Answer: a. 0.4667
b. 0.4667 and C 0.0667
Explanation:
Given Data:
N = population size (10)
n = random selection (2)
r = number of observations = 7
Therefore
f(y) = ( r/y ) ( N - r / n - y ) / ( N /n )
When y = 1
f(1) = ( 7/1 ) ( 10 - 7 / 2 -1 ) / ( 10/2 )
= 7 / 15
= 0.4667
When y = 2
f(2) = ( 7/2 ) ( 10 - 7 / 2 -2 ) / ( 10/2 )
= 7 / 15
= 0.4667
When y = 0
f(0) = ( 7/0 ) ( 10 - 7 / 2 -0) / ( 10/2 )
= 1 / 15
= 0.0667
Steam enters an adiabatic condenser (heat exchanger) at a mass flow rate of 5.55 kg/s where it condensed to saturated liquid water at P = 20 kPa. The change in enthalpy of the steam is – 2,491 kJ/kg. The steam is cooled by water from a river. Environmental regulations require that the maximum increase in the river water temperature is 10°C.
a) What is the minimum mass flow rate of river water through the condenser to cool the steam? The Cp for water = 4.184 kJ/kg/K.
Answer:
The minimum mass flow rate will be "330 kg/s".
Explanation:
Given:
For steam,
[tex]m_{s}=5.55 \ kg/s[/tex]
[tex]\Delta h=2491 \ kg/kj[/tex]
For water,
[tex]\Delta T=10^{\circ}C[/tex]
[tex](Cp)_{w}=4.184 \ kJ/kg^{\circ}C[/tex]
They add energy efficiency as condenser becomes adiabatic, with total mass flow rate of minimal vapor,
⇒ [tex]m_{s}\times (\Delta h)=M_{w}\times(Cp)_{w}\times \Delta T[/tex]
On putting the estimated values, we get
⇒ [tex]5.55\times 2491=M_{w}\times 4.184\times 10\\[/tex]
⇒ [tex]13825.05=M_{w}\times 41.84[/tex]
⇒ [tex]M_{w}=330 \ kg/s[/tex]
Find the requested quantities for the circuit. We used the mesh-current method to identify the meshes. We then identified the mesh currents and wrote a KVL equation for each mesh and a constraint equation for the dependent source that defines its controlling variable in terms of the mesh currents. We solved these equations simultaneously for the unknown mesh currents and constrained current, and we checked the solution by verifying that the power in the circuit balances Now use the mesh-current values to calculate the voltage v0 and the total power generated in the circuit. Enter your answers directly on the figure.
Answer:
Explanation:
The image that is supposed to be attached to the question is displayed in the diagram below.
Applying Nodal Analysis at node 1;
[tex]\dfrac{V_o -50}{12.5*10^{-3}} + \dfrac{V_o}{50*10^3}+\dfrac{V_o-7500 \ in}{10*10^3}=0[/tex]
where;
[tex]in = \dfrac{V_o}{50*10^3}[/tex] (from the circuit)
= [tex]\dfrac{V_o-50}{12.5}+\dfrac{V_o}{50} + V_o -\dfrac{7500 *V_o }{\frac{50*10^3}{10}}=0[/tex]
= [tex]V_o [ \dfrac{1}{12.5}+\dfrac{1}{50}+\dfrac{1}{10}-\dfrac{75}{500}] = \dfrac{50}{12.5}[/tex]
= [tex]V_o[ \dfrac{500*500+12.5*5000+12.5*5000*5-75*12.5*500}{12.5*50*10*500}]= \dfrac{50}{12.5}[/tex]
= [tex]V_o = 80 \ volts[/tex]
[tex]in = \frac{80}{50*10^3}= 1.6 mA \\ \\ 7500*in = 120 volts \\ \\ I = \frac{120-80}{10(10^3} =4*10^{-3} Amps \\ \\ \\ \\ P_{generated} = 75000*in*I \\ \\ P_{generated} = 120*4*10^{-3} \\ \\ P_{generated} = 480 \ MW[/tex]
A. A solidified lava flow containing zircon mineral crystals is present in a sequence of rock layers that are exposed in a hillside.A mass spectrometer analysis was used to count the atoms of uranium-235 and lead-207 isotopes in zircon samples from thelava flow. The analysis revealed that 71% of the atoms were uranium-235, and 29% of the atoms were lead-207. Refer toFIGURE 8.11 to help you answer the following questions.1. About how many half-lives of the uranium-235to lead-207 decay pair have elapsed in the zircon crystals? ______________2. What is the absolute age of the lava flow based onits zircon crystals? Show your calculations.3. What is the age of the rocklayers above the lava flow? _______________4. What is the age of the rocklayers beneath the lava flow? _______________B. Astronomers think that Earth probably formed at the same time as
Answer:
1. 0.494
2. = 352.299 million years
3. It is between 0 years to 352.299 million years
4. greater than the 352.299 million years.
Explanation:
Given
Uranium 235 ———— lead 207( zircon sample)
t= 0 100%. 0%
t= t. 71%. 29%
1)- half lives elapsed = n
(1/2)n = 0.71
By taking log and solving n = 0.494
No of half lives = 0.494
2) Calculating the age of the larva
- age of lava flow = half life of uranium 235 x n
= 713 x 0.494
= 352.299 million years
3)- The rock layers was created above this lava flow is of later occurrence so the age will be lower than that of the age of the lava flow so the age of rock over the lava flow will lower than the 352.299 million years it will be in between 0 years to 352.299 million years.
4)- The rock layers created underneath this lava flow is of earlier occurrence so its age is more than that of the age of the lava flow so the age of rock will more than the lava flow is greater than the 352.299 million years.
B. The astronomers think the earth was created as all of the other rock materials in our solar system, including the oldest meteorites. The oldest meteorites ever found on Earth contain nearly equal amounts of both Uranium-238 and Lead-206.
The number of half-lives of the uranium-235to lead-207 decay pair have elapsed in the zircon crystals is; 0.494
What is the number of half lives?We are given;
The atoms of Uranium-235 and lead-207 which made up the zircon sample.
At t = 0; Uranium-235 is 100% while lead-207 is 0%
At t = t; Uranium-235 is 71% while lead-207 is 29%
1) Let the half lives that elapsed be n. Thus;
(¹/₂)ⁿ = 0.71
n*log0.5 = log 0.71
n = (log 0.71)/(log 0.5)
n = 0.494
Thus;
Number of half lives = 0.494
2) Formula to get the absolute age of the larva is;
Absolute age age of lava flow = half life of uranium-235 * n
The half life of uranium-235 is 713 million years. Thus;
Absolute age age of lava flow = 713 * 0.494
Absolute age of lava flow = 352.22 million years
3) The rock layers above the lava flow were created after the lava flow and so the age will be lower than that of the age of the lava flow. Thus, it's age will be between 0 years and 352.299 million years.
4) A: The rock layers beneath the lava flow were in existence earlier than the lava flow and as such, the age of rock layers beneath the lava flow will be greater than 352.22 million years.
B; The astronomers think the earth was created as all of the other rock materials in our solar system.
Read more about Half life at; https://brainly.com/question/26148784
A certain printer requires that all of the following conditions be satisfied before it will send a HIGH to la microprocessor acknowledging that it is ready to print: 1. The printer's electronic circuits must be energized. 2. Paper must be loaded and ready to advance. 3. The printer must be "on line" with the microprocessor. As each of the above conditions is satisfied, a HIGH is generated and applied to a 3-input logic gate. When all three conditions are met, the logic gate produces a HIGH output indicating readiness to print. The basic logic gate used in this circuit would be an): A) NOR gate. B) NOT gate. C) OR gate. D) AND gate.
Answer:
D) AND gate.
Explanation:
Given that:
A certain printer requires that all of the following conditions be satisfied before it will send a HIGH to la microprocessor acknowledging that it is ready to print
These conditions are:
1. The printer's electronic circuits must be energized.
2. Paper must be loaded and ready to advance.
3. The printer must be "on line" with the microprocessor.
Now; if these conditions are met the logic gate produces a HIGH output indicating readiness to print.
The objective here is to determine the basic logic gate used in this circuit.
Now;
For NOR gate;
NOR gate gives HIGH only when all the inputs are low. but the question states it that "a HIGH is generated and applied to a 3-input logic gate". This already falsify NOR gate to be the right answer.
For NOT gate.
NOT gate operates with only one input and one output device but here; we are dealing with 3-input logic gate.
Similarly, OR gate gives output as a high if any one of the input signals is high but we need "a HIGH that is generated and applied to a 3-input logic gate".
Finally, AND gate output is HIGH only when all the input signal is HIGH and vice versa, i.e AND gate output is LOW only when all the input signal is LOW. So AND gate satisfies the given criteria that; all the three conditions must be true for the final signal to be HIGH.
Why does a BJT transistor require detailed calculations for its base resistor value to operate?
Answer: Because of the role the base region play in the transistor.
Explanation:
The base region of BJT transistor - an opposite polarity charge carrier from emitter region to collector region, plays a vital role in triggering for a sufficient emiter - to - collector current.
The current received by the base region of BJT determines the effect of the continue flow of current into the collector region which will eventually determine the output current.
Indicate the correct statement about the effect of Reynolds number on the character of the flow over an object.
If Reynolds number is high enough the effect of viscosity is negligible and the fluid flows over the plate without sticking to the surface.
If Reynolds number increases the extent of the region around the object that is affected by viscosity decreases.
If Reynolds number is low enough the effect of viscosity is so high that there is a region near the plate where the fluid is stationary.
If Reynolds number increases the size of the region around the object that is affected by viscosity increases.
Answer:
If Reynolds number increases the extent of the region around the object that is affected by viscosity decreases.
Explanation:
Reynolds number is an important dimensionless parameter in fluid mechanics.
It is calculated as;
[tex]R_e__N} = \frac{\rho vd}{\mu}[/tex]
where;
ρ is density
v is velocity
d is diameter
μ is viscosity
All these parameters are important in calculating Reynolds number and understanding of fluid flow over an object.
In aerodynamics, the higher the Reynolds number, the lesser the viscosity plays a role in the flow around the airfoil. As Reynolds number increases, the boundary layer gets thinner, which results in a lower drag. Or simply put, if Reynolds number increases the extent of the region around the object that is affected by viscosity decreases.
Mr. auric goldfinger, criminal mastermind, intends to smuggle several tons of gold across international borders by disguising it as lumps of iron ore. he commands his engineer minions to form the gold into little spheres with a diameter of exactly and paint them black. however, his chief engineer points out that customs officials will surely notice the unusual weight of the "iron ore" if the balls are made of solid gold (density ). he suggests forming the gold into hollow balls instead (see sketch at right), so that the fake "iron ore" has the same density as real iron ore one of the balls of fake "iron ore," sliced in half. calculate the required thickness of the walls of each hollow lump of "iron ore." be sure your answer has a unit symbol, if necessary, and round it to significant digits.
Answer:
The thickness of the walls of each hollow lump of "iron ore" is 2.2 cm
Explanation:
Here we have that the density of solid gold = 19.3 g/cm³
Density of real iron ore = 5.15 g/cm³
Diameter of sphere of gold = 4 cm
Therefore, volume of sphere = 4/3·π·r³ = 4/3×π×2³ = 33.5 cm³
Mass of equivalent iron = Density of iron × Volume of iron = 5.15 × 33.5
Mass of equivalent iron = 172.6 cm³
∴ Mass of gold per lump = Mass of equivalent iron = 172.6 cm³
Volume of gold per lump = Mass of gold per lump/(Density of the gold)
Volume of gold per lump = 172.6/19.3 = 8.94 cm³
Since the gold is formed into hollow spheres, we have;
Let the radius of the hollow sphere = a
Therefore;
Total volume of the hollow gold sphere = Volume of gold per lump - void sphere of radius, a
Therefore;
[tex]33.5 = 8.94 - \frac{4}{3} \times \pi \times a^3[/tex]
[tex]\frac{4}{3} \times \pi \times a^3 = 33.5 - 8.94[/tex]
[tex]a^3 = \frac{24.6}{\frac{3}{4} \pi } = 5.9[/tex]
a = ∛5.9 = 1.8
The thickness of the walls of each hollow lump of "iron ore" = r - a = 4 - 1.8 = 2.2 cm.
42. A vehicle has sagged rear springs and reduced rear curb riding height. This problem results
in?
A. Excessive positive camber on the front wheels
B. Excessive toe-out on the front wheels
C. Excessive positive caster on the front wheels
D. Excessive toe-in on the front wheels
43. On a vehicle equipped with rear parallel leaf springs and a solid rear axle, customer is
complaining that the vehicle reacts erratically (it darts) during turns. What is the most likely
cause of this complaint?
A. Incorrect ride height
B. Incorrect driveline angle
C. Loose rear axle U-bolts
D. Missing jounce/rebound bumpers
44. A power steering pump is being tested with a pressure gauge for maximum output pressure.
Which of the following statements is correct?
A. The pressure gauge should be attached to the pump return port
B. The steering wheel should be held in the right lock position
C. A maximum output pressure dead heading test should last no longer than 5 seconds
D. One should check output pressure with engine speed above 4000rpm
Answer:1. Driving under the influence of any drug that makes you drive unsafely is:
a. Permitted if it is prescribed by a doctor
b. Against the law
c. Permitted if it is a diet pill or cold medicine
2. Which fires can you put out with water:
a. Tire fires
b. Gasoline fires
c. Electrical fires
3. How far should a driver look ahead of the vehicle while driving:
a. 9-12 seconds
b. 12-15 seconds
c. 18-21 seconds
4. To prevent shifting, there should be at least one tie-down for ever ____feet of cargo: a. 10
b. 15 c. 18
5. Which of these statements about downshifting is true:
a. When you downshift for a curve, you should do so before you enter the curve
b. When you downshift for a hill, you should do so after you start down the hill
c. When you downshift for a curve, you should do so after you enter the curve
6. How do you test hydraulic brakes for a leak:
a. Move the vehicle slowly and see if it stops when the brake is applied
b. With the vehicle stopped, pump the pedal three time, apply pressure then hold
For five seconds and see it the pedal moves.
c. Step on the brake pedal and accelerator at the same time and see if the vehicle moves
7. For an average driver, driving 55 MPH on dry pavement, it will take about _____ to bring The vehicle to a stop:
a. Twice the length of the vehicle
b. Half the length of a football field
c. The length of a football field
8. You are driving a vehicle with a light load, traffic is moving at 35 MPH in a 55 MPH zone. The safest speed for your vehicle in this situation is most likely:
a. 30 MPH
b. 35 MPH
c. 40 MPH
9. Which of these is a good rule to follow when driving at night:
a. Keep your speed slow enough to stop within the range of your headlights
b. Look directly at oncoming headlights
c. Keep your instrument lights bright
10. A moving vehicle ahead of you has a red triangle with an orange center on the rear. What does this mean?
a. The vehicle is hauling hazardous materials
b. It may be a slow-moving vehicle
c. It may be oversized
11. You wish to turn right form a two lane two way street to make the turn. Which of these Drawings show how the turn should be made.
12. You are driving a heavy vehicle and must exit a highway using an offramp that curves downhill:
a. Use the posted speed limit for the offramp
b. Slow down to a safe speed before the turn
c. Wait until you are in the turn before downshifting
13. Which of these statements about using mirrors is true:
a. You should look at a mirror for several seconds at a time
b. There are “blind spots” that your mirror cannot show you
c. A lane change requires you to look at the mirrors twice
14. You must park on the side of a level, straight, two-lane road. Where should you place the three reflective triangles?
a. one within 10 feet of the rear of the vehicle, one about 100 feet to the rear and one about 200 feet to the rear.
b. One with 10 feet of the rear of the vehicle, one about 100 feet to the rear and one about 100 feet from the front of the vehicle
c. One about 50 feet from the rear of the vehicle, one about 100 feet to the rear and one about 100 feet from the front of the vehicle
15. Your vehicle is in a traffic emergency and may collide with another vehicle if you do not take action. Which of these is a good rule to remember at such a time?
a. Stopping is always the safest action in a traffic emergency
b. Heavy vehicles can almost always turn more quickly than they can stop
c. Leaving the road is always more risky than hitting another vehicle
16. The most important reason for being alert to hazards is:
a. Law enforcement personnel can be called
b. You will have time to plan your escape if the hazard becomes an emergency
c. You can help impaired drivers
17. You are traveling down a long, steep hill. Your brakes begin to fade and then fail. What should you do?
a. Downshift
b. Pump the brake pedal
c. Look for an escape ramp or escape route
18. The most common cause of serious vehicle skids is:
a. Driving too fast for road conditions
b. Poorly adjusted brakes
c. Bad tires
19. To avoid a crash, you had to drive onto the right shoulder. You are now driving at 40 MPH on the shoulder. How should you move back onto the pavement?
a. If clear, come to a complete stop before steering back onto the pavement
b. Brake had to slow the vehicle, then steer sharply onto the pavement
c. Keep moving at the present speed and steer very gently back onto the pavement
20. If
a. Slide sideways and spin out
b. Go straight ahead but will turn if you turn the steering wheel
c. Go straight ahead even if the steering wheel is turned
A small family home in Tucson, Arizona, has a rooftop area of 1967 square feet, and it is possible to capture rain falling on about 56% of the roof. A typical annual rainfall is about 14 inches. If the family wanted to install a tank to capture the rain for an entire year, without using any of it, what would be the required volume of the tank in m3 and in gallons? How much would the water weigh when the tank was full (in N and in lbf)?
Answer:
V = 36.4 m³ = 4.86 gallons
W = 80193.88 lbf = 356720 N = 356.72 KN
Explanation:
We have the following data given in the question:
At = Total area of roof = 1967 ft²
h = Annual Rainfall = 14 inches = 1.17 ft
V = Volume of tank in m³ and gallons = ?
W = Weight of water in N and lbf = ?
So, for volume we know that the area of roof that receives rainfall is 56% of total area and 14 inches of annual rainfall means that there is a standing height of 14 inches of rain water for a given area, for 1 year.
Area to receive rain = A = 0.56*1967 ft² = 1101.52 ft²
Now,
Volume = V = A * h = 1101.52 ft²)(1.17 ft)
V = 1285.11 ft³
Converting to m³:
V = (1285.11 ft³)(1 m³/35.3147 ft³)
V = 36.4 m³
Converting to gallons:
V = (1285.11 ft³)(1 m³/264.172 gal)
V = 4.86 gal
Now, for the weight of water, we use formula:
W = ρVg
where,
W = weight of water = ?
ρ = Density of water = 1000 kg/m³
V = Volume of tank = 36.4 m³
g = 9.8 m/s²
Therefore,
W = (1000 kg/m³)(36.4 m³)(9.8 m/s²)
W = 356720 N = 356.72 KN
Converting to lbf:
W = (356720 N)(1 lbf/4.44822 N)
W = 80193.88 lbf
Wheels A and B have weights of 150 lb and 100 lb , respectively. Initially, wheel A rotates clockwise with a constant angular velocity of 100 / A rad s and wheel B is at rest. If A is brought into contact with B, determine the time required for both wheels to attain the same angular velocity. The coefficient of kinetic friction between the two wheels is 0.3 k and the radii of gyration of A and B about their respective centers of mass are 1 A k ft and 0.75 B k ft . Neglect the weight of link AC.
The image attached that is supposed to be attached to the question is shown in the first file below.
Answer:
t = 2.19 seconds
Explanation:
The free body diagram showing the center of mass A and B is attached in the second diagram below.
NOTE : that from the second diagram; Mass A and B do not have any acceleration
Taking the moment about wheel A:
[tex]\sum M_A = I_A \alpha _A[/tex]
[tex]-f(r_A) = I_A \alpha _A ----- (1)[/tex]
The equilibrium forces in the y-direction is 0
i.e
[tex]F_y = 0[/tex]
So;
[tex]N +T sin 30^0 -W_A = 0 ----- (2)[/tex]
The equilibrium forces in the x-direction is as follows:
[tex]\sum F_x = 0[/tex]
[tex]Tcos 30^0 + f= 0 -----(3)[/tex]
The kinetic friction f can be expressed as :
[tex]f = \mu _k N[/tex]
From above equation (2) and equation (3);
[tex]N + [\dfrac{-f}{cos 30^0}]sin 30^0 -150 =0[/tex]
[tex]N - \mu _k N \ tan 30^0 -150 =0[/tex]
[tex]N = \dfrac{150}{1-0.3 \ tan 30^0}[/tex]
N = 181.423 lb
Similarly; from equation(1)
[tex]\alpha_A = - \dfrac{f(r_A)}{I_A}[/tex]
[tex]\alpha _A = \dfrac{-\mu_k N(r_A)}{I_A}[/tex]
[tex]\alpha _A = \dfrac{-0.3*181.423*1.25}{\frac{150}{32.2}*I^2}[/tex]
[tex]\alpha _A =-14.6045 \ rad/s^2[/tex]
However; from the kinematics ; as moments are constant ; so is the angular acceleration is constant )
Thus;
[tex]\omega _A - \omega_o^A = \alpha_A t[/tex]
[tex]\omega _A = \omega_o^A + \alpha_A t[/tex]
[tex]\omega _A = 100 -14.6045 \ t ---- (4)[/tex]
Let's take a look at wheel B now;
Taking the moment about wheel B from the equation of motion:
[tex]\sum M_B = I_B \alpha _B[/tex]
[tex]f(r_B) = I_B \alpha _B[/tex]
[tex]\mu_k N (r_B) = I_B \alpha_B[/tex]
[tex]\mu_k N (r_B) = \dfrac{W_B}{g}* k^2_B \alpha_B[/tex]
[tex]\alpha_B = \dfrac{0.3*181.423*1}{\frac{100}{32.2}*0.75^2}[/tex]
[tex]\alpha = 31.1563 \ rad/s^2[/tex]
Again; from the kinematics; as the moments are constant which lead to the angular accleration;
[tex]\omega _B = \omega _o^B + \alpha _B \ t[/tex]
[tex]\omega _B =0 + 31.156 \ t-----(5)[/tex]
From equation 4 and 5 which attain the same angular velocity; we have;
[tex]\omega^A = \omega^B[/tex]
100 - 14.6045 t = 31.1563 t
100 = 31.1563 t + 14.6045 t
100 = 45.761 t
t = 100/45.761
t = 2.19 seconds
Experiment: With the battery voltage set to 15 volts, measure the current in a parallel circuit with 1, 2, 3, and 4 light bulbs. (In each case, place the ammeter next to the battery.) Use Ohm’s law to calculate the total resistance of the circuit. Record results below. Is this right?
Answer:
No
Explanation:
We expect current to be proportional to the number of identical bulbs. The total resistance is the ratio of voltage to current, so will be inversely proportional to the number of bulbs.
The current readings look wrong in that the first bulb caused the current to be 1 A, but each additional bulb increased it by 2 A. If that is what happened, the bulbs were not identical. That may be OK, but we expect the point of the experiment is to let you see the result described above.
In any event, the total resistance is not calculated properly. It should be the result of dividing voltage (15 V) by current.
Answer:
No, it is not right.
Explanation:
Your table is not consistent with bulbs of the same resistance.
Current comes from a measurement, but resistance comes from a calculation.
I presume that the measured currents are correct.
Ohm's Law states that the current flowing in a circuit is directly proportional to the voltage.
We usually write it as
V/I = R
1. One bulb in circuit
[tex]R = \dfrac{V}{I} = \dfrac{\text{15 V}}{\text{1 A}}= \mathbf{15 \, \Omega}[/tex]
2. Two bulbs
[tex]R = \dfrac{V}{I} = \dfrac{\text{15 V}}{\text{3 A}} = \mathbf{5 \, \Omega}[/tex]
3. Three bulbs
[tex]R = \dfrac{V}{I} = \dfrac{\text{15 V}}{\text{5 A}} = \mathbf{3 \, \Omega}[/tex]
4. Four bulbs
[tex]R = \dfrac{V}{I} = \dfrac{\text{15 V}}{\text{7 A}} = \mathbf{2.1 \, \Omega}[/tex]
Aerotron Electronics is considering purchasing a water filtration system to assist in circuit board manufacturing. The system costs $32,000. It has an expected life of 7 years at which time its salvage value will be $5,000. Operating and maintenance expenses are estimated to be $2,000 per year. If the filtration system is not purchased, Aerotron Electronics will have to pay Bay City $13,000 per year for water purification. If the system is purchased, no water purification from Bay City will be needed. Aerotron Electronics must borrow 1/2 of the purchase price, but they cannot start repaying the loan for 2 years. The bank has agreed to 3 equal annual payments, with the 1st payment due at the end of year 2. The loan interest rate is 8.5 % compounded annually. Aerotron electronics’ MARR is 12.5 % compounded annually.
Required:
a. What is the present worth of this investment? (Carry all interim calculations to 5 decimal places and then round your final answer to the nearest dollar. The tolerance is £10.)
b. What is the decision rule for judging the attractiveness of investments based on present worth?
c. Should Aerotron Electronics buy the water filtration system?
Answer:
Explanation:
a) Present worth of the system:
First step :
Calculation of bank installment:
We are given:
·nitial costs = $32,000
Borrow amount = ½ of purchase price
Payment of borrow amount = EOY 2 to EOY 4 (3 Equal installments)
Bank loan interest = 8.5% = 8.5/100 = 0.085
Assume the installment amount is F. They will be paid at end of year 2 to end of year 4. Their present value must be equal to borrow amount.
Present value of cost to be incurred in future can be calculated by below formula:
F P (1 + i)
F= Future cost
i = Rate of interest
n = time (in years)
Therefore,
F F $32,000 2 F (1 +0.085)2 (1 +0.085) 3 (1 +0.085) + +
.: 2.35394F = 16,000 or F = 16,000 2.35394 - $6,797.11
Step 2: Present worth of the system:
Given Data:
Initial costs = $32,000
Expected life = 7 years
Salvage value = $5,000
O&M Costs = $2,000 per year
MARR = 12.5%
= 12.5/100
= 0.125
Present value of uniform recurring payments is given by below formula:
P=A (1 + i) - 1 i(1+i)n 72
Where,
P = Present Value
A = Recurring payments per annum
i = rate of interest
n = time (in years)
Hence present value of O&M costs,
P1 = -2,000 x (1 + 0.125) 7-1 0.125 x (1 + 0.125) 7 -$8,984.60
Present worth of the system calculated in below table:
Description
F ($)
MARR (i)
per year
n (years)
P ($)
Initial investment (1/2 of purchase price)
-16,000.00
0.125
0
-16,000.00
Bank installment EOY2
-6,797.11
0.125
2
-5,370.56
Bank installment EOY3
-6,797.11
0.125
3
-4,773.83
Bank installment EOY4
-6,797.11
0.125
4
-4,243.40
Salvage value
5,000.00
0.125
7
2,192.31
O&M Costs
-8,984.60
0.125
0
-8,984.60
The current worth of the new system
-37,180.07
Part b) Decision rule of judgment:
Assuming current value of costs is lower than current value of benefit, an alternative is known to be economic to use based on current worth analysis.
Part c) Decision for the water filtration system:
Given Data:
· Annual savings from filtration system (A) = $13,000 per year
· Expected life (n) = 7 years
· MARR = 12.5%
= 12.5/100
= 0.125
Present value of benefits = 13,000 X (1 + 0.125)7 - 1 0.125 x (1 + 0.125) 7 $58,399.91
Since current value of costs is less than current value of benefit, this is an worthwhile system and has to be be purchased.
The steady-state data listed below are claimed for a power cycle operating between hot and cold reservoirs at 1200 K and 400 K, respectively. For each case, evaluate the net power developed by the cycle, in kW, and the thermal efficiency. Also in each case apply the equation below on a time-rate basis to determine whether the cycle operates reversibly, operates irreversibly, or is impossible.
(a) Qh(dot)=600 kW, Qc(dot)=400 kW
(b) Qh(dot)=600 kW, Qc(dot)=0 kW
(c) Qh(dot)=600 kW, Qc(dot)=200 kW
∮ (δQ/T)_b = -σ_cycle
Answer:
(a) Qh(dot)=600 kW, Qc(dot)=400 kW is an irreversible process.
(b) Qh(dot)=600 kW, Qc(dot)=0 kW is an impossible process.
(c) Qh(dot)=600 kW, Qc(dot)=200 kW is a reversible process.
Explanation:
T(hot) = 1200k, T(cold) = 400
efficiency n = (Th - Tc ) / Tc
n = (1200 - 400) / 1200 = 0.667 (this will be the comparison base)
(a)
Qh = 600 kW, Qc = 400 kW
n = (Qh - Qc) / Qh ⇒ (600 - 400) / 600
n = 0.33
0.33 is less than efficiency value from temperature 0.67
∴ it is irreversible process
(b)
Qh = 600 kW, Qc = 0
n = (Qh - Qc) / Qh ⇒ (600 - 0) / 600 = 1
efficiency in any power cycle can never be equal to one.
∴ it is an impossible process.
(c)
Qh = 600 kW, Qc = 200 kW
n = (Qh - Qc) / Qh = (600 - 200) / 600
n = 0.67 (it is equal to efficiency value from temperature)
∴ it is a reversible process
– A cloud customer has asked you to do a forensics analysis of data stored in on CSP’s server. The customer’s attorney explains that the CSP offers little support for data acquisition and analysis will help you with data collection for a fee. The attorney asks you to prepare a memo with detailed questions of what you need to know to perform the task .She plans to use this memo to negotiate for services you will provide in collecting and analyzing evidence .Write a one –to two page menu with questions to ask the CSP .
Answer:
A one -two pages menu was written with questions directed to the CSP which is stated below in the explanation section
Explanation:
Solution
If CSP has no team or limited staff, you will need to ask the following questions to understand how the CSP is set up:
Is detailed knowledge of cloud topology, storage devices is available ?Are there any restrictions in taking digital evidence from a cloud storage?For e-discovery demands on multi tenant cloud systems, is the data of investigation local or remote?Does the investigator have the power to make use of cloud staff conduct an investigation? What is the relationship of CSP's with cloud users?What are the SLA's and what are the guidelines to define them ? SLAs should also specify support options, penalties for services not provided, system performance,fees, provided software/hardware. CSP must explain who has the right to access the data ? and limitations for conducting acquisitions for an investigation.For guidelines of operations, digital forensics should review CSP's policies, and standards..What are the CSP's business continuity and disaster recovery plans.Are there Any plans to revise current laws ?Are there Any cases involving data commingling with other customer's data?Ask What law controls data stored in the cloud is a challenge?To access evidence in the cloud :
What is the configuration of the CSP?Is the data storage location secretly kept or it is open ?Are there any court orders, subpoenas with prior notice, search warrants etc?What are the procedures for log keeping ? so that complications we not arise in the investigations chain of evidence.What is the configuration of the CSP?What is the right key of encryption to read the data if at all the CSP has provided encryption to the data.Is there any threat from hackers so that they will not use any malware an modify the file meta data?Does CSP have a personnel trained to respond to network incidents?Who are the data owners, identity protection, users and access controls for a better role management.It is required to design and implement: 1. A counter which counts from 0 to 255 with seven segment display.
2. A logic function y=∑( 0,3,5,10,16,20,30 35).
3. Summing and subtracting circuit of 8 digit numbers.
4. ROM (a,b,c,d) to 60 to 75 in binary.
5. A timing module which counts time in us or ms or seconds.
Answer:
A counter which counts from 0 to 255 with seven segment display
Timer Mode Control (TMOD)
Explanation:
A spherical ball of solid, nonporous naphthalene, a "moth ball," is suspended in still air. The naphthalene ball slowly sublimes, releasing the naphthalene into the surrounding air by a diffusion limited process.
1. Estimate the time required to reduce the diameter from 2 cm to 0.5 cm. when the surrounding air is at 347 K and 1.013 x 10^5 Pa. Naphthalene has a molecular weight of 128 g/mole, a solid density of 1.145 g/cm^3, a diffusion coefficient in air of 8.19 x 10^-6 m^2/s, and exerts a vapor pressure of 5 torr (670 Pa) at 347 K.
Answer:
61.6 hours will be needed to reduce the diameter of solid spherical ball from 2 cm to 0.50 cm.
Explanation:
Find the given attachments
A circuit-switching scenario in whichNcs users, each requiring a bandwidth of 25 Mbps, must share a link of capacity 150 Mbps.
A packet-switching scenario withNps users sharing a 150 Mbps link, where each user again requires 25 Mbps when transmitting, but only needs to transmit 10 percent of the time.
What is the probability that a given (specific) user is transmitting, and the remaining users are not transmitting?
Answer:
0.09
Explanation:
Packet switching involves breaking a message into packets and sending them independently. Since the user only needs to transmit 10 percent of the time, the probability that a given (specific) user is transmitting = 10% = 0.1
The probability that a user is not transmitting = 100% - 10% = 90% = 0.9
Therefore, the probability that a given (specific) user is transmitting, and the remaining users are not transmitting = 0.1 * 0.9 = 0.09
Q#3:(A)Supose we extend the circular flow mode to add imports and export copy the circular flow digram onto a sheet paper and then add a foreign country as athird agent.Draw a through sketch of the flows of imports exports and the payment for each on your digrams?
Answer & Explanation:
Circular Flow model denotes how goods & services, factor incomes & prices move within sectors of economy.
A closed economy has two sectors - households & firms, having following features of circular flow between them:
Households provide factor services to firms , & get factor payments from firms in returnFirms provide goods & services to households, & get prices for households in returnIn case of open economy - with rest of world & foreign country, exports & imports also come in circular flow.
Firms export to foreign ROW, receive export payments from them. Households, firms import from foreign ROW, pay their import payments to them.What can you do to protect your hands from any hazards on your worksite? Select the 2 answer options that apply. Use your gloves correctly Avoid jobs that could hurt your hands Always wear gloves Inspect your gloves for any damage before wearing
Answer:
"Use your gloves correctly" and "Inspect your gloves for any damage before wearing."
Explanation:
If you don't use your gloves correctly you could risk your hands being infected/burned. If your gloves are damaged they will provide minimum protection but there is still a potential safety hazard.
Answer:
Use your gloves correctly
Inspect your gloves for any damage before wearing
Calibrations on a recent version of an operating system showed that on the client side, there is a delay of at least 0.5 ms for a packet to get from an application to the network interface and a delay of 1.4 ms for the opposite path (network interface to application buffer). The corresponding minimum delays for the server are 0.20 ms and 0.30 ms, respectively.
What would be the accuracy of a run of the Cristian's algorithm between a client and server, both running this version of Linux, if the round trip time measured at the client is 6.6 ms?
Answer:
4.2ms
Explanation:
Calibrated time= 0.3+0.2+0.5+1.4= 2.4
Measured time= 6.6ms
Accuracy is closeness of measurement to an observed or true value
Accuracy= 6.6-2.4= 4.2ms
Because A-B=A+ (-B), the subtraction of signed numbers can be accomplished by adding the complement. Subtract each of the following pairs of 5-bit binary numbers by adding the complement of the subtrahend to the minuend. Indicate when an overflow occurs. Assume that negative number are represented in 1’s complement. Then repeat using 2’s complement.
a) 01001-11010b) 11010-11001c) 10110-01101d) 11011-00111e) 11100-10101
Answer:
Using 1's complement
a)
Therefore the difference is -10001
b)
Therefore the difference is 00001
c)
Therefore the difference is 01001
d)
Therefore the difference is 10100
e)
Therefore the difference is 00111
Explanation:
Using 1's complement
a) The 1's complement of the subtrahend 11010 = 00101.
Therefore 01001-11010 = 01001 + 00101 = 01110
Since no overflow, we take the 1's complement of the result and it is negative.
Therefore the difference is -10001
b) The 1's complement of the subtrahend 11001 = 00110.
Therefore 11010-11001 = 11010 + 00110 =1 00000
Since there is an overflow, we add the overflow to the result
Therefore the difference is 00001
c) The 1's complement of the subtrahend 01101 = 10010
Therefore 10110-01101 = 10110 + 10010 = 1 01000
Since there is an overflow, we add the overflow to the result
Therefore the difference is 01001
d) The 1's complement of the subtrahend 00111 = 11000
Therefore 11011-00111= 11011 + 11000 = 1 10011
Since there is an overflow, we add the overflow to the result
Therefore the difference is 10100
e) The 1's complement of the subtrahend 10101 = 01010
Therefore 11100-10101= 11100 + 01010 = 1 00110
Since there is an overflow, we add the overflow to the result
Therefore the difference is 00111
Using 2's complement
a) The 2's complement of the subtrahend 11010 = 00110.
Therefore 01001-11010 = 01001 + 00110 = 01111
Since no overflow, we take the 2's complement of the result and it is negative.
Therefore the difference is -10001
b) The 2's complement of the subtrahend 11001 = 00111.
Therefore 11010-11001 = 11010 + 00111 =1 00001
Since there is an overflow, we drop the overflow
Therefore the difference is 00001
c) The 1's complement of the subtrahend 01101 = 10011
Therefore 10110-01101 = 10110 + 10011 = 1 01001
Since there is an overflow, we drop the overflow
Therefore the difference is 01001
d) The 1's complement of the subtrahend 00111 = 11001
Therefore 11011-00111= 11011 + 11001 = 1 10100
Since there is an overflow, we drop the overflow
Therefore the difference is 10100
e) The 1's complement of the subtrahend 10101 = 01011
Therefore 11100-10101= 11100 + 01011 = 1 00111
Since there is an overflow, we drop the overflow
Therefore the difference is 00111
It is proposed to absorb acetone from air using water as a solvent. Operation is at 10 atm and is isothermal at 20°C. The total flow rate of entering gas is 10 kmol /h. The entering gas is 1.2 mol% acetone. Pure water is used as the solvent. The water flow rate is 15 kmol/h. The desired outlet gas concentration should be 0.1 mol % acetone. For this system, Henry's law holds and Ye = 1.5 X where Ye is the mol fraction of acetone in the vapour in equilibrium with a mol fraction X in the liquid.
KGa = 0.4 kmol*m^-3*s^-1
1. Draw a schematic diagram to represent the process.
2. Determine the mole fraction of acetone in the outlet liquid.
Answer:
The meole fraction of acetone in the outlet liquid is [tex]x_1 = 0.0072[/tex]
Explanation:
1.
The schematic diagram to represent this process is shown in the diagram attached below:
2.
the mole fraction of acetone in the outlet liquid is determined as follows:
solute from Basis Gas flow rate [tex]G_s = 10(1-0.012) =9.88 kmol/hr[/tex]
Let the entering mole be :[tex]y_1 = 1.2[/tex] % = 0.012
[tex]y_1 =(\dfrac{y_1}{1-y_1})[/tex]
[tex]y_1 =(\dfrac{0.012}{1-0.012})[/tex]
[tex]y_1 =0.012[/tex]
Let the outlet gas concentration be [tex]y_2[/tex] = 0.1% = 0.001
[tex]y_2 = 0.001[/tex]
Thus; the mole fraction of acetone in the outlet liquid is:
[tex]G_s y_1 + L_s x_2 = y_2 L_y + L_s x_1[/tex]
[tex]9.88(0.012-0.001)=15*x_1[/tex]
[tex]9.88(0.011) = 15x_1[/tex]
[tex]x_1 = \dfrac{0.10868}{15}[/tex]
[tex]x_1 = 0.0072[/tex]
The mole fraction of acetone in the outlet liquid is [tex]x_1 = 0.0072[/tex]
If someone told you that a certain AC circuit was a capacitive, you would know that in that circuit the current
A) current and voltage are zero
B) leads the voltage
C) and voltage are in phrase
D) lags the voltage
Answer:
B) leads the voltage
Explanation:
One way to think about it is that the current causes charge to be accumulated on the capacitor, changing its voltage. The current must be non-zero before the voltage can change. Hence current leads voltage.
Helium is used as the working fluid in a Brayton cycle with regeneration. The pressure ratio of the cycle is 8, the compressor inlet temperature is 300 K, and the turbine inlet temperature is 1800 K. The effectiveness of the regenerator is 75 percent. Determine the thermal efficiency and the required mass flow rate of helium for a net power output of 60 MW, assuming both the compressor and the turbine have an isentropic efficiency of (a) 100 percent and (b) 80 percent. The properties of Helium are cp = 5.1926 kJ/kg.K and k = 1.667.
Answer:
Explanation:
Find the temperature at exit of compressor
[tex]T_2=300 \times 8^{\frac{1.667-1}{1.667} }\\=689.3k[/tex]
Find the work done by the compressor
[tex]\frac{W}{m} =c_p(T_2-T_1)\\\\=5.19(689.3-300)\\=2020.4kJ/kg[/tex]
Find the actual workdone by the compressor
[tex]\frac{W}{m} =n_c(\frac{W}{m} )\\\\=1 \times 2020.4kJ/kg[/tex]
Find the temperature at exit of the turbine
[tex]T_4=\frac{1800}{8^{\frac{1.667-1}{1.667} }} \\\\=787.3k[/tex]
Find the actual workdone by the turbine
[tex]1 \times 5.19 (1800-783.3)\\=5276.6kJ/kg[/tex]
Find the temperature of the regeneration
[tex]\epsilon = \frac{T_5-T_2}{T_4-T_2} \\\\0.75=\frac{T_5-689.3}{783.3-689.3} \\\\T_5=759.8k[/tex]
Find the heat supplied
[tex]Q_i_n=c_p(T_3-T_5)\\\\=5.19(1800-759.8)\\\\=5388.2kJ/kg[/tex]
Find the thermal efficiency
[tex]n_t_h=\frac{W_t-W_c}{Q_i_n} \\\\=\frac{5276.6-2020.4}{5388.2} \\\\n_t_h=60.4[/tex]
60.4%
Find the mass flow rate
[tex]m=\frac{W_net}{P} \\\\\frac{60 \times 10^3}{5276.6-2020.4} \\\\=18.42[/tex]
Find the actual workdone by the compressor
[tex]\frac{W_c}{m} =\frac{(\frac{W}{m} )}{n_c} \\\\=\frac{2020.4}{0.8} \\\\=2525.5kg[/tex]
Find the actual workdone by the turbine
[tex]\frac{W_t}{m} =n_t(\frac{W}{m} )\\\\=0.8 \times5.19(1800-783.3)\\\\=4221.2kJ/kg[/tex]
Find the temperature of the compressor exit
[tex]\frac{W_t}{m} =c_p(T_2_a-T_1)\\2525.5=5.18(T_2_a-300)\\T_2_a=787.5k[/tex]
Find the temperature at the turbine exit
[tex]4221.2=5.18(1800-T_4_a)\\\\T_4_a=985k[/tex]
Find the temperature of regeneration
[tex]\epsilon =\frac{T_5-T_2}{T_4-T_2}\\\\0.75=\frac{T_5-787.5}{985-787.5}\\\\T_5=935.5k[/tex]
Answer:
a) 60.4%; 18.42 kg/s
b) 37.8% ; 35.4 kg/s
Explanation:
a) at an isentropic efficiency of 100%.
Let's first find the exit temperature of the compressor T2, using the formula:
[tex](r_p) ^k^-^1^/^k = \frac{T_2}{T_1}[/tex]
Solving for T2, we have:
[tex] T_2 = 300 * (8)^1^.^6^6^7^-^1^/^1^.^6^6^7 = 689.3 K [/tex]
Let's now find the work dine by the compressor.
[tex] \frac{W_c}{m} = c_p(T_2 - T_1) [/tex]
[tex] \frac{W_c}{m} = 5.19(689.3 - 300) = 2020.4 KJ/kg[/tex]
The actual work done by the compressor =
[tex] W_c = 1 * 2020.4 = 2020.4 KJ/kg [/tex]
Let's find the temperature at the exit of the turbine, T4
[tex](r_p) ^k^-^1^/^k = \frac{T_3}{T_4}[/tex]
Solving for T4, we have:
[tex]T_4 = \frac{1800}{(8)^1^.^6^6^7^-^1^/^1^.^6^6^7} = 783.3 K[/tex]
Let's find the work done by the turbine.
[tex]\frac{W_t}{m} = c_p(T_3 - T_4)[/tex]
[tex]\frac{W_t}{m} = 5.19(1800 - 783.3) = 5276.6 KJ/kg[/tex]
The actual work done by the turbine:
= 1 * 5276.6 = 5276.6 KJ/kg
Let's find the regeneration temperature, using the formula:
[tex] e = \frac{T_r - T_2}{T_4 - T_2}[/tex]
Substituting figures, we have:
[tex] 0.75 = \frac{T_r - 689.3}{783.3 - 689.3} [/tex]
[tex] T_r = [0.75(783.3 - 689.3)] + 689.3 = 759.8 [/tex]
Let's calculate the heat supplied.
[tex]Q = c_p(T_3 - T_r)[/tex]
[tex] Q = 5.19(1800 - 759.8) [/tex]
Q = 5388.2 kJ/kg
For thermal efficiency, we have:
[tex] n = \frac{W_t - W_c}{Q} [/tex]
Substituting figures, we have:
[tex] n = \frac{5276.6 - 2020.4}{5388.2} = 0.604 [/tex]
0.604 * 100 = 60.4%
For mass flow rate:
Let's use the formula:
[tex] m = \frac{W_n_e_t}{P} [/tex]
Wnet = 60MW = 60*1000
[tex] m = \frac{60*10^3}{5276.6 - 2020.4} = 18.42 [/tex]
b) at an isentropic efficiency of 80%.
Let's now find the work done by the compressor.
[tex] \frac{W_c}{m} = c_p(T_2 - T_1) [/tex]
[tex] \frac{W_c}{m} = 5.19(689.3 - 300) = 2020.4 KJ/kg[/tex]
The actual work done by the compressor =
[tex] W_c = \frac{2020.4}{0.8}= 2525.5 KJ/kg [/tex]
Let's find the work done by the turbine.
[tex] \frac{W_t}{m} = c_p(T_3 - T_4) [/tex]
[tex] \frac{W_t}{m} = 5.19(1800 - 787.5) = 5276.6 KJ/kg[/tex]
The actual work done by the turbine:
= 0.8 * 5276.6 = 4221.2 KJ/kg
Let's find the exit temperature of the compressor T2, using the formula:
[tex]\frac{W_c}{m} = c_p(T_2 - T_1) [/tex]
[tex] 2525.5 = 5.19(T_2 - 300) [/tex]
Solving for T2, we have:
[tex] T_2 = \frac{2525.5 + 300}{5.19} = 787.5 [/tex]
Let's find the temperature at the exit of the turbine, T4
[tex] \frac{W_t}{m} = c_p(T_3 - T_4) [/tex]
[tex] 4221.2 = 5.19(1800 - T_4) [/tex]
Solving for T4 we have:
[tex] T_4 = 958 K[/tex]
Let's find the regeneration temperature, using the formula:
[tex] e = \frac{T_r - T_2}{T_4 - T_2}[/tex]
Substituting figures, we have:
[tex] 0.75 = \frac{T_r - 787.5}{985 - 787.5} [/tex]
[tex] T_r = [0.75(958 - 787.5)] + 787.5 = 935.5 K [/tex]
Let's calculate the heat supplied.
[tex]Q = c_p(T_3 - T_r)[/tex]
[tex] Q = 5.19(1800 - 935.5) [/tex]
Q = 4486.2 kJ/kg
For thermal efficiency, we have:
[tex] n = \frac{W_t - W_c}{Q} [/tex]
Substituting figures, we have:
[tex] n = \frac{4221.2 - 2525.2}{4486.2} = 0.378 [/tex]
0.378 * 100 = 37.8%
For mass flow rate:
Let's use the formula:
[tex] m = \frac{W_n_e_t}{P} [/tex]
Wnet = 60MW = 60*1000
[tex] m = \frac{60*10^3}{4221.2 - 2525.2} = 35.4 kg/s [/tex]
define sheer stress
Every one deals with stress. Stress or anxiety is usually when your scared or anxious because you know you could have done better then what you did. For example on a test, you could be scared about what grade you get. You can also have stress when you feel something bad is going to happen.
Answer:
it is the component of stress coplanar with a material cross section. It arises from the shear force, the component of force vector parallel to the material cross section
Explanation:
hope this helps and have a good day :-)
You wonder why Andy acted in this fashion, and you guess that, because the door was unlocked, he must be afraid that someone broke into their home. Which reading strategy did you use to understand the character's behavior? A. Inferring B. Visualizing C. Summarizing D. Asking questions
Answer:
option A. Inferring
Explanation:
inferring/ inference as reading strategy simply is the process by which one uses what he/she knows to make a guess about what you don't know or reading between the lines. Readers in making inferences uses clues found inside text along with their own views or experiences to help them figure out what is not directly said,thereby causing a personal and memorable text. for one to draw an inference from the passage via reading, Identify if its an Inference Question.inferring involves Trusting the Passage or what you are seeing, then you start Hunting for Clues thereafter you Narrow Down the Choices. and then come to a conclusion or Practice.
A 50-cm x 50-cm circuit board that contains 121 square chips on one side is to be cooled by combined natural convection and radiation by mounting it on a vertical surface in a room at 25 °C. Each chip dissipates 0.18 W of power, and the emissivity of the chip surfaces is 0.7. Assuming the heat transfer from the back side of the circuit board to be negligible, and the temperature of the surrounding surfaces to be the same as the air temperature of the room, determine the surface temperature of the chips. Evaluate air properties at a film temperature of 30 °C and 1 atm pressure. Is this a good assumption?
Answer:
Ts = 311.86 K = 38.86°C
Explanation:
The convection heat transfer coefficient for vertical orientation of the board is given by the formula:
[tex]h = 1.42(\frac{T_{s} - T_{f} }{L})^{0.25}[/tex]
where,
h = heat transfer coefficient
[tex]T_{s}[/tex] = surface temperature
[tex]T_{f}[/tex] = Temperature of fluid (air) = 30°C + 273 = 303 K
L = Characteristic Length = 50 cm = 0.5 m
Since the heat transfer through convection is given as:
[tex]Q_{conv} = hA_{s}(T_{s} - T_{f})[/tex]
using value of h, we get:
[tex]Q_{conv} = 1.42(\frac{T_{s} - T_{f} }{L})^{0.25} A_{s} (T_{s} - T_{f} )[/tex]
[tex]Q_{conv} = 1.42 A_{s} \frac{(T_{s} - T_{f} )^{1.25} }{L^{0.25} }[/tex]
where,
[tex]A_{s}[/tex] = Surface Area = (0.5 m)(0.5 m) = 0.25 m²
Now, the radiation heat transfer is given by:
[tex]Q_{rad} =[/tex] εσ[tex]A_{s} [(T_{s})^{4} - (T_{surr})^{4}][/tex]
where,
ε = emissivity of surface = 0.7
σ = Stefan Boltzman Constant = 5.67 x 10⁻⁸ W/m².k⁴
[tex]T_{surr}[/tex] = Temperature of surroundings = 25°C +273 = 298 k
Now, the total heat transfer rate will be:
[tex]Q_{total} = Q_{conv} + Q_{rad}[/tex]
using values:
[tex]Q_{total} =[/tex] [tex]1.42 A_{s} \frac{(T_{s} - T_{f} )^{1.25} }{L^{0.25} } +[/tex] εσ[tex]A_{s} [(T_{s})^{4} - (T_{surr})^{4}][/tex]
we know that the total heat transfer from the board can be found out by:
[tex]Q_{total} = (0.18 W) (121) = 21.78 W[/tex]
using values in the equation:
21.78 = (1.42)(0.25)[tex](T_{s} - 303)^{1.25}/0.5^{0.25}[/tex] + (0.7)(5.67 x 10⁻⁸)(0.25)[tex][(T_{s})^{4} - 298^{4}][/tex]
21.78 = (0.4222)[tex](T_{s} - 303)^{1.25}[/tex] + 9.922 x 10⁻⁹[tex](T_{s} )^{4}[/tex] - 78.25
100.03 = (0.4222)[tex](T_{s} - 303)^{1.25}[/tex]+ 9.922 x 10⁻⁹[tex](T_{s} )^{4}[/tex]
Solving this equation numerically by Newton - Raphson Method (Here, any numerical method or an equation solver can be used), we get the value of Ts to be:
Ts = 311.86 K = 38.86°C
The film temperature is the average of surface temperature and surrounding temperature. Therefore,
Film Temperature = (25°C + 38.86°C)/2 = 31.93°C
Since, this is very close to 30°C.
Hence, the assumption is good.
Makine yüzeysel gemi ünvan değişikliği teknolojisi
I inferred you meant emerging technologies we see today.
Explanation:
1. 3D Printing
A three dimensional printing allows a digital model to be printed (constructed) into a physical object.
A box, an industrial design and many more could be printed within minutes.
2. AI voice recognition devices
Another trend in the tech world is the rise in artificial intelligence been used in voice recognition devices that not only recognize one's voice but also take commands from the user.
This technology allows one to listen to the internet, such as listening to music online.
The tech world is ever changing with new technology beyond one's consumption on the increase daily.
A peasant finds himself on a riverbank with a wolf, a goat, and a head of cabbage. He needs to transport all three to the other side of the river in his boat. However, the boat has room for only the peasant himself and one other item (either the wolf, the goat, or the cabbage). In his absence, the wolf would eat the goat, and the goat would eat the cabbage.
a) Solve this problem for the peasant or prove it has no solution. (Note: The peasant is a vegetarian but does not like cabbage and hence can eat neither the goat nor the cabbage to help him solve the problem. And it goes without saying that the wolf is a protected species.)
Answer:
The solution is presented in explanation
Explanation:
This problem can be solved in following steps:
1) In the first round the peasant will take the goat to the other side.
2) Now, the peasant will come back alone.
3) The peasant will now take the wolf with him to other side.
4) The peasant will return with the goat to riverbank.
5) Now, he will take cabbage to the other side of the river, where the wolf is already present.
6) Peasant will leave cabbage and wolf on other side and come back to riverbank alone. Since, wolf does not eat cabbage.
7) Now, finally the peasant will take goat to the other side of river.
In this way, all three of them shall be transported to the other side of the river without eating each other.
Oxygen combines with nitrogen in the air to form NOx at about
A. 1500
B. 2500
C.50
D.000
Oxygen combines with nitrogen in the air to form NO at about 2500 celsius.
What combines nitrogen and oxygen in the air?
The enormous energy of lightning breaks nitrogen molecules and enables their atoms to combine with oxygen in the air forming nitrogen oxides. These dissolve in rain, forming nitrates, that are carried to the earth.
At these high temperatures, nitrogen and oxygen from the air combine to produce nitrogen monoxide. One nitrogen molecule (N2) reacts with one oxygen molecule (O2) to make two nitrogen monoxide molecules (NO).
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