Answer:
The water of the saturated clayed soil is 66.67 %.
Explanation:
Given;
mass of saturated clayed soil, [tex]M_s[/tex] = 600 g
mass of dry soil sample, [tex]M_d[/tex] = 200 g
mass of water content, [tex]M_w[/tex] = [tex]M_s[/tex] - [tex]M_d[/tex] = 600 g - 200 g = 400 g
The water content is determined as;
[tex]M_w(\%) = \frac{M_s - M_d}{M_s} *100\%\\\\M_w(\%) = \frac{600-200}{600} *100 \% \\\\M_w(\%) = 66.67 \%[/tex]
Therefore, the water of the saturated clayed soil is 66.67 %.
1. Why does condensed water drip from the air-conditioning system?
Answer:
Your answer is: Evaporator, the condensed water drip comes from something called the evaporator. When warm air touches the cold coil ( or the cold pipe) the water vapor condensed while the cold coil absorbs.
Explanation:
Hope this helped : )
Consider the open-loop plant:
P(S) = 10/(s2 + 2s + 5) and a controller: k(s +1) /(s) = s(s +2) structured in a unity feedback control architecture: = w r u y C(s) P(s) (f) (numerical) Suppose the maximum control effort the actuator can provide is Umar = 0.52. Run a separate iteration, this time beginning with k = 0.01, and incrementing by 0.01 to determine the largest value of k such that the actuator does not exceed its limit. Simulate for 30 seconds during each iteration. With this k, display a 1 x 2 subplot showing the control command u(t) and the output y(t). On the u(t) plot, show the maximum value Umax being less than 0.52. Also display the k value that yields this response. Note: the loop will probably take a long time to run since it's running the simulation each time.
(a) (numerical) Next, run the simulation for 60 seconds, this time introducing a unit step distur- bance w(t) beginning at 30 seconds. With the value of k found in (f), plot the output y(t) and the control command u(t). Is the system capable of rejecting a constant disturbance? E(S) (h) (by hand) As a follow-up to
(b), compute the transfer function T(s) in terms of k, and W(s) note the number of zeros at the origin. Keep this number in mind as we move into the next topic, system TYPE & steady-state error. It can tell us definitively what types of disturbances we can reject and what types of references we can track.
Answer:
Explanation:
r the open-loop plant:
P(S) = 10/(s2 + 2s + 5) and a controller: k(s +1) /(s) = s(s +2) structured in a unity feedback control architecture: = w r u y C(s) P(s) (f) (numerical) Suppose the maximum control effort the actuator can provide is Umar = 0.52. Run a separate iteration, this time beginning with k = 0.01, and incrementing by 0.01 to determine the largest value of k such that the actuator does not exceed its limit. Simulate for 30 seconds during each iteration. With this k, display a 1 x 2 subplot showing the control command u(t) and the output y(t). On the u(t) plot, show the maximum value Umax being less than 0.52. Also display the k value that yields this response. Note: the loop will probably take a long time to run since it's running the simulation each time.
(a) (numerical) Next, run the simulation for 60 seconds, this time introducing a unit step distur- bance w(t) beginning at 30 seconds. With the value of k found in (f), plot the output y(t) and the control command u(t). Is the system capable of rejecting a constant disturbance? E(S) (h) (by hand) As a follow-up to
(b), compute the transfer function T(s) in terms of k, and W(s) note the number of zeros at the origin. Keep this number in mind as we move into the next topic, system TYPE & steady-state error. It can tell us definitively what types of disturbances we can reject and what types of references we can track.
Multiple Choice
Which of the following best describes the role of engineers?
A) Engineers must clearly determine exactly what needs to be solved or designed.
B) Engineers must earn a bachelor’s and a master’s degree before getting hired.
C) Engineers must be sure every change they make is a functional design improvement.
D) Engineers must be content working in an office setting.
Answer:
A
Explanation:
Answer:
A) Engineers must clearly determine exactly what needs to be solved or designed.
Explanation:
1. The area(in square centimeters) of a square coaster can be represented by
d2 + 8d + 16 cm2.How long is it’s side? What is its perimeter?
Answer: (d−4) 2
Explanation: Factoring d2-8d+16
The first term is, d2 its coefficient is 1 .
The middle term is, -8d its coefficient is -8 .
The last term, "the constant", is +16
Step-1 : Multiply the coefficient of the first term by the constant 1 • 16 = 16
Step-2 : Find two factors of 16 whose sum equals the coefficient of the middle term, which is -8 .
-16 + -1 = -17
-8 + -2 = -10
-4 + -4 = -8 That's it
Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -4 and -4
d2 - 4d - 4d - 16
Step-4 : Add up the first 2 terms, pulling out like factors :
d • (d-4)
Add up the last 2 terms, pulling out common factors :
4 • (d-4)
Step-5 : Add up the four terms of step 4 :
(d-4) • (d-4)
Which is the desired factorization
Multiply (d-4) by (d-4)
The rule says : To multiply exponential expressions which have the same base, add up their exponents.
In our case, the common base is (d-4) and the exponents are :
1 , as (d-4) is the same number as (d-4)1
and 1 , as (d-4) is the same number as (d-4)1
The product is therefore, (d-4)(1+1) = (d-4)2
Please mark me brainlyest
Consider a well-insulated horizontal rigid cylinder that is divided into two compartments by a piston that is free to move but does not allow either gas to leak into the other side. Initially, one side of the piston contains 1m3 of N2 gas at 500 kPa and 160 degrees C while the other side contains 1m3 of He gas at 500 kPa and 60 degrees C. Now thermal equilibrium is established in the cylinder as a result of heat transfer through the piston. Using constant specific heats at room temperature, determine the final equilibrium temperature in the cylinder. What would your answer be if the piston were not free to move?
Answer:
- the final equilibrium temperature in the cylinder is 85.67 °C
- Also the answer would be the same if the piston were not free to move since it would effect only pressure and not the specific heats.
Explanation:
Given That;
Compartment Nitrogen:
Volume V1 = 1 m³, Pressure P1 = 500 kPa , Temperature T = 120°C
Compartment of helium:
Volume V1H = 1 m³, Pressure P1H = 500 kPa, Temperature T1H = 40°C
From the ideal specific heat of gases
-Nitrogen
The gas k (constant) and the k (constant) volume specific heats are;
R = 0.2968 kJ
/kg.K
Cv = 0.743 kJ/kg.K
also From the ideal specific heat of gases
-helium
R = 2.0769 kJ /kg.K
Cv = 3.1156 kJ/kg.K
we know that
PV = mRT
Mass of the nitrogen
mN2 = (P1V1 /RT1)_N2
mN2 = (500)(1) / (0.2968)(393)
= 4.29kg
mass of helium
mHe = (P1HVIH /RT1H)_He
mHe = (P1V1 /RT1)_N2
mHe = (500)(1) / (2.0769)(313)
= 0.769kg
Taking the whole contents of the cylinder,
the 1st law relation can be expressed as;
Ein - Eout = ΔEsystem
0 = ΔU = (ΔU)_N2 + (ΔU)_He
0 = [mcV(T2 -T1)]_N2 + [mcV(T2 - Ti)]He
Since T2 =T_F
0 = [mcV(T_F -T1)]_N2 + [mcV(T_F - Ti)]He
we Substitute
(4.29)(0.743)(T_F - 120) + (0.769)(3.1156)(T_F - 40) = 0
T_F = 85.67 °C
therefore the final equilibrium temperature in the cylinder is 85.67 °C
Also the answer would be the same if the piston were not free to move since it would effect only pressure and not the specific heats.
Driving next to another vehicle can sometimes take away the option to _____.
A. steer around a hazard
B. improve your mileage
C. conduct your scan
D. manage your tire wear
Answer: A. steer around a hazard
Explanation: say there was a vehicle next to you on your left and there was a pile of rocks it would prevent you from steer from going straight but instead have you steer around the rocks and cross into another lane that’s what it said in the lesson
Driving next to another vehicle can sometimes take away the option to A. steer around a hazard
What are Traffic Rules?This refers to the rules and regulations that guide road users and pedestrians to ensure their safety.
Hence, we can see that when driving on the road next to another vehicle, the option to steer around a hazard is taken away, and hence, it is not advisable to do this.
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A single-phase 50-kVA, 2400/240-volt, 60-Hz distribution transformer is used as a step-down transformer at the load end of a 2400-volt feeder whose series impedance is (1.0 + j2.0) ohms. The equivalent series impedance of the transformer is (1.0 + j2.5) ohms referred to the highvoltage (primary) side. The transformer is delivering rated load at a 0.8 power factor lagging and at a rated secondary voltage. Neglecting the transformer exciting current.
Determine:
a. The voltage at the transformer primary terminals
b. The voltage at the sending end of the feeder.
c. The real and reactive power delivered to the sending end of the feeder.
What components should you inspect if the crankshaft end play is out of specifications?
Answer:
gdyc ddxtfvytg4dgtfxdwcftcd3rcby
If the crankshaft end play is out of specifications, check for:
Thrust Bearings
Main Bearings
Crankshaft
Crankshaft Thrust Washers
Engine Block
To understand the crankshaft when it is out of specifications, check for:
Thrust Bearings: Check the condition of the thrust bearings, which are located at the front and/or rear of the engine block. Excessive wear or damage to the thrust bearings can contribute to increased crankshaft end play.
Main Bearings: Inspect the main bearings, which support the crankshaft within the engine block. Worn or damaged main bearings can cause excessive movement of the crankshaft.
Crankshaft: Check the crankshaft itself for any signs of damage, such as scoring or bending. A damaged crankshaft may not sit properly within the bearings, leading to increased end play.
Crankshaft Thrust Washers: Some engines use thrust washers to control the end play of the crankshaft. Inspect these washers for wear, damage, or improper installation.
Engine Block: Check the engine block for any signs of damage or distortion that may affect the alignment and support of the crankshaft.
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C+ Write a program that converts degrees Fahrenheit to Celsius using the following formula. degreesC = 5(degreesF – 32)/9
Prompt the user to enter a temperature in degrees Fahrenheit.Convert the temperature to Farenheit if the Celsius is entered,or to Celsius if Farenheit is entered.Display the result in readable format.If anything other than C,c,F and f is entered,print an error mssage and stop.
Answer:
Written in C++
#include<iostream>
#include<cmath>
using namespace std;
int main()
{
float degreeC, degreeF;
cout<<"Degree Fahrenheit: ";
cin>>degreeF;
degreeC = 5 * (degreeF - 32)/9;
cout<<"Degree Celsius: "<<degreeC<<" C";
return 0;
}
Explanation:
The question requests that input should be in degree Fahrenheit
Declare all necessary variables
float degreeC, degreeF;
Prompt user for input in degrees Fahrenheit as stated in the question
cout<<"Degree Fahrenheit: ";
Get User Input
cin>>degreeF;
Convert degree Fahrenheit to Celsius
degreeC = 5 * (degreeF - 32)/9;
Display output
cout<<"Degree Celsius: "<<degreeC<<" C";
a line passes through (1,9) is parallel to the line through (-2,2) and (4,5) find its equation.
What is the full form of AWM
Answer:
its a gun
Explanation:
The Accuracy International AWM (Arctic Warfare Magnum or AI-Arctic Warfare Magnum) is a bolt-action sniper rifle manufactured by Accuracy International designed for magnum rifle cartridges. ...
Effective firing range: 1,100 m (1,203 yd) (.300 ...
Manufacturer: Accuracy International
In service: 1996–present
The next day at SLS found everyone in technical support busy restoring computer systems to their former state and installing new virus and worm control software. Amy found herself learning how to install desktop computer operating systems and applications as SLS made a heroic effort to recover from the attack of the previous day.
Required:
Do you think this event was caused by an insider or outsider?
Answer:
Yes, it was caused by an insider or outsider.
Explanation:
Yes, I believe that this event was caused by an insider or outsider. This is because, if a USB removable flash drive is attached to all the office computers by an insider and now an outsider sends a mail to the company’s General group mail; if the mail has a link that contains some infected worms and virus; once anybody clicks on the link in the mail then, all the computers will be affected by the virus.
Therefore, this event may be caused by the insider who clicked the infected link or by the outsider sent the infected link in the mail.
what was the code phrase the japanese used to initiate the attack on pearl harbor
Answer:Itaka
Explanation:
Answer:
Tora!
Explanation:
It was an expression that illustrated the complete surprise of the attack.
list five properties of most transition metals
Which AWS specification covers electrodes used for welding low-alloy steels?
Answer:
American Welding Society filler metal specification A5. 5-96
Explanation:
You prepare low alloy electrodes by adding the correct alloying elements to the electrode coating. This specification covers the following;
requirements for size and lengths of requirements for marking and manufacturingrequirements for chemical composition and mechanical properties of weld metalmetal soundness test and moisture testsusability test for electrodesChapter 1:
What story his father would tell William about Chief Mwase and the Battle of Kasungu. What did
The Ngoni fail to appreciate about the Chewa? What was the moral to the story?
Answer: At least I’m not the only one looking for answers lol
Explanation:
Find the slope of the line containing the points (6, 4) and (6,8).
A. -1
B. 1
C. 0
D. undefined
Oxygen initially is at 35°F and 16.8 lbf/in2. It fills a closed, rigid 5 ft3 tank filled with a paddle wheel. During the process, a paddle wheel provides 4 Btu of energy transfer by work to the gas. During the process, the gas temperature increases up to 90°F. Assuming ideal gas behavior and ignoring K.E and P.E effects, determine the mass of the oxygen in the tank, the amount of heat transfer in Btu, and the final pressure of the tank.
Answer:
m = 228 gm
ΔQ = 4.35 Btu
P₂ = 43.2 lbf/in² = 2.98 x 10⁵ Pa
Explanation:
FOR MASS OF OXYGEN:
We will use ideal gas equation for initial conditions:
P₁V₁ = nRT₁
P₁V₁ = mRT₁/M
where,
P₁ = Initial Pressure = (16.8 lbf/in²)(6894.76 Pa/1 lbf/in²) = 1.15 x 10⁵ Pa
V₁ = Initial Volume = (5 ft³)(0.028316 m³/1 ft³) = 0.1415 m³
m = mass of oxygen = ?
R = Universal Gas Constant = 8.314 J/mol.k
T₁ = Initial Temperature = 35°F = 274.67 k
M = Molecular Mass of Oxygen = 32 gm
Therefore,
(1.15 x 10⁵ Pa)(0.1415 m³) = m(8.314 J/mol.k)(274.67 k)/(32 gm)
m = (1.15 x 10⁵ Pa)(0.1415 m³)(32 gm)/(8.314 J/mol.k)(274.67 k)
m = 228 gm
FOR AMOUNT OF HEAT TRANDFER:
Using First Law of Thermodynamics:
ΔQ = ΔU +W
Since, this is a constant volume process. Therefore, the work done in this process must be equal to zero:
ΔQ = ΔU + 0
ΔQ = ΔU
and,
ΔU = m Cv ΔT
Therefore,
ΔQ = m Cv ΔT
where,
ΔQ = Heat Transfer Amount = ?
m = mass of oxygen = 228 g = 0.228 kg
Cv = Molar Specific Heat of Oxygen at Constant Volume = 0.659 KJ/kg.k
ΔT = Change in temperature = 305.22 k - 274.67 k = 30.55 k
Therefore,
ΔQ = (0.228 kg)(0.659 KJ/kg.k)(30.55 k)
ΔQ = (4.6 KJ)(0.9478 Btu/1 KJ)
ΔQ = 4.35 Btu
FOR FINAL PRESSURE IN TANK:
Using equation of state:
P₁V₁/T₁ = P₂V₂/T₂
Here, the volume will be constant due to rigid tank. Hence,
V₁ = V₂ = V
Therefore,
P₁V/T₁ = P₂V/T₂
P₁/T₁ = P₂/T₂
(16.8 lbf/in²)/(35°F) = P₂/(90°F)
P₂ = (16.8 lbf/in²)(90°F)/(35°F)
P₂ = 43.2 lbf/in² = 2.98 x 10⁵ Pa
Which is currently the most supported theory about the future of the universe?
a.The universe could remain exactly as it is today.
b.The universe may continue to expand.
c.Stars could burn out, causing the universe to become dark and cold.
d.Gravity could pull galaxies back together, causing a reverse of the big bang.
Answer:
b. The universe may continue to expand.
Explanation:
The most supported theory about the future of the universe is that, it may continue to expand up to such point that it will become too cold to allow any forms of life to pull through. This is popularly known as the "Big Freeze."
Such acceleration of expansion is caused by the so-called, "dark energy." This also explains why the acceleration of expansion is not constant. This thereby causes the distance between galaxies to grow farther apart.
Answer:
b
Explanation:
edge 2021
Technician A says that the unitized structure of a hybrid vehicle is considerably different when compared to the same conventional model.
Technician B says that hybrid vehicles have 12-volt and high voltage batteries.
Who is right?
Technician A says that the unitized structure of a hybrid vehicle is considerably different when compared to the same conventional model is right.
What are hybrid vehicle?Hybrid vehicle are defined as a powered by a combustion engine and/or a number of electric motors that draw power from batteries. A gas-powered car simply has a traditional gas engine, but a hybrid car also features an electric motor.
One important advantage of hybrid cars is their capacity to reduce the size of the main engine, which improves fuel efficiency. Many hybrid vehicles employ electric motors to accelerate slowly at first until they reach higher speeds. They then use gasoline-powered engines to increase fuel efficiency.
Thus, technician A says that the unitized structure of a hybrid vehicle is considerably different when compared to the same conventional model is right.
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How much would you spend on gasoline each year if you drove 10,000 miles over the year and your vehicle achieves 15 miles per gallon with gasoline prices at $4.00 a gallon? Now substitute your vehicle with a hybrid-electric automobile that achieves 60 miles per gallon. Calculate the yearly cost for fuel with this vehicle
Answer:
a. [tex]Total\ Cost = \$2667[/tex]
b. [tex]Total\ Cost = \$667[/tex]
Explanation:
(a)
Given
[tex]Distance = 10000\ miles[/tex]
[tex]Rate = 15\ miles/gallon[/tex]
[tex]Price = \$4.00/gallon[/tex]
Required
Determine the total amount spent in a year
First, we need to determine the number of gallons used in a year;
[tex]Total\ Gallons = Distance/Rate[/tex]
[tex]Total\ Gallons = 10000miles /\frac{15miles}{gallon}[/tex]
[tex]Total\ Gallons = 10000miles * \frac{1\ gallon}{15\ miles}[/tex]
[tex]Total\ Gallons = 10000 * \frac{1\ gallon}{15}[/tex]
[tex]Total\ Gallons = \frac{10000\ gallons}{15}[/tex]
[tex]Total\ Gallons = \frac{10000}{15}\ gallons[/tex]
Next, is to determine the total cost:
[tex]Total\ Cost = Total\ Gallons * Price[/tex]
[tex]Total\ Cost = \frac{10000}{15}\ gallon * \frac{\$4}{gallon}[/tex]
[tex]Total\ Cost = \frac{10000}{15} * \$4[/tex]
[tex]Total\ Cost = \frac{10000 * \$4}{15}[/tex]
[tex]Total\ Cost = \frac{\$40000}{15}[/tex]
[tex]Total\ Cost = $2666.66666667[/tex]
[tex]Total\ Cost = \$2667[/tex] (approximated)
(b)
Given
[tex]Rate = 60\ miles/gallon[/tex]
Required
Determine the total amount spent in a year
First, we need to determine the number of gallons used in a year;
[tex]Total\ Gallons = Distance/Rate[/tex]
[tex]Total\ Gallons = 10000miles /\frac{60\ miles}{gallon}[/tex]
[tex]Total\ Gallons = 10000miles * \frac{1\ gallon}{60\ miles}[/tex]
[tex]Total\ Gallons = 10000 * \frac{1\ gallon}{60}[/tex]
[tex]Total\ Gallons = \frac{10000\ gallons}{60}[/tex]
[tex]Total\ Gallons = \frac{10000}{60}\ gallons[/tex]
Next, is to determine the total cost:
[tex]Total\ Cost = Total\ Gallons * Price[/tex]
[tex]Total\ Cost = \frac{10000}{60}\ gallon * \frac{\$4}{gallon}[/tex]
[tex]Total\ Cost = \frac{10000}{60} * \$4[/tex]
[tex]Total\ Cost = \frac{10000 * \$4}{60}[/tex]
[tex]Total\ Cost = \frac{\$40000}{60}[/tex]
[tex]Total\ Cost = \$666.666666667[/tex]
[tex]Total\ Cost = \$667[/tex] (approximated)
two technicians are discussing catalytic converters technician a says that the exhaust mixture must fluctuate between rich and lean mixture for the best l efficiently technician b says that the air fuel mixture must be leaner than 14:7:1 for best performance from a three way catalytic converter who is correct
Answer:
A
Explanation:
The popularity of orange juice, especially as a breakfast drink, makes it an important factor in the economy of orange-growing regions.Marketed juice has either gone through a process in which it was concentrated, or it may be a not-from-concentrate juice. Frozen concentrated juice is reconstituted before consumption. Although concentrated juices are less popular in the United States than at one time, they still have a major segment of the market.The approaches to concentrating orange juice include evaporation, freeze concentration, and reverse osmosis. Here we examine the evaporation process by focusing only on two constituents in the juice: solids and water.Fresh orange juice contains approximately 10.0 wt% solids (sugar, citric acid and other indigenous ingredients) and frozen concentrate contains approximately 45.0 wt% solids. The frozen concentrate is obtained by evaporating water from the fresh juice to produce a mixture that is approximately 67.0 wt% solids. However, so that the flavor of the concentrate will closely approximate that of fresh juice, the concentrate from the evaporator is blended with fresh orange juice (and other minor additives) to produce a final concentrate that is approximately 45.0 wt% solids.Assume a basis of 100.0 kg of fresh juice fed to the process.a) What is the mass of final concentrate produced?b) What is the fraction of fresh juice that bypasses the evaporator?
Answer:
A) 22.22 kg
B) 0.1364
Explanation:
When we perform a degree of freedom (DOF) analysis on the bypass subsystem, we will have 2 unknown masses which are; (m1, m2).l with degree of freedom as 1.
Secondly, when we perform a degree of freedom analysis on the overall system, we will have 2 unknown masses namely (m3, m5) with degree of freedom as 0.
Thirdly, when we perform a degree of freedom analysis on the evaporator, we will have 3 unknown masses namely (m1, m3, m4) with Degree of freedom as 1.
Lastky,when we perform a degree of freedom analysis on mixing point, we will have 3 unknown masses namely, (m2, m4, m5) with degree of freedom as 1.
We are given;
Fresh orange = 10.0 wt% solids
Frozen juice concentrate = 45.0wt% solids
Mixture produced = 67.0 wt% solids
Final concentrate = 45.0 wt% solids
Now, making use of the overall solids balance, we have;
0.1 × 100 = 0.45(m5)
m5 = (0.1 × 100)/0.45
m5 = 22.22 kg
Making use of overall mass balance, we have;
100 kg = m3 + m5
m3 = 100 - 22.22
m3 = 77.78 kg
Now, making use of the Mixing point mass balance, we have;
m4 + m2 = m5
So; m4 + m2 = 22.22 - - - (eq 1)
Also, making use of the Mixing point solids balance, we have;
0.67m4 + 0.1m2 = 0.45m5
0.67m4 + 0.1m2 = 9.999 - - - (eq 2)
Solving eq 1 and eq 2 simultaneously, we have;
m2 = 13.64 kg
m4 = 8.58 kg
Final concentrate is m5 = 22.22 kg
Now, fraction = m2/100 = 13.64/100 = 0.1364
Consider a regenerative gas-turbine power plant with two stages of compression and two stages of expansion. The overall pressure ratio of the cycle is 9. The air enters each stage of the compressor at 300 K and each stage of the turbine at 1200 K. Accounting for the variation of specific heats with temperature, determine the minimum mass flow rate of air needed to develop a net power output of 105 MW.
Answer: the minimum mass flow rate of air required to generate a power output of 105 MW is 238.2 kg/s
Explanation:
from the T-S diagram, we get the overall pressure ratio of the cycle is 9
Calculate the pressure ratio in each stage of compression and expansion. P1/P2 = P4/P3 = √9 = 3
P5/P6 = P7/P8 = √9 =3
get the properties of air from, "TABLE A-17 Ideal-gas properties of air", in the text book.
At temperature T1 =300K
Specific enthalpy of air h1 = 300.19 kJ/kg
Relative pressure pr1 = 1.3860
At temperature T5 = 1200 K
Specific enthalpy h5 = 1277.79 kJ/kg
Relative pressure pr5 = 238
Calculate the relative pressure at state 2
Pr2 = (P2/P1) Pr5
Pr2 =3 x 1.3860 = 4.158
get the two values of relative pressure between which the relative pressure at state 2 lies and take the corresponding values of specific enthalpy from, "TABLE A-17 Ideal-gas properties of air", in the text book.
Relative pressure pr = 4.153
The corresponding specific enthalpy h = 411.12 kJ/kg
Relative pressure pr = 4.522
The corresponding specific enthalpy h = 421.26 kJ/kg
Find the specific enthalpy of state 2 by the method of interpolation
(h2 - 411.12) / ( 421.26 - 411.12) =
(4.158 - 4.153) / (4.522 - 4.153 )
h2 - 411.12 = (421.26 - 411.12) ((4.158 - 4.153) / (4.522 - 4.153))
h2 - 411.12 = 0.137
h2 = 411.257kJ/kg
Calculate the relative pressure at state 6.
Pr6 = (P6/P5) Pr5
Pr6 = 1/3 x 238 = 79.33
Obtain the two values of relative pressure between which the relative pressure at state 6 lies and take the corresponding values of specific enthalpy from, "TABLE A-17 Ideal-gas properties of air", in the text book.
Relative pressure Pr = 75.29
The corresponding specific enthalpy h = 932.93 kJ/kg
Relative pressure pr = 82.05
The corresponding specific enthalpy h = 955.38 kJ/kg
Find the specific enthalpy of state 6 by the method of interpolation.
(h6 - 932.93) / ( 955.38 - 932.93) =
(79.33 - 75.29) / ( 82.05 - 75.29 )
(h6 - 932.93) = ( 955.38 - 932.93) ((79.33 - 75.29) / ( 82.05 - 75.29 )
h6 - 932.93 = 13.427
h6 = 946.357 kJ/kg
Calculate the total work input of the first and second stage compressors
(Wcomp)in = 2(h2 - h1 ) = 2( 411.257 - 300.19 )
= 222.134 kJ/kg
Calculate the total work output of the first and second stage turbines.
(Wturb)out = 2(h5 - h6) = 2( 1277.79 - 946.357 )
= 662.866 kJ/kg
Calculate the net work done
Wnet = (Wturb)out - (Wcomp)in
= 662.866 - 222.134
= 440.732 kJ/kg
Calculate the minimum mass flow rate of air required to generate a power output of 105 MW
W = m × Wnet
(105 x 10³) kW = m(440.732 kJ/kg)
m = (105 x 10³) / 440.732
m = 238.2 kg/s
therefore the minimum mass flow rate of air required to generate a power output of 105 MW is 238.2 kg/s
Under the right conditions, it is possible, due to surface tension, to have metal objects float on water. Consider placing a short length of a small diameter steel (sp. wt.=500 lb/ft3) rod on a surface of water. What is the maximum diameter, Dmax, that the rod can have before it will sink? Assume that the surface tension forces act vertically upward. Note: A standard paper clip has a diameter of 0.036 in. Partially unfold a paper clip and see if you can get it to float on water. Do the results of this experiment support your analysis?
Answer:
D = 0.060732 in
Explanation:
given data
sp. wt. = 500 lb/ft³
diameter = 0.036 in
solution
we get here maximum diameter of rod that is express as
D = [tex]\sqrt{\frac{8 \sigma }{\pi y}}[/tex] ......................1
here [tex]\sigma[/tex] surface tension of water at 60⁰f = 5.03 × [tex]10^{-3}[/tex] lb/ft and y = 500 lb/ft³
so put here value and we will get
D = [tex]\sqrt{\frac{8 \times 5.03 \times 10^{-3} }{\pi \times 500}}[/tex]
D = 0.005061 ft
D = 0.060732 in
If, for a particular junction, the acceptor concentration is 1017/cm3 and the donor concentration is 1016/cm3, find the junction built-in voltage. Assume ni = 1.5 × 1010/cm3. Also, find the width of the depletion region (W) and its extent in each of the p and n regions when the junction terminals are left open. Calculate the magnitude of the charge stored on either side of the junction. Assume that the junction area is 10 µm2.
Answer:
A) V_o = 0.7543 V
B) W = 328.42 nm
C) X_n = 298.56 nm and X_p = 29.86 nm
D) Q_j = 47.767 × 10^(-15)
Explanation:
We are given;
Acceptor concentration; N_A = 10^(17) /cm³
donor concentration; N_D = 10^(16) /cm³
ni = 1.5 × 10^(10) /cm3
A) The formula for the built - in - voltage at the junction is given by;
V_o = V_T(In (N_A × N_D/ni²))
Where V_T is thermal voltage at room temperature with a value from online sources as 25.9 mV
Thus;
V_o = 25.9(In (10^(17) × 10^(16))/(1.5 × 10^(10))²
V_o = 0.7543 V
B) Now, formula for the width of the depletion region (W)is given as;
W = √(2ε_s/q[(1/N_A) + (1/N_D)]V_o)
Where;
ε_s is the permittivity in the semiconductor with a constant value of 1.04 × 10^(-12) F/cm
q is the electron charge = 1.6 × 10^(-19) C
Thus;
W = √(2 × 1.04 × 10^(-12)/(1.6 × 10^(-19)) [(1/10^(17)) + (1/10^(16)]0.7543)
W = 32.842 × 10^(-6) cm
Converting to m gives;
W = 328.42 nm
C) Formula for extent of the n region is;
X_n = W × (N_A/(N_A + N_D))
X_n = 32.842 × 10^(-6) × (10^(17)/(10^(17) + 10^(16))
X_n = 29.856 × 10^(-6) cm
Converting to m gives;
X_n = 298.56 nm
Formula for extent of the p region is;
X_p = W - X_n
X_p = 328.42 nm - 298.56 nm
X_p = 29.86 nm
D) The charge stored on either side of the junction is given by the formula;
Q_j = Aq[(N_A × N_D)/(N_A + N_D)]W
Where A is junction Area and we are given junction area as 10 µm² = 100 × 10^(-8) cm
Thus;
Q_j = (100 × 10^(-8) × 1.6 × 10^(-19))[(10^(17) × 10^(16))/(10^(17) + 10^(16)] × 32.842 × 10^(-6)
Q_j = 47.767 × 10^(-15)
Steam at 20 bars is in the saturated vapor state (call this state 1) and contained in a pistoncylinderdevice with a volume of 0.03 m3. Assume the steam is cooled at constantvolume (i.e. the piston is held fixed in place) until the temperature reaches 200 C (callthis state 2). Then the steam is expanded isothermally until its volume is three times theinitial value (state 3).
Required:
a. Determine the pressures at state 2 and 3. ans. 15.5 bar, ~10 bar
b. Determine the change in specific internal energy, u, for each of the two processes.
-389 kJ/kg, 410 kJ/kg
c. Make qualitatively correct sketches of the processes on a T-v plot.
Answer:
Explanation:
Given that:
At state 1:
Pressure P₁ = 20 bar
Volume V₁ = 0.03 [tex]\mathbf{m^{3}}[/tex]
From the tables at saturated vapour;
Temperature T₁ = 212.4⁰ C ; [tex]v_1 = vg_1[/tex] = 0.0996 [tex]\mathbf{m^{3}}[/tex] / kg
The mass inside the cylinder is m = 0.3 kg, which is constant.
The specific internal energy u₁ = ug₁ = 2599.2 kJ/kg
At state 2:
Temperature T₂ = 200⁰ C
Since the 1 - 2 occurs in an isochoric process v₂ = v₁ = 0.099 [tex]\mathbf{m^{3}}[/tex] / kg
From temperature T₂ = 200⁰ C
[tex]v_f_2 = 0.0016 \ m^3/kg[/tex]
[tex]vg_2 = 0.127 \ m^3/kg[/tex]
Since [tex]vf_2 < v_2<vg_2[/tex] , the saturated pressure at state 2 i.e. P₂ = 15.5 bar
Mixture quality [tex]x_2 = \dfrac{v_2-vf_2}{vg_2 -vf_2}[/tex]
[tex]x_2 = \dfrac{(0.099-0.0016)m^3/kg}{(0.127 -0.0016) m^3/kg}[/tex]
[tex]x_2 = \dfrac{(0.0974)m^3/kg}{(0.1254) m^3/kg}[/tex]
[tex]\mathsf{x_2 =0.78}[/tex]
At temperature T₂, the specific internal energy [tex]u_f_2 = 850.6 \ kJ/kg[/tex] , also [tex]ug_2 = 2594.3 \ kJ/kg[/tex]
Thus,
[tex]u_2 = uf_2 + x_2 (ug_2 -uf_2)[/tex]
[tex]u_2 =850.6 +0.78 (2594.3 -850.6)[/tex]
[tex]u_2 =850.6 +1360.086[/tex]
[tex]u_2 =2210.686 \ kJ/kg[/tex]
At state 3:
Temperature [tex]T_3=T_2 = 200 ^0 C ,[/tex]
[tex]V_3 = 2V_1 = 0.06 \ m^3[/tex]
Specific volume [tex]v_3 = 0.2 \ m^3/kg[/tex]
Thus; [tex]vg_3 =vg_2 = 0.127 \ m^3/kg[/tex] ,
SInce [tex]v_3 > vg_3[/tex], therefore, the phase is in a superheated vapour state.
From the tables of superheated vapour tables; at [tex]v_3 = 0.2 \ m^3/kg[/tex] and T₃ = 200⁰ C
The pressure = 10 bar and v =0.206 [tex]\ m^3/kg[/tex]
The specific internal energy [tex]u_3[/tex] at the pressure of 10 bar = 2622.3 kJ/kg
The changes in the specific internal energy is:
[tex]u_2-u_1[/tex]
= (2210.686 - 2599.2) kJ/kg
= -388.514 kJ/kg
≅ - 389 kJ/kg
[tex]u_3-u_2[/tex]
= (2622.3 - 2210.686) kJ/kg
= 411.614 kJ/kg
≅ 410 kJ/kg
We can see the correct sketches of the T-v plot showing the diagrammatic expression in the image attached below.
Steam heated at constant pressure in a steam generator enters the first stage of a supercritical reheat cycle at 28 MPa, 5208C. Steam exiting the first-stage turbine at 6 MPa is reheated at constant pressure to 5008C. Each turbine stage has an isentropic efficiency of 78% while the pump has an isentropic efficiency of 82%. Saturated liquid exits the condenser that operates at constant pressure of 6 kPa. Determine the quality of the steam exiting the second stage of the turbine and the thermal efficiency.
This question is incomplete, the complete question is;
Steam heated at constant pressure in a steam generator enters the first stage of a supercritical reheat cycle at 28 MPa, 520°C. Steam exiting the first-stage turbine at 6 MPa is reheated at constant pressure to 500°C. Each turbine stage has an isentropic efficiency of 78% while the pump has an isentropic efficiency of 82%. Saturated liquid exits the condenser that operates at constant pressure of 6 kPa.
Determine the quality of the steam exiting the second stage of the turbine and the thermal efficiency.
Answer:
- the quality of the steam exiting the second stage of the turbine is 0.9329
- the thermal efficiency is 36.05%
Explanation:
get the properties of steam at pressure p1 = 28 MPa and temperature T2 = 520°C .
Specific enthalpy h1= 3192.3 kJ/kg
Specific entropy s1 = 5.9566 kJ/kg.K
Process 1 to 2s is isentropic expansion process in the turbine
S1 = S2s
get the enthalpy at state 2s at pressure p2 = 6 MPa and S2s = 5.9566 kJ/kg.K
h2s = 2822.2 kJ/kg
get the enthalpy at state 2 using isentropic turbine efficiency of the turbine. nT1 = (h1 - h2) / (h1 - h2s)
0.78 = (3192.3 - h2) / (3192.3 - 2822.2)
h2 = 2903.6 kJ/kg
get the enthalpy at state 3 at pressure p2 = p3 = 6 MPa and T3 = 500°C
h3 = 3422.2 kJ/kg
s3 = 6.8803 kJ/kg.K
Process 3 to 4s is isentropic expansion process in the turbine
S3 = S4s
get the enthalpy at state 4s at pressure p4s = p4 = 6 kPa and S4s = 6.8803 kJ/kg.K
h4s = 2118.8 kJ/kg
get the enthalpy at state 4 using isentropic turbine efficiency of the turbine. nT2 = (h3 - h4) / (h3 - h4s)
0.78 = (3422.2 - h4) / ( 3422.2 - 2118.8 )
h4 = 2405.5 kJ/kg
get the properties at pressure, p5 = 6 kPa
h5 = hf
= 151.53 kJ/kg
v5 = Vf
= 0.0010064 m³/kg
get the enthalpy at state 6 using isentropic pump efficiency of the turbine, at
p6 = p1 = 28 MPa
np = v5( p6 - p5) / (h6 - h5)
0.82 = ((0.0010064)( 28000 - 6)) / (h6 - 151.53)
h6 = 185.89 kJ/kg
Now to find the quality of the steam at the exit of the second stage of the turbine
At stat4, p4 = 6kPa
h4f = 151.53 kJ/kg
h4fg = 2415.9 kJ/kg
h4 = h4f + x4h4fg
2405.5 = 151.53 + (x4 (2415.9))
x4 = 0.9329
the quality of the steam exiting the second stage of the turbine is 0.9329
Also to find the efficiency of the power plant, we use the following equation;
n = Wnet / Qin
= (Wt1 + Wt2 - Wp) / (Q61 + Q23)
= [(h1 - h2) + (h3 - h4) - (h6 - h5)] / [(h1 - h6) + (h3 - h2)]
[(3192.3 - 2903.6) + (3422.2 - 2405.5) - (185.89 - 151.53)] / [(3192.3 - 185.89) + (3422.2 - 2903.6)]
= 0.3605
n = 36.05%
therefore the thermal efficiency is 36.05%
A house is maintained at a comfortable temperature by means of an electrical resistor heater during winter. The heater is operated at a constant current (I) under an applied voltage of 110 V. For simplicity, let us assume that the heater operates at a steady state, with 100% of conversion efficiency from the electrical energy to the internal energy of air in the house.
Required:
Find if the instructor pays $8.80/day for heating, with electricity cost $0.09/kWh.
b. Calculate the heat produced by such a heater hourly.
Answer:
$8.89
Explanation:
$8.80/a day for heat and $0.09kwh for electricity
A process engineer performed jar tests for a water in order to determine the optimal pH and dose using alum. A test was conducted by first dosing each jar with the same alum dose of 10 mg/L and varying pH in each jar from 5.0 to 7.5 with an increment of 0.5 unit of pH. After the first test, he/she plotted the results of remaining turbidity versus pH and found the optimal pH was 6.25. He/she continuously perform a second set of jar tests by holding the optimal pH of 6.25 constant and varying alum doses from 10 to 15 mg/L. Here are the results:
Results of Jar Tests for raw water at optimal pH of 6.25
Turbidity of raw water = 15 NTU
Alkalinity of raw water = 5 mg/L expressed as CaCO3.
Alum Dose 10 11 12 13 14 15
(mg/L)
Turbidity 5.0 4.6 4.5 3.0 5.0 6.0
Remaining
(NTU)
Determine:
1) Plot turbidity versus dose and find the optimal dose of alum (mg/L) with the water lowest remaining turbidity.
2) The theoretical amount of alkalinity consumed at the optimal dose expressed as CaCO3, mg/L.
3) Compared with the theoretical alkalinity from the above results, is the raw water alkalinity sufficient for the coagulation? If not, what kinds of chemical do you need in order to enhance the alkalinity?
Answer:
1) 13 mg/liters
2) 72.22 mg/lit
3) The Alkanity in raw water is not sufficient enough and the kinds of chemicals needed to enhance its acidity are
CaOKOHNa2CO3NaOHCO2NaHCO3Explanation:
1) plot of turbidity versus dose and optimal dose of Alum ( mg/L )
Optimal dose of Alum = 13 mg/liters from the graph attached below
2) Theoretical amount of alkalinity consumed at the optimal dose can be calculated as follows
Alkanity is due to [tex]HCO^-_{3}[/tex]
given optimal dose of Alum = 13 mg/liters for question 1
I mole of alum = 2 moles of AL(OH)3
666 grams of alum = 2*27 = 54 grams of AL(OH)3
hence 1 mole of [tex]AL^{+3}[/tex] = (13/54 ) mMole / lit
The moles of HCO3 = 6 * [tex]\frac{13}{54}[/tex] because 1 mole of Alum reacts with 6 moles of HC03
[HCO3] as CaCO3 = 6 * (13/54) * 50
= 72.22 mg/lit (theoretical amount of alkalinity consumed)
3) The Alkanity in raw water is not sufficient enough and the kinds of chemicals needed to enhance its acidity are
CaOKOHNa2CO3NaOHCO2NaHCO3