PWM input and output signals are often converted to analog voltage signals using low-pass filters. Design and simulate the following: a PWM signal source with 1 kHz base frequency and adjustable pulse-width modulation(PWM source), an analog filter with time constant of 0.01 s. Simulate the input and output of the low pass filter for a PWM duty cycle of 25, 50, and 100%
Answer:
Attached below
Explanation:
PWM signal source has 1 KHz base frequency
Analog filter : with time constant = 0.01 s
low pass transfer function = [tex]\frac{1}{0.01s + 1 }[/tex]
PWM duty cycle is a constant block
Attached below is the design and simulation into Simulink at 25% , 50% and 100% respectively
A site is underlain by two layers of normally consolidated clayey sand. The unit weights of the top layer are 19 and 21 kN/m2 and the layer is 6 meters thick. The unit weights of the bottom layer are 20 and 22 kN/m, and the layer is 8 meters thick. Below the bottom layer lies bedrock. The water table is located 2 meters below the ground surface. The top layer soil has a friction angle of 38 degrees, a cohesion intercept of 20 kPa, and an undrained strength of 160 kPa. The bottom layer soil has a friction angle of 33 degrees, a cohesion intercept of 10 kPa, and an undrained strength of 120 kPa. Point A is located 9 meters underground. All layers have a lateral stress ratio of 0.5.
Required:
a. Draw the profile neatly, add dimensions, and draw a tree on the ground surface.
b. Determine the geostatic vertical effective stress and the geostatic horizontal effective stress at point A.
c. Draw the Mohr Circle to determine the effective stress that acts at point A on a plane inclined 10 degrees counter-clockwise from the horizontal plane. Make sure the Mohr Circle is a well-drawn circle
Answer:
A) attached below
B) Geostatic vertical effective stress ( бv )
= 119.33 KN/m^2
Geostatic horizontal effective stress ( бn )
= 59.66 KN/m^2
C) attached below
Explanation:
attached below is a detailed solution
A) attached below
B) Determine the geostatic vertical effective stress and the geostatic horizontal effective stress at point A
Geostatic vertical effective stress ( бv )
= 119.33 KN/m^2
Geostatic horizontal effective stress ( бn )
= 59.66 KN/m^2
C) attached below
A river has an average rate of water flow of 59.6 M3/s. This river has three tributaries, tributary A, B and C, which account for 36%, 47% and 17% of water flow respectively. How much water is discharged in 30 minutes from tributary B?
Answer:
50421.6 m³
Explanation:
The river has an average rate of water flow of 59.6 m³/s.
Tributary B accounts for 47% of the rate of water flow. Therefore the rate of water flow through tributary B is:
Flow rate of water through tributary B = 47% of 59.6 m³/s = 0.47 * 59.6 m³/s = 28.012 m³/s
The volume of water that has been discharged through tributary B = Flow rate of water through tributary B * time taken
time = 30 minutes = 30 minutes * 60 seconds / minute = 1800 seconds
The volume of water that has been discharged through tributary B in 30 seconds = 28.012 m³/s * 1800 seconds = 50421.6 m³
Axial forces in a column due to service loads are as follows (assume the live load used in these calculationsis less than 100 psf):Dead:150 kcompressionLive:280 kcompressionRoof Live:40 kcompressionSnow:50 kcompressionWind:120 kcompression or tensionEarthquake:200 kcompression or tension1. Compute the required axial strength,TuandPu, for this column in tension and compression usingLRFD load combinations. (Neglect self-weight.)2. Describe the loading scenario that represents the worst case tension and compression loading for thecolumn (remember, wind and earthquake loading can act in either direction).
Answer:
a) attached below
b) The worst case in tension is case ( 9-7 ) b which is
= -65k
The worst case in compression is case ( 9-5 ) a which is
= 670 k
Explanation:
Given data :
D = 150k , L = 280k , Lr = 40k , s = 50k , w = ± 120k
E = ± 200k
attached below is a detailed solution to the given problem ( problem 1 )
A) attached below
b) The worst case in tension is case ( 9-7 ) b which is
= -65k
The worst case in compression is case ( 9-5 ) a which is
= 670 k
Perform the following unit conversions. Please do not use an on-line unit converter since this problem is given to you as practice in preparation for what you need to be proficient in:
a. 180 in^3 to L
b. 750 ft-lbf to kJ
c. 75.0 hp to kW
d. 2500.0 lb/h to kg/s
e. 120 psia to kPa
f. 120 psig to kPa
g. 300 ft/min to m/s
h. 125 km/h to miles/h
i. 6000 N to Ibf
j. 6000 N to ton
Answer:
The answers are below
Explanation:
a. 180 in^3 to L
1 in³ = 0.0164L
180 in³ = [tex]180\ in^3*\frac{0.0164\ L}{1\ in^3}= 2.95\ L[/tex]
b. 750 ft-lbf to kJ
1 ft-lbf = 0.00136 kJ
750 ft-lbf = [tex]750\ ft-lbf *\frac{0.00136\ kJ}{1\ ft-lbf} =1.02\ kJ[/tex]
c. 75.0 hp to kW
1 hp = 0.746 kW
75 hp = [tex]75\ hp*\frac{0.746\ kW}{1\ hp}=55.95\ kW[/tex]
d. 2500.0 lb/h to kg/s
1 lb/h = 0.000126 kg/s
2500.0 lb/h = [tex]2500.0\ lb/h*\frac{0.000126\ kg/s}{1\ lb/h} =0.315\ kg/s[/tex]
e. 120 psia to kPa
1 psia = 6.89 kPa
120 psia = [tex]120\ psia*\frac{6.89\ kPa}{1\ psia} =826.8\ kPa[/tex]
f. 120 psig to kPa
1 psig = 6.89 kPa
120 psig = [tex]120\ psia*\frac{6.89\ kPa}{1\ psig} =826.8\ kPa[/tex]
g. 300 ft/min to m/s
1 ft/min = 0.005 m/s
300 ft/min = [tex]300\ ft/min*\frac{0.005\ m/s}{1\ ft/min} = 1.5\ m/s\\[/tex]
h. 125 km/h to miles/h
1 km/h = 0.62 mph
125 km/h = [tex]125\ km/h*\frac{0.62\ mph}{1\ km/h} =77.5\ mph[/tex]
i) 6000 N to Ibf
1 N = 0.2248 lbf
6000 N = [tex]6000\ N*\frac{ 0.2248\ lbf}{1\ N}=1348.8\ N[/tex]
j. 6000 N to ton
1 N = 0.000102 Ton-force
6000 N = [tex]6000\ N*\frac{ 0.000102\ Ton-force}{1\ N}=0.612\ N[/tex]
A Russian rocket engine (RD-110 with LOX-kerosene) consists of four thrust chambers supplied by a single turbopump. The exhaust from the turbine of the turbopump then is ducted to four vernier nozzles (which can be rotated to provide some control of the flight path). Using the information below, determine the thrust and mass flow rate of the four vernier gas nozzles. For individual thrust chambers (vacuum): F= 73.14 kN, c = 3279 m/sec Overall engine with verniers (vacuum): F= 297.93 kN, c = 3197 m/sec.
Answer:
- Vernier thrust is 5.37 kN
- mass flow rate of the four Vernier gas nozzles is 0.4048 kg/s
Explanation:
Given that;
For individual thrust chambers (vacuum);
Fc = 73.14 kN , Cc = 3279 m/sec
For Overall engine with Vernier (vacuum);
Foa = 297.93 kN = , Coa = 3197 m/sec.
- determine the Vernier thrust
Vernier thrust Fv = Foa - ( 4 × Fc )
Vernier thrust Fv = 297.93 - ( 4 × 73.14)
Vernier thrust Fv = 297.93 - 292.56
Vernier thrust Fv = 5.37 kN
Therefore, Vernier thrust is 5.37 kN
-
Vernier mass flow rate;
we know that
[tex]Co_{a}[/tex] = Fc + Fv / mc + mv
mv = Foa/Coa - Fc/Cc
we convert kilonewton to kilograms
1 kn = 102 kg
Fc = 73.14 kN = 73.14 × 102 = 7460.28 kg
Foa = 297.93 kN = 297.93 × 102 = 30388.86 kg
we substitute
mv = (30388.86 / 3197) - (( 4 × 7460.28) / 3279)
mv = 9.5054 - 9.1006
mv = 0.4048 kg/s
Therefore, mass flow rate of the four Vernier gas nozzles is 0.4048 kg/s
Engineer drawing:
How can i draw this? Any simple way?
Please calculate the energy stored in a flying wheel. The steel flying wheel is 10 metric tons, in adiameter of 10m, and it rotates at a speed of 1000 rpm. The steel density is 8050 kg/m^3. When energyis fully released, the flying wheel stops its rotation.
Answer:
685.38 MJ
Explanation:
Given that:
mass = 10 tons = 1.0 × 10 ⁴ kg
diameter D = 10 m
radius R = 5 m
speed N = 1000 rpm
Using the formula for K.E = [tex]\dfrac{1}{2}I \omega^2[/tex] to calculate the energy stored
where;
[tex]= \dfrac{2 \pi \times N}{60}[/tex]
[tex]= \dfrac{2 \pi \times 1000}{60}[/tex]
= 104.719 rad/s
Hence, the energy stored is;
[tex]= \dfrac{1}{2}\times (\dfrac{MR^2}{2}) \times \omega^2[/tex]
[tex]= \dfrac{1}{2}\times (\dfrac{10^4\times 5^2}{2}) \times 104.719^2[/tex]
= 685379310.1
= 685.38 MJ
2.) Technician A says that milky colored ATF could indicate a leaking transmission cooler in the radiator.
Technician B says that milky colored ATF could indicate the presence of leak detection dye.
Who is right?
OA. A only
OB. B only
OC. Both A and B
OD. Neither A nor B
A brass alloy rod having a cross sectional area of 100 mm2 and a modulus of 110 GPa is subjected to a tensile load. Plastic deformation was observed to begin at a load of 39872 N. a. Determine the maximum stress that can be applied without plastic deformation. b. If the maximum length to which a specimen may be stretched without causing plastic deformation is 67.21 mm, what is the original specimen length
Answer:
a) the maximum stress that can be applied without plastic deformation is 398.72 N/mm²
b) length of the specimen is 66.97 mm
Explanation:
Given the data in the question;
a) Determine the maximum stress that can be applied without plastic deformation
when know that; maximum stress σ[tex]_{max}[/tex] = F / A
where F is the force in the rod ( 39872 N )
A is the cross-sectional area of the rod ( 100 mm² )
so we substitute;
σ[tex]_{max}[/tex] = 39872 N / 100 mm²
σ[tex]_{max}[/tex] = 398.72 N/mm²
Therefore, the maximum stress that can be applied without plastic deformation is 398.72 N/mm²
b)
strain in the members can be calculated using the expression
ε = σ / E
where σ is the stress in the rod
E is the module of elasticity ( 110 GPa = 110000 N/mm² )
(Sl-L) / L = σ/E
where Sl-L is the change in length of the member
L is the original length of the specimen
so we substitute
(67.21 - L) / L = 398.72 / 110000
110000( 67.21 - L) = 398.72L
7393100 - 110000L = 398.72L
7393100 = 398.72L+ 110000L
7393100 = 110398.72L
L = 7393100 / 110398.72
L = 66.97 mm
Therefore; length of the specimen is 66.97 mm
Determine the voltage drop from the top terminal to the bottom terminal, vab, in the right hand branch and, vcd, in the left hand branch of the circuit. Determine each voltage drop based on the elements in the corresponding branch.
Answer:
Hello your question is incomplete attached below is the missing part of the question
answer ;
voltage drop in the Vcd branch = 30 V
Voltage drop in the middle branch = 40v - 30v = 10 volts
voltage drop in AB = 60 + ( -600 * 0.05 ) = 60 - 30 = 30 volts
Explanation:
Determine voltage drop from top terminal to bottom terminal ( Vab ) in the right hand branch and Vcd in left hand branch
40v and 50mA are in series hence; Ix = 50mA
also Vcd = 30V
CD is parallel to AB hence; Vcd = Vab = 30 V
Vab = ∝*Ix + 60 v
30v = ∝ ( 50mA ) + 60
therefore ∝ = -600
voltage drop in the Vcd branch = 30 V
Voltage drop in the middle branch = 40v - 30v = 10 volts
voltage drop in AB = 60 + ( -600 * 0.05 ) = 60 - 30 = 30 volts
3.) Technician A says that a scan tool can be used to verify engine operating temperature,
Technician B says that a refractometer can be used to verify engine operating temperature.
Who is right?
OA. A only
OB. B only
OC. Both A and B
OD. Neither A nor B
