Answer:
The intrinsic Fermi energy is slightly below the center of the bandgap ( option A )
Explanation:
Given that the effective mass of holes is > than the effective mass of electrons in a semiconductor the Intrinsic Fermi energy is slightly below the center of the bandgap.
Intrinsic Fermi energy is experienced/present in the occupation of energy levels in the valence or conduction bands of an intrinsic conductor
A center-point bending test was performed on a 2 in. x d in. wood lumber according to ASTM D198 procedure with a span of 4 ft and the 4 in. side is positioned vertically. If the maximum load was 240 kips and the modulus of rupture was 940.3 ksi, what is the value of d
Answer:
3.03 INCHES
Explanation:
According to ASTM D198 ;
Modulus of rupture = ( M / I ) * y ----- ( 1 )
M ( bending moment ) = R * length of span / 2
= (120 * 10^3 ) * 48 / 2 = 288 * 10^4 Ib-in
I ( moment of inertia ) = bd^3 / 12
= ( 2 )*( d )^3 / 12 = 2d^3 / 12
b = 2 in , d = ?
length of span = 4 * 12 = 48 inches
R = P / 2 = 240 * 10^3 / 2 = 120 * 10^3 Ib
y ( centroid distance ) = d / 2 inches
back to equation ( 1 )
( M / I ) * y
940.3 ksi = ( 288 * 10^4 / 2d^3 / 12 ) * d / 2
= ( 288 * 10^4 * 12 ) / 2d^3 ) * d / 2
940300 = 34560000* d / 4d^3
4d^3 ( 940300 ) = 34560000 d ( divide both sides with d )
4d^2 = 34560000 / 940300
d^2 = 9.188 ∴ Value of d ≈ 3.03 in
In a ground-water basin of 12 square miles, there are two aquifers: an upper unconfined aquifer 500 ft in thickness and a lower confined aquifer with an available hydraulic head drop of 150 ft. Hydraulic tests have determined that the specific yield of the upper unit is 0.12 and the storativity of the lower unit is 4x10-4. What is the amount of recoverable ground water in the basin
Answer:
0.1365 m^3
Explanation:
thickness of upper aquifer = 500 ft
lower aquifer head drop = 150 ft
area of ground water basin = 12 m^2
specific yield of upper unit = 0.12
Storativity of lower unit = 4 * 10^-4
determine the amount of recoverable ground water
first step : calculate volume of unconfined aquifer
= 12 * 500/5280 = 1.1364 miles^3
The recoverable volume of water from unconfined aquifer
= 1.1364 * 0.12 = 0.1364 miles^3
next : calculate volume of confined aquifer
= 12 * 150/5250 = 0.341 miles^3
The recoverable volume of water from confined aquifer
= 0.341 * ( 4 * 10^-4 ) = 1.364 * 10^-4 miles^3
Hence the amount of recoverable ground water in the basin
= ∑ recoverable ground water from both aquifer
= 0.1365 m^3
Please help ASAP and show all work. Thanks
1. An industrial water shutoff valve is designed to operate with 10 lb of effort force. The valve will encounter 100 lb of resistance force applied to a 2 in. diameter axle.
--a. What is the required actual mechanical advantage of the system?
--b. What is the required wheel diameter to overcome the resistance force?
2. A worker on a zip line crew lifts participants weighing approximately 200 lb several times a day from the ground 20 feet below. A block and tackle system is designed with 50 lb of effort force designed to lift the materials.
--a. What is the required actual mechanical advantage?
--b. How many supporting strands will be needed in the pulley system?
3. A simple gear train is composed of three gears. Gear A is the driver and has 10 teeth, gear B has 8 teeth, and gear C has 20 teeth.
--a. If the output is at C, what is the gear ratio?
--b. If gear A rotates at 60 rpm, how fast is gear C rotating?
--c. If the output of torque at gear C is 150 ftlb, what is the input torque at gear A?
4. Look at the picture provided.
--a. What class of lever is shown in the figure? (How do you know?)
--b. How much effort force is needed to balance the 100 lb load?
Answer:
1. --a 10
--b 100 in.
2. --a 4
--b 4
3. --a 2
--b 30 rpm
--c 75 ft.·lb.
4. --a Second class lever
--b 50 lbs.
Explanation:
The Actual Mechanical Advantage, AMA, is given as follows;
[tex]AMA = \dfrac{F_R}{F_E}[/tex]
Where;
[tex]F_R[/tex] = The resistance force magnitude
[tex]F_E[/tex] = The effort force magnitude
1. a. We have;
[tex]F_R[/tex] = 10 lb.
[tex]F_E[/tex] = 100 lb.
[tex]AMA = \dfrac{100 \ lb}{10 \ lb} = 10[/tex]
b. [tex]Mechanical \ advantage, \ M.A. = \dfrac{Distance \ moved \ by \ load}{Distance \ moved \ by \ effort}[/tex]
The diameter of the axle, d = 2 in.
Let 'D' represent the diameter of the wheel, we have;
The distance moved by the axle, c = π·d
The distance moved by the load, C = π·D
[tex]M.A. = 10 = \dfrac{\pi \cdot D}{\pi \times 2}[/tex]
∴ 2 × 10 = D
D = 20 in.
The required wheel diameter to overcome the resistance force, D = 100 in.
2. --a The mass of the participants, m = 200 lb.
The depth of the ground of the participants = 20 feet
The effort force = 50 lb
Actual Mechanical Advantage, AMA = 200 lb./(50 lb.) = 4
--b. The number of strands of pulley needed ≈ The mechanical advantage = 4
3. The number of gears on Gear A = 10 teeth
The number of gears on Gear B = 8 teeth
The number of gears on Gear C = 20 teeth
-a. Given that the driver gear = Gear A
The output gear = Gear C
[tex]The \ gear \ ratio = \dfrac{The \ number \ of \ teeth \ on \ the \ driven \ gear}{The \ number \ of \ teeth \ on \ the \ driver \ gear}[/tex]
The driver gear = The input gear
Therefore, we have;
[tex]The \ gear \ ratio = \dfrac{20 \ teeth}{10 \ teeth} = 2[/tex]
The gear ratio = 2
-b [tex]The \ gear \ ratio = \dfrac{The \ driver \ gear\ speed}{The \ driven \ gear\ speed}[/tex]
Therefore, we have;
[tex]The \ gear \ ratio = 2 = \dfrac{60 \ rpm}{Gear\ C \, speed}[/tex]
[tex]Gear\ C \, speed = \dfrac{60 \ rpm}{2} = 30 \ rpm[/tex]
-c The output (driven) gear torque at Gear C = 150 ft.·lb.
[tex]The \ gear \ ratio = \dfrac{Driven \ gear \ torque}{Driver \ gear \ torque}[/tex]
Therefore;
[tex]2 = \dfrac{150 \ ft \cdot lb}{Driver \ gear \ torque}[/tex]
[tex]Driver \ gear \ torque = \dfrac{150 \ ft \cdot lb}{2} = 75 \ ft \cdot lb[/tex]
The input (driver) torque at Gear A = 75 ft·lb
4. -a Given that the load is between the effort and the fulcrum, we have;
The type of lever is a second class lever
-b The distance between the load and the fulcrum = 4 feet
The distance between the effort and the fulcrum = 8 feet
We have;
100 lbs × 4 ft. = Effort × 8 ft.
∴ Effort = 100 lbs × 4 ft./(8 ft.) = 50 lbs.
The effort = 50 lbs.
Air at 308C, 1 bar, 50% relative humidity enters an insulated chamber operating at steady state with a mass flow rate of 3 kg/min and mixes with a saturated moist air stream entering at 58C, 1 bar with a mass flow rate of 5 kg/min. A single mixed stream exits at 1 bar. Determine (a) the relative humidity and temperature, in 8C, of the exiting stream. (b) the rate of exergy destruction, in kW, for T0 5 208C. Neglect kinetic and potential energy effects.
Answer: the question of whether or is that a new place for a person to come in the morning and then the day that we have a good day at school or to come home with us to go to the church or we could meet up at my place in about a week or so to get the rest of the kids and I can go out to the school to go to the gym to go to the doctor to pick them out or not I have a good time to come over to get the stuff out of the car so I’m going out of the house to go to the store to pick it out I don’t have any money for that
Explanation:
rosbel or Janette lol baakkaaa
Answer:
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Explanation:gt555555555555555555555555555555555555555555555555
Answer:
dawbkjbjwwjhjfbfjewfaekfhawkjndwkja
Explanation: dum*as*
What is the importance of food handling machines in food industry (Explain in points).
Answer:
Explanation:
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Rosie Milojevic, Business Development at iComplied
Answered 3 years ago
For those who run or work in a business that handles food, you understand the importance of food safety and regulation compliance as it pertains to not only to certification and licensing of a company, but also the life or death of those who consume the products. Because food safety is such an important issue, we want to ensure that all companies who work in the production, preparation, or processing of food have the tools and information they need to ensure they are in full compliance 100% of the time. One slip in food safety compliance could cost someone their health or life, and this information spreads mistrust in the general public toward the company who sold the product, and also the entire product across companies throughout an entire country and beyond. So how does each company working with food ensure they comply with each and every food safety regulation? Through excellent auditing tools.
Auditing in Food Safety Must Change
Auditing in food safety compliance is essential in ensuring that all regulations are being complied with. Most auditing practices for major companies are severely out of date. Many rely on several different forms, stacks of paperwork, and data stored on multiple different computers and databases. This is incredibly inefficient, as it can be easy to miss something when comparing one sheet to the next, and paper can easily be lost or destroyed. It is also time consuming for companies to share audits over separate computers and databases, as it must be passed down from one employee to the next. Technology has granted all food safety compliance managers a simple solution. Auditing apps are the new wave solutions for companies who want all of their data stored on one cloud database that is instantly accessible with real-time data for all employees. This means that no one ever gets left out of the loop, and no new information will fall through the cracks.
8. What is the density of an object with a mass of 290.5 g and volume of 83 cm 3?
Great amounts of electro-magnetic energy from our son and other bodies n space travel through space. Who's is a logical conclusion about these electro-magnetic waves?
Answer:
Logical conclusion : there are more electromagnetic waves than sunlight
Explanation:
The traveling of electromagnetic energy from the sun and other bodies through space leads to Electromagnetic radiation.
Hence the logical conclusion about Electromagnetic waves is that there are more electromagnetic waves than sunlight
While the travelling of electromagnetic waves through space is described as gliding through space
A rectangular channel 3.0 m wide has a flow rate of 5.0 m3/s with a normal depth of 0.50 m. The flow then encounters a dam that rises 0.25 m above the channel bottom. Will a hydraulic jump occur?
Answer:
The hydraulic will jump since the flow is subcritical ( i.e. Y2 > Yc )
Explanation:
width of channel = 3.0 m
Flow rate = 5 m^3/s
Normal depth = 0.50 m
Flow encounters a dam rise of 0.25 m
To know if the hydraulic jump will occur
we will Determine the new normal depth
Y2 = 3.77m
Yc ( critical depth )= 0.66m
Attached below is the detailed solution
Could anyone answer this, please? It's about solid mechanics. I will give you 100 points!!! It's due at midnight.
Answer:
sorry i don't know
Explanation:
A continuous and aligned fiber-reinforced composite is manufactured using 80 vol% aramid fiber (a kevlar-like compound) embedded nylon 6-6. A part for a high-performance aircraft utilizes this composite. If the part experiences 953 lb-f (pounds force) along the fiber alignment axis, what is the force conveyed by the fibers ?
Answer:
the force conveyed by the fibers is 947.93 lb-f
Explanation:
Given the data in the question;
V_f = 80% = 0.8
V_m = 1 - V_f = 1 - 0.8 = 0.2
Now,
length of fibre L_f = length of Nylon L_n
V_f = A_f × L_f = 0.8
V_m = A_n × L_n = 0.2
so
V_f/V_m = A_f/A_n = 0.8/0.2
A_f/A_n = 4
now, the strains in fibre is equal to strains in nylon
(P/AE)f = (P/AE)n
P_f/A_fE_f = P_n/A_nE_n
P_f = (A_f/A_n)(E_f/E_n)(P_n)
P_f = ( 4 )( 131 / 2.8 )(Pn)
P_f = 187.14Pn
and P_n = Pf / 187.14
Hence
given that P_total = 953 lb-f
P_f + P_n = 953
P_f + ( P_f / 187.14 ) = 953
P_f( 1 + ( 1 / 187.14 ) ) = 953
P_f( 1.00534359 = 953
P_f = 953 / 1.00534359
P_f = 947.93 lb-f
Therefore, the force conveyed by the fibers is 947.93 lb-f
An aircraft engine operates on a simple ideal Brayton Cycle with a pressure ratio of 10. Heat is added to the cycle at a rate of 500 kW; air passes through the engine at a rate of 1 kg/s; and the air at the beginning of the compression is at 70 kPa and 0oC. Determine the power produced by this engine and its thermal efficiency. Use constant specific heats at room temperature.
Answer: look at the screenshot
The power and thermal efficiency of this engine is equal to 241 Kilowatts and 48.2% respectively.
How to calculate the power and thermal efficiency?
First of all, we would determine the thermal efficiency of this engine by applying the following formula:
[tex]\eta= 1-\frac{1}{r_p^{k-1/k}} \\\\\eta= 1-\frac{1}{10^{1.4-1/1.4}} \\\\\eta= 1-\frac{1}{10^{0.4/1.4}}\\\\\eta= 1-\frac{1}{10^{0.4/1.4}}\\\\\eta= 1-\frac{1}{10^{0.2857}}\\\\\eta = 0.482[/tex]
Thermal efficiency = 48.2%.
Now, we can determine net power output as follows:
[tex]W_{out}=nq_{in}\\\\W_{out}= 0.482 \times 500\\\\W_{out}=241\;kW.[/tex]
Power = 241 Kilowatts.
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A cylindrical rod of copper (E = 110 GPa) having a yield strength of 240 MPa is to be subjected
to a load of 6660 N. If the length of the rod is 380 mm, what must be the diameter to allow an
elongation of 0.50 mm?
Answer:
"7.654 mm" is the correct solution.
Explanation:
According to the question,
[tex]E=110\times 10^3 \ N/mm^2[/tex][tex]\sigma_y = 240 \ mPa[/tex][tex]P = 6660 \ N[/tex][tex]L = 380 \ mm[/tex][tex]\delta = 0.5 \ mm[/tex]Now,
As we know,
The Elongation,
⇒ [tex]E=\frac{\sigma}{e}[/tex]
[tex]=\frac{\frac{P}{A} }{\frac{\delta}{L} }[/tex]
or,
⇒ [tex]\delta=\frac{PL}{AE}[/tex]
By substituting the values, we get
[tex]0.5=\frac{6660\times 380}{(\frac{\pi}{4}D^2)(110\times 10^3)}[/tex]
then,
⇒ [tex]D^2=58.587[/tex]
[tex]D=\sqrt{58.587}[/tex]
[tex]=7.654 \ mm[/tex]
3. When mixing repair adhesive, how do you know when the material is ready?
A) O The mix is uniform in color
B) O The mix has set for 2 minutes
C)The mix has no lumps
D)The mix turns blue
Answer:
O The mix is uniform in color
The blue and white “Alfa” flag signifies what?
A) A tow boat is towing another vessel to safety
B) A ski boat that has a participant in the water
C) A dive boat that is restricted in its ability to maneuver
D) A Coast Guard boat that is on patrol
The blue and white "Alfa" flag signifies: C. a dive boat that is restricted in its ability to maneuver.
What is a dive boat?A dive boat can be defined as a type of boat that is commonly used by scuba divers or recreational divers to reach a deep dive site which cannot be conveniently accomplished by swimming from the bank or sea shore.
As a safety precaution, the blue and white "Alfa" flag is designed and developed to signify a dive boat that is restricted in its ability to maneuver.
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Two objects labeled K and L have equal mass but densities 0.95Do and Do, respectively. Each of these objects floats after being thrown into a deep swimming pool. Which is true about the buoyant forces acting on these objects?
a. The buoyant force is greater on Object K since it has lower density and lower density objects always float "higher" in the fluid.
b. The buoyant force is greater on Object L since it is denser than K and therefore "heavier."
c. The buoyant forces are equal on the objects since they have equal mass.
d. Without knowing the specific gravity of the objects, nothing can be determined.
Answer:
C. The buoyant forces are equal on the objects since they have equal mass.
Explanation:
Correct option (C) The buoyant forces are equal on the objects since they have equal mass. For floating objects, the buoyant force equals the weight of the objects. Since each object has the same weight, they must have the same buoyant force to counteract that weight and make them float.
forty gal/min of a hydrocarbon fuel having a spesific gravity of 0.91 flow into a tank truck with load limit of 40,000 lb of fuel. How long will it takee to fill the tank in the truck?
Answer: 131.75minutes
Explanation:
First if all, we've to find the density of liquid which will be:
= Specific gravity × Density to pure water
= 0.91 × 8.34lb/gallon
= 7.59lb/gallon
Then, the volume that's required to fill the tank will be:
= Load limit/Density of fluid
= 40000/7.59
= 5270.1gallon
Now, the time taken will be:
= V/F
= 5270.1/40
= 131.75min
It'll take 131.75 minutes to fill the tank in the truck.
You do a simple experiment at home with the plastic body of a syringe, where you close the exit with one thumb, and push in the plunger with your other thumb. You are able to compress the air inside the syringe from 5ml to 1ml. Assume that you hold the plunger for long enough such that the temperature equalizes to ambient conditions.
Required:
a. Given that the circular plunger's diameter is 1.5cm, how much force is being exerted to hold the plunger in the compressed state?
b. Given that the opening of the syringe has a diameter of 2mm, how much force is exerted on the thumb used to trap the air from escaping?
Answer:
a. 89.5 N b. 1.59 N
Explanation:
a. Given that the circular plunger's diameter is 1.5cm, how much force is being exerted to hold the plunger in the compressed state?
Using Boyle's law, we find the final pressure at the compressed state given that the initial pressure is atmospheric pressure
So, P₁V₁ = P₂V₂ where P₁ = initial atmospheric pressure in syringe = 1 atm = 1.013 × 10⁵ N/m²,V₁ = initial volume of syringe = 5 ml, P₂ = final pressure in syringe at compression and V₂ = final volume of syringe = 1 ml
So, making P₂ subject of the formula, we have
P₂ = P₁V₁/V₂
Substituting the values of the variables into the equation, we have
P₂ = P₁V₁/V₂
P₂ = 1.013 × 10⁵ N/m² × 5 ml/1 ml
P₂ = 1.013 × 10⁵ N/m² × 5
P₂ = 5.065 × 10⁵ N/m²
Since pressure, P = F/A where F = force and A = cross-sectional area of syringe = πd²/4 where d = diameter of syringe = 1.5 cm = 1.5 × 10⁻² m.
So, F = PA
F = P₂πd²/4
substituting the values of the variables into the equation, we have
F = P₂πd²/4
F = 5.065 × 10⁵ N/m²π(1.5 × 10⁻² m)²/4
F = 5.065 × 10⁵ N/m²π(2.25 × 10⁻⁴ m²)/4
F = 35.8 × 10/4 N
F = 8.95 × 10
F = 89.5 N
b. Given that the opening of the syringe has a diameter of 2mm, how much force is exerted on the thumb used to trap the air from escaping?
Since the pressure in the syringe after compression is constant, we have
P₂ = F₁/A₁ where F₁ = force exerted on thumb and A₁ = cross-sectional area of opening of syringe = πd₁²/4 where d = diameter of opening of syringe = 2 mm = 2 × 10⁻³ m.
So, F₁ = P₂A₁
F = P₂πd₁²/4
substituting the values of the variables into the equation, we have
F = P₂πd²/4
F = 5.065 × 10⁵ N/m²π(2 × 10⁻³ m)²/4
F = 5.065 × 10⁵ N/m²π(4 × 10⁻⁶ m²)/4
F = 15.91 × 10⁻¹
F = 1.591 N
F ≅ 1.59 N
Air is compressed from 100 kPa, 300 K to 1200 kPa in a two-stage compressor with intercooling between the stages. The intercooler pressure is 400 kPa. The air is cooled back to 300 K in the intercooler before entering the second compressor stage. Each compressor stage is isentropic. For steady-state operation and negligible changes in kinetic and potential energy from inlet to exit, determine (a) the temperature at the exit of the seco
Answer:
410.7 K
Explanation:
Given data:
P1 ( initial air pressure ) = 100 kPa
Pi ( intercooler pressure ) = 400 kPa
P2( compressed air pressure ) = 1200 kPa
T1 = 300 k
Ti = 300 k
Calculate for the temperature at the exit of the second stage ( T2 )
we can calculate this using the relation below
T2 / Ti = ( P2/Pi )^(r-1/r)
∴ T2 = 300 ( 1200 / 400 )^( 1.4 - 1 / 1.4 )
= 300 ( 3 )^0.286
= 300 * 1.369 = 410.7 K
describe five tools used in suspension system service and repair
Answer:
Explanation:
You'll still need wrenches, sockets and screwdrivers, but there are other things that it may be necessary to buy to complete the work.
Spring Compressor. One part of suspension repair is replacing coil springs.
Hydraulic Puller. -Hydraulic pullers are used to remove shaft-fitted parts (bearings or couplings). Pullers use a controlled hydraulic force in an effective way and can quickly separate (especially compared to the manual alternative) the parts.
CV Boot Tool. he CV Boot is a ribbed, rubber flexible boot that keeps water and dirt out of the joint and the special grease inside the joint.
Torque Wrench. .A torque wrench is a tool used to apply a specific torque to a fastener such as a nut, bolt, or lag screw. ... A torque wrench is used where the tightness of screws and bolts is crucial.
Ball Joint Separator. This tool is used to separate the ball joint from the spindle support arm. It works on many domestic and import front wheel drive vehicles and is adjustable up to 2" for different size ball joints.
Strut Nuts.
Tie Rod Puller.
etc....
In a certain balanced three phase system each line current is a 5a and each line voltage is 220v . What is the approximate real power if the power factor is 0.7
Answer:
1,334
Explanation:
In a certain balanced three phase system each line current is a 5a and each line voltage is 220v . 3.87 kilowatts is the approximate real power if the power factor is 0.7.
The idea of real power in a balanced three-phase system is critical in electrical power systems. It denotes the actual power transferred and used by the system, which is usually measured in kilowatts (kW) or megawatts (MW). Understanding real power is critical for evaluating electrical system efficiency and performance. Real power is the component of power in a balanced three-phase system that does useful work, such as operating motors, generating heat, or powering appliances.
Real Power (P) = √3 × Line Current (I) × Line Voltage (V) × Power Factor (PF)
P = √3 × 5 A × 220 V × 0.7
P = 3.87 kilowatts (kW)
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Consider a turbofan engine installed on an aircraft flying at an altitude of 5500m. The CPR is 12 and the inlet diameter of this engine is 2.0m The bypass ratio of this engine 8. The bypass ratio (BPR) of a turbofan engine is the ratio between the mass flow rate of the bypass stream to the mass flow rate entering the core. The inlet temperature is 253K and the outlet temperature is 233K. Determine the thrust of this engine in order to fly at the velocity of 250 m/s. Assume cold air approach. The engine is ideal.
Answer:
The thrust of the engine calculated using the cold air is 34227.35 N
Explanation:
For the turbofan engine, firstly the overall mass flow rate is considered. The mass flow rate is given as
[tex]\dot{m}=\rho AV_a[/tex]
Here
ρ is the density which is given as [tex]\dfrac{P}{RT}[/tex]P is the pressure of air at 5500 m from the ISA whose value is 50506.80 PaR is the gas constant whose value is 286.9 J/kg.KT is the temperature of the inlet which is given as 253 KA is the cross-sectional area of the inlet which is given by using the diameter of 2.0 mV_a is the velocity of the aircraft which is given as 250 m/sSo the equation becomes
[tex]\dot{m}=\rho AV_a\\\dot{m}=\dfrac{P}{RT} AV_a\\\dot{m}=\dfrac{50506.80}{286.9\times 253} \times (\dfrac{\pi}{4}\times 2^2)\times 250\\\dot{m}=546.4981\ kgs^{-1}[/tex]
Now in order to find the flow from the fan, the Bypass ratio is used.
[tex]\dot{m}_f=\dfrac{BPR}{BPR+1}\times \dot{m}[/tex]
Here BPR is given as 8 so the equation becomes
[tex]\dot{m}_f=\dfrac{BPR}{BPR+1}\times \dot{m}\\\dot{m}_f=\dfrac{8}{8+1}\times 546.50\\\dot{m}_f=485.77\ kgs^{-1}[/tex]
Now the exit velocity is calculated using the total energy balance which is given as below:
[tex]h_4+\dfrac{1}{2}V_a^2=h_5+\dfrac{1}{2}V_e^2[/tex]
Here
h_4 and h_5 are the enthalpies at point 4 and 5 which could be rewritten as [tex]c_pT_4[/tex] and [tex]c_pT_5[/tex] respectively.The value of T_4 is the inlet temperature which is 253 KThe value of T_5 is the outlet temperature which is 233KThe value of c_p is constant which is 1005 J/kgKV_a is the inlet velocity which is 250 m/sV_e is the outlet velocity that is to be calculated.So the equation becomes
[tex]h_4+\dfrac{1}{2}V_a^2=h_5+\dfrac{1}{2}V_e^2\\c_pT_4+\dfrac{1}{2}V_a^2=c_pT_5+\dfrac{1}{2}V_e^2[/tex]
Rearranging the equation gives
[tex]\dfrac{1}{2}V_e^2=c_pT_4-c_pT_5+\dfrac{1}{2}V_a^2\\\dfrac{1}{2}V_e^2=c_p(T_4-T_5)+\dfrac{1}{2}V_a^2\\V_e^2=2c_p(T_4-T_5)+V_a^2\\V_e=\sqrt{2c_p(T_4-T_5)+V_a^2}\\V_e=\sqrt{2\times 1005\times (253-233)+(250)^2}\\V_e=320.46 m/s[/tex]
Now using the cold air approach, the thrust is given as follows
[tex]T=\dot{m}_f(V_e-V_a)\\T=485.77\times (320.46-250)\\T=34227.35\ N[/tex]
So the thrust of the engine calculated using the cold air is 34227.35 N
Q2 [45 marks] Consider Ibra region where the installed solar panels cost on average 2 OMR /W.
[10 marks] What is the cost to install a 5kW PV system for a residence?
[10 marks] If the solar irradiance in Ibra is on average 800W/m2 and the installed panels have efficiency of 18%. How many panels are required if the panel’s area is 2m2?
[15 marks] Assume Ibra has an average of 10 day-hours, dusty environment which causes the efficiency of the solar system to drop by 10% on average, and 30 cloudy days/year which cause the efficiency of the solar panels drops by 50%. If electrical power cost per kWh is 0.05 OMR determine the break-even time for the 5kW PV system.
[10 marks] If the system to be off-grid, what would be the backup time if three 12-V batteries were selected each with a capacity of 200Ah. Assume that you can discharge the batteries up to 80% of their capacities.
Answer:
so hard it is
Explanation:
I don't know about this
please mark as brainleast
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A liquid propellant engine has the following characteristics: chamber pressure of 7 MPa, constant ratio of specific heats of 1.3, and a characteristic velocity of 1600 m/s. The nozzle has the following characteristics: throat area of 0.010 m2 and an expansion ratio of 10. Calculate the following:
1. Thrust coefficient at sea level
2. Specific impulse at sea level
3. Altitude at optimal expansion
4. Thrust coefficient at optimal expansion
5. Mass flux through the throat
Answer:
1.55260 N.s3370 m1.643.75 kg/sExplanation:
1) Thrust coefficient at sea level.
Cfsl = TSL / Pca
TSL = Mp * Vc + ( Pc - Pa )Ac
Mp = mass flux = 43.75 kg/s
∴ Cfsl = Mp Vc / Pca + ( Pc - Pa )/Pc * ( Ac / A* )
= 1.6 - 0.04923 = 1.55
2) Specific impulse at sea
Isp = Vc / g = 2549.75 / 9.81
= 260 N.s
3) Altitude at optimal expansion
H = 3370 m
4) thrust coefficient at optimal expansion
CF = 1.6
attached below is the detailed solution
5) Mass flux through the throat
Mass flux = P1 * At / Cc
= ( 7*10^6 * 0.01 ) / 1600
= 43.75 kg/s
The volume fraction of particles in a WC-particle Cu-matrix CERMET is 0.84 Calculate the minimum expected elastic modulus, in GPa, of this particle reinforced composite. The moduli of the two components are 682 GPa and 110 GPa, respectively. Use what you know about these composite phases to discern which modulus is which.
Answer:
the minimum expected elastic modulus is 372.27 Gpa
Explanation:
First we put down the data in the given question;
Volume fraction [tex]V_f[/tex] = 0.84
Volume fraction of matrix material [tex]V_m[/tex] = 1 - 0.84 = 0.16
Elastic module of particle [tex]E_f[/tex] = 682 GPa
Elastic module of matrix material [tex]E_m[/tex] = 110 GPa
Now, the minimum expected elastic modulus will be;
[tex]E_{CT[/tex] = ([tex]E_f[/tex] × [tex]E_m[/tex] ) / ( [tex]E_f[/tex][tex]V_m[/tex] + [tex]E_m[/tex] [tex]V_f[/tex] )
so we substitute in our values
[tex]E_{CT[/tex] = (682 × 110 ) / ( [ 682 × 0.16 ] + [ 110 × 0.84] )
[tex]E_{CT[/tex] = ( 75,020 ) / ( 109.12 + 92.4 )
[tex]E_{CT[/tex] = 75,020 / 201.52
[tex]E_{CT[/tex] = 372.27 Gpa
Therefore, the minimum expected elastic modulus is 372.27 Gpa
An ocean thermal energy conversion system is being proposed for electric power generation. Such a system is based on the standard power cycle for which the working fluid is evaporated, passed through a turbine, and subsequently condensed. The system is to be used in very special locations for which the oceanic water temperature near the surface is approximately 300 K, while the temperature at reasonable depths is approximately 280 K. The warmer water is used as a heat source to evaporate the working fluid, while the colder water is used as a heat sink for condensation of the fluid. Consider a power plant that is to generate 2 MW of electricity at an efficiency (electric power output per heat input) of 3%. The evaporator is a heat exchanger consisting of a single shell with many tubes executing two passes. If the working fluid is evaporated at its phase change temperature of 290 K, with ocean water entering at 300 K and leaving at 292 K.
Required:
a. What is the heat exchanger area required for the evaporator?
b. What flovw rate must be maintained for the water passing through the evaporator?
Answer:
a) the heat exchanger area required for the evaporator is 11178.236 m²
b) the required flow rate is 1993630.38 kg/s
Explanation:
Given the data in the question;
Water temperature near the surface = 300 K
temperature at reasonable depths ( cold ) = 280 K
power plant output W' = 2 MW
efficiency η = 3% = 0.03
we know that; efficiency η = W'[tex]_{power-out[/tex] / Q[tex]_{supplied[/tex]
we substitute
0.03 = 2 / Q[tex]_{supplied[/tex]
Q[tex]_{supplied[/tex] = 2 / 0.03
Q[tex]_{supplied[/tex] = 66.667 MW = 66.667 × 10⁶ Watt
T[tex]h_{in[/tex] = 300 K T[tex]h_{out[/tex] = 292 K
T[tex]c_{in[/tex] = 290 K T[tex]c_{out[/tex] = 290 K
Now, Heat transfer in evaporator;
Q = UA( LMTD )
so
LMTD = (ΔT₁ - ΔT₂) / ln( ΔT₁ / ΔT₂ )
first we get ΔT₁ and ΔT₂
ΔT₁ = T[tex]h_{in[/tex] - T[tex]c_{out[/tex] = 300 - 290 = 10 K
ΔT₂ = T[tex]h_{out[/tex] - T[tex]c_{in[/tex] = 292 - 290 = 2 K
so we substitute into our equation;
LMTD = (10 - 2) / ln( 10 / 2 )
LMTD = 8 / ln( 5 )
LMTD = 8 / 1.6094379
LMTD = 4.97
a) Heat transfer Area will be;
Q[tex]_H[/tex] = UA( LMTD )
we substitute
66.667 × 10⁶ = 1200 × A × 4.97
66.667 × 10⁶ = 5964 × A
A = (66.667 × 10⁶) / 5964
A = 11178.236 m²
Therefore, the heat exchanger area required for the evaporator is 11178.236 m²
b) Flow rate
we know that;
Q[tex]_H[/tex] = m'C[tex]_P[/tex]( [tex]T_{in[/tex] - [tex]T_{out[/tex] )
specific heat capacity of water Cp = 4.18 (kJ/kg∙°C)
we substitute
66.667 × 10⁶ = m' × 4.18 × ( 300 - 292 )
66.667 × 10⁶ = m' × 33.44
m' = ( 66.667 × 10⁶ ) / 33.44
m' = 1993630.38 kg/s
Therefore, the required flow rate is 1993630.38 kg/s
An engineer sets up an experiment to determine the coefficient of static friction "us" for an unknown material. She cuts the material in to a disc and places a test mass on top, L = .75m from the center, and she proceeds to spin the disc with an angular acceleration of theta-double-dot = 40t rad/s^2 counterclockwise. The engineer notes that the test mass slips at t=.2s. What is "us" (friction)?
Answer:
HF I am writing this email with your company is it a try and make them easier for us and we will need a new job and then email
A 2.00 kg piece of aluminum metal at 75.0 °C is placed in 6.00 liters (= 6.00 kg) of water at 35.0 °C. Determine the final temperature .
Answer:
T = 46.97 °C
Explanation:
Applying the law of conservation of energy, we get:
Heat Energy Lost by Aluminum Piece = Heat Energy Gained by Water
[tex]m_aC_a\Delta T_a=m_wC_w\Delta T_w[/tex]
where,
[tex]m_w[/tex] = mass of water = (Density)(Volume) = (1000 kg/m³)(1 L)(0.001 m³/1 L)
[tex]m_w[/tex] = 1 kg
[tex]m_a[/tex] = mass of auminum piece = 2 kg
[tex]C_w[/tex] = specific heat capacity of water = 4200 J/kg.°C
[tex]C_a[/tex] = specific heat capacity of aluminum = 897 J/kg.°C
[tex]\Delta T_w[/tex] = Change in Temperature of Water = T - 35°C
[tex]\Delta T_a[/tex] = Change in Temperature of Aluminum Piece = 75°C - T
T = Final Temperature = ?
Therefore,
[tex](2\ kg)(897\ J/kg.^oC)(75^oC - T)=(1\ kg)(4200\ J/kg.^oC)(T - 35^oC)\\\\134550\ J - (1794\ J/^oC)T = (4200\ J/^oC)T - 147000\ J\\281550\ J = (5994\ J/^oC)T\\\\T = \frac{281550\ J }{5994\ J/^oC}[/tex]
T = 46.97 °C
Explain when it is appropriate to use Tier II?
????? what do u mean tier 2
It is desired to filter a cell broth at a rate of 3000 liters/h on a rotary vacuum filter at a vacuum pressure of 70kPa. The cycle time for the drum will be 60 s and the cake formation time (filtering time) will be 20 s. The broth to be filtered has a viscosity of 2.0 cP and a cake solids (dry basis) per volume of filtrate of 10 g/liter. From laboratory test, the specific cake resistance has been determined to be 9 x 1010 cm/g. Determine the area of the filter that is required.
Answer:
0.5667 m^2
Explanation:
Given data :
Flow rate = 3000 liters/h
vacuum pressure = 70 kPa
cycle time for drum = 60 s
cake formation time ( filtering time ) = 20 s
Broth viscosity = 2.0 cP
Cake solids per volume = 10 g/liter
specific cake resistance = 9 * 10^10 cm/g
Calculate the area of filter required
Area of filter required = 0.5667 m^2
Attached below is the detailed solution