Answer:
a. impossible
b. possible and reversible
c. possible and irreversible
Explanation:
a. 1000kw
Qh - Wnet
we have
QH = 1500
wnet = 1000
1500 - 1000
= 500kw
σcycle = [tex]-[\frac{QH}{TH} -\frac{QC}{TC} ][/tex]
Qh = 1500
Th = 1000
Tc = 500
Qc = 500
[tex]-[\frac{1500}{1000} -\frac{500}{500} ][/tex]
solving this using LCM
= -0.5
the cycle is impossible since -0.5<0
b. 750Kw
Qc = 1500 - 750
=750Kw
Qh = 1500
Th = 1000
Tc = 500
Qc = 750
σ-cycle
[tex]-[\frac{1500}{1000} -\frac{750}{500} ]\\= 1.5 -1.5\\= 0[/tex]
This cycle is possible and it is also reversible
c. 0 kw
Qc = 1500-0
= 1500
Qh = 1500
Th = 1000
Tc = 500
Qc = 1500
σ- cycle
[tex]-[\frac{1500}{1000} -\frac{1500}{500} ]\\-(1.5-3)\\-(-1.5)\\= 1.5[/tex]
1.5>0
so this cycle is possible and irreversible
A network has three independent file servers, each with 90 percent reliability. The probability that the network will be functioning correctly (at least one server is working) at a given time is:
Answer:
The correct answer is "99.9%".
Explanation:
According to the information given in the question,
[tex]P(1 \ fail) = 0.1[/tex]
The probability of all fail will be:
[tex]P(all \ fail) = (0.1)^3[/tex]
[tex]=0.001[/tex]
hence,
[tex]P(not \ all \ fail)= 1-P(all \ fail)[/tex]
[tex]=0.999[/tex]
[tex]=99.9[/tex] (%)
Thus the above is the right answer.
A balanced three-phase inductive load is supplied in steady state by a balanced three-phase voltage source with a phase voltage of 120 V rms. The load draws a total of 10 kW at a power factor of 0.85 (lagging). Calculate the rms value of the phase currents and the magnitude of the per-phase load impedance. Draw a phasor diagram showing all tlme voltages and currents.
Answer:
Following are the solution to the given question:
Explanation:
Line voltage:
[tex]V_L=\sqrt{3}V_{ph}=\sqrt{3}(120) \ v[/tex]
Power supplied to the load:
[tex]P_{L}=\sqrt{3}V_{L}I_{L} \cos \phi[/tex]
[tex]10\times 10^3=\sqrt{3}(120 \sqrt{3}) I_{L}\ (0.85)\\\\I_{L}= 32.68\ A[/tex]
Check wye-connection, for the phase current:
[tex]I_{ph}=I_L= 32.68\ A[/tex]
Therefore,
Phasor currents: [tex]32.68 \angle 0^{\circ} \ A \ ,\ 32.68 \angle 120^{\circ} \ A\ ,\ and\ 32.68 -\angle 120^{\circ} \ A[/tex]
Magnitude of the per-phase load impedance:
[tex]Z_{ph}=\frac{V_{ph}}{I_{ph}}=\frac{120}{32.68}=3.672 \ \Omega[/tex]
Phase angle:
[tex]\phi = \cos^{-1} \ (0.85) =31.79^{\circ}[/tex]
Please find the phasor diagram in the attached file.
4.3
While a train is standing still, its smoke blows 12 m/s north.
What will the resulting velocity be of the smoke relative to the train if the train
is moving at 25 m/s south?
(3)
Cho thanh có tiết diện thay đổi chịu tải trọng dọc trục (hình 1).
Biết d1 = 5 cm, d2 = 8 cm, a= 15 cm, b=10cm, P1 =400kN, P2 =200kN, E= 2.104 kN/cm2.
a) Vẽ biểu đồ lực dọc.
b) Kiểm tra bền của thanh AC, [ϭ] =10 (kN/cm2).
c) Xác định chuyển vị theo phương dọc trục của tâm tiết diện C
Answer:
saay in English language
Car B is traveling a distance dd ahead of car A. Both cars are traveling at 60 ft/s when the driver of B suddenly applies the brakes, causing his car to decelerate at ft/s^2. It takes the driver of car A 0.75 s to react (this is the normal reaction time for drivers). When he applies his brakes, he decelerates at 18 ft/s^2.
Required:
Determine the minimum distance d between the cars so as to avoid a collision.
Answer:
Explanation:
Using the kinematics equation [tex]v = v_o + a_ct[/tex] to determine the velocity of car B.
where;
[tex]v_o =[/tex] initial velocity
[tex]a_c[/tex] = constant deceleration
Assuming the constant deceleration is = -12 ft/s^2
Also, the kinematic equation that relates to the distance with the time is:
[tex]S = d + v_ot + \dfrac{1}{2}at^2[/tex]
Then:
[tex]v_B = 60-12t[/tex]
The distance traveled by car B in the given time (t) is expressed as:
[tex]S_B = d + 60 t - \dfrac{1}{2}(12t^2)[/tex]
For car A, the needed time (t) to come to rest is:
[tex]v_A = 60 - 18(t-0.75)[/tex]
Also, the distance traveled by car A in the given time (t) is expressed as:
[tex]S_A = 60 * 0.75 +60(t-0.75) -\dfrac{1}{2}*18*(t-0.750)^2[/tex]
Relating both velocities:
[tex]v_B = v_A[/tex]
[tex]60-12t = 60 - 18(t-0.75)[/tex]
[tex]60-12t =73.5 - 18t[/tex]
[tex]60- 73.5 = - 18t+ 12t[/tex]
[tex]-13.5 =-6t[/tex]
t = 2.25 s
At t = 2.25s, the required minimum distance can be estimated by equating both distances traveled by both cars
i.e.
[tex]S_B = S_A[/tex]
[tex]d + 60 t - \dfrac{1}{2}(12t^2) = 60 * 0.75 +60(t-0.75) -\dfrac{1}{2}*18*(t-0.750)^2[/tex]
[tex]d + 60 (2.25) - \dfrac{1}{2}(12*(2.25)^2) = 60 * 0.75 +60((2.25)-0.75) -\dfrac{1}{2}*18*((2.25)-0.750)^2[/tex]
d + 104.625 = 114.75
d = 114.75 - 104.625
d = 10.125 ft
Calculate the number of 12 V batteries (capacity 120 Ah) needed to run a 3 kW DC motor that operates in 240 V. How many hours the motor will run with 20 of such batteries connected in series?
Answer:
20 batteries9.6 hoursExplanation:
To obtain 240 V from 12 V batteries they must be connected in series. The number needed is ...
240/12 = 20 . . . batteries needed
__
The current draw will be ...
(3000 W)/(240 V) = 12.5 A
Then the time available from the battery stack is ...
(120 Ah)/(12.5 A) = 9.6 h
The motor can run 9.6 hours from the series connection.
A steel rod, which is free to move, has a length of 200 mm and a diameter of 20 mm at a temperature of 15oC. If the rod is heated uniformly to115 oC, determine the length and the diameter of this rod to the nearest micron at the new temperature if the linear coefficient of thermal expansion of steel is 12.5 x 10-6 m/m/oC. What is the stress on the rod at 115oC.
Explanation:
thermal expansion ∝L = (δL/δT)÷L ----(1)
δL = L∝L + δT ----(2)
we have δL = 12.5x10⁻⁶
length l = 200mm
δT = 115°c - 15°c = 100°c
putting these values into equation 1, we have
δL = 200*12.5X10⁻⁶x100
= 0.25 MM
L₂ = L + δ L
= 200 + 0.25
L₂ = 200.25mm
12.5X10⁻⁶ *115-15 * 20
= 0.025
20 +0.025
D₂ = 20.025
as this rod undergoes free expansion at 115°c, the stress on this rod would be = 0
Assignment Using Perman's equation, estimate the potential evapotranspiration for the month of August in locality with the following data
latitude 20 degrees north
Elevation 200m
• Mean monthly temperature 20 degreeso
• Mean relative humidity: 75%
• Mean sunshine hour= 9hrs
Wind at 2 m height equal 85 km per day
nature of surface cover =green grass
Answer:
what do you need help with
Explanation:
The value of universal gas constant is same for all gases?
a) yes
b)No
Answer:
The answer of these questions is
Explanation:
b) NO
a basketball player pushes down with a force of 50N on a basketball that is indlated to a gage pressure of 8.0x10^4 Pa. What is the diameter of comtact between the ball nad the floor
Answer:
The diameter of the contact area between the ball and the floor is approximately 28 milimeters.
Explanation:
The basketball experiments a normal stress ([tex]\sigma[/tex]), in pascals, due to normal force from the floor ([tex]N[/tex]). By definition of normal stress, we have the following equation:
[tex]\sigma = \frac{4\cdot N}{\pi\cdot D^{2}}[/tex] (1)
Where [tex]D[/tex] is the diameter of the contact area between the ball and the floor, in meters.
Please notice that magnitude of the normal force equals the magnitude of external force given by the basketball player and weight is negligible in comparison with normal and external forces.
If we know that [tex]N = 50\,N[/tex] and [tex]\sigma = 8.0\times 10^{4}\,Pa[/tex], then the diameter of the contact area is:
[tex]\sigma = \frac{4\cdot N}{\pi\cdot D^{2}}[/tex]
[tex]D^{2} = \frac{4\cdot N}{\pi\cdot \sigma}[/tex]
[tex]D = 2\cdot \sqrt{\frac{N}{\pi\cdot \sigma} }[/tex]
[tex]D = 2\cdot \sqrt{\frac{50\,N}{\pi\cdot (8\times 10^{4}\,Pa)} }[/tex]
[tex]D\approx 0.028\,m[/tex] [tex](28\,mm)[/tex]
The diameter of the contact area between the ball and the floor is approximately 28 milimeters.
Future solution for air pollution in new zealand
Answer:
New Zealand may use some of these solutions to prevent air pollution
Explanation:
Using public transports.
Recycle and Reuse
No to plastic bags
Reduction of forest fires and smoking
Use of fans instead of Air Conditioner
Use filters for chimneys
Avoid usage of crackers
4) A steel tape is placed around the earth at the equator when the temperature is 0 C. What will the clearance between the tape and the ground (assumed to be uniform) be if the temperature of the tape rises to 30 C. Neglect the expansion of the earth (the radius of the earth is 6.37 X 106 m)
Answer:
2102.1 m
Explanation:
Temperature at the equator = 0⁰
Radius of the earth = 6.37x10⁶
Required:
We how to find out what the clearance between tape and ground would be if temperature increases to 30 degrees.
Final temperature = ∆T = 303-273 = 30
S = 11x10^-6
The clearance R = Ro*S*∆T
=6.37x10⁶x 11x10^-6x30
= 2102.1m
Or 2.102 kilometers
Thank you
What is the built-in pollution control system in an incinerator called
Explanation:
hbyndbnn☝️
Refrigerant-134a enters an adiabatic compressor at -30oC as a saturated vapor at a rate of 0.45 m3 /min and leaves at 900 kPa and 55oC. Determine (a) the power input to the compressor, (b) the isentropic efficiency of the compressor, and (c) the rate of exergy destruction and the second-law efficiency of the compressor. Take T0
Answer:
a) 1.918 kw
b) 86.23%
c) 0.26 kw
Explanation:
Given data:
T1 = -30°C = 243 k , T0 = 27°C
using steam tables
h1 = 232.19 KJ/kg
s1 = 0.9559 Kj/Kgk
T2 = 55°C P2 = 900 kPa
Psat = 1492 kPa, h2 = 289.95 Kj/Kg, s2 = 0.9819 Kj/kgk , m = 0.0332 kg/s
a) Determine the power input to the compressor
power input = 1.918 kw
b) Determine isentropic efficiency of compressor
Isentropic efficiency = 86.23%
c) Determine rate of exergy destruction
rate = 0.26 kw
Attached below is the detailed solution of the given problems
Bài 3: Cho cơ cấu culít (hình 3.5) với các kích thước động lAB = 0,5lAC = 0,1m. Khâu 3 chịu tác dụng của mô men M3 = 500 N. Cơ cấu ở trạng thái cân bằng. Tại thời điểm khâu 1 ở vị trí υ1 = 900 hãy tính áp lực tại các khớp động tại B, C và A.
You are working for a company that creates special magnetic environments. Your new supervisor has come from the financial side of the organization rather than the technical side. He has promised a client that the company can provide a device that will create a magnetic field inside a cylindrical chamber that is directed along the cylinder axis at all points in the chamber and increases in the axial direction as the square of the value of y, where y is in the axial direction and y = 0 is at the bottom end of the cylinder. Prepare a calculation to show that the field requested by your supervisor and promised to a client is impossible.
Answer:
Following are the responses to the given question:
Explanation:
When we take the entire cylinder as a surface that is:
[tex]B=B.y^2\ j\\\\[/tex]
for the magnetic filed existance
[tex]\bigtriangledown . \underset{b}{\rightarrow}=0[/tex]
[tex]\therefore[/tex]
the flub by this cyclinder is zero implies there theaming flubs = outgoing flubs
[tex]\bigtriangledown . \underset{b}{\rightarrow}\\\\=\frac{\delta }{\delta x} B_x+\frac{\delta }{\delta y } B_y+ \frac{\delta }{\delta z} B_z\\\\=\frac{\delta }{\delta x} B_0+\frac{\delta }{\delta y } B_{y^2}+ 0\\\\=\frac{\delta }{\delta x} B_0\ {y^2}=2 B_0\ y\\\\So,\\\\\bigtriangledown . \underset{b}{\rightarrow}\neq 0[/tex]
that's why it is impossible field.
Design a ductile iron pumping main carrying a discharge of 0.35 m3/s over a distance of 4 km. The elevation of the pumping station is 140 m and that of the exit point is 150 m. The required terminal head is 10 m. Estimate the pipe diameter and pumping head using the explicit design procedure g
Answer:
[tex]D=0.41m[/tex]
Explanation:
From the question we are told that:
Discharge rate [tex]V_r=0.35 m3/s[/tex]
Distance [tex]d=4km[/tex]
Elevation of the pumping station [tex]h_p= 140 m[/tex]
Elevation of the Exit point [tex]h_e= 150 m[/tex]
Generally the Steady Flow Energy Equation SFEE is mathematically given by
[tex]h_p=h_e+h[/tex]
With
[tex]P_1-P_2[/tex]
And
[tex]V_1=V-2[/tex]
Therefore
[tex]h=140-150[/tex]
[tex]h=10[/tex]
Generally h is give as
[tex]h=\frac{0.5LV^2}{2gD}[/tex]
[tex]h=\frac{8Q^2fL}{\pi^2 gD^5}[/tex]
Therefore
[tex]10=\frac{8Q^2fL}{\pi^2 gD^5}[/tex]
[tex]D=^5\frac{8*(0.35)^2*0.003*4000}{3.142^2*9.81*10}[/tex]
[tex]D=0.41m[/tex]
What statement about the print() function is true?
print() has a variable number of parameters.
print() can have only one parameter.
print() can be used to obtain values from the keyboard.
print() does not automatically add a line break to the display.
Explanation:
print() has a variable number of parameters. this is the answer.
hope this helps you
have a nice day
A(n) ____ combines two planetary gearsets to provide more gear ratio possibilities. A)compound planetary gearset B)orifice C)detent D)lock-up torque converter
Answer:
The answer is A. Compound Planetary Gearset.
Explanation:
The Compound Planetary Gear block represents a planetary gear train with composite planet gears. Each composite planet gear is a pair of rigidly connected and longitudinally arranged gears of different radii. One of the two gears engages the centrally located sun gear while the other engages the outer ring gear.
Compound planetary gear sets have at least two planet gears attached in line to the same shaft, rotating and orbiting at the same speed while meshing with different gears. Compounded planets can have different tooth numbers, as can the gears they mesh with.
Refrigerant 134a enters an insulated compressor operating at steady state as saturated vapor at 2208C with a mass flow rate of 1.2 kg/s. Refrigerant exits at 7 bar, 708C. Changes in kinetic and potential energy from inlet to exit can be ignored. Determine (a) the volumetric flow rates at the inlet and exit, each in m3 /s, and (b) the power input to the compressor, in kW.'
Answer:
a)[tex]V_1=4.88m^2/s[/tex]
[tex]V_2=4.88m^2/s[/tex]
b)[tex]P=-119.18kW[/tex]
Explanation:
From the question we are told that:
Steady State Saturated vapor [tex]T_1= -20C=>253k[/tex]
Mass Flow rate [tex]M=1.2kg/s[/tex]
Exit Pressure [tex]P_2=7bar[/tex]
Exit Temperature [tex]T_2=70C=>373k[/tex]
From Refrigerant 134a Properties
[tex]T_1= -20C =>P_1=1.399 bar[/tex]
Generally the equation for Volumetric Flow rate is mathematically given by
For Inlet
[tex]V_1=m\frac{RT_1}{P_1}[/tex]
[tex]V_1=m\frac{8314*253}{1.399*10^3}[/tex]
[tex]V_1=18.97m^2/s[/tex]
For outlet
[tex]V_2=m\frac{RT_2}{P_2}[/tex]
[tex]V_2=1.2*\frac{8314*343}{7*10^3}[/tex]
[tex]V_2=4.88m^2/s[/tex]
b)
Generally the equation for Steady state mass and energy equation is mathematically given by
[tex]P=m(h_1-h_2)[/tex]
From Refrigerant 134a Properties
[tex]T_1= -20C =>h_1=24.76kJ/kg[/tex]
[tex]T_2= 70C =>h_2=124.08kJ/kg[/tex]
Therefore
[tex]P=1.2(12.76-124.08)[/tex]
[tex]P=-119.18kW[/tex]
Therefore
Power input into the compressor is
[tex]P=-119.18kW[/tex]
Which of the following is not a part sympathetic activation during the fight or flight response?
Answer:
Digestion functions become more active
Explanation:
I just took the text!
Unfiltered full wave rectifier with a 120 V 60 Hz input produces an output with a peak of 15V. When a capacitor-input filter and a 1k ohm load are connected the DC output voltage is 14V. What is... The value of the capacitor? The value of the peak to peak ripple voltage?
Answer:
[tex]V_{pp}=2V[/tex]
Explanation:
Source Voltage [tex]V= 120V[/tex]
Frequency [tex]f=60Hz[/tex]
Peak output voltage [tex]Vp=15V[/tex]
Peak Output Voltage with filter [tex]V_p'=14V[/tex]
Generally the equation for Peak to peak voltage is mathematically given by
[tex]V_p'=V_p-\frac{V_{pp}}{2}[/tex]
Therefore
[tex]V_{pp}=2(V_p-v_p')[/tex]
[tex]V_{pp}=2(15-14)[/tex]
[tex]V_{pp}=2V[/tex]
In the construction of a large reactor pressure vessel, a new steel alloy with a plane strain fracture toughness of 55 MPa-m1/2 and a Y value of 1.0. An in-service stress level of 200 MPa has been calculated. What is the length of a surface crack (in mm) that will lead to fracture
Answer:
[tex]l=24mm[/tex]
Explanation:
From the question we are told that:
Plane strain fracture toughness of [tex]T=55 MPa-m1/2[/tex]
Y value [tex]Y=1.0[/tex]
Stress level of[tex]\sigma =200 MPa[/tex]
Generally the equation for length of a surface crack is mathematically given by
[tex]l=\frac{1}{\pi}(\frac{T}{Y*\sigma})^2[/tex]
[tex]l=\frac{1}{3.142}(\frac{55}{1*200})^2[/tex]
[tex]l=0.024m[/tex]
Therefore
in mm
[tex]l=24mm[/tex]
Given : x² + 200x = 166400 The current park is a square, and the addition will increase the width by 200 meters to give the expanded park a total area of 166,400 square meters To Find : the side length of the current square park. Solution: x² + 200x = 166400 => x(x + 200) = 166400 166400 = 320 * 520 => (320)(320 + 200) = 166400 => x = 320 side length of the current square park. = 320 m Learn More: Which expression is a possible leading term for the polynomial ... brainly.In/question/13233517
Answer:
320 m
Explanation:
To find the side length of the current park, x, we solve the quadratic equation for the area of the park
x² + 200x = 166400
x² + 200x - 166400 = 0
We multiply -166400 by x² to get -166400x². We now find the factors of 166400x² that will add up to 200x. These factors are -320x and 520x
So, we re-write the expression as
x² + 200x - 166400 = 0
x² + 520x - 320x - 166400 = 0
We write out the factors of the expression,
x² + 520x - 320x - 320 × 520 = 0
Factorizing the expression, we have
x(x + 520) - 320(x + 520) = 0
(x + 520)(x - 320) = 0
x + 520 = 0 or x - 320 = 0
x = -520 or x = 320
Since x is not negative, we take the positive answer.
So, x = 320 m
A rod that was originally 100-cm-long experiences a strain of 82%. What is the new length of the rod?
122 cm
182 cm
82 cm
22 cm
108.2 cm
Answer: (b)
Explanation:
Given
Original length of the rod is [tex]L=100\ cm[/tex]
Strain experienced is [tex]\epsilon=82\%=0.82[/tex]
Strain is the ratio of the change in length to the original length
[tex]\Rightarrow \epsilon =\dfrac{\Delta L}{L}\\\\\Rightarrow 0.82=\dfrac{\Delta L}{100}\\\\\Rightarrow \Delta L=82\ cm[/tex]
Therefore, new length is given by (Considering the load is tensile in nature)
[tex]\Rightarrow L'=\Delta L+L\\\Rightarrow L'=82+100=182\ cm[/tex]
Thus, option (b) is correct.
For each function , sketch the Bode asymptotic magnitude and asymptotic phase plots.
a. G(s)= 1/s(s+2)(s+4)
b. G(s)= (s+5)/(s+2)(s+4)
c. G(s)= (s+3)(s+5)/s(s+2)(s+4)
Answer:
attached below
Explanation:
a) G(s) = 1 / s( s+2)(s + 4 )
Bode asymptotic magnitude and asymptotic phase plots
attached below
b) G(s) = (s+5)/(s+2)(s+4)
phase angles = tan^-1 w/s , -tan^-1 w/s , tan^-1 w/4
attached below
c) G(s)= (s+3)(s+5)/s(s+2)(s+4)
solution attached below
A route for a proposed 8-m-wide highway crosses a region with a 4-m-thick saturated, soft, normally consolidated clay (CH) above impermeable rock. Groundwater level is 1 m below the surface. The geotechnical data available during the preliminary design stage consist of Atterberg limits (LL 5 68% and PL 5 32%) and the natural water content (w 5 56%). Based on experience, the geotechnical engineer estimated the coefficient of consolidation at 8 m2 per year. To limit settlement, a 4-m-high embankment will be constructed as a surcharge from fill of unit weight 16 kN/m3.
(a) Estimate the compression and recompression indices.
(b) Estimate the total primary consolidation settlement under the center of the embankment.
(c) Plot a time–settlement curve under the center of the embankment.
(d) How many years will it take for 50% consolidation to occur?
(e) Explain how you would speed up the consolidation.
(f) Estimate the rebound (heave) when the surcharge is removed.
Sketch a settlement profile along the base of the embankment. Would the settlement be uniform along the base? Explain your answer.A route for a proposed 8-m-wide highway crosses a
Compute the minimum length of vertical curve that will provide 220 m stopping sight distance for a design speed of 110 km/h at the intersection of a -3.50% grade and a +2.70% grade.
i have made notes and saved it as a pdf u can take it to answer question and make ur concept good
The minimum length of vertical curve that will provide 220 m stopping sight distance is; 458.8 m
We are given;
Stopping sight distance; S = 220 m
Design Speed; V = 110 km/h
Intersection grade 1; G1 = +2.7
Intersection Grade 2; G2 = -3.5
From the AASHTO Table attached, we can trace the value of the radius of vertical curvature for the given stopping sight distance and design speed.From the table, at S = 220 m and V = 110 km/h, we can see that;
Radius of vertical curvature; K = 74
Now, the difference in grade given is;A = G1 - G2
A = 2.7 - (-3.5)
A = 2.7 + 3.5
A = 6.2
Formula for the minimum length of vertical curve is;L = KA
Thus;
L = 74 × 6.2
L = 458.8 m
Read more about stopping sight distance at; https://brainly.com/question/2087168
Why water parameters of Buriganga river vary between wet and dry seasons?
Explain.
Select the correct statement(s) regarding IEEE 802.16 WiMAX BWA. a. WiMAX BWA describes both 4G Mobile WiMAX and fixed WiMax b. DSSS and CDMA are fundamental technologies used with WiMAX BWA c. OFDM is implemented to increase spectral efficiency and to improve noise performance d. all of the statements are correct
Answer:
d. all of the statements are correct.
Explanation:
WiMAX Broadband Wireless Access has the capacity to provide service up to 50 km for fixed stations. It has capacity of up to 15 km for mobile stations. WiMAX BWA describes both of 4G mobile WiMAX and fixed stations WiMAX. OFMD is used to increase spectral efficiency of WiMAX and to improve noise performance.
