Copy of Lab 4 _ The Human Arm
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Apr 3, 2024
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1. What is the mass of the arm and an estimate of its uncertainty? Explain how you
estimated the uncertainty.
Angle ( )
◦
M
l
(kg)
M
t
(kg)
M
T
(kg)
δ
90
0
1.885
0.01
50
0
2.200
0.01
50
100
3.970
0.02
130
0
2.050
0.02
Table 4.1:
The table shows the values for the angle of the arm when lifted at a certain angle, the
mass of the load (M
l
), mass applied (M
t
), and the uncertainty of the mass applied ( M
T
).
δ
Using equation 11.4 m
T
d
1
sin(θ) - m
L
(d
1
+ d
2
+ d
3
) - m
A
(d
1
+ d
2
) = 0 where m
T
is the mass applied
(kg), m
l
is the mass of the load (kg), θ is the angle of the metal arm (
°),
and d
1
, d
2
, d
3
are the
given distances of the arm given in the lab manual. By isolating for m
A
, the mass of the arm is
obtained:
m
A
(1.885)(2.35𝑠𝑖?(90))
14.5
=
m
A
= 0.3055 kg
We calculate the value of the mass applied by taking the average of the maximum and minimum
values:
. We then obtain its uncertainty by taking the difference
1.885 + 1.895
2
=
1. 890 𝑘𝑔
between the two m
T
values and averaging them:
For that reason, m
T
1.895 − 1.885
2
= ± 0. 005.
=
. Accordingly, the uncertainty of the mass of the arm (m
A
) is:
(1. 890 ± 0. 005)
.
(0. 3055 ± 0. 005)
2. Using Eq. 11.4, what is the theoretical value of
𝑚
T
(no uncertainty required) that will
balance the arm for
𝜃
= 50
°
? Show your work. What is the experimental value (estimate its
uncertainty) and is it consistent with the theoretical value?
m
T
d
1
sin(θ) - m
L
(d
1
+ d
2
+ d
3
) - m
A
(d
1
+ d
2
) = 0
m
T
=
?𝐴(?1 + ?2) ?1𝑠𝑖?(50)
= 0.3055 * 14.5
2.35*𝑠𝑖?(50)
=
2. 460 𝑘𝑔
When isolating for m
T
, the mass applied is neglected because there was no mass attached to the
arm. Therefore, the theoretical mass value of m
T
that will balance the arm for
𝜃
= 50
°
is 2.460
kg. The experimental value for the mass applied is found to be 2.200 kg and its uncertainty is
obtained by averaging the difference between its maximum and minimum values:
Its uncertainty is then obtained by averaging the difference between
2.200 + 2.210
2
=
2. 205 𝑘𝑔.
the maximum and minimum values:
As a result, the experimental
2.210 − 2.200
2
= ± 0. 005.
value for m
T
is
Comparing the experimental and theoretical values of m
T
, it
(2. 205 ± 0. 005).
can be observed that there is a moderate agreement between them.The difference between the
values (2.460-2.205 = 0.255) which falls within the standard deviations of the theoretical value
proving there is an agreement between the values. Hence, the experimental value of m
T
is
consistent with its theoretical value with no mass load applied.
3. Show your work for calculating the force at the elbow joint (
??
) for
𝜃
= 50
°
and
𝑚𝐿
= 0
g. Do not include uncertainty.
To calculate the force at the elbow joint, equation 11.5 must be used:
F
x
= 0 and
F
y
= 0.
Σ
Σ
𝑇??𝑠(θ) = 𝐹???𝑠(θ)
gm
T
cos( ) = F
E
cos(
θ
ϕ)
1.889*9.81*cos(50
°
= F
Ex
)
F
Ex
= 11.90 N, to the left
F
y
= 0
-sin( )m
T
g - m
A
g - m
L
g - F
E
sin( ) = 0
θ
θ
F
Ey
= [-sin(50)*
2.2
9*9.81] -[ 0.3055*9.81 + 0*9.81]
F
Ey
= 16.50 N, downwards
Thus, F
E
=
𝐹?
2
+ 𝐹?
2
= (− 11. 90)
2
+ (− 16. 50)
2
= 20. 34𝑁. 4. Calculate and measure
𝑚𝑇
for 50
°
and a load (either 100 g or 150 g depending on your
setup). Calculate
??
(no uncertainty). You should include an estimate of the uncertainty for
the measured
𝑚𝑇
. How do these values compare to the answers without a load? Show your
work for all parts of this question.
Using equation 11.4, the theoretical mass applied can be isolated from the equation as follows:
m
T
=
?𝐿(?1+?2+?3) + ?𝐴(?1 + ?2) ?1𝑠𝑖?(50)
=
[0.1*0.29]+[0.3055*0.145]
0.0235*𝑠𝑖?(50)
=
4. 072 𝑘𝑔 However, the experimental mass obtained from the lab experiment is 3.970 kg. To calculate the
experimental value of m
T
, the average mass values are taken:
The
3.970+3.990
2
=
3. 980 𝑘𝑔.
uncertainty is then found by:
Hence, the experimental value of m
T
when
3.990−3.970
2
= ± 0. 01.
m
L
is 150 grams is
Comparing the theoretical and experimental values of m
T
,
(3. 980 ± 0. 01).
it is observed that the difference between the values (4.072 - 3.980 = 0.092) and the uncertainty
is (0.01 - 0 = 0.01), which does not fall within the standard deviation of the theoretical value,
showing that there is a slightly poor agreement between them. The experimental value of m
T
is
not consistent with its theoretical value due to the human error of miscalculating the weights
applied and not properly balancing the arm when the apparatus is at 50° with a load of 0.100 kg
applied.
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Related Questions
Dr. Brown has built a 94 cm pendulum and pulled the pendulum 25° from vertical and then
measured the period of 30 consecutive swings. The data is shown below. How do you interpret
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Period of Dr. Brown's pendulum
Period of swing (s)
1.98
1.978
1.976
1.974
1.972
1.97
1.968
1.966
1.964
0
10
20
Number of swings since release
30
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O The random uncertainty starts large (approximately 0.03 s) but decreases quickly until it is immeasureably
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Total Uncertainty
P
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V
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145.0
L
0
d.
Q
Joom
ма 8-42.00
で
how
equal to I'm in
Thermal conductivity.
materi di
L
M
Heat Flow ?
b. 3.24 kW
N
0
d
२
R
I. thermal Resistance 17
a. 0.80 KW
b. none
OF
c. 9.25 KIKW
d. 1.08
e.
K| kw
0.68
44.12 KW
37.5 KW
M
R
1.0m
the above
d
mall
a. decrease
b. unchanged
c. increase
the walls
side is exposed to temperatures of 5000
upper
while its lower siche is exposed to
unidirectional heat flow in
Assume steady
state,
radiation and Convectim heat effects.
20°C.
y direction Neglect
none of
IF antract reere tance
heat
N
S
1.0m
op
the above
a row of bricks, an taking
thicknees
Bricks
k (kw/m-k)
Flow?
0.50
0.25
0-50
0-80
0.4⁰
0.50
0.25
0.50
•
not neglighe, what happen to overall
d. inereact
then decrease
e increase / decrease depending
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B
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T (°C)
v(m³/kg)
T (°C)
v(m³/kg)
200
0.2060
200
0.1325
0.1483
0.1627
0.2275
240
280
240
280
0.2480
Data encountered in solving problems often do not fall exactly on the grid of values
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becomes necessary. Using the data provided here, estimate
i.
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iii.
11.
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v(m³/kg)
p = 1.0 MPa
T ("C)
v(m³/kg)
T ("C)
200
0.2060
200
0.1325
240
280
0.2275
0.2480
240
280
0.1483
0.1627
Data encountered in solving problems often do not fall exactly on the grid of
values provided by property tables, and linear interpolation between adjacent
table entries becomes necessary. Using the data provided here, estimate
i. the specific volume at T= 240 °Č, p = 1.25 MPa, in m/kg
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Areas Under the Standard Normal Curve-The Values Were Generated Using the Standard Normal
Distribution Function of Excel
Note that the standard normal curve is symmetrical about the mean.
z
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
0.1
0.11
0.12
1
0.95
0.96
0.97
0.98
0.99
1.01
1.02
1.03
1.04
1.05
Mean - 0
1.06
1.07
1.08
1.09
A
0.0000
0.0040
0.0080
0.0120
0.0160
0.0199
0.0239
0.0279
0.0319
0.0359
0.0398
0.0438
0.0478
A
0.3186
0.3212
0.3238
0.3264
0.3289
0.3315
0.3340
0.3365
0.3389
Z
0.3413
0.3438
0.3461
0.3485
0.3508
0.3531
0.3554
0.3577
0.3599
0.3621
0.13
0.14
0.15
0.16
0.17
0.18
0.19
0.2
0.21
0.22
0.23
0.24
0.25
1.12
1.13
1.14
1.15
1.16
1.17
A
z
0.0517
0.0557
0.26
0.27
0.28
0.29
0.0596
0.0636
0.0675 0.3
0.0714 0.31
0.0753 0.32
0.0793 0.33
0.0832 0.34
0.0871 0.35
0.0910
0.0948
0.0987
1.18
1.19
1.2
1.21
1.22
1.23
1.24
1.25
1.26
1.27
1.28
A
0.3643
0.3665
0.3686
0.3708
0.3729
0.3749
0.3770
0.3790
0.3810
0.36
0.3830
0.3849
0.3869
0.3888
0.3907
0.3925
0.3944
0.3962
0.3980
0.3997
0.37…
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a temperature of T (in degrees Fahrenheit) and pressure P (in pounds per square inch).
T//P
480
500
520
540
20 22
24
26
27.85 25.31 23.19 21.39
28.46 25.86 23.69 21.86
29.06 26.41 24.20 22.33
29.66 26.95 24.70 22.79
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Use the tables to approximate your partial derivatives and recall the equation of our tangent
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b) Use the tangent plane to estimate the volume of a pound of steam at a temperature of
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16000
1.440
14000
1.435
12000
1.430
10000
1.425
8000
1.420
6000
1.415
4000
1.410
2000
1.405
1.400
45
15
20
25
30
35
40
Temperature ["C]
+dynamic viscosity (mPas]
+ density (gicm")
Figure 1
(a) As shown in Figure 1, viscosity and density of honey was found to decrease with increase
in temperature. Explain the reasons behind these phenomenon related to Fluid Mechanics
in microscopic view.
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100
80
60
40
20
0.002
0.004
0.006
0.008
0.01
0.012
Strain, in/in.
FIGURE P1.17
1.18 Use Problem 1.17 to graphically determine the following:
a. Modulus of resilience
b. Toughness
Hint: The toughness (u) can be determined by calculating the area under
the stress-strain curve
u =
de
where & is the strain at fracture. The preceding integral can be approxi-
mated numerically by using a trapezoidal integration technique:
u, = Eu, = o, + o e, - 6)
%3D
c. If the specimen is loaded to 40 ksi only and the lateral strain was found to
be -0.00057 in./in., what is Poisson's ratio of this metal?
d. If the specimen is loaded to 70 ksi only and then unloaded, what is the
permanent strain?
Stress, ksi
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As an engineer working for a water bottling company, you collect the following
data in order to test the performance of the bottling systems. Assume normal
distribution.
Milliliters of Water in the Bottle Frequency
485
490
milliliters
495
500
505
510
515
What is the mean (in milliliters)?
milliliters
What is the standard deviation (in milliliters)?
What is the z value corresponding to 490 milliliters?
Z =
6
12
20
33
18
11
00
8
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a) The mass of a test specimen of metal alloy is measured to be m = 674.4 g. The procedure has a standard uncertainty of 1.0 g.
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density of the specimen and its uncertainty.
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uncertainty to within 2% and its radius R can be measured with a standard uncertainty within 1%.
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b)
The variation in the experimental density of water, p, with temperature T, in the range
of 20°C
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30. In a typical tension test a dog-bone
shaped specimen is pulled in a
machine. During the test, the force F
needed to pull the specimen and the
length L of a gauge section are measured. This data is used for plotting a
stress-strain diagram of the material. Two definitions, engineering and true,
exist for stress and strain. The engineering stress
F
de
and strain
e are
defined by oe
F
%3D
and
A0
L-Lo where Lo and Ao are the initial gauge
6, =
Lo
length and the initial cross-sectional area of the specimen, respectively. The
true stress o, and strain e are defined by o,
는는 and q
%3D
= In
A0 Lo
The following are measurements of force and gauge length from a ten-
sion test with an aluminum specimen. The specimen has a round cross sec-
tion with a radius of 0.25 in. (before the test). The initial gauge length is 0.5
in. Use the data to calculate and generate the engineering and true stress-
strain curves, both on the same plot. Label the axes and use a legend to
identify the curves.
Units:…
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Creep exercises
The following data in the table below was obtained a from rupture test.
Stress (MPa)
Temperature (°C)
Rupture time (hr)
T 550
580
0.75
550
555
5.5
550
540
19.2
550
525
62.0
70
760
2.05
70
730
8.2
70
700
36.5
70
675
134
Assume: C = 46
Establish whether the Larson-Miller parameter is valid for both stresses (550 and 70
MPa).
Was there a change in creep mechanism when changing the stress from 550 to 70 MPa
during rupture test? Motivate your answer with the aid of calculation.
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In measuring the surface tension of a
liquid (drop weight method), 20 drops of
the liquid (r = 0.2cm) falling apart from
the tip whose diameter is 0.4 cm were
found to weight 0.95 gram. What is the
surface tension of the liquid?
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A Pitot tube measures a pressure difference Pa = Pt- Ps (dynamic = total - static)
between the front and side of the Pitot tube. The dynamic pressure is given by the
kinetic energy due to the airspeed v, allowing airspeed to be calculated from the
measured pressure difference.
(a)
P₁ = ² pv² ⇒ v=
When designing a Pitot tube, a manufacturer wants to make it precise enough to measure
airspeed to within 0.5 % uncertainty, i.e. = 0.005, at sea level.
Av
p=(1.225 +0.0005) kg/m³ at sea level.
(b)
2Pd
Av
Using error propagation, derive an expression for Simplify as much as
possible!
ΔΡΑ
Pd
Calculate the maximum permissible %-uncertainty in Pa, i.e.
allow the Pitot tube to meet the manufacturer's requirements.
Hint: Answer 1%
x 100, that will
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A temperature measurement system has the following specifications:
-128 to 781°C
Range
Linearity error
0.29% FSO
Hysteresis error
0.12% FSO
Sensitivity error
0.04% FSO
Zero drift
0.32% FSO
FSO stands for "Full Scale Output". Estimate the overall instrument
uncertainty for this system based on the available information. Use the
maximum possible output range over the FSO in your computations.
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QUESTION 1
You find that statistical uncertainty is your largest measurement uncertainty and that the V value is your largest propagated uncertainty. How can you try to improve both uncertainties in the simplest. but most effective way?
O use a new V instrument and a new DV instrument to collect more measurements at a wicer variety of V data points
O collect DV data for new IV data points
O use a new V instrument but the same DV instrument to collect the same numtber of measurements at a wicer variety of IV data points
o take more measurements for the DV value at the IV data points aiready collected and add more IV data points
O use a new DV instrument to take more DV measurements for the Ivdata points already collected and collect additional DV data for more IV data points not previcusly measured
o start over to collect the same number of DV measurements as before. but al a greater variety of IV data points than before
o take more measurements for the DV values at the IV data points…
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Related Questions
- Dr. Brown has built a 94 cm pendulum and pulled the pendulum 25° from vertical and then measured the period of 30 consecutive swings. The data is shown below. How do you interpret the uncertainties involved in this data? Period of Dr. Brown's pendulum Period of swing (s) 1.98 1.978 1.976 1.974 1.972 1.97 1.968 1.966 1.964 0 10 20 Number of swings since release 30 O Some random effect lowers the period of the pendulum by about 0.01 s in between the first and last data point. O The random uncertainty starts large (approximately 0.03 s) but decreases quickly until it is immeasureably small by the end of the data set. O Some systematic effect lowers the period of the pendulum by about 0.01 s in between the first and last data point. O The systematic uncertainty starts large (approximately 0.01 s) but decreases quickly until it immeasurably small by the end of the data set.arrow_forward1- Consider the measurement of pressure in a gas through measurement of its temperature T and density p Let the uncertainties of T and p be ôT and õp where p is determined from the ideal gas p=pRT. Derive an expression of uncertainty (error) in P and determine ôP/P. Suppose the uncertainty in ôp is 1% and in ôT is 2° at T=27°.arrow_forward3. Examine your data in Data Table 2 and in Data Table 3. For the data with the smallest percentage difference, compare the total energy at each point. Calculate the sum U₁+Ug at x₁ as ½kx-mgx₁. Calculate that sum at x₂ as ½kx²-mgx₂. Do you expect them to agree reasonably well? Explain why they should or should not be the same. 4. Consider the same data as used in Question 3. Calculate the value of x halfway between x₁ and x₂. Calculate U₁+Ug=½kx² − mgx_for_that point. Do you expect them to agree with the energy calculated in Question 3? If they agree reasonably well, explain why they do. If they do not agree, explain why they do not agree.arrow_forward
- 1. To determine the dynamic viscosity of a liquid you take repeated measurements of the liquid density and kinematic viscosity. These data are summarized in the following table: Avg. measurement Total Uncertainty P 999 kg/m³ ±0.5% V 1.23 mm²/s ±0.01 mm²/s kg Ns What is the dynamic viscosity? Give dynamic viscosity in units of; which is the same as ms 2. Determine the absolute and relative uncertainties in the dynamic viscosity.arrow_forwardThe following diagram illustrates measurements that 145.0 L 0 d. Q Joom ма 8-42.00 で how equal to I'm in Thermal conductivity. materi di L M Heat Flow ? b. 3.24 kW N 0 d २ R I. thermal Resistance 17 a. 0.80 KW b. none OF c. 9.25 KIKW d. 1.08 e. K| kw 0.68 44.12 KW 37.5 KW M R 1.0m the above d mall a. decrease b. unchanged c. increase the walls side is exposed to temperatures of 5000 upper while its lower siche is exposed to unidirectional heat flow in Assume steady state, radiation and Convectim heat effects. 20°C. y direction Neglect none of IF antract reere tance heat N S 1.0m op the above a row of bricks, an taking thicknees Bricks k (kw/m-k) Flow? 0.50 0.25 0-50 0-80 0.4⁰ 0.50 0.25 0.50 • not neglighe, what happen to overall d. inereact then decrease e increase / decrease depending on value of Contact resistancearrow_forwardThe heat transfer conducted through material is calculated from the equation: Q = KX AXTD/L Where K: Conductivity of material A: Area of heat transfer TD: Temperature difference across material L: Thickness of material A student measures the area, thickness and temperature difference and assumes that the error in conductivity is negligible. The student also estimates the uncertainty range for each variable. In estimating the maximum possible value of Q, the student should use the following formula: A B Q max= K x A max x TD max / L max Q max= K x A max x TD max / L nom Q max= Q nominal + dQ/dLmin Q max= K x A max x TD max / L minarrow_forward
- The following table lists temperatures and specific volumes of water vapor at two pressures: p = 1.0 MPa p = 1.5 MPa T (°C) v(m³/kg) T (°C) v(m³/kg) 200 0.2060 200 0.1325 0.1483 0.1627 0.2275 240 280 240 280 0.2480 Data encountered in solving problems often do not fall exactly on the grid of values provided by property tables, and linear interpolation between adjacent table entries becomes necessary. Using the data provided here, estimate i. the specific volume at T = 240 °C, p = 1.25 MPa, in m³/kg the temperature at p = 1.5 MPa, v = 0.1555 m³/kg, in °C the specific volume at T = 220 °C, p = 1.4 MPa, in m³/kg ii. iii. 11.arrow_forwardThe following table lists temperatures and specific volumes of water vapor at two pressures: p = 1.5 MPa v(m³/kg) p = 1.0 MPa T ("C) v(m³/kg) T ("C) 200 0.2060 200 0.1325 240 280 0.2275 0.2480 240 280 0.1483 0.1627 Data encountered in solving problems often do not fall exactly on the grid of values provided by property tables, and linear interpolation between adjacent table entries becomes necessary. Using the data provided here, estimate i. the specific volume at T= 240 °Č, p = 1.25 MPa, in m/kg ii. the temperature at p = 1.5 MPa, v = 0.1555 m/kg, in °C ii. the specific volume at T = 220 °C, p = 1.4 MPa, in m'/kgarrow_forwardAreas Under the Standard Normal Curve-The Values Were Generated Using the Standard Normal Distribution Function of Excel Note that the standard normal curve is symmetrical about the mean. z 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1 0.11 0.12 1 0.95 0.96 0.97 0.98 0.99 1.01 1.02 1.03 1.04 1.05 Mean - 0 1.06 1.07 1.08 1.09 A 0.0000 0.0040 0.0080 0.0120 0.0160 0.0199 0.0239 0.0279 0.0319 0.0359 0.0398 0.0438 0.0478 A 0.3186 0.3212 0.3238 0.3264 0.3289 0.3315 0.3340 0.3365 0.3389 Z 0.3413 0.3438 0.3461 0.3485 0.3508 0.3531 0.3554 0.3577 0.3599 0.3621 0.13 0.14 0.15 0.16 0.17 0.18 0.19 0.2 0.21 0.22 0.23 0.24 0.25 1.12 1.13 1.14 1.15 1.16 1.17 A z 0.0517 0.0557 0.26 0.27 0.28 0.29 0.0596 0.0636 0.0675 0.3 0.0714 0.31 0.0753 0.32 0.0793 0.33 0.0832 0.34 0.0871 0.35 0.0910 0.0948 0.0987 1.18 1.19 1.2 1.21 1.22 1.23 1.24 1.25 1.26 1.27 1.28 A 0.3643 0.3665 0.3686 0.3708 0.3729 0.3749 0.3770 0.3790 0.3810 0.36 0.3830 0.3849 0.3869 0.3888 0.3907 0.3925 0.3944 0.3962 0.3980 0.3997 0.37…arrow_forward
- Designing safe boilers depends on knowing how steam behaves under certain changes in temperature and pressure. Steam tables, such as the one below, are published giving values of the function V = f(T, P) where V is the volume (in cubic feet) of one pound of steam at a temperature of T (in degrees Fahrenheit) and pressure P (in pounds per square inch). T//P 480 500 520 540 20 22 24 26 27.85 25.31 23.19 21.39 28.46 25.86 23.69 21.86 29.06 26.41 24.20 22.33 29.66 26.95 24.70 22.79 a) Find the tangent plane to V = f(T, P) for T near 520°F and P near 24 lb/in². [Hint: Use the tables to approximate your partial derivatives and recall the equation of our tangent plane is: V = VT(T − To) + Vp(P - Po) + f(T, P) ] b) Use the tangent plane to estimate the volume of a pound of steam at a temperature of 525°F and P near 24.3 lb/in².arrow_forward16000 1.440 14000 1.435 12000 1.430 10000 1.425 8000 1.420 6000 1.415 4000 1.410 2000 1.405 1.400 45 15 20 25 30 35 40 Temperature ["C] +dynamic viscosity (mPas] + density (gicm") Figure 1 (a) As shown in Figure 1, viscosity and density of honey was found to decrease with increase in temperature. Explain the reasons behind these phenomenon related to Fluid Mechanics in microscopic view. Dynamic viscolsty [mPa.s] Density [g/cm³]arrow_forward100 80 60 40 20 0.002 0.004 0.006 0.008 0.01 0.012 Strain, in/in. FIGURE P1.17 1.18 Use Problem 1.17 to graphically determine the following: a. Modulus of resilience b. Toughness Hint: The toughness (u) can be determined by calculating the area under the stress-strain curve u = de where & is the strain at fracture. The preceding integral can be approxi- mated numerically by using a trapezoidal integration technique: u, = Eu, = o, + o e, - 6) %3D c. If the specimen is loaded to 40 ksi only and the lateral strain was found to be -0.00057 in./in., what is Poisson's ratio of this metal? d. If the specimen is loaded to 70 ksi only and then unloaded, what is the permanent strain? Stress, ksiarrow_forward
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