necessary for treatment. Benefits of membership in clinical or hospital settings can help the patient in frequent visits for treatment, concession in laboratory tests, and medication.Learn more about customer:
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A non-inductive load takes a current of 15 A at 125 V. An inductor is then connected in series in order that the same current shall be supplied from 240 V, 50 Hz mains. Ignore the resistance of the inductor and calculate: i. the inductance of the inductor; ii. the impedance of the circuit; iii. the phase difference between the current and the applied voltage.
Answer:
(i) The inductance of the inductor is = 43.43 mH (ii) the impedance of the circuit is = 16∠58.61° Ω (iii) the phase difference for current and the voltage applied is Q = 58.61°
Explanation:
Solution
Given that:
I= 5 A
V = 125V
Resistance R= Not known yet
Thus
To find the resistance we have the following formula which is shown below:
R = V/I
=125/15
R =8.333Ω
Now,
Voltage = 240
Frequency = 50Hz
Current (I) remain at = 15A
Z= not known (impedance)
so,
To find the impedance we have the formula which is shown below:
Z = V/I =240/15
Z= 16Ω⇒ Z = R + jXL
Z = 8.333 + jXL = 16
Thus
√8.333² + XL² = 16²
8.333² + XL² = 16²
XL² = 186.561
XL = 13.658Ω
Now
We find the inductance of the Inductor and the impedance of the circuit.
(i) In solving for the inductance of the inductor, a formula is applied here, which is shown below:
L = XL/w
=13.658/ 2π * 50
=13.658/314.15 = 0.043 = 43.43 mH
Note: w= 2πf
(ii) For the impedance of the circuit we have the following:
z = 8.333 + j 13.658
z = 16∠58.61° Ω
(iii) The next step is to find the phase difference between the applied voltage and current.
Q = this is the voltage across the inductor in a series of resonant circuit.
Q can also be called the applied voltage
Thus,
Q is described as an Impedance angle
Therefore, Q = 58.81°
An isentropic steam turbine processes 5.5 kg/s of steam at 3 MPa, which is exhausted at 50 kPa and 100°C. Five percent of this flow is diverted for feedwater heating at 500 kPa. Determine the power produced by this turbine. Use steam tables.
Answer:
The answer is 1823.9
Explanation:
Solution
Given that:
m = 5.5 kg/s
= m₁ = m₂ = m₃
The work carried out by the energy balance is given as follows:
m₁h₁ = m₂h₂ +m₃h₃ + w
Now,
By applying the steam table we have that<
p₃ = 50 kPa
T₃ = 100°C
Which is
h₃ = 2682.4 kJ/KJ
s₃ = 7.6953 kJ/kgK
Since it is an isentropic process:
Then,
p₂ = 500 kPa
s₂=s₃ = 7.6953 kJ/kgK
which is
h₂ =3207.21 kJ/KgK
p₁ = 3HP0
s₁ = s₂=s₃ = 7.6953 kJ/kgK
h₁ =3854.85 kJ/kg
Thus,
Since 5 % of this flow diverted to p₂ = 500 kPa
Then
w =m (h₁-0.05 h₂ -0.95 )h₃
5.5(3854.85 - 0.05 * 3207.21 - 0.95 * 2682.4)
5.5( 3854.83 * 3207.21 - 0.95 * 2682.4)
5.5 ( 123363249.32 -0.95 * 2682.4)
w=1823.9
Find the largest number. The process of finding the maximum value (i.e., the largest of a group of values) is used frequently in computer applications. For example, an app that determines the winner of a sales contest would input the number of units sold by each salesperson. The sales person who sells the most units wins the contest. Write pseudocode, then a C# app that inputs a series of 10 integers, then determines and displays the largest integer. Your app should use at least the following three variables:
Counter: Acounter to count to 10 (i.e., to keep track of how many nimbers have been input and to determine when all 10 numbers have been processed).
Number: The integer most recently input by the user.
Largest: The largest number found so far.
Answer:
See Explanation
Explanation:
Required
- Pseudocode to determine the largest of 10 numbers
- C# program to determine the largest of 10 numbers
The pseudocode and program makes use of a 1 dimensional array to accept input for the 10 numbers;
The largest of the 10 numbers is then saved in variable Largest and printed afterwards.
Pseudocode (Number lines are used for indentation to illustrate the program flow)
1. Start:
2. Declare Number as 1 dimensional array of 10 integers
3. Initialize: counter = 0
4. Do:
4.1 Display “Enter Number ”+(counter + 1)
4.2 Accept input for Number[counter]
4.3 While counter < 10
5. Initialize: Largest = Number[0]
6. Loop: i = 0 to 10
6.1 if Largest < Number[i] Then
6.2 Largest = Number[i]
6.3 End Loop:
7. Display “The largest input is “+Largest
8. Stop
C# Program (Console)
Comments are used for explanatory purpose
using System;
namespace ConsoleApplication1
{
class Program
{
static void Main(string[] args)
{
int[] Number = new int[10]; // Declare array of 10 elements
//Accept Input
int counter = 0;
while(counter<10)
{
Console.WriteLine("Enter Number " + (counter + 1)+": ");
string var = Console.ReadLine();
Number[counter] = Convert.ToInt32(var);
counter++;
}
//Initialize largest to first element of the array
int Largest = Number[0];
//Determine Largest
for(int i=0;i<10;i++)
{
if(Largest < Number[i])
{
Largest = Number[i];
}
}
//Print Largest
Console.WriteLine("The largest input is "+ Largest);
Console.ReadLine();
}
}
}
A walrus loses heat by conduction through its blubber at the rate of 220 W when immersed in −1.00°C water. Its internal core temperature is 37.0°C, and it has a surface area of 2.23 m2. What is the average thickness of its blubber? The conductivity of fatty tissue without blood is 0.20 (J/s · m · °C).
Answer:
The average thickness of the blubber is 0.077 m
Explanation:
Here, we want to calculate the average thickness of the Walrus blubber.
We employ a mathematical formula to calculate this;
The rate of heat transfer(H) through the Walrus blubber = dQ/dT = KA(T2-T1)/L
Where dQ is the change in amount of heat transferred
dT is the temperature gradient(change in temperature) i.e T2-T1
dQ/dT = 220 W
K is the conductivity of fatty tissue without blood = 0.20 (J/s · m · °C)
A is the surface area which is 2.23 m^2
T2 = 37.0 °C
T1 = -1.0 °C
L is ?
We can rewrite the equation in terms of L as follows;
L × dQ/dT = KA(T2-T1)
L = KA(T2-T1) ÷ dQ/dT
Imputing the values listed above;
L = (0.2 * 2.23)(37-(-1))/220
L = (0.2 * 2.23 * 38)/220 = 16.948/220 = 0.077 m
Under normal operating conditions, the electric motor exerts a torque of 2.8 kN-m.on shaft AB. Knowing that each shaft is solid, determine the maximum shearing stress in a) shaft AB b) shaft BC c) shaft CD (25 points) Given that the torque at B
Answer:
Explanation:
The image attached to the question is shown in the first diagram below.
From the diagram given ; we can deduce a free body diagram which will aid us in solving the question.
IF we take a look at the second diagram attached below ; we will have a clear understanding of what the free body diagram of the system looks like :
From the diagram; we can determine the length of BC by using pyhtagoras theorem;
SO;
[tex]L_{BC}^2 = L_{AB}^2 + L_{AC}^2[/tex]
[tex]L_{BC}^2 = (3.5+2.5)^2+ 4^2[/tex]
[tex]L_{BC}= \sqrt{(6)^2+ 4^2}[/tex]
[tex]L_{BC}= \sqrt{36+ 16}[/tex]
[tex]L_{BC}= \sqrt{52}[/tex]
[tex]L_{BC}= 7.2111 \ m[/tex]
The cross -sectional of the cable is calculated by the formula :
[tex]A = \dfrac{\pi}{4}d^2[/tex]
where d = 4mm
[tex]A = \dfrac{\pi}{4}(4 \ mm * \dfrac{1 \ m}{1000 \ mm})^2[/tex]
A = 1.26 × 10⁻⁵ m²
However, looking at the maximum deflection in length [tex]\delta[/tex] ; we can calculate for the force [tex]F_{BC[/tex] by using the formula:
[tex]\delta = \dfrac{F_{BC}L_{BC}}{AE}[/tex]
[tex]F_{BC} = \dfrac{ AE \ \delta}{L_{BC}}[/tex]
where ;
E = modulus elasticity
[tex]L_{BC}[/tex] = length of the cable
Replacing 1.26 × 10⁻⁵ m² for A; 200 × 10⁹ Pa for E ; 7.2111 m for [tex]L_{BC}[/tex] and 0.006 m for [tex]\delta[/tex] ; we have:
[tex]F_{BC} = \dfrac{1.26*10^{-5}*200*10^9*0.006}{7.2111}[/tex]
[tex]F_{BC} = 2096.76 \ N \\ \\ F_{BC} = 2.09676 \ kN[/tex] ---- (1)
Similarly; we can determine the force [tex]F_{BC}[/tex] using the allowable maximum stress; we have the following relation,
[tex]\sigma = \dfrac{F_{BC}}{A}[/tex]
[tex]{F_{BC}}= {A}*\sigma[/tex]
where;
[tex]\sigma =[/tex] maximum allowable stress
Replacing 190 × 10⁶ Pa for [tex]\sigma[/tex] ; we have :
[tex]{F_{BC}}= 1.26*10^{-5} * 190*10^{6} \\ \\ {F_{BC}}=2394 \ N \\ \\ {F_{BC}}= 2.394 \ kN[/tex] ------ (2)
Comparing (1) and (2)
The magnitude of the force [tex]F_{BC} = 2.09676 \ kN[/tex] since the elongation of the cable should not exceed 6mm
Finally applying the moment equilibrium condition about point A
[tex]\sum M_A = 0[/tex]
[tex]3.5 P - (6) ( \dfrac{4}{7.2111}F_{BC}) = 0[/tex]
[tex]3.5 P - 3.328 F_{BC} = 0[/tex]
[tex]3.5 P = 3.328 F_{BC}[/tex]
[tex]3.5 P = 3.328 *2.09676 \ kN[/tex]
[tex]P =\dfrac{ 3.328 *2.09676 \ kN}{3.5 }[/tex]
P = 1.9937 kN
Hence; the maximum load P that can be applied is 1.9937 kN
Two blocks of rubber (B) with a modulus of rigidity G = 14 MPa are bonded to rigid supports and to a rigid metal plate A. Knowing that c = 80 mm and P = 46 kN, determine the smallest allowable dimensions a and b of the blocks if the shearing stress in the rubber is not to exceed 1.4 MPa and the deflection of the plate is to be at least 7 mm.
Answer:
a = 0.07m or 70mm
b = 0.205m or 205mm
Explanation:
Given the following data;
Modulus of rigidity, G = 14MPa=14000000Pa.
c = 80mm = 0.08m.
P = 46kN=46000N.
Shearing stress (r) in the rubber shouldn't exceed 1.4MPa=1400000Pa.
Deflection (d) of the plate is to be at least 7mm = 0.007m.
From shearing strain;
[
[tex]Modulus Of Elasticity, E = \frac{d}{a} =\frac{r}{G}[/tex]
Making a the subject formula;
[tex]a = \frac{Gd}{r}[/tex]
Substituting into the above formula;
[tex]a = \frac{14000000*0.007}{1400000}[/tex]
[tex]a = \frac{98000}{1400000}[/tex]
[tex]a = 0.07m or 70mm[/tex]
a = 0.07m or 70mm.
Also, shearing stress;
[tex]r = \frac{P}{2bc}[/tex]
Making b the subject formula;
[tex]b = \frac{P}{2cr}[/tex]
Substituting into the above equation;
[tex]b = \frac{46000}{2*0.08*1400000}[/tex]
[tex]b = \frac{46000}{224000}[/tex]
[tex]b = 0.205m or 205mm[/tex]
b = 0.205m or 205mm
You are tasked with designing a thin-walled vessel to contain a pressurized gas. You are given the parameters that the inner diameter of the tank will be 60 inches and the tank wall thickness will be 5/8" (0.625 inches). The allowable circumferential (hoop) stress and longitudinal stresses cannot exceed 30 ksi.
(1) What is the maximum pressure that can be applied within the tank before failure? = psi(2) If you had the opportunity to construct a spherical tank having an inside diameter of 60 inches and a wall thickness of 5/8" (instead of the thin-walled cylindrical tank as described above), what is the maximum pressure that can be applied to the spherical tank? = psi
Answer:
Explanation:
For cylinder
Diameter d = 60 inches
thickness t = 0.625 inches
circumferential (hoop) stress = 30 ksi
[tex]hoop \ \ stress =\sigma_1=\frac{P_1d}{2t}\\\\\sigma_1=30ksi\\\\30000=\frac{P_1\times 60}{2\times0.625}\\\\P_1=624psi[/tex]
[tex]longitudinal \ \ stress =\sigma_2=\frac{P_2d}{2t}\\\\\sigma_2=30ksi\\\\30000=\frac{P_2\times 60}{4\times0.625}\\\\30000=\frac{P_2\times 60}{2.5}\\\\75000=P_2\times60\\\\P_2=\frac{75000}{60} \\\\P_1=1250psi[/tex]
Therefore maximum pressure without failure is P₁ = 625 psi
ii) For Sphere
[tex]\sigma_1=\sigma_2=\frac{Pd}{4t} \\\\P=\frac{30000\times 4 \times 0.625}{60} \\\\=\frac{75000}{60}\\\\=1250\ \ psi[/tex]
Scheduling can best be defined as the process used to determine:
Answer:
Overall project duration
Explanation:
Scheduling can best be defined as the process used to determine a overall project duration.
Solid spherical particles having a diameter of 0.090 mm and a density of 2002 kg/m3 are settling in a solution of water at 26.7C. The volume fraction of the solids in the water is 0.45. Calculate the settling velocity and the Reynolds number.
Answer:
Settling Velocity (Up)= 2.048*10^-5 m/s
Reynolds number Re = 2.159*10^-3
Explanation:
We proceed as follows;
Diameter of Particle = 0.09 mm = 0.09*10^-3 m
Solid Particle Density = 2002 kg/m3
Solid Fraction, θ= 0.45
Temperature = 26.7°C
Viscosity of water = 0.8509*10^-3 kg/ms
Density of water at 26.7 °C = 996.67 kg/m3
The velocity between the interface, i.e between the suspension and clear water is given by,
U = [ ((nf/ρf)/d)D^3] [18+(1/3)D^3)(1/2)]
D = d[(ρp/ρf)-1)g*(ρf/nf)^2]^(1/3)
D = 2.147
U = 0.0003m/s (n = 4.49)
Up = 0.0003 * (1-0.45)^4.49 = 2.048*10^-5 m/s
Re=0.09*10^-3*2.048*10^-5*996.67/0.0008509 = 2.159*10^-3
An amplifier which needs a high input resistance and a high output resistance is : Select one: a. A voltage amplifier b. None of these c. A transresistance amplifier d. A current amplifier e. A transconductance amplifier Clear my choice
Answer:
None of these
Explanation:
There are different types of amplifiers, and each has different characteristics.
Voltage amplifier needs high input and low output resistance.Current amplifier needs Low Input and High Output resistance.Trans-conductance amplifier Low Input and High Output resistance.Trans-Resistance amplifier requires High Input and Low output resistance.Therefore, the correct answer is "None of these "
Consider a series RC circuit at the left where C = 6 µ F, R = 2 MΩ, and ε = 20 V. You close the switch at t = 0. Find (a) the time constant for the circuit, (b) the half-life of the circuit, (c) the current at t = 0, (d) the voltage across the capacitor at t = 0, and (e) the voltage across the resistor after a very long time.
Answer:
(a) 12 seconds (b) t = 8.31 seconds (c) 10µ A (d) V = 20 V (e) V =0
Explanation:
Solution
Given that:
C = 6 µ which is = 6 * 10^ ⁻6
R = 2 MΩ, which is = 2 * 10^ 6
ε = 20 V
(a) When it is at the time constant we have the following:
λ = CR
= 6 * 10^ ⁻6 * 2 * 10^ 6
λ =12 seconds
(b) We solve for the half life of the circuit which is given below:
d₀ = d₀ [ 1- e ^ ⁺t/CR
d = decay mode]
d₀/2 = d₀ 1- e ^ ⁺t/12
2^⁻1 = e ^ ⁺t/12
Thus
t/12 ln 2
t = 12 * ln 2
t = 12 * 0.693
t = 8.31 seconds
(c) We find the current at t = 0
So,
I = d₀/dt
I = d₀/dt e ^ ⁺t/CR
= CE/CR e ^ ⁺t/CR
E/R e ^ ⁺t/CR
Thus,
at t = 0
I E/R = 20/ 2 * 10^ 6
= 10µ A
(d) We find the voltage across the capacitor at t = 0 which is shown below:
V = IR
= 10 * 10^ ⁻6 * 2 * 10^ 6
V = 20 V
(e) We solve for he voltage across the resistor.
At t = 0
I = 0
V =0
While having a discussion about O-rings at the bottom of filters, Technician A says that the Automotive Filter Manufacturers Council recommends that the filter O-ring be lubricated with oil after installing the filter. Technician B says that the filter O-ring should be lubricated before installation. Who is correct
Answer:
Technician B is correct
Explanation:
O- rings are used with oil transmission filters to avoid transmission failures but some people use lip seals as well. either of them is inserted onto the outer part of the transmission system i.e it is inserted/found in-between Transmission filters and the transmission systems and it main purpose is to avoid leaks and transmission failure in the short and long term.
0-rings should be lubricated before installation this is because the o-rings are usually super tight when installing and would require lubrication to ease the installation process else the rubber might get ruptured and this would lead to instant transmission failure.
cubical tank 1 meter on each edge is filled with water at 20 degrees C. A cubical pure copper block 0.46 meters on each edge with an initial temperature of 100 degrees C is quickly submerged in the water, causing an amount of water equal to the volume of the smaller cube to spill from the tank. An insulated cover is placed on the tank. The tank is adiabatic. Estimate the equilibrium temperature of the system (block + water). Be sure to state all applicable assumptions.
Answer:
final temperature = 26.5°
Explanation:
Initial volume of water is 1 x 1 x 1 = 1 [tex]m^{3}[/tex]
Initial temperature of water = 20° C
Density of water = 1000 kg/[tex]m^{3}[/tex]
volume of copper block = 0.46 x 0.46 x 0.46 = 0.097 [tex]m^{3}[/tex]
Initial temperature of copper block = 100° C
Density of copper = 8960 kg/[tex]m^{3}[/tex]
Final volume of water = 1 - 0.097 = 0.903 [tex]m^{3}[/tex]
Assumptions:
since tank is adiabatic, there's no heat gain or loss through the wallsthe tank is perfectly full, leaving no room for cooling airtotal heat energy within the tank will be the summation of the heat energy of the copper and the water remaining in the tank.mass of water remaining in the tank will be density x volume = 1000 x 0.903 = 903 kg
specific heat capacity of water c = 4186 J/K-kg
heat content of water left Hw = mcT = 903 x 4186 x 20 = 75.59 Mega-joules
mass of copper will be density x volume = 8960 x 0.097 = 869.12 kg
specific heat capacity of copper is 385 J/K-kg
heat content of copper Hc = mcT = 869.12 x 385 x 100 = 33.46 Mega-joules
total heat in the system = 75.59 + 33.46 = 109.05 Mega-joules
this heat will be distributed in the entire system
heat energy of water within the system = mcT
where T is the final temperature
= 903 x 4186 x T = 3779958T
for copper, heat will be
mcT = 869.12 x 385 = 334611.2T
these component heats will sum up to the final heat of the system, i.e
3779958T + 334611.2T = 109.05 x [tex]10^{6}[/tex]
4114569.2T = 109.05 x [tex]10^{6}[/tex]
final temperature T = (109.05 x [tex]10^{6}[/tex])/4114569.2 = 26.5°
A spherical tank for storing gas under pressure is 25 m in diameter and is made of steel 15 mm thick. The yield point of the material is 240 MPa. A factor of safety of 2.5 is desired. The maximum permissible internal pressure is most nearly: 90 kPa 230 kPa 430 kPa D. 570 kPa csauteol psotolem here Pcr 8. A structural steel tube with a 203 mm x 203 mm square cross section has an average wall thickness of 6.35 mm. The tube resists a torque of 8 N m. The average shear flow is most nearly
A. 100 N/m
B. 200 N/m
C. 400 N/m
D. 800 N/m
Answer:
1) 2304 kPa
2) B. 200 N/m
Explanation:
The internal pressure of the of the tank can be found from the following relations;
Resisting wall force F = p×(1/4·π·D²)
σ×A = p×(1/4·π·D²)
Where:
σ = Allowable stress of the tank
A = Area of the wall of the tank = π·D·t
t = Thickness of the tank = 15 mm. = 0.015 m
D = Diameter of the tank = 25 m
p = Maximum permissible internal pressure pressure
∴ σ×π·D·t = p×(1/4·π·D²)
p = 4×σ×t/D = 4 × 240 ×0.015/2.5 = 5.76 MPa
With a desired safety factor of 2.5, the permissible internal pressure = 5.76/2.5 = 2.304 MPa
2) The formula for average shear flow is given as follows;
[tex]q = \dfrac{T}{2 \times A_m}[/tex]
Where:
q = Average shear flow
T = Torque = 8 N·m
[tex]A_m[/tex] = Average area enclosed within tube
t = Thickness of tube = 6.35 mm = 0.00635 m
Side length of the square cross sectioned tube, s = 203 mm = 0.203 m
Average area enclosed within tube, [tex]A_m[/tex] = (s - t)² = (0.203 - 0.00635)² = 0.039 m²
[tex]\therefore q = \dfrac{8}{2 \times 0.039} = 206.9 \, N/m[/tex]
Hence the average shear flow is most nearly 200 N/m.
Following are the solution to the given question:
Calculating the allowable stress:
[tex]\to \sigma_{allow} = \frac{\sigma_y}{FS} \\\\[/tex]
[tex]= \frac{240}{2.5} \\\\= 96\\\\[/tex]
Calculating the Thickness:
[tex]\to t =15\ mm = \frac{15\ }{1000}= 0.015\ m\\\\[/tex]
The stress in a spherical tank is defined as
[tex]\to \sigma = \frac{pD}{4t}\\\\\to 96 = \frac{p(25)}{4(0.015)}\\\\\to p = 0.2304\;\;MPa\\\\\to p = 230.4\;\;kPa\\\\\to p \approx 230\;\;kPa\\\\[/tex]
[tex]\bold{\to A= 203^2= 41209\ mm^2} \\\\[/tex]
Calculating the shear flow:
[tex]\to q=\frac{T}{2A}[/tex]
[tex]=\frac{8}{2 \times 41209 \times 10^{-6}}\\\\=\frac{8}{0.082418}\\\\=97.066\\\\[/tex]
[tex]\to q=97 \approx 100 \ \frac{N}{m}\\[/tex]
Therefore, the final answer is "".
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An Ideal gas is being heated in a circular duct as while flowing over an electric heater of 130 kW. The diameter of duct is 500 mm. The gas enters the heating section of the duct at 100 kPa and 27 deg C with a volume flow rate of 15 m3/s. If heat is lost from the gas in the duct to the surroundings at a rate of 80 kW, Calculate the exit temperature of the gas in deg C. (Assume constant pressure, ideal gas, negligible change in kinetic and potential energies and constant specific heat; Cp =1000 J/kg K; R = 500 J/kg K)
Answer:
Exit temperature = 32 °C
Explanation:
We are given;
Initial Pressure;P1 = 100 KPa
Cp =1000 J/kg.K = 1 KJ/kg.k
R = 500 J/kg.K = 0.5 Kj/Kg.k
Initial temperature;T1 = 27°C = 273 + 27K = 300 K
volume flow rate;V' = 15 m³/s
W = 130 Kw
Q = 80 Kw
Using ideal gas equation,
PV' = m'RT
Where m' is mass flow rate.
Thus;making m' the subject, we have;
m' = PV'/RT
So at inlet,
m' = P1•V1'/(R•T1)
m' = (100 × 15)/(0.5 × 300)
m' = 10 kg/s
From steady flow energy equation, we know that;
m'•h1 + Q = m'h2 + W
Dividing through by m', we have;
h1 + Q/m' = h2 + W/m'
h = Cp•T
Thus,
Cp•T1 + Q/m' = Cp•T2 + W/m'
Plugging in the relevant values, we have;
(1*300) - (80/10) = (1*T2) - (130/10)
Q and M negative because heat is being lost.
300 - 8 + 13 = T2
T2 = 305 K = 305 - 273 °C = 32 °C
13000 + 300 - 8000 = T2
A cylinder of metal that is originally 450 mm tall and 50 mm in diameter is to be open-die upset forged to a final height of 100 mm. The strength coefficient is 230 MPa and the work hardening exponent is 0.15 while the coefficient of friction of the metal against the tool is 0.1. If the maximum force that the forging hammer can deliver is 3 MN, can the forging be completed
Answer:
Yes, the forging can be completed
Explanation:
Given h = 100 mm, ε = ㏑(450/100) = 1.504
[tex]Y_f = 230 \times 1.504^{0.15} = 244.52[/tex]
V = π·D²·L/4 = π × 50²×450/4 = 883,572.93 mm³
At h = 100 mm, A = V/h = 883,572.93 /100 = 8835.73 mm²
D = √(4·A/π) = 106.07 mm
[tex]K_f[/tex] = 1 + 0.4 × 0.1 × 106.07/100 = 1.042
F = 1.042 × 244.52 × 8835.73 = 2252199.386 N =2.25 MN
Hence the required force = 2.25 MN is less than the available force = 3 MN therefore, the forging can be completed.
Sensors are used to monitor the pressure and the temperature of a chemical solution stored in a vat. The circuitry for each sensor produces a HIGH voltage when a specified maximum value is exceeded. An alarm requiring a LOW voltage input must be activated when either the pressure or the temperature is excessive. Design a circuit for this application
g A rectangular bar of length L has a slot in the central half of its length. The bar has width b, thickness t, and elastic modulus E. The slot has width b/3. The overall length of the bar is L = 570 mm, and the elastic modulus of the material is 77 GPa. If the average normal stress in the central portion of the bar is 200 MPa, calculate the overall elongation δ of the bar.
Answer:
the overall elongation δ of the bar is 1.2337 mm
Explanation:
From the information given :
According to the principle of superposition being applied to the axial load P of the system; we have:
[tex]\delta = \delta_{AB} +\delta_{BC} + \delta_{CD}[/tex]
where;
δ = overall elongation
[tex]\delta _{AB}[/tex] = elongation of bar AB
[tex]\delta _{BC}[/tex] = elongation of bar BC
[tex]\delta _{CD} =[/tex] elongation of bar CD]
If we replace; [tex]\dfrac{PL}{AE}[/tex] for δ and bt for area;
we have:
[tex]\delta = \dfrac{P_{AB}L_{AB}}{(b_{AB}t)E} +\dfrac{P_{BC}L_{BC}}{(b_{BC}t)E}+\dfrac{P_{CD}L_{CD}}{(b_{CD}t)E}[/tex]
where ;
P = load
L = length of the bar
A = area of the cross-section
E = young modulus of elasticity
Let once again replace:
P for [tex]P_{AB}, P_{BC} , P_{CD}[/tex] (since load in all member of AB, BC and CD will remain the same )
[tex]\dfrac{L}{4}[/tex] for [tex]L_{AB}[/tex],
[tex]\dfrac{L}{2}[/tex] for [tex]L_{BC}[/tex] and
[tex]\dfrac{L}{4}[/tex] for [tex]L_{CD}[/tex]
[tex]2\dfrac{b}{3}[/tex] for [tex]b_{BC}[/tex]
b for [tex]b_{CD}[/tex]
[tex]\delta = \dfrac{P (\dfrac{L}{4})}{btE}+ \dfrac{P (\dfrac{L}{2})}{2 \dfrac{b}{3}tE}+\dfrac{P (\dfrac{L}{4})}{btE}[/tex]
[tex]\delta = \dfrac{PL}{btE}[\dfrac{1}{4}+ \dfrac{1}{2}*\dfrac{3}{2}+ \dfrac{1}{4}][/tex]
[tex]\delta = \dfrac{5}{4}\dfrac{PL}{btE} --- \ (1)[/tex]
The stress in the central portion can be calculated as:
[tex]\sigma = \dfrac{P}{A}[/tex]
[tex]\sigma = \dfrac{P}{\dfrac{2}{3}bt}[/tex]
[tex]\sigma = \dfrac{3P}{2bt}[/tex]
So; Now:
[tex]\delta = \dfrac{5}{4}* \dfrac{2 * \sigma}{3}*\dfrac{L}{E}[/tex]
[tex]\delta= \dfrac{5}{4}* \dfrac{2 * 200}{3}*\dfrac{570}{77*10^3 \ MPa}[/tex]
δ = 1.2337 mm
Therefore, the overall elongation δ of the bar is 1.2337 mm
(a) Consider a message signal containing frequency components at 100, 200, and 400 Hz. This signal is applied to a SSB modulator together with a carrier at 100 kHz, with only the upper sideband retained. In the coherent detector used to recover the local oscillator supplies a sinusoidal wave of frequency 100.02 kHz. Determine the frequency components of the detector output. (b) Repeat your analysis, assuming that only the lower sideband is transmitted.
Answer:
Explanation:
The frequency components in the message signal are
f1 = 100Hz, f2 = 200Hz and f3 = 400Hz
When amplitude modulated with a carrier signal of frequency fc = 100kHz
Generates the following frequency components
Lower side band
[tex]100k - 100 = 99.9kHz\\\\100k - 200 = 99.8kHz\\\\100k - 400 = 99.6kHz\\\\[/tex]
Carrier frequency 100kHz
Upper side band
[tex]100k + 100 = 100.1kHz\\\\100k + 200 = 100.2kHz\\\\100k + 400 = 100.4kHz[/tex]
After passing through the SSB filter that filters the lower side band, the transmitted frequency component will be
[tex]100k, 100.1k, 100.2k\ \texttt {and}\ 100.4kHz[/tex]
At the receive these are mixed (superheterodyned) with local ocillator frequency whichh is 100.02KHz, the output frequencies will be
[tex]100.02 - 100.1k = 0.08k = 80Hz\\\\100.02 - 100.2k = 0.18k = 180Hz\\\\100.02 - 100.4 = 0.38k = 380Hz[/tex]
After passing through the SSB filter that filters the higher side band, the transmitted frequency component will be
[tex]100k, 99.9k, 99.8k\ \ and \ \99.6kHz[/tex]
At the receive these are mixed (superheterodyned) with local oscillator frequency which is 100.02KHz, and then fed to the detector whose output frequencies will be
[tex]100.02 - 99.9k = 0.12k = 120Hz\\\\100.02 - 99.8k = 0.22k = 220Hz\\\\100.02 - 99.6k = 0.42k = 420Hz[/tex]
A) The frequency Components of the Detector Output are;
80 Hz, 120 Hz and 380 Hz
B) The frequency Components if only the lower sideband is transmitted are; 120 Hz, 220 Hz and 420 Hz
Message SignalsA) We are given the frequency components in the message signal as;
f1 = 100Hzf2 = 200Hzf3 = 400HzWe are told that the carrier signal has a frequency; fc = 100kHz
Thus, the frequency components generated are;
Lower side band:
100 kHz - 100 Hz = 99.9 kHz100 kHz - 200 Hz = 99.8 kHz100 kHz - 400 Hz = 99.6 kHzUpper side band:
100 kHz + 100 Hz = 100.1 kHz100 kHz + 200 Hz = 100.2 kHz100 kHz + 400 Hz = 100.4 kHzWe are told that the local oscillator now supplies a sinusoidal wave of frequency 100.02 kHz.
Thus, the output frequencies are;
100.02 kHz - 100.1 kHz = 80 Hz
100.02 kHz - 100.2 kHz = 180 Hz
100.02 kHz - 100.4 kHz = 380 Hz
B) Repeating the analysis assuming only the lower sideband is repeated gives us the frequencies as;
100.02 kHz - 99.9 kHz = 120 Hz
100.02 kHz - 99.8 kHz = 220 Hz
100.02 kHz - 99.6 kHz = 420 Hz
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Technician A says that one planetary gear set can provide gear reduction, overdrive, and reverse. Technician B says that most transmissions today use compound (multiple) planetary gear sets. Which technician is correct?
Answer:
Both technician A and technician B are correct
Explanation:
A planetary gearbox consists of a gearbox with the input shaft and the output shaft that is aligned to each other. It is used to transfer the largest torque in the compact form. A planetary gearbox has a compact size and low weight and it has high power density.
One planetary gear set can provide gear reduction, overdrive, and reverse. Also, most transmissions today use compound (multiple) planetary gears set.
So, both technician A and technician B are correct.
A motor vehicle has a mass of 1.8 tonnes and its wheelbase is 3 m. The centre of gravity of the vehicle is situated in the central plane 0.9 m above the ground and 1.7 m behind the front axle. When moving on the level at 90 km/h the brakes applied and it comes to a rest in a distance of 50 m.
Calculate the normal reactions at the front and rear wheels during the braking period and the least coefficient of friction required between the tyres and the road. (Assume g = 10 m/s2)
Answer:
1) The normal reactions at the front wheel is 9909.375 N
The normal reactions at the rear wheel is 8090.625 N
2) The least coefficient of friction required between the tyres and the road is 0.625
Explanation:
1) The parameters given are as follows;
Speed, u = 90 km/h = 25 m/s
Distance, s it takes to come to rest = 50 m
Mass, m = 1.8 tonnes = 1,800 kg
From the equation of motion, we have;
v² - u² = 2·a·s
Where:
v = Final velocity = 0 m/s
a = acceleration
∴ 0² - 25² = 2 × a × 50
a = -6.25 m/s²
Force, F = mass, m × a = 1,800 × (-6.25) = -11,250 N
The coefficient of friction, μ, is given as follows;
[tex]\mu =\dfrac{u^2}{2 \times g \times s} = \dfrac{25^2}{2 \times 10 \times 50} = 0.625[/tex]
Weight transfer is given as follows;
[tex]W_{t}=\dfrac{0.625 \times 0.9}{3}\times \dfrac{6.25}{10}\times 18000 = 2109.375 \, N[/tex]
Therefore, we have for the car at rest;
Taking moment about the Center of Gravity CG;
[tex]F_R[/tex] × 1.3 = 1.7 × [tex]F_F[/tex]
[tex]F_R[/tex] + [tex]F_F[/tex] = 18000
[tex]F_R + \dfrac{1.3 }{1.7} \times F_R = 18000[/tex]
[tex]F_R[/tex] = 18000*17/30 = 10200 N
[tex]F_F[/tex] = 18000 N - 10200 N = 7800 N
Hence with the weight transfer, we have;
The normal reactions at the rear wheel [tex]F_R[/tex] = 10200 N - 2109.375 N = 8090.625 N
The normal reactions at the front wheel [tex]F_F[/tex] = 7800 N + 2109.375 N = 9909.375 N
2) The least coefficient of friction, μ, is given as follows;
[tex]\mu = \dfrac{F}{R} = \dfrac{11250}{18000} = 0.625[/tex]
The least coefficient of friction, μ = 0.625.
Use a delta-star conversion to simplify the delta BCD (40 , 16 , and 8 ) in the
bridge network in Fig. f and find the equivalent resistance that replaces the network
between terminals A and B, and hence find the current I if the source voltage is 52 V.
Answer:
Current, I = 4A
Explanation:
Since the connection is in delta, let's convert to star.
Simplify BCD:
[tex] R1 = \frac{40 * 8}{40 + 16 + 8} = \frac{320}{64} = 5 ohms [/tex]
[tex] R2 = \frac{16 * 8}{40 + 16 + 8} = \frac{128}{64} = 2 ohms [/tex]
[tex] R3 = \frac{40 * 16}{40 + 16 + 8} = \frac{640}{64} = 10 ohms [/tex]
From figure B, it can be seen that 6 ohms and 6 ohms are connected in parallel.
Simplify:
[tex] \frac{6 * 6}{6 + 6} = \frac{36}{12} = 3 \ohms [/tex]
Req = 10 ohms + 3 ohms
Req = 13 ohms
To find the current, use ohms law.
V = IR
Where, V = 52volts and I = 13 ohms
Solve for I,
[tex] I = \frac{V}{R} = \frac{52}{13} = 4A[/tex]
Current, I = 4 A
What's the "most common" concern with using variable frequency drives (VFDs)? 1) carrier frequency 2) harmonic distortion 3) hertz modulation
The common" concern with using variable frequency drives (VFDs) is C. hertz modulation.
What is variable frequency drive?It should be noted that a variable frequency drive simply means a type of motor drive that us used in mechanical drive system.
In this case, common" concern with using variable frequency drives (VFDs) is hertz modulation
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A 10-mm-diameter Brinell hardness indenter produced an indentation 1.55 mm in diameter in a steel alloy when a load of 500 kg was used. Calculate the Brinell hardness (in HB) of this material. Enter your answer in accordance to the question statement HB
Answer:
HB = 3.22
Explanation:
The formula to calculate the Brinell Hardness is given as follows:
[tex]HB = \frac{2P}{\pi D\sqrt{D^{2}- d^{2} } }[/tex]
where,
HB = Brinell Hardness = ?
P = Applied Load in kg = 500 kg
D = Diameter of Indenter in mm = 10 mm
d = Diameter of the indentation in mm = 1.55 mm
Therefore, using these values, we get:
[tex]HB = \frac{(2)(500)}{\pi (10)\sqrt{10^{2}- 1.55^{2} } }[/tex]
HB = 3.22
The basic behind equal driving is to
Follow traffic signs , Keep distance between cars , Be patient in traffic.
2) Consider schedules S3, S4, and S5 below. Determine whether each schedule is strict, cascadeless, recoverable, or non-recoverable. You need to explain your reason.
S3: r1(x), r2(z), r1(z), r3(x), r3(y), w1(x), c1, w3(y), c3, r2(y), w2(z),w2(y),c2
S4: r1(x), r2(z), r1(z), r3(x), r3(y),w1(x),w3(y), r2(y),w2(z),w2(y), c1,c2, c3
S5: r1(x), r2(z), r3(x), r1(z), r2(y), r3(y), w1(x), c1, w2(z), w3(y), w2(y), c3, c2
Answer:
Explanation:
Consider schedules S3, S4, and S5 below. Determine whether each schedule is strict, cascadeless, recoverable, or non-recoverable. You need to explain your reason.
S3: r1(x), r2(z), r1(z), r3(x), r3(y), w1(x), c1, w3(y), c3, r2(y), w2(z),w2(y),c2
S4: r1(x), r2(z), r1(z), r3(x), r3(y),w1(x),w3(y), r2(y),w2(z),w2(y), c1,c2, c3
S5: r1(x), r2(z), r3(x), r1(z), r2(y), r3(y), w1(x), c1, w2(z), w3(y), w2(y), c3, c2
Strict schedule:
A schedule is strict if it satisfies the following conditions:
Tj reads a data item X after Ti has written to X and Ti is terminated means aborted or committed.
Tj writes a data item X after Ti has written to X and Ti is terminated means aborted or committed.
S3 is not strict because In a strict schedule T3 must read X after C1 but here T3 reads X (r3(X)) before Then T1 has written to X (w1(X)) and T3 commits after T1.
S4 is not strict because In a strict schedule T3 must read X after C1, but here T3 reads X (r3(X)) before T1 has written to X (w1(X)) and T3 commits after T1.
S5 is not strict because T3 reads X (r3(X)) before T1 has written to X (w1(X))
but T3 commits after T1. In a strict schedule T3 must read X after C1.
Cascadeless schedule:
Cascadeless schedule follows the below condition:
Tj reads X only? after Ti has written to X and terminated means aborted or committed.
S3 is not cascadeless schedule because T3 reads X (r3(X)) before T1 commits.
S4 is not cascadeless schedule because T3 reads X (r3(X)) before T1 commits.
S5 is not cascadeless schedule because T3 reads X (r3(X)) before T1 commits or T2 reads Y (r2(Y)) before T3 commits.
But while come to the definition of cascadeless schedules S3, S4, and S4 are not cascadeless, and T3 is not affected if T1 is rolled back in any of the schedules, that is,
T3 does not have to roll back if T1 is rolled back. The problem occurs because these
schedules are not serializable.
Recoverable schedule:
Schedule that follows the below condition:
-----Tj commits after Ti if Tj has?read any data item written by Ti.
Ci > Cj means that Ci happens before Cj. Ai denotes abort Ti. To test if a schedule is
recoverable one has to include abort operations. Thus in testing the recoverability abort
operations will have to used in place of commit one at a time. Also the strictest condition is
------where a transaction neither reads nor writes to a data item, which was written to by a transaction that has not committed yet.
If A1?>C3>C2, then schedule S3 is recoverable because rolling back of T1 does not affect T2 and
T3. If C1>A3>C2. schedule S3 is not recoverable because T2 read the value of Y (r2(Y)) after T3 wrote X (w3(Y)) and T2 committed but T3 rolled back. Thus, T2 used non- existent value of Y. If C1>C3>A3, then S3 is recoverable because roll back of T2 does not affect T1 and T3.
Strictest condition of schedule S3 is C3>C2.
If A1?>C2>C3, then schedule S4 is recoverable because roll back of T1 does not affect T2 and T3. If C1>A2>C3, then schedule S4 is recoverable because the roll back of T2 will restore the value of Y that was read and written to by T3 (w3(Y)). It will not affect T1. If C1>C2>A3, then schedule S4 is not recoverable because T3 will restore the value of Y which was not read by T2.
The Rappahannock River near Warrenton, VA, has a flow rate of 3.00 m3/s. Tin Pot Run (a pristine stream) discharges into the Rappahannock at a flow rate of 0.05 m3/s. To study mixing of the stream and river, a conserva- tive tracer is to be added to Tin Pot Run. If the instruments that can mea- sure the tracer can detect a concentration of 1.0 mg/L, what minimum concentration must be achieved in Tin Pot Run so that 1.0 mg/L of tracer can be measured after the river and stream mix? Assume that the 1.0 mg/L of tracer is to be measured after complete mixing of the stream and Rappa- hannock has been achieved and that no tracer is in Tin Pot Run or the Rap- pahannock above the point where the two streams mix. What mass rate (kg/d) of tracer must be added to Tin Pot Run?
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