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California State University Long Beach
Department of CECEM
CE 336 Fluid Mechanics Laboratory
Bernoulli’s Theorem Demonstration
Experiment #4
Instructor: Dr. Loan Miller
Group 3
Group Members:
Major(s)
Andy Nicky Sok
Mechanical Engineering
Huu Tran
Mechanical Engineering
Kevin Alvarez
Mechanical Engineering
Date: September 29th, 2023
Fall 2023
2
Table of Contents
1.
Purpose of Study…………………………………………………… 3
2.
Introduction………………………………………………………… 3
3.
Theory……………………………………………………………… 4
4.
Equipment and Experimental Set-up………………………………. 6
5.
Discussion…………………………………………………………. 8
6.
Conclusion………………………………………………………… 14
7.
References…………………………………………………………. 16
3
Purpose of Study
Bernoulli's theorem is tested in this experiment using a venturi meter. Through the venturi meter and the measurement of volumetric flow rates, static head and kinetic head are measured, recorded, and calculated. Thus, the total amount of energy that passes via the venturi pipe can be calculated.
Introduction To verify Bernoulli’s Theorem, in this experiment a venturi meter is used to demonstrate the discharge of the fluid which then can be calculated to see if the law of conservation of energy is applied to the steady flow of the fluid. To do this, a series of manometers connected by a venturi tube and equipped with a piezometric head were observed. As the experiment is running, the system needs to be regulated to have a consistent and reliable reading. Check for air bubbles that may be trapped in the system and to remove them by bleeding which is located on top of the venturi meters. Adjust the flow rate of the system by the hydraulic bench to 0.105 and measure the height of the heads on each manometer. Repeat the process but increase
the flow rate by 0.05 until the flow rate reaches 0.55. The diameter of the ducts can be used to determine the area of the duct, and from that area, the flow rate and velocity can be calculated along with the velocity and velocity head. Static head and dynamic head can be added together at
the end to get the total head.
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4
Theory Bernoulli’s equation:
P is the pressure (Pa) ?
is the density of fluid (kg/m) v is the velocity of water (m/s) z is the distance from datum to a specific point (m)
g is gravity constant (m/s^2)
Flow rate:
Q is the flow rate (m^3 /s)
t is the time to collect water (s)
V is the volume of water collected (m^3)
Flow Velocity:
v is the velocity of flow (m/s)
A is the cross-sectional area at specific points (m^2 )
Q is the flow rate (m^3 /s)
5
Dynamic head:
ℎ
?
is dynamic head (m)
g is gravity constant (m/s^2)
v is the velocity of flow (m/s)
Total head:
H is total head (m)
h is static head (m) g is gravity constant (m/s^2)
v is the velocity of flow (m/s)
6
Equipment & Experimental Set-up Figure 1: Hydraulic Bench to adjust the flow rate of the system.
Figure 2: Venturi Meter used for its different pressure areas for each piezometer.
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7
Figure 3: Manometer used to measure the height of each tubing.
Figure 4: Venturi Meter Description of each Piezometer.
8
Discussion ●
Table of Recorded Data and Results
Flow
Rate
Distance
into Duct
Area of
Duct
Static
Head
Static
Head
Velocity
Dynamic
Head
Total
Head
Q
A
h
h
v
H
m^3/s
(m)
m^2
(mm)
(m)
(m/s)
(m)
(m)
0.000105
h1
0.000
0.0005309
242
0.242
0.198
0.00199
0.24399
h2
0.020
0.0004227
242
0.242
0.248
0.00314
0.24514
h3
0.032
0.0002659
238
0.238
0.395
0.00795
0.24595
h4
0.046
0.0002011
230
0.230
0.522
0.01390
0.24390
h5
0.061
0.0002214
230
0.230
0.474
0.01146
0.24146
h6
0.076
0.0002679
234
0.234
0.392
0.00783
0.24183
h7
0.091
0.0003192
238
0.238
0.329
0.00551
0.24351
h8
0.106
0.0003746
240
0.240
0.280
0.00400
0.24400
h9
0.121
0.0004348
240
0.240
0.241
0.00297
0.24297
h10
0.136
0.0004992
242
0.242
0.210
0.00226
0.24426
h11
0.156
0.0005309
242
0.242
0.198
0.00199
0.24399
0.000155
h1
0.000
0.0005309
248
0.248
0.292
0.00434
0.25234
h2
0.020
0.0004227
246
0.246
0.367
0.00685
0.25285
h3
0.032
0.0002659
236
0.236
0.583
0.01732
0.25332
h4
0.046
0.0002011
222
0.222
0.771
0.03029
0.25229
h5
0.061
0.0002214
224
0.224
0.700
0.02498
0.24898
h6
0.076
0.0002679
232
0.232
0.579
0.01706
0.24906
h7
0.091
0.0003192
238
0.238
0.486
0.01202
0.25002
h8
0.106
0.0003746
240
0.240
0.414
0.00873
0.24873
h9
0.121
0.0004348
242
0.242
0.356
0.00648
0.24848
h10
0.136
0.0004992
244
0.244
0.311
0.00491
0.24891
h11
0.156
0.0005309
246
0.246
0.292
0.00434
0.25034
0.000205
h1
0.000
0.0005309
257
0.257
0.386
0.00760
0.26460
h2
0.020
0.0004227
254
0.254
0.485
0.01199
0.26599
h3
0.032
0.0002659
237
0.237
0.771
0.03030
0.26730
h4
0.046
0.0002011
210
0.210
1.020
0.05299
0.26299
h5
0.061
0.0002214
214
0.214
0.926
0.04369
0.25769
h6
0.076
0.0002679
230
0.230
0.765
0.02984
0.25984
h7
0.091
0.0003192
238
0.238
0.642
0.02102
0.25902
h8
0.106
0.0003746
243
0.243
0.547
0.01526
0.25826
h9
0.121
0.0004348
246
0.246
0.471
0.01133
0.25733
h10
0.136
0.0004992
249
0.249
0.411
0.00860
0.25760
h11
0.156
0.0005309
251
0.251
0.386
0.00760
0.25860
9
Flow Rate
Distance
into Duct
Area of
Duct
Static
Head
Static
Head
Velocit
y
Dynamic
Head
Total
Head
Q
A
h
h
v
H
m^3/s
(m)
m^2
(mm)
(m)
(m/s)
(m)
(m)
0.000255
h1
0.000
0.0005309
267
0.267
0.480
0.01176
0.27876
h2
0.020
0.0004227
263
0.263
0.603
0.01855
0.28155
h3
0.032
0.0002659
237
0.237
0.959
0.04688
0.28388
h4
0.046
0.0002011
196
0.196
1.268
0.08198
0.27798
h5
0.061
0.0002214
203
0.203
1.152
0.06761
0.27061
h6
0.076
0.0002679
226
0.226
0.952
0.04617
0.27217
h7
0.091
0.0003192
239
0.239
0.799
0.03253
0.27153
h8
0.106
0.0003746
247
0.247
0.681
0.02362
0.27062
h9
0.121
0.0004348
251
0.251
0.586
0.01753
0.26853
h10
0.136
0.0004992
254
0.254
0.511
0.01330
0.26730
h11
0.156
0.0005309
257
0.257
0.480
0.01176
0.26876
0.000301
h1
0.000
0.0005309
280
0.280
0.567
0.01638
0.29638
h2
0.020
0.0004227
275
0.275
0.712
0.02584
0.30084
h3
0.032
0.0002659
238
0.238
1.132
0.06531
0.30331
h4
0.046
0.0002011
180
0.180
1.497
0.11423
0.29423
h5
0.061
0.0002214
190
0.190
1.359
0.09420
0.28420
h6
0.076
0.0002679
224
0.224
1.123
0.06433
0.28833
h7
0.091
0.0003192
241
0.241
0.943
0.04532
0.28632
h8
0.106
0.0003746
252
0.252
0.803
0.03290
0.28490
h9
0.121
0.0004348
259
0.259
0.692
0.02442
0.28342
h10
0.136
0.0004992
262
0.262
0.603
0.01853
0.28053
h11
0.156
0.0005309
266
0.266
0.567
0.01638
0.28238
0.000355
h1
0.000
0.0005309
294
0.294
0.669
0.02279
0.31679
h2
0.020
0.0004227
288
0.288
0.840
0.03594
0.32394
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10
h3
0.032
0.0002659
238
0.238
1.335
0.09085
0.32885
h4
0.046
0.0002011
160
0.160
1.766
0.15889
0.31889
h5
0.061
0.0002214
176
0.176
1.603
0.13103
0.30703
h6
0.076
0.0002679
220
0.220
1.325
0.08948
0.30948
h7
0.091
0.0003192
242
0.242
1.112
0.06304
0.30504
h8
0.106
0.0003746
258
0.258
0.948
0.04577
0.30377
h9
0.121
0.0004348
266
0.266
0.816
0.03397
0.29997
h10
0.136
0.0004992
272
0.272
0.711
0.02578
0.29778
h11
0.156
0.0005309
278
0.278
0.669
0.02279
0.30079
Flow
Rate
Distance
into Duct
Area of
Duct
Static
Head
Static
Head
Velocity
Dynamic
Head
Total
Head
Q
A
h
h
v
H
m^3/s
(m)
m^2
(mm)
(m)
(m/s)
(m)
(m)
0.000405
h1
0.000
0.0005309
310
0.310
0.763
0.02966
0.33966
h2
0.020
0.0004227
300
0.300
0.958
0.04678
0.34678
h3
0.032
0.0002659
234
0.234
1.523
0.11824
0.35224
h4
0.046
0.0002011
136
0.136
2.014
0.20680
0.34280
h5
0.061
0.0002214
160
0.160
1.829
0.17054
0.33054
h6
0.076
0.0002679
214
0.214
1.512
0.11646
0.33046
h7
0.091
0.0003192
244
0.244
1.269
0.08205
0.32605
h8
0.106
0.0003746
263
0.263
1.081
0.05957
0.32257
h9
0.121
0.0004348
275
0.275
0.931
0.04421
0.31921
h10
0.136
0.0004992
282
0.282
0.811
0.03355
0.31555
h11
0.156
0.0005309
289
0.289
0.763
0.02966
0.31866
0.000451
h1
0.000
0.0005309
330
0.330
0.849
0.03678
0.36678
h2
0.020
0.0004227
317
0.317
1.067
0.05801
0.37501
11
h3
0.032
0.0002659
237
0.237
1.696
0.14663
0.38363
h4
0.046
0.0002011
110
0.110
2.243
0.25645
0.36645
h5
0.061
0.0002214
141
0.141
2.037
0.21148
0.35248
h6
0.076
0.0002679
207
0.207
1.683
0.14441
0.35141
h7
0.091
0.0003192
244
0.244
1.413
0.10174
0.34574
h8
0.106
0.0003746
270
0.270
1.204
0.07387
0.34387
h9
0.121
0.0004348
283
0.283
1.037
0.05483
0.33783
h10
0.136
0.0004992
293
0.293
0.904
0.04161
0.33461
h11
0.156
0.0005309
301
0.301
0.849
0.03678
0.33778
0.000500
h1
0.000
0.0005309
350
0.350
0.942
0.04520
0.39520
h2
0.020
0.0004227
333
0.333
1.183
0.07130
0.40430
h3
0.032
0.0002659
234
0.234
1.880
0.18022
0.41422
h4
0.046
0.0002011
80
0.080
2.487
0.31520
0.39520
h5
0.061
0.0002214
123
0.123
2.258
0.25992
0.38292
h6
0.076
0.0002679
200
0.200
1.866
0.17750
0.37750
h7
0.091
0.0003192
247
0.247
1.566
0.12505
0.37205
h8
0.106
0.0003746
277
0.277
1.335
0.09079
0.36779
h9
0.121
0.0004348
294
0.294
1.150
0.06739
0.36139
h10
0.136
0.0004992
306
0.306
1.002
0.05114
0.35714
h11
0.156
0.0005309
315
0.315
0.942
0.04520
0.36020
Flow
Rate
Distance
into Duct
Area of
Duct
Static
Head
Static
Head
Velocity
Dynamic
Head
Total
Head
Q
A
h
h
v
H
m^3/s
(m)
m^2
(mm)
(m)
(m/s)
(m)
(m)
0.000550
h1
0.000
0.0005309
374
0.374
1.036
0.05470
0.42870
h2
0.020
0.0004227
355
0.355
1.301
0.08628
0.44128
h3
0.032
0.0002659
231
0.231
2.068
0.21807
0.44907
12
h4
0.046
0.0002011
48
0.048
2.736
0.38140
0.42940
h5
0.061
0.0002214
100
0.100
2.484
0.31451
0.41451
h6
0.076
0.0002679
192
0.192
2.053
0.21477
0.40677
h7
0.091
0.0003192
249
0.249
1.723
0.15131
0.40031
h8
0.106
0.0003746
285
0.285
1.468
0.10986
0.39486
h9
0.121
0.0004348
306
0.306
1.265
0.08154
0.38754
h10
0.136
0.0004992
320
0.320
1.102
0.06188
0.38188
h11
0.156
0.0005309
332
0.332
1.036
0.05470
0.38670
Table 1
: Recorded Data and Results
●
Sample Calculations
Using the first flow rate for the sample calculation:
Figure 5. The venturi meter section with Area, Distance from Datum and corresponding
diameters of the section (all measurements are in mm)
All millimeter is converted to meter.
⮚
Average velocity through the pipe section, at each location of the piezometric tube:
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13
v
=
Q
A
=
0.000105
0.00053093
=
¿
0.198
(
m
s
)
⮚
Dynamic Head (Velocity Head) at each location of the piezometric tube:
v
2
2
g
=
v
2
2
∗
9.81
=
0.198
2
2
∗
9.81
=
¿
0.00199
(
m
)
⮚
Total energy (H) at each location of the piezometric tube:
H
=
P
γ
+
z
+
v
2
2
g
=
static head
+
0
+
dynamichead
H
=
h
+
v
2
2
g
=
0.242
+
0.00199
=
0.24399
(
m
)
See Table 1 for results of calculations.
0.000 0.020 0.040 0.060 0.080 0.100 0.120 0.140 0.160 0.180 0.200
0.03
0.08
0.13
0.18
0.23
0.28
0.33
0.38
HGL vs Distance x Q1
Q2
Q3
Q4
Q5
Q6
Q7
Q8
Q9
Q10
Distance x (m)
Static Head (m)
Figure 6. HGL vs. distance along the duct for the recorded flows.
14
0.000
0.020
0.040
0.060
0.080
0.100
0.120
0.140
0.160
0.180
0.2
0.25
0.3
0.35
0.4
0.45
0.5
EGL vs Distance x
Q1
Q2
Q3
Q4
Q5
Q6
Q7
Q8
Q9
Q10
Distance x (m)
Total Head (m)
Figure 7. EGL vs. distance along the duct for the recorded flows.
Conclusion
15
The purpose of this laboratory experiment is to empirically verify Bernoulli's theorem using a Venturi meter, demonstrating the principle of conservation of energy in fluid flow and its practical applications in fluid dynamics and engineering.
Questions
1.
What is a venturi meter? What is advantage of using a venturi meter over a weir (exp #2) or orifice (exp #3) to measure discharge?
A Venturi meter is a precision flow measurement device with a gradually narrowing tube that allows for accurate flow rate determination by measuring pressure differences. Using a Venturi meter offers advantages over weirs and orifices for measuring discharge because it provides higher accuracy across a broader range of flow rates, reduces head loss, is less sensitive to changes in fluid properties, and is less prone to cavitation. 2.
What are the assumptions of Bernoulli’s equation?
Bernoulli's equation assumes the fluid is incompressible, with a constant density, and that the flow is steady, devoid of unsteadiness over time. Additionally, it presumes an inviscid and irrotational flow, meaning no internal friction or rotation occurs within the fluid. The equation also relies on the conservation of energy principle, asserting that the total mechanical energy of a fluid particle remains constant along its path, given no external forces apart from gravity.
Discussion of Results
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1.
How HGL and EGL changed along the length of the duct?
At the outset, at the Venturi meter's entrance, the HGL stood at a higher level, indicating the fluid's pressure head. As the fluid gained speed while passing through the narrow throat of the Venturi meter, the HGL declined due to the intensified fluid velocity, consistent with Bernoulli's principle. Upon exiting the throat and entering the diverging section, the HGL experienced a partial rebound but stayed below its initial level. This alteration in HGL showcased the transformation of pressure energy into kinetic energy as the fluid accelerated, followed by a partial restoration of pressure energy.
The Energy Grade Line (EGL) exhibits an initial increase followed by a decrease along the length of the duct. This phenomenon reflects energy transformations within the flowing fluid. Initially, as the fluid accelerates in the constricted throat, pressure energy is
converted into kinetic energy, causing the EGL to rise. However, in the diverging section,
only a partial conversion of kinetic energy back into pressure energy occurs, resulting in a
net energy loss and causing the EGL to decrease. This pattern demonstrates the principles
of energy conservation.
2.
Discuss on the validity of Bernoulli's equation for Convergent flow and Divergent flow.
Bernoulli's equation demonstrated its accuracy for convergent flow within the Venturi meter's throat. The equation's assumptions aligned well with the conditions in this
section, where fluid velocity increased, leading to a decrease in pressure. Precise pressure
measurements affirmed the equation's validity, confirming Bernoulli's theorem..
17
In the divergent flow section, Bernoulli's equation continued to be relevant but needed specific attention. It assumed adiabatic conditions and took into account the fluid's deceleration as it expanded. Although the pressure decrease was not as dramatic as
in the throat, the experiment still showcased a measurable reduction in pressure energy along the diverging section, consistent with the principles of Bernoulli's theorem.
References
Sultana, R. (2017). Lab # 4: Bernoulli’s Theorem Demonstration. In CE 336 Fluid Mechanics Lab Manual
(pp. 21–30). essay, CSULB Fluid Mechanics Lab Coordinator. Armfield, 2012, “Bernoulli’s Theorem Demonstration”, Instruction Manual.
Munson, B. R., T. H. Okiishi, W. W. Huebsch, A. P. Rothmayer, 2012, “Fundamentals of Fluid Mechanics”, 7th edition, John Wiley, Chapter 8. CE 336 Fluid Mechanics Student Manual, 1993, CSULB.
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4. In the obove problem, how long, in minutes, after its removal from the room will
the reading be 6 deg C?
5. In the obove problem, ofter 30 seconds, whot will be the thermometer reading?
5.2
dT(body temp)/dt = k(Temp of medium - 5)
dr(body temp)/dt = k(Temp of body - Temp of medium)
Temp of body = Cenkt +5
2.5
13.7
dr(body temp)/dt = k(Temp of medium - Temp of body)…
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As the tempering temperature of a heat-treated steel increases, the associated values of Brinell hardness also increases, while the values of tensile strength and yield strength decrease.
True
False
The correct definition of Dimensional Analysis is:
A procedure to find the yield point of a material when no definitive yield point can be determined graphically.
A procedure to calculate the Elastic Modulus.
A procedure to verify the size of a mass of material.
A methodical process of converting from one set of units to another set of units.
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4.
Find the width of the belt, necessary to transmit 7.5 kW to a pulley 300 mm diameter, if the pulley
makes 1600 r.p.m and the coefficient of friction between the belt and the pulley is 0.22. Assume the
angle of contact as 210° and the maximum tension in the belt is not to exceed 8 N/mm width.
[Ans. 67.4 mm]
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APPLICATIONS OF FIRST ORDER DIFFERENTIAL EQUATIONS
Newton’s Law of Cooling and Joint Proportions
At 4:00pm, a thermometer reading of 28 deg C is taken outside where the ambient temperature is –11 deg C. At 4:05 pm, the thermometer reads 8 deg above zero. After a while, the thermometer is returned to the room maintained at 28 deg C. At 4:12pm, the thermometer reads 15 deg C. When was the thermometer returned to the room?
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PHYS X PHYS X
印 PHYS X
PHYS X
POTPHYS X
PHYS X
E PHYS X E PHYS
top/semester2/physics%20for%20engineers/PHYS220_CH15_Lecture%20Notes_Problems%2015 19,15.29 S
(D Page view
A Read aloud
V Draw
Problem-15-19: page-475
A 0.500-kg object attached to a spring with a force constant of 8.00 N/m vibrates in simple harmonic motion
with an amplitude of 10.0 cm. Calculate the maximum value of its
(a) speed, and acceleration.
(b) the speed and the acceleration when the object is 6.00 em from the equilibrium position, and
(c) the time interval required for the object to move from.r50 to r5 8.O0 cm.
Solution:
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Topics:
Statics of Rigid Bodies, Equilibrium of Rigid Bodies, Equilibrium in Two Dimensions, etc.
I will rate you with “LIKE/UPVOTE," if it is COMPLETE STEP-BY-STEP SOLUTION.
If it is INCOMPLETE SOLUTION and there are SHORTCUTS OF SOLUTION, I will rate you with “DISLIKE/DOWNVOTE.”
THANK YOU FOR YOUR HELP.
PS: If you have answered this already, don’t answer it again; give chance to other experts to answer it. I want to verify if all of you will arrive in the same final answer; thats why I ask it multiple times. If you answer it again, i'll dislike all your entries/answers.
arrow_forward
Topics:
Statics of Rigid Bodies, Equilibrium of Rigid Bodies, Equilibrium in Two Dimensions, etc.
I will rate you with “LIKE/UPVOTE," if it is COMPLETE STEP-BY-STEP SOLUTION.
If it is INCOMPLETE SOLUTION and there are SHORTCUTS OF SOLUTION, I will rate you with “DISLIKE/DOWNVOTE.”
THANK YOU FOR YOUR HELP.
PS: If you have answered this already, don’t answer it again; give chance to other experts to answer it. I want to verify if all of you will arrive in the same final answer; thats why I ask it multiple times. If you answer it again, i'll dislike all your entries/answers.
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i need the answer quickly
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dear tutor please provide neat and clean and detailed answer.
dont copy from google
adress both questions well
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A new electronic component for aircraft is tested to withstand temperatures as low as -50°F. One engineer suggests that the device cannot actually stand temperatures that low. What are the Type I and Type II errors?
Type I: The device CAN NOT withstand temperatures as low as -50F but the engineer incorrectly finds that it CAN.Type II: The device CAN withstand temperatures as low as -50F but the engineer incorrectly finds that it CAN NOT.
Type I: The device CAN withstand temperatures as low as -50F but the engineer incorrectly finds that it CAN NOT.Type II: The device CAN NOT withstand temperatures as low as -50F but the engineer incorrectly finds that it CAN.
Is the answer the first or second bullet point?
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Thermodynamics
Please READ THE PROBLEM! I’m tired of posting the same problem over and over. please Read.
This is what the asking for please:
-1. sketch and label all your answers on the nozzle
-2. Sketch and label all your answers on the process on a P-V diagram....
Thank you
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Instrumentation & Measurements
This homework measures your capability to design/analyze various components/variables of ameasurement system based on what you have studied.
Question is Attached in image. Thank you.
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The rapid progress of engineering design and information technology has caused difficulties in analyzing system reliability. Because of the increased complexity in system reliability structure (component/subsystem interfaces), many unexpected failure modes could occur, and their behaviors are interdependent. At a system’s design and development stage, the main challenge in analyzing a complex system is the failure uncertainty introduced by the incomplete knowledge of the system. This makes it hard to decompose system reliability into subsystem/component reliability in a deterministic manner, such as series or parallel systems. As a result, some common reliability analysis tools such as fault tree (FT) and reliability block diagram (RBD) become inadequate. Do you agree, why or why not? Are there any other approaches to system reliability assessment beside these tools at the early system’s design and development stage (what are these approaches)?
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- dear tutor please provide neat and clean and detailed answer. dont copy from google adress both questions wellarrow_forwardA new electronic component for aircraft is tested to withstand temperatures as low as -50°F. One engineer suggests that the device cannot actually stand temperatures that low. What are the Type I and Type II errors? Type I: The device CAN NOT withstand temperatures as low as -50F but the engineer incorrectly finds that it CAN.Type II: The device CAN withstand temperatures as low as -50F but the engineer incorrectly finds that it CAN NOT. Type I: The device CAN withstand temperatures as low as -50F but the engineer incorrectly finds that it CAN NOT.Type II: The device CAN NOT withstand temperatures as low as -50F but the engineer incorrectly finds that it CAN. Is the answer the first or second bullet point?arrow_forwardThermodynamics Please READ THE PROBLEM! I’m tired of posting the same problem over and over. please Read. This is what the asking for please: -1. sketch and label all your answers on the nozzle -2. Sketch and label all your answers on the process on a P-V diagram.... Thank youarrow_forward
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