The question asks for pump pressure calculations in a pipeline, requiring applications of fluid dynamics principles, such as Bernoulli's equation, and considering factors like elevation changes, viscosity, and pipe characteristics.
Explanation:The student's question relates to the field of fluid dynamics, specifically the calculation of required pump pressure in a pipeline system using principles such as Bernoulli's equation and concepts like pressure, flow rate, pipe diameter, density, viscosity, and elevation change. To solve this type of problem, you typically apply the Bernoulli equation, along with any additional head losses due to viscosity, known as the head loss due to friction, which can be calculated using the Darcy-Weisbach equation or other empirical formulas. You must then calculate the total head needed, convert this to pressure, taking into account the fluid's density, gravitational acceleration, and add any other relevant pressures, such as atmospheric pressure if the pipeline opens to the atmosphere.
A hollow aluminum alloy [G = 3,800 ksi] shaft having a length of 12 ft, an outside diameter of 4.50 in., and a wall thickness of 0.50 in. rotates at 3 Hz. The allowable shear stress is 6 ksi, and the allowable angle of twist is 5°. What horsepower may the shaft transmit?
Answer:
Horse power = 167.84 hp
Explanation:
Horsepower is calculated using the formula;
P = T * w
See the attached file for the calculation
Researchers compared protein intake among three groups of postmenopausal women: (1) women eating a standard American diet (STD), (2) women eating a lacto-ovo-vegetarian diet (LAC), and (3) women eating a strict vegetarian diet (VEG).GroupMean protein intake (mg)SDnSTD75910LAC571310VEG47176a.Without using MATLAB ANOVA functions, determine if there is a significant difference in mean protein intake between these groups.
Answer:
see explaination
Explanation:
The statistical procedure for comparing the 3 groups was the F test with 2 degrees of freedom in the numerator (groups - 1) and 23 df in the denominator (N total - groups)
It is calculated as SSB/2/(SSW/23) = 10.22754
As Fcrit =
from Excel, finv(.05,2,23) = 3.4221, we can reject the null hypothesis of no difference between groups.
SSW = the sum of std i ^ 2 * (n i - 1)
SSB = the sum of ni (mean i - mean)^2
2. From Excel, we get the pvalue from fdist(10.22754,2,23) = .000664
3. For LSD, we calculate (mean 1 - mean 2)/(s * sqrt(1/n1+1/n2))
This is the pooled s = sqrt(SSW/23)
Then, we found t crit from tinv(.05,23) = 2.068
Making use of the chart i made, We found significant differences between std and lac as well as std and veg, but no significant difference between lac and veg.
Consider an oil-to-oil double-pipe heat exchanger whose flow arrangement is not known. The temperature measurements indicate that the cold oil enters at 20°C and leaves at 55°C, while the hot oil enters at 80°C and leaves at 45°C.
a. Do you think this is a parallel-flow or counter-flow heat exchanger? Why?
b. Assuming the mass flow rates of both fluids to be identical, determine the effectiveness of this heat exchanger.
a. This is a counter flow heat exchanger.
b. The effectiveness of heat exchanger is 0.615
A counter-flow heat exchanger is one in which the direction of the flow of one of the working fluids is opposite to the direction to the flow of the other fluid. In a parallel flow exchanger, both fluids in the heat exchanger flow in the same direction.Counter flow heat exchanger distributes the heat more evenly across the heat exchanger and allows for maximum efficiency.The effectiveness of heat exchanger is defined as ratio of actual heat transfer to the maximum possible heat transfer.
Given that the cold oil enters at 20°C and leaves at 55°C, while the hot oil enters at 80°C and leaves at 45°C.
[tex]Effectiveness=\frac{55-15}{80-15}=\frac{40}{65}=0.615[/tex]
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You wish to filter out 60 Hz noise (which arises from electrical interference at the frequency of AC current in our electrical grid) using a simple RC circuit, which is one useful form has a transfer function:
G(s) = 1/RCS + 1
a. What value of the product RC should you choose so that amplitude at 60 Hz is attenuated by 90%?
b. For that value of RC, what is the largest frequency that is attenuated by less than 5%?
c. Turn in a Bode plot for your proposed system generated in Matlab.
Answer:
a. 0.02639
b. 1.98H
Explanation:
Please see attachment
A 2-m-long and 3-m-wide horizontal rectangular plate is submerged in water. The distance of the top surface from the free surface is 5 m. The atmospheric pressure is 95 kPa. Considering the atmospheric pressure, the hydrostatic force acting on the top surface of this plate is _____. Solve this problem using appropriate software.
Answer:
864 KN
Explanation:
Atmospheric pressure is defined as the force per unit area exerted against a surface by the weight of the air above that surface.
Please kindly check attachment for the step by step solution of the given problem.
the upper surface of a 1x1 ASTM B152 copper plate is being cooled by air at 20 c. if the rate of convection heat transfer from the plate surface is 700 W, would the use of ASTM B152 plate be in compliance with tht ASME Code for process piping?
Answer:
Since the Reynolds number is below 3 x 10⁵ the use of ASTM B152 plate WILL NOT be in complaince with ASME Code for process piping.
Explanation:
Detailed Explanation is given in the attached document.
A supply fan is operating at 30000 cfm and 4 inch of water with an efficiency of 50%. (a) Calculate the fan power at the current operating condition. (b) Calculate the pressure and power if the supply fan reduces its speed to deliver 20000 cfm.
Answer:
The pressure and power of fan is 1.77 and 11.18 Hp respectively.
Explanation:
Given:
Discharge [tex]Q_{1} = 30000[/tex] cfm
Pressure difference [tex]\Delta P = 4[/tex] inch
Efficiency [tex]\eta = 50\%[/tex]
(A)
From the formula of fan power,
[tex]P _{1} = \frac{Q \Delta P}{6356 \eta}[/tex]
[tex]P_{1} = \frac{30000 \times 4}{6356 \times 0.5}[/tex]
[tex]P_{1} = 37.76[/tex] Hp
(B)
Fan power and pressure is given by,
We know that pressure difference is proportional to the square of discharge.
[tex]\frac{\Delta p_{2} }{\Delta P_{1} } = (\frac{Q_{2} }{Q_{1} } ) ^{2}[/tex]
[tex]\Delta P_{2} = (\frac{20000}{30000} ) \times 4[/tex]
[tex]\Delta P_{2} = 1.77[/tex]
Fan power proportional to the cube of discharge.
[tex]\frac{P_{2} }{P^{1} } = (\frac{Q_{2} }{Q_{1} } )^{3}[/tex]
[tex]P_{2} = \ (\frac{20000}{30000} ) ^{3} \times 37.76[/tex]
[tex]P_{2} = 11.18[/tex] Hp
Therefore, the pressure and power of fan is 1.77 and 11.18 Hp respectively.
What evidence indicates that a reaction has occurred? (Select all that apply.)
The temperature decreased.
A solid brown product formed.
The temperature increased.
A gas formed.
An explosion occurred.
The evidence that indicates that a reaction has occurred include the following:
B. A solid brown product formed.
C. The temperature increased.
D. A gas formed.
E. An explosion occurred.
A chemical reaction is a chemical process that involves the continuous transformation (rearrangement) of the ionic, atomic or molecular structure of a chemical element by breaking down and forming chemical bonds, in order to produce a new chemical compound while new bonds are formed.
This ultimately implies that, a chemical change would give rise to the chemical properties of matter by causing the transformation of one chemical substance into one or more different chemical substances.
Estimate pressure drop for an estimate of pipe diameter Pressure drop is a function of flow rate, length, diameter, and roughness. Either iterative methods OR equation solvers are necessary to solve implicit problems. For a first guess of a 1 ft diameter pipe, what is the fluid velocity? V = 5.67 ft/s What is the Reynolds number? Re = 96014 What is the pipe relative roughness?
Answer:
Explanation:
By using Bernoulli's Equation:
[tex]\frac{P_1}{P_g}+\frac{v_1^2}{2g}+z_1=\frac{P_2}{P_g}+\frac{v_2^2}{2g}+z_2+f\frac{L}{D}\frac{v^2}{2g}[/tex]
where;
[tex]z_1 = z_2 \ and \ v_1 = v_2[/tex]
[tex]P_1 - P_2 = f \frac{L}{D}\frac{1}{2}\rho v^2[/tex]
[tex]P_1-P_2 = \frac{5 \ lb}{in^2}( 144 \frac{in^2}{ft^2})[/tex]
[tex]P_1-P_2 = 720 \frac{lb}{ft^2}[/tex]
[tex]V = \frac{Q}{A} \\ \\ V = \frac{6.684 \ ft^2/s}{\frac{\pi}{4}D^2} \\ \\V = \frac{8.51}{D^2}[/tex]
Density of gasoline [tex]\rho = 1.32 \ slug/ft^3[/tex]
Dynamic Viscosity [tex]\mu[/tex] = [tex]6.5*10^{-6} \frac{lb.s}{ft}[/tex]
[tex]P_1-P_2 = f \frac{L \rho V^2}{2D}[/tex]
[tex]720 = f \frac{L(100)(1.32)}{2D}(\frac{8.51}{D^2})^2[/tex]
D = 1.46 f
[tex]Re, = \frac{\rho VD}{\mu} = \frac{1.32 *\frac{8.51}{D^2} D}{6.5*10^{-6}}[/tex]
[tex]= \frac{1.72*10^6}{D}[/tex]
[tex]\frac{E}{D}= \frac{0.00015}{D}[/tex]
However; the trail and error is as follows;
Assume ; f= 0.02 → D = 0.667ft[tex]\left \{ {{Re=2.576*10^6} \atop {\frac{E}{D}=0.000225}} \right.[/tex] [tex]\right \{ {{f=0.014} \atop {\neq 2}}[/tex]
f = 0.0145 → D = 0.0428 ft [tex]\left \{ {{Re=4.018*10^6} \atop {\frac{E}{D}=0.00035}} \right.[/tex] [tex]\right \{ {{f=0.015} \atop {\neq 0.0145}}[/tex]
f = 0.0156 → D = 0.43 ft [tex]\left \{ {{Re=4.0*10^6} \atop {\frac{E}{D}=0.000348}} \right.[/tex] [tex]f = 0.0156[/tex]
∴ pipe diameter d = 0.43 ft
Given that:
D = 1 ft
[tex]V = \frac{Q}{A} \\ \\ V = \frac{6.684}{\frac{\pi}{4}(1)^2} \\ \\ V = 8.51 \ ft/s[/tex]
[tex]Re = \frac{\rho \ V \ D}{\mu } \\ \\ Re = \frac{1.32 *8.51 *1 }{6.5*10^{-6}}[/tex]
[tex]Re = 1.72 *10^6[/tex]
[tex]\frac{E}{D} = \frac{0.00015}{1} \\ \\ = 0.00015[/tex]
[tex]f = 0.0136[/tex]
[tex]P_2-P_1 = \frac{fL \rho V^2 }{2D}[/tex]
[tex]P_2-P_1 = \frac{0.036(100)(1.32)(8.51)^2 }{2*1}[/tex]
[tex]P_2-P_1 = 65 \frac {lb}{ft^2}[/tex] to psi ; we have:
[tex]P_2-P_1 = 0.45 \ psi[/tex]
A spherical gas container made of steel has a(n) 17-ft outer diameter and a wall thickness of 0.375 in. Knowing that the internal pressure is 60 psi, determine the maximum normal stress and the maximum shearing stress in the container.
Answer:
Maximum Normal Stress σ = 8.16 Ksi
Maximum Shearing Stress τ = 4.08 Ksi
Explanation:
Outer diameter of spherical container D = 17 ft
Convert feet to inches D = 17 x 12 in = 204 inches
Wall thickness t = 0.375 in
Internal Pressure P = 60 Psi
Maximum Normal Stress σ = PD / 4t
σ = PD / 4t
σ = (60 psi x 204 in) / (4 x 0.375 in)
σ = 12,240 / 1.5
σ = 8,160 P/in
σ = 8.16 Ksi
Maximum Shearing Stress τ = PD / 8t
τ = PD / 8t
τ = (60 psi x 204 in) / (8 x 0.375 in)
τ = 12,240 / 3
τ = 4,080 P/in
τ = 4.08 Ksi
For a very rough pipe wall the friction factor is constant at high Reynolds numbers. For a length L1 the pressure drop over the length is Δp1. If the length of the pipe is then doubled, what is the relation of the new pressure drop Δp2 to the original pressure drop Δp1 at the original mass flow rate?
delta P2=?
Answer:
[tex]\Delta p_{2} = 2\cdot \Delta p_{1}[/tex]
Explanation:
The pressure drop is directly proportional to the length of the pipe. Then, the new pressure drop is two time the previous one.
[tex]\Delta p_{2} = 2\cdot \Delta p_{1}[/tex]
A doubly drained specimen, 2.54 cm in height, is consolidated in the lab under an applied stress. The time for 50 % overall (or average) consolidation is 12 min. (a) Compute the cv value for the lab specimen. (b) How long will it take for the specimen to consolidate to an average consolidation of 90 %? (c) If the final consolidation settlement of the specimen is expected to be 0.43 cm, how long will it take for 0.18 cm of settlement to occur? (d) After 14 minutes, what percent consolidation has occurred at the middle of the specimen?
Answer:
Cv = 0.026 cm²/min
t = 52.60 min
v% = 41.86 %
tv = 0.1375
t = 8.53 min
v = 53.61 %
Explanation:
given data
height = 2.54 cm
50 % consolidation = 12 min
solution
we get here first Cv value that is express as
Tv = [tex]\frac{Cv\times t}{d^2}[/tex] .................1
here Tv for 50% is 0.196
put here value and we get
0.196 = [tex]\frac{Cv\times 12}{\frac{2.54}{2}^2}[/tex]
solve it we get
Cv = 0.026 cm²/min
and
for tv for 90 % consolidation is 0.848
put value in equation 1
0.848 = [tex]\frac{0.026\times t}{\frac{2.54}{2}^2}[/tex]
solve it we get t
t = 52.60 min
and
v% will be here is
v% = [tex]\frac{0.18}{0.43} \times 100[/tex]
v% = 41.86 %
and
tv = [tex]\frac{\pi }{4}\times \frac{4}{100}^2[/tex]
tv = 0.1375
so now put value in equation 1 we get
0.1375 = [tex]\frac{0.026 \times t}{\frac{2.54}{2}^2}[/tex]
solve it we get
t = 8.53 min
and
now put value of t 14 min in equation 1 will be
tv = [tex]\frac{0.026 \times 14}{\frac{2.54}{2}^2}[/tex]
t = 0.225 min
and v will be after 14 min
0.0225 = [tex]\frac{\pi }{4}\times \frac{v}{100}^2[/tex]
v = 53.61 %
A field sample of an unconfined aquifer is packed in a test cylinder. The length and diameter of the cylinder are 50 cm and 6 cm respectively. The field sample is tested for a period of three minutes under a constant head difference of 16.3 cm. As a result, 45.2 cm3 of water is collected at the outlet. Determine the hydraulic conductivity of the aquifer sample.
Answer:
0.09cm/sec
Explanation:
We are going to describe Hydraulic conductivity as a measure of the ease with which water flows through sediments, determining renewal rates of water, dissolved gases, and nutrients.
See attachment for a detailed solution.
The hydraulic conductivity of the given aquifer sample packed in a test cylinder is; 0.027 cm/s
What is the hydraulic Conductivity?
We are given;
Head Loss; H = 16.3 cm
Length of cylinder; L = 50 cm
Diameter of cylinder; d = 6 cm
radius; r = 6/2 = 3 cm
time; t = 3 minutes = 180 s
Volume; V = 45.2 cm³
Formula for the hydraulic gradient is;
i = H/L
i = 16.3/50
i = 0.326
Formula for volumetric rate of flow is;
Q = V/t
Q = 45.2/180
Q = 0.25 cm³/s
Formula for area is;
A = πd²/4
A = π × 6²/4
A = 9π
Formula for hydraulic conductivity is;
K = Q/(A * i)
K = 0.25/(9π * 0.326)
K = 0.027 cm/s
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Technician A says a limited slip differential can redirect power from a drive wheel that is slipping to the wheel that has traction. Technician B says traction control can redirect power by applying the brake on a drive wheel that is slipping. Who is correct?
Answer:
Both Technician A, and Technician B are correct
Explanation:
The Traction control are found in those modern automobile, it's a part of the electronic stability control and it becomes active once the automobile get acceleration. It helps the tired of the car not to slip when the car speed up.
It functions by making the car wheel to stop spinning through the reduction of power that is transferred to the wheel i.e application of traction on the wheels of the car. when car is moving with acceleration on a road with with little friction, the Traction is used.
During raining or snow when the road become slippery , In the old cars that doesn't have traction control, the gas pedal is feathered. Which helps to function as traction control
Assume the following LTI system where the input signal is an impulse train (i.e.,x(t)=∑????(t−nT0)[infinity]n=−[infinity].a)Find the Fourier series coefficient of x(t). Then find its Fourier transform and sketch the magnitude and phase spectra.b)Sketch the magnitude and phase spectra of the output (i.e., |Y(????)|and∡Y(????)) if the system is a low-pass filter with H(????)={1|????|<3????020other????ise, where ????0=2πT0.c)Sketch the magnitude and phase spectra of the output(|Y(????)|and∡Y(????)) if the system is a high-pass filter with H(????)={1|????|>5????020other????ise, where ????0=2πT0.d)Sketch the magnitude and phase spectra of the outputif the system is a filter with H(????)=11+j????.
Answer:
See explaination
Explanation:
The Fourier transform of y(t) = x(t - to) is Y(w) = e- jwto X(w) . Therefore the magnitude spectrum of y(t) is given by
|Y(w)| = |X(w)|
The phase spectrum of y(t) is given by
<Y(w) = -wto + <X(w)
please kindly see attachment for the step by step solution of the given problem.
A simply supported beam spans 20 ft and carries a uniformly distributed dead load of 0.8 kip/ft, including the beam self-weight, and a live load of 2.3 kip/ft. Determine the minimum required plastic section modulus (Zx) and select the lightest-weight W-shape to carry the moment. Consider only the limit state of yielding and use A992 steel.
Answer:
Bending stress = 32.29ksi ∠ 33.0 ksi
Explanation:
CHECK BELOW FOR THE ANSWER IN THE FILE ATTACHED.
A single lane highway has a horizontal curve. The curve has a super elevation of 4% and a design speed of 45 mph. The PC station is 105+00 and the PI is at 108+75.
What is the station of the PT?
Answer: 112 + 19.27
Explanation:
Super elevation is an inward transverse slope provided through out the length of the horizontal curve which ends up serving as a counteract to the centrifugal force and checks tendency of overturning. It changes from infinite radius to radius of a transition curve.
Super curve elevation (e) = 4%
4/100= 0.04
e= V^2/gR
Make R the subject of the formula.
egR= V^2
R= V^2/eg
V= 45mph
=45 × 0.44704m/s
=20.1168m/s
g (force due to gravity) =9.81
Therefore,
R= (20.1168)^2/9.81 × 0.04
= 1031.31m
Tangent Length( T) = PI - PC
Tangent Length= 10875 - 10500
=375m
T= R Tan(I/2)
375= 1031.31 × Tan(I/2)
I= 39.96
Also,
L= πRI/180
= 719.27m
Station PT= Stat PC+ L
10500 + 719.27
=11219.27
=112 + 19.27
Very thin films are usually deposited under vacuum conditions to prevent contamination and ensure that atoms can fly directly from the source to the depositing surface without being scattered along the way.
a. To get an idea of how few and far between the air molecules are in a thin-film deposition chamber, determine the mean free path of a generic "air" molecule with an effective diameter of 0.25 nm at a pressure of 1.5 x 10-6 Pa and temperature of 300 K.
b. If the chamber is spherical with a diameter of 10 cm, estimate how many times a given molecule will collide with the chamber before colliding with another air molecule.
c. How many air molecules are in the chamber (treating "air" as an ideal gas)?
Answer:
a. 9947 m
b. 99476 times
c. 2*10^11 molecules
Explanation:
a) To find the mean free path of the air molecules you use the following formula:
[tex]\lambda=\frac{RT}{\sqrt{2}\pi d^2N_AP}[/tex]
R: ideal gas constant = 8.3144 Pam^3/mol K
P: pressure = 1.5*10^{-6} Pa
T: temperature = 300K
N_A: Avogadros' constant = 2.022*10^{23}molecules/mol
d: diameter of the particle = 0.25nm=0.25*10^-9m
By replacing all these values you obtain:
[tex]\lambda=\frac{(8.3144 Pa m^3/mol K)(300K)}{\sqrt{2}\pi (0.25*10^{-9}m)^2(6.02*10^{23})(1.5*10^{-6}Pa)}=9947.62m[/tex]
b) If we assume that the molecule, at the average, is at the center of the chamber, the times the molecule will collide is:
[tex]n_{collision}=\frac{9947.62m}{0.05m}\approx198952\ times[/tex]
c) By using the equation of the ideal gases you obtain:
[tex]PV=NRT\\\\N=\frac{PV}{RT}=\frac{(1.5*10^{-6}Pa)(\frac{4}{3}\pi(0.05m)^3)}{(8.3144Pa\ m^3/mol\ K)(300K)}=3.14*10^{-13}mol\\\\n=(3.14*10^{-13})(6.02*10^{23})\ molecules\approx2*10^{11}\ molecules[/tex]
5) Initially, the pressure and temperature of steam inside a solid capsule is at 100-pound force per square inch absolute (psia), and 600 degrees Fahrenheit (°F), respectively. Because heat is gradually removed from this container, the pressure inside the capsule drops by the amount of 10-pound force per square inch absolute (psia). Answer the following questions, A. (10 points) The change of entropy per unit mass between the initial and final states B. (5 points) The amount of heat transfer per unit mass for the process C. (5 points) Sketch the T-s diagram for the process, showing the associated values of the thermodynamic properties for states and 2 on your sketch.
Answer:
Check the explanation
Explanation:
Kindly check the attached image below for the full step by step explanation to your question.
A 45-kg iron block initially at 280°C is quenched in an insulated tank that contains 100 kg of water at 18°C. Assuming the water that vaporizes during the process condenses back in the tank, determine the total entropy change during this process. The specific heat of water at 25°C is cp = 4.18 kJ/kg·K. The specific heat of iron at room temperature is cp = 0.45 kJ/kg·K
Answer: −3.46kJ/K
Explanation:
From the question above, we have:
The mass of the block (m) = 45kg
The initial temperature of the block (T1) = 280∘C
The weight of the water (mw) = 100kg
The temperature of water (Tw) = 18∘C
Recall the energy balance equation,
ΔUI = −ΔUw
In this case ΔUI is the internal energy of the iron, while ΔUw is the internal energy of water.
[mcp (T2 − T1)]I = −[mcp (T2 − T1)]w
Here cp is the specific heat at constant pressure.
The specific heat of iron is (cp)I = 0.45kJ/kg⋅K, and the specific heat of water is (cp)w = 4.18kJ/kg⋅K.
Now, we substitute the values in above equation,
[45 × 0.45(T2 − 280)]I = −[100 × 4.18(T2 − 18)]w
[20.25(T2 − 280)] = −[418(T2 − 18)]
20.25T2 − 5,670 = −[418T2 − 7,524]
20.25T2 − 5,670 = −418T2 + 7,524
20.25T2 + 418T2 = 7,524 + 5,670
438.35T2 = 13,194
T2 = 30.1K
Recall, the expression to calculate the total entropy change is given as:
ΔStotal = ΔSI + ΔSw
ΔStotal = [mcpln(T2/T1)]I + [mcpln(T2/T1)]w
Now, we substitute the values in above equation,
ΔStotal = [45 × 0.45ln(297.6/553)]I + [100 × 4.18ln(297.6/291)]w
ΔStotal = −12.55 + 9.09
ΔStotal = −3.46kJ/K
Thus the total entropy change is −3.46kJ/K.
The aerodynamic behavior of a flying insect is to be investigated in a wind tunnel using a ten-times scale model. It is known that the insect’s velocity depends on its size (characteristic length L), wing flapping frequency ω, surrounding fluid’s density rho and viscosity μ. If the insect flaps its wings 50 times a second when flying at 1.25 m/s, determine the wind tunnel air speed and wing oscillation frequency required for dynamic similarit
Answer:
Explanation:
Write the equation for Reynolds number as follows:
Re = VL/v
For dynamic similarity,
(VL/v)m + (VL/v)p…… (1)
Since, the model and prototype are in same medium, the kinematic viscosity remains same.
From equation (1), we can write
(VL)m = (VL)p
Here, L represents length, and V is the velocity.
Re-write the equation as follows:
Vm = Lp/Lm x Vp
Substitute 1/8 for Lp/Lm and 1.5m/s for Vp .
Vm = 1/8 x 1.5
Vm = 0.1875m/s
Therefore, the wind tunnel air speed is 0.1875m/s.
Water flows with a velocity of 3 m/s in a rectangular channel 3 m wide at a depth of 3 m. What is the change in depth and in water surface elevation produced when a gradual contraction in the channel to a width of 2.6 m takes place? Determine the greatest contraction allowable without altering the specified upstream conditions.
Answer: new depth will be 3.462m and the water elevation will be 0.462m.
The maximum contraction will be achieved in width 0<w<3
Explanation:detailed calculation and explanation is shown in the image below
You are designing a vascular prosthesis made of woven Dacron and you would like to reinforce it with thin metal wires. The prosthesis has a cylindrical shape. The principal directions of stress are axial and circumferential. What orientation would you choose for the metal wires to avoid rupture/delamination?
Answer:
In place of spiral configuration, we can also use metal wire rings which will provide better strength against circumferential stress. But this spiral configuration will provide strength against circumferential along with axial stress also. If the vascular prosthesis gets elongated along axial direction the corresponding radius of the rings will become short and it will prohibit the expansion of the vascular prosthesis along the circumferential direction.
Explanation:
The orientation is attached below.
Bulk wind shear is calculated by finding the vector difference between the winds at two different heights. Using the supercell wind profile you identified, calculate the 0-1 km and 0-6 km bulk wind shear values. This means we will find the difference between the surface wind (lowest wind barb on the sounding) and the speed of the wind at 1 km and 5 km. The atmospheric pressure at 1 km above sea level is typically very close to 850 mb. The pressure at 6 km above sea level is very close to 500 mb. Please calculate the 0-1 km and 0-6 km wind shear values in knots (kts). For simplicity, assume that the surface winds are due south easterly, the 850 mb winds are due southerly, and the 500 mb winds are due westerly. Show your work.
Answer:
See explaination
Explanation:
2. 0-1 km shear value: taking winds at 1000mb and 850 mb
15 kts south easterly and 50 kts southerly
Vector difference 135/15 and 180/50 will be 170/61 or southerly 61 kts
3. 0-6 km shear value: taking winds at 1000 mb and 500 mb
15 kts south easterly and 40 kts westerly
Vector difference 135/15 and 270/40 will be 281/51 kts
please see attachment
An element representing maximum in-plane shear stress with the associated ________ normal stresses is oriented at an angle of _______ from an element representing the ________ stresses.
Answer: 1) tensile
2) 45 degree orientation
3) principal shear stress
Explanation:
An element representing maximum in-plane shear stress with the associated _tensile normal stresses is oriented at an angle of ____45_degree__ from an element representing the ____principal shear____ stresses.
A car hits a tree at an estimated speed of 10 mi/hr on a 2% downgrade. If skid marks of 100 ft. are observed on dry pavement (F=0.33) followed by 200 ft. on an unpaved shoulder (F=0.28), what is the initial speed of the vehicle just before the pavement skid was begun?
Answer:
[tex]v_{o} = 22.703\,\frac{m}{s}[/tex] [tex]\left(50.795\,\frac{m}{s}\right)[/tex]
Explanation:
The deceleration of the car on the dry pavement is found by the Newton's Law:
[tex]\Sigma F = -\mu_{k,1}\cdot m\cdot g \cdot \cos \theta + m\cdot g \cdot \sin \theta = m\cdot a_{1}[/tex]
Where:
[tex]a_{1} = (-\mu_{k,1}\cdot \cos \theta + \sin \theta)\cdot g[/tex]
[tex]a_{1} = (-0.33\cdot \cos 1.146^{\textdegree}+\sin 1.146^{\textdegree})\cdot \left(9.807\,\frac{m}{s^{2}} \right)[/tex]
[tex]a_{1} = -3.040\,\frac{m}{s^{2}}[/tex]
Likewise, the deceleration of the car on the unpaved shoulder is:
[tex]a_{2} = (-\mu_{k,2}\cdot \cos \theta + \sin \theta)\cdot g[/tex]
[tex]a_{2} = (-0.28\cdot \cos 1.146^{\textdegree}+\sin 1.146^{\textdegree})\cdot \left(9.807\,\frac{m}{s^{2}} \right)[/tex]
[tex]a_{2} = -2.549\,\frac{m}{s^{2}}[/tex]
The speed just before the car entered the unpaved shoulder is:
[tex]v_{o} = \sqrt{\left(4.469\,\frac{m}{s} \right)^{2}-2\cdot \left(-2.549\,\frac{m}{s^{2}} \right)\cdot (60.88\,m)}[/tex]
[tex]v_{o} = 18.175\,\frac{m}{s}[/tex]
And, the speed just before the pavement skid was begun is:
[tex]v_{o} = \sqrt{\left(18.175\,\frac{m}{s} \right)^{2}-2\cdot \left(-3.040\,\frac{m}{s^{2}} \right)\cdot (30.44\,m)}[/tex]
[tex]v_{o} = 22.703\,\frac{m}{s}[/tex] [tex]\left(50.795\,\frac{m}{s}\right)[/tex]
The initial speed of the vehicle just before the pavement skid was begun is 5284.65 ft/hr.
Dry pavement friction coefficient Fdry = 0.33
Length of skid marks on dry pavement ddry = 100 ft
Friction coefficient on unpaved shoulder Fshoulder = 0.28
Length of skid marks on unpaved shoulder = 200 ft
First, let's calculate the work done on dry pavement:
Work on dry pavement = Fdry × ddry = [tex]0.33 *100[/tex]
= 33 ft·lbf
Work on unpaved shoulder = Fshoulder × dshoulder
= [tex]0.28 * 200[/tex]
= 56 ft·lbf
Total work done = Work on dry pavement + Work on unpaved shoulder = 33 + 56
= 89 ft·lbf
Assuming the car's mass remains constant, and the final speed is 0, we have:
89 ft·lbf = (1/2)m × (10 mi/hr)²
Convert the final speed to feet per hour:
[tex]10 mi/hr = 10 × 5280/3600 = 5280 ft/hr[/tex]
Now, solve for the initial speed:
v = √((2 × 89 ft·lbf) / m)
v ≈ √((2 × 89) / (6.38 × 10⁻⁶)) ft/hr
v ≈ [tex]\sqrt{(27889055.9}[/tex] ft/hr
v ≈ 5284.65 ft/hr
A plate in the shape of an isosceles triangle 3 feet high and 4 feet wide is submerged vertically in water, base doward, with the base 5 ft bellow the surface. Find the force exerted by the water on one side of the plate.
Answer:
The force exerted by the water on one side of the plate is F = 24*pg
Explanation:
From the given question, the first step to take is to find he force exerted by the water on one side of the plate.
Solution
Given that:
Let the pressure the at a depth of y ft be = pgy lb/Pa
the area of the atrip is given as = f(y)*delta(y) = 4/3*(y-2)delta(y)
Then
we combine with the range for y as = y E [2 , 5]
Thus,
F = 4/3*pg * integral from (2 , 5) [y(y-2)] dy
Recall that,
p = water density
g= gravity of acceleration
so,
F = 4/3*pg * integral from (2 , 5) [y^2 - 2y]dy]
F = 4/3*pg * [y^3/3 - y^2] [2 , 5]
F = 4/3*pg * [18]
Finally, F = 24*pg
A water-filled manometer is used to measure the pressure in an air-filled tank. One leg of the manometer is open to atmosphere. For a measured manometer deflection of 5 m of water, determine the tank static pressure if the barometric pressure is 101.3 kPa absolute.
(a) 140 kPa absolute
(b) 150 kPa absolute
(c) 160 kPa absolute
(d) 170 kPa absolute
Answer:
[tex]P = 150.335\,kPa[/tex] (Option B)
Explanation:
The absolute pressure of the air-filled tank is:
[tex]P = 101.3\,kPa + \left(1000\,\frac{kg}{m^{3}} \right)\cdot \left(9.807\,\frac{kg}{m^{3}} \right)\cdot (5\,m)\cdot \left(\frac{1\,kPa}{1000\,Pa} \right)[/tex]
[tex]P = 150.335\,kPa[/tex]
An aluminum metal rod is heated to 300oC and, upon equilibration at this temperature, it features a diameter of 25 mm. If a tensile force of 1 kN is applied axially to this heated rod, what is the expected mechanical response
Answer: the metal will experience a strain of approximately 2.037Mpa
This strain is lesser than if the force was applied at room temperature.
This will reduce internal stress and increase some mechanical properties of the aluminum such as mechanical hardening.
Explanation:
Detailed explanation and calculation and comparison with equivalent tensile stretching at ordinary room temperature is compared.
Create a shell script (utilities1.sh) that will print a menu of commands to execute. (a) The script will prompt the user for a number as input. The input will be read, the corresponding command will be executed, and you will see the output of the executed command on your screen. The script will then exit normally, and return a value of 0 (exit 0). See the sample run below.
Answer:
Explanation:
check the attached files for solution.