A rectangular channel 6 m wide with a depth of flow of 3 m has a mean velocity of 1.5 m/s. The channel undergoes a smooth, gradual contraction to a width of 4.5 m.
(a) Calculate the depth and velocity in the contracted section.
(b) Calculate the net fluid force on the walls and floor of the contraction in the flow direction.
In each case, identify any assumptions made.
Answers
Answer:
Depth in the contracted section = 2.896m
Velocity in the contracted section = 2.072m/s
Explanation:
Please see that attachment for the solving.
Assumptions:
1. Negligible head losses
2. Horizontal channel bottom
Related Questions
A particular NMOS device has parameters VT N = 0.6 V, L = 0.8µm, tox = 200 Å, and µn = 600 cm2 /V–s. A drain current of ID = 1.2 mA is required when the device is biased in the saturation region at VGS = 3 V. Determine the required channel width of the device.
Answers
Answer:
[tex]W= 3.22 \mu m[/tex]
Explanation:
the transistor In saturation drain current region is given by:
[tex]i_D}=K_a(V_{GS}-V_{IN})^2[/tex]
Making [tex]K_a[/tex] the subject of the formula; we have:
[tex]K_a=\frac {i_D} {(V_{GS} - V_{IN})^2}[/tex]
where;
[tex]i_D = 1.2m[/tex]
[tex]V_{GS}= 3.0V[/tex]
[tex]V_{TN} = 0.6 V[/tex]
[tex]K_a=\frac {1.2m} {(3.0 - 0.6)^2}[/tex]
[tex]K_a = 208.3 \mu A/V^2[/tex]
Also;
[tex]k'_n}=\frac{\mu n (\frac{cm^2}{V-s} ) \epsilon _{ox}(\frac{F}{cm} ) }{t_{ox}(cm)}[/tex]
where:
[tex]\mu n (\frac{cm^2}{V-s} ) = 600[/tex]
[tex]\epsilon _{ox}=3.9*8.85*10^{-14}[/tex]
[tex]{t_{ox}(cm)=200*10^{-8}[/tex]
substituting our values; we have:
[tex]k'_n}=\frac{(600)(3.988.85*10^{-14})}{(200*10^{-8})}[/tex]
[tex]k'_n}=103.545 \mu A/V^2[/tex]
Finally, the width can be calculated by using the formula:
[tex]W= \frac{2LK_n}{k'n}[/tex]
where;
L = [tex]0.8 \mu m[/tex]
[tex]W= \frac{2*0.8 \mu m *208.3 \mu}{103.545 \mu}[/tex]
[tex]W= 3.22 \mu m[/tex]
Develop a simulation model for a square-wave inverter connected to a dc source of 96 V and an output frequency of 60 Hz. The load is a series RL load with R = 5 Ohm and L = 100 mH.
Answers
Answer:
The answer to this question is attached fully with the explanation.
3. Suppose that a class named Bicycle contains a private nonstatic integer named height, a public nonstatic String named model, and a public static integer named wheels. Which of the following are legal statements in a class named BicycleDemo that has instantiated an object as Bicycle myBike new Bicycle C);? f. Bicycle. model Hurricane a. myBike height 26; b. my Bike model Cyclone g. Bicycle. int 3 3; c. myBike Wheels 3 d. my Bike .model 108; i. Bicycle wheels 2 e. Bicycle height 24; j. Bicycle yourBike myBike
Answers
Answer:
The solution to the given problem is provided below.
Explanation:
a.) myBike.height = 26; Not Legal statement
b.) myBike.model = “Cyclone”: Legal statement
c.) myBike.wheels = 3; Legal statement
d.) myBike.model = 108; Not legal statement
e.) Bicycle.height = 24; Not Legal statement
f.) Bicycle.model = “Hurricane”; Not legal statement
g.) Bicycle.int = 3; Not Legal statement
h.) Bicycle.model = 108; Not Legal Statement
i.) Bicycle.wheels = 2; Legal Statement
j.) Bicycle yourBike = myBike; Legal Statement
Before entering an underground utility vault to do repairs, a work crew analyzed the gas in the vault and found that it contained 29 mg/m3of hydrogen sulfide. Because the allowable exposure level is 14 mg/m3, the work crew began ventilating the vault with a blower. If the volume of the vault is 160 m3and the flow rate of contaminant-free air is 10 m3/min, how long will it take to lower the hydrogen sulfide level to a level that will allow the work crew to enter? Assume the manhole behaves as a CMFR and that hydrogen sulfide is nonreactive in the time period considered.
Answers
Answer:
11.65 minutes.
Explanation:
See the attached picture for detailed explanation.
Answer:
11.65 min
Explanation:
Hydrogen sulfide is very poisonous and as a result, it is essential to reduce the concentration of the gas to the lowest possible value to minimize its effects. The time taken to reduce the amount of hydrogen sulfide in the system to the allowable limit can be estimated as shown below:
t = (V/Q)*ln(Ci/Co) = (160/10)*ln(14/29) = 16*0.728 = 11.65 min
"A horizontal curve is being designed for a new two-lane highway (12-ft lanes). The PI is at station 250 + 50, the design speed is 65 mi/h, and a maximum superelevation of 0.07 ft/ft is to be used. If the central angle of the curve is 38 degrees, design a curve for the highway by computing the radius and stationing of the PC and PT."
Answers
Answer:
Radius = 1565ft ; PC = 245 + 11.13 ; PT = 255 + 48.88
Explanation:
1. Accordingly to the law of mechanics;
Centrifugal factor =S+F = V²/15R
Where; S = Super-elevation slope = 0.07ft/ft
F= Slide friction factor
V= Design speed=65mi/h
R= Radius
From the graph (see attached), at design speed of 65mi/h, coefficient of slide friction factor, F =0.11
Applying the figures in the equation above;
0.07+0.11 = 65²/15R
R=281.67/0.18
R=1564.8
Approximately Radius = 1565ft
2. Stationing PC = Stationing PI - T
where T = Tangent distance
T= Rtan(Δ/2) where Δ = central angle = 38° & Stationing PI = 250 + 50
T=1565tan19°
T=538.87ft
Stationing PC = 250 + 50 - (5 + 38.87)
PC = 245 + 11.13
3. Stationing PT = Stationing PC + L
Where L = Length of the circular curve
L = π/180*(RΔ)
L=0.01745*1565*38
L=1037.75
Therefore;
Stationing PT= 245 + 11.13 + (10 +37.75)
PT = 255 + 48.88
An earthen trapezoidal channel (n = 0.025) has a bottom width of 5.0 m, side slopes of 1.5 horizontal: 1 vertical and a uniform flow depth of 1.10 m. In an economic study to remedy excessive seepage from the canal two proposals, (a) to line the sides only and, (b) to line the bed only are considered. If the lining is of smooth concrete (n = 0.012), calculate the equivalent roughness in the above two cases for a bottom slope of 0.005
Answers
Answer:
A. 0.020
B. 0.018
Explanation: check the attached file
Answer:
a. n=0.020 b. n=0.018
Explanation:
a.
Case 1: To line the sides only
n=(Σn₁¹°⁵P₁)^2/3/P^2/3
n = equivalent roughness
n₁=corresponding roughness coefficients
P=length
At the bed: n₁=0.025 and P₁=5m
At the sides: n₂=0.012 and P₂= 2*1.1*√1+1.5²=2.2*1.8=3.96m
P = P₁+P₂=8.96m
Equivalent roughness, n = [5*(0.025)^1.5+3.96*(0.012)^1.5]^2/3/(8.96)^2/3
n= [(5*0.00395)+(3.96*0.0013)]^2/3/4.317
n=0.0249^2/3/4.317
n=0.0842/4.317
n=0.0195
n=0.020
b.
Case 2: To line the bed only
P₁=5m n₁=0.012
P₂=3.96 n₂=0.025
P=8.96
Equivalent roughness n= [5*(0.012)^1.5+3.96*(0.025)^1.5]^2/3/(8.96)^2/3
n=[(5*0.0013)+(3.96*0.00395)]^2/3/4.317
n=0.0221^2/3/4.317
n=0.078/4.317
n=0.018
For each of the characteristic equations of feedback control systems given, determine the range of K so that the system is asymptotically stable. Determine the value of Kso that the system is marginally stable and the frequency of sustained oscillation if applicable. s4 + Ks3 + 2s2 + (K + 1)s + 10 = 0
Answers
Answer:
Explanation:
The method or principle applied is the Routh- hurtwitz criterion for solving characteristics equation.
The steps by step analysis and appropriate substitution is carefully shown in the attached file.
With reference to the NSPE Code of Ethics, which one of the following statements is true regarding the ethical obligations of the engineers involved in the VW Emissions Cheating Scandal.
a. The VW engineers involved conducted themselves honorably, responsibly, ethically, and lawfully so as to enhance the honor, reputation, and usefulness of the profession.
b. As faithful agents and trustees of Volkswagen the engineers involved could not ethically report the emissions cheating violations to public authorities.
c. The VW engineers involved were ethically obligated to hold paramount the health, welfare and safety of the public even if their supervisors directed them to implement software and hardware that enabled cheating on the emissions testing software.
d. The VW engineers involved were not ethically obligated to report to their supervisors and to upper management the emissions cheating violations being implemented in the control system hardware and software.
Answers
Answer: c. The VW engineers involved were ethically obligated to hold paramount the health, welfare and safety of the public even if their supervisors directed them to implement software and hardware that enabled cheating on the emissions testing software.
Explanation: The National Society of professional Engineers, NSPE define the code of ethics which must guide engineers in their duty. These codes act as principles of personal conduct, towards the public and their employers.
One of the areas covered by these codes is overriding importance of the safety and health of the public to any other factor. In addition, engineers are to avoid deception and maintain the reputation of their profession. These cannot be sacrificed for the financial gain of their employers or explained away by saying they are following the direction of their employers. While they have certain responsibilities to their employers, the health welfare and safety of the public is more important.
Silicon carbide nanowires of diameter 15 nm can be grown onto a solid silicon carbide surface by carefully depositing droplets of catalyst liquid onto a flat silicon carbide substrate. Silicon carbide nanowires grow upward from the deposited drops, and if the drops are deposited in a pattern, an array of nanowire fins can be grown, forming a silicon carbide nano-heat sink. Consider finned and unfinned electronics packages in which an extremely small, 10 μm × 10 μm electronics device is sandwiched between two 100-nm-thick silicon carbide sheets. In both cases, the coolant is a dielectric liquid at 20°C. A heat transfer coefficient of 1.0 × 105 W/m2·K exists on the top and bottom of the unfinned package and on all surfaces of the exposed silicon carbide fins, which are each 300 nm long. Each nano-heat sink includes a 50 × 50 array of nanofins. Determine the maximum allowable heat rate that can be generated by the electronic device so that its temperature is maintained at 95°C for (a) the unfinned and (b) the finned packages
Answers
Answer:
Please find attached file for complete answer.
Explanation:
Which statement is not correct regarding the deformation of a circular shaft in torsion? a. Cross sections remain flat. b. Longitudinal lines remain straight. c. Circular sections remain circular. d. Radial lines on the sections remain straight.
Answers
Answer:
B
Explanation:
Torsion is application of torque to a shaft to turn it about its longitudinal axis. When torque is applied to a shaft the circle remains unchanged in a circular state, its cross section does not warp but remains flat with a straight radial lines but its longitudinal lines changes into an helix intersecting the circular shaft
A bridge hand consists of 13 cards. One way to evaluate a hand is to calculate the total high point count (HPC) where an ace is worth four points, a king is worth three points, a
Answers
Answer: Let us use the pickled file - DeckOfCardsList.dat.
Explanation: So that our possible outcome becomes
7♥, A♦, Q♠, 4♣, 8♠, 8♥, K♠, 2♦, 10♦, 9♦, K♥, Q♦, Q♣
HPC (High Point Count) = 16
A ship tows a submerged cylinder, 1.5 m in diameter and 22 m long, at U = 5 m/s in fresh water at 20°C. Estimate the towing power in kW if the cylinder is (a) parallel, and (b) normal to the tow direction.
Answers
Based on the available information, the the estimated towing power required is:
(a) For the parallel orientation: 655 kW
(b) For the normal orientation: 4,116.75 kW
Given:
- Diameter of the cylinder: 1.5 m
- Length of the cylinder: 22 m
- Towing speed (U): 5 m/s
- Water temperature: 20°C
(a) Cylinder parallel to the tow direction:
Step 1: Calculate the drag force for the parallel orientation.
The drag force for a submerged cylinder in parallel orientation is given by the formula:
F_D = 0.5 × ρ × C_D × A × U^2
Where:
- ρ is the density of fresh water at 20°C, which is approximately 998 kg/m³.
- C_D is the drag coefficient for a cylinder in parallel orientation, which is approximately 0.82.
- A is the cross-sectional area of the cylinder, which is π × D × L = π × 1.5 m × 22 m = 103.87 m².
Calculating the drag force:
F_D = 0.5 × 998 kg/m³ × 0.82 × 103.87 m² × (5 m/s)² = 131,000 N
Step 2: Calculate the towing power for the parallel orientation.
Towing power = Drag force × Towing speed
Towing power = 131,000 N × 5 m/s = 655,000 W = 655 kW
(b) Cylinder normal to the tow direction:
Step 3: Calculate the drag force for the normal orientation.
The drag force for a submerged cylinder in normal orientation is given by the formula:
F_D = 0.5 × ρ × C_D × A × U^2
Where:
- C_D is the drag coefficient for a cylinder in normal orientation, which is approximately 1.2.
- A is the cross-sectional area of the cylinder, which is π × D × D/4 = π × 1.5 m × 1.5 m/4 = 1.77 m².
Calculating the drag force:
F_D = 0.5 × 998 kg/m³ × 1.2 × 1.77 m² × (5 m/s)² = 823,350 N
Step 4: Calculate the towing power for the normal orientation.
Towing power = Drag force × Towing speed
Towing power = 823,350 N × 5 m/s = 4,116,750 W = 4,116.75 kW
Therefore, the estimated towing power required is:
(a) For the parallel orientation: 655 kW
(b) For the normal orientation: 4,116.75 kW
A single axial load of magnitude P = 15 kips is applied at end C of thesteel rod ABC. Knowing that E = 30 × 106 psi, determine the diameterd of portion BC for which the deflection of point C will be 0.05 in.
Answers
Answer:
d = 0.868 in
Explanation:
note:
solution is attached due to error in mathematical equation. please find the attachment
This question involves solving for the diameter d at the portion BC of a steel rod under axial load. It involves utilizing Hooke's Law and expressing cross-sectional area in terms of diameter. The final calculation will depend on the length of the portion BC.
Explanation:The question appears to involve the concept of stress and strain in the area of mechanical engineering. In this problem, we have a steel rod ABC, where end C, experiencing a single axial load P = 15 kips, deflections to 0.05 inches. We are tasked to find the diameter, denoted as d, of portion BC.
To solve this, we could utilize Hooke's Law of elasticity that connects stress, strain, and the Young's modulus (E). The formula is represented as δ = PL/AE. In the equation, P is the load, L is the length, A is the cross-sectional area and E is Young's modulus. We are given P, E, and δ (deflection), leaving the cross-sectional area and length as variables to be found. Since we are looking for the diameter (d), we need to express the area (A) in terms of the diameter using the formula A = πd²/4.
The length L is dependent on the specific conditions of the problem and are not given directly. Depending on the conditions, L might need to be solved differently. First, solve the equation for diameter, afterwards, plug the known values in and evaluate, giving the diameter in inches of the steel rod at portion BC.
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Bridge A is the longest suspension bridge in a Country. Bridge B is 5555 feet shortershorter than Bridge A. If the length of Bridge A is m feet, express the length of Bridge B as an algebraic expression in m. Write an expression representing the length of Bridge B in terms of m.
Answers
The length of Bridge B is expressed as m - 5555 feet, where m represents the length of Bridge A. For example, if Bridge A is 10,000 feet long, then Bridge B would be 4445 feet long.
Explanation:Bridge B is 5555 feet shorter than Bridge A, so we can represent the length of Bridge B as m - 5555 feet. This expression represents the length of Bridge B in terms of m, where m represents the length of Bridge A. For example, if Bridge A is 10,000 feet long, then Bridge B would be 10,000 - 5555 = 4445 feet long.
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Electric heater wires are installed in a solid wall having a thickness of 8 cm and k=2.5 W/m.°C. The right face is exposed to an environment with h=50 W/m2°C and k'=30°C, while the left face is exposed to h=75 W/m2°C and T[infinity]=50°C. What is the maximum allowable heat generation rate such that the maximum temperature in the solid does not exceed 300°C.
Answers
Answer:
2.46 * 10⁵ W/m³
Explanation:
See attached pictures for detailed explanation.
Answer:
[tex]q^.=2.46*10^5W/m^3[/tex]
Explanation:
[tex]Given\\k=2.5W/m\\h_{1} =75(left)\\h_{2} =50(right)\\T_{1} =50^oC\\T_{2} =30^oC[/tex]
so
[tex]T=-\frac{q^.x^2}{2k} +c_{1}x+ c_{2} \\T=T_{1} \\at \\x=-0.04\\T=T_{2} \\at\\x=+0.04[/tex]
[tex]dT/dx=-q^.x/k+c_{1} \\T=T_{max} =300\\at\\x=c_{1} \frac{k}{q^.} (1)[/tex]
[tex]h_{1}(T_{1infinity} -T_{1} )=-k\frac{dT}{dx} |_{x=0.04} (2)\\-k\frac{dT}{dx} |_{x=0.04} =h_{2} (T_{2}-T_{2infinity} (3)[/tex]
[tex]300=-\frac{q^.}{2k} [c_{1} \frac{k}{q} ]^2+c_{1} [c_{1} \frac{k}{q} ]+c_{2} (1)[/tex]
[tex]75[50+\frac{q^2}{2k} (0.04)^2+c_{1} (0.04)-c_{2} ]=-k[\frac{+q^2(0.04)}{2k} ](2)[/tex]
[tex]-k[\frac{-q^.(0.04)}{2k} ]=50[\frac{-q^.(0.04)}{2k} +c_{1} (0.04)+c_{2} -30](3)[/tex]
solving above 3 equations for 3 unknowns c1,c2,q
we get [tex]q^.=2.46*10^5W/m^3[/tex]
A large tower is to be supported by a series of steel wires. It is estimated that the load on each wire will be 11,100 N (2500 lb f ). Determine the minimum required wire diameter assuming a factor of safety of 2 and a yield strength of 1030 MPa (150,000 psi).
Answers
Answer:
5.24m
Explanation:
Data given
force, F= 11,100N
safety factor,N=2
yield strength, =1030MPa
To determine the minimum diameter, we first determine the working strength which is expressed as
[tex]working strength=\frac{yield strength}{safty factor}\\ Working strength =1030/2=515MPa\\[/tex]
Since also the working strength is define as the ration of the force to the area, we have
[tex]515=\frac{11100}{A}\\ A=21.55m^{2}[/tex]
hence the required diameter is given as
[tex]A=\pi d^2/4\\d=\sqrt{\frac{4*21.55 }{\pi }} \\d=5.24m[/tex]
What is the purpose of the following algorithm? input somenum Repeat the following steps for 14 times input variable1 if variable1 < somenum then somenum = variable1 print somenum
Answers
Answer:
The purpose of the algorithm is to print the least digit among a total of 15 digita
Explanation:
input somenum
Repeat the following steps for 14 times
input variable1
if variable1 < somenum then
somenum = variable1
print somenum
On line 1, the algorithm takes an input through variable1
An iteration is started on line 2 and ends on line 6
Line 3,4,5 re performed repeatedly;
On line 3, the algorithm accepts another input through somenum and it keep accepting it till the end of the iteration.
On line 4, the algorithm tests if variable1 is lesser than somenum.
If yes, line 5 is executed and the value of variable1 is assigned to somenum
Else, line 5 is skipped; the iteration moves to line 3 as long as the condition is still valid.
At the end of the iteration, the least value stored in somenum is printed through
. A steam turbine operates between 500°C and 3.5 MPa to 200°C and 0.3 MPa. If the turbine generates 750 kW and the heat loss is 100 kW, what is the flow rate of steam through the turbine?
Answers
Answer:
1.757 kg/s
Explanation:
According to the First Law of Thermodynamics, the physical model for a turbine working at steady state is:
[tex]-\dot Q_{out} - \dot W_{out} + \dot m \cdot (h_{in}-h_{out})=0[/tex]
The flow rate of steam is:
[tex]\dot m = \frac{\dot Q_{out}+\dot W_{out}}{h_{in}-h_{out}}[/tex]
Water enters and exits as superheated steam. After looking for useful data in a property table for superheated steam, specific enthalpies at inlet and outlet are presented below:
[tex]h_{in} = 3451.7 \frac{kJ}{kg} \\h_{out} = 2967.9 \frac{kJ}{kg} \\[/tex]
Finally, the flow rate is calculated:
[tex]\dot m = \frac{100 kW + 750 kW}{3451.7 \frac{kJ}{kg} - 2967.9 \frac{kJ}{kg}}\\\dot m =1.757 \frac{kg}{s}[/tex]
15 points) A horizontal curve is being designed for a new two-lane highway (12-ft lanes). The PI is at station 250 00, design speed is 65 mph, and a maximum superelevation of 0.08 ft/ft is to be used. If the central angle of the curve is 35 degrees, design a curve for the highway by computing the radius and stationing of the PC and PT.
Answers
Answer:
59.78 m
Explanation:
Data:
PI station = 250 will allow a speed of 65 mph
The maximum superelevation will be = 0.08 ft/ft
Central angle of the curve = 35 degrees
A figure will be used to present the information. This gives the height of 59.78 m as the maximum elevation for the safe speed of the vehicle.
(CLO 3—Boolean/Comb. Logic) It is desired to multiplex four different input data lines, a-d, onto one output. Three address lines, ("x" [MSB] through "z" [LSB]) control input-to-output selection. The three-bit address can be stated as a decimal number ranging from 0 to 7. Input a is MUXed out on address 3, b on address 4, c on 6, d on 7. Draw the MUX circuit below.
Answers
Answer:
Please find attached file for complete answer solution and explanation of same question.
Explanation:
In the casting of steel under certain mold conditions, the mold constant in Chvorinov's Rule is known to be 4.0 min/cm2, based on previous experience. The casting is a flat plate whose length = 35 cm, width = 10 cm, and thickness = 15 mm. Determine how long it will take for the casting to solidify.
Answers
Answer:
it will take for the casting to solidify 2.55 min
Explanation:
given data
mold constant = 4 min/cm²
length = 35 cm
width = 10 cm
thickness = 15 mm
solution
we use here Chvorinov's Rule that is
Chvorinov's Rule = mold constant × [tex](\frac{V}{A})^{1.9}[/tex] ..............1
put here value
Chvorinov's Rule = 4 × [tex](\frac{600}{760})^{1.9}[/tex]
Chvorinov's Rule = 2.55 min/in
so heer unit flow become [tex]min/in^{1.9}[/tex]
The portable lighting equipment for a mine is located 100 meters from its dc supply source. The mine lights use a total of 5 kW and operate at 120 V dc. Determine the required cross-sectional area of the copper wires used to connect the source to the mine lights if we require that the power lost in the copper wires be less than or equal to 5 percent of the power required by the mine lights.
Answers
To calculate the required cross-sectional area of copper wires needed to connect the DC supply source to the mine lights with less than 5% power loss, calculate the permissible power loss, and then use the relationship between power loss, current, and wire resistance. Considering the resistivity of copper and the round trip length of the wire, the formula A = ρL/(δP/I²) determines the necessary wire thickness.
Explanation:To determine the required cross-sectional area of copper wires for the lighting equipment, we need to ensure the power loss (δP) is no more than 5% of the lights' power usage (P), which is 5 kW (5000 W). As the power loss in wires is given by δP = I²R, where I is the current and R is the resistance of the wire, we must first calculate the current using P = IV, giving I = P/V = 5000W/120V = 41.67 A. The power loss allowed is thus 5% of 5000 W, equating to 250 W. Given δP, we can find R using δP = I²R, which gives R = δP/I².
The resistance of a copper wire is also given by R = ρL/A, where ρ is the resistivity of copper (1.68 x 10^-8 Ωm), L is the length of the wire (200 meters round trip), and A is the cross-sectional area we need to find. Equating the two expressions for R and solving for A gives A = ρL/(δP/I²). Substituting the given and calculated values yields the required cross-sectional area. Finally, as resistance depends on the entire length of the circuit, remember to double the distance to account for both the outgoing and return paths.
Prob. 4.2.1. A well that pumps at a constant rate of 0.5 m3/s fully penetrates a confined aquifer of 34-m thickness. After a long period of pumping, steady-state drawdowns are measured at two observation wells 50 and 100 m from the pumping well as 0.9 m and 0.4 m, respectively. Determine: (a) the hydraulic conductivity and transmissivity of the aquifer; (b) the radius of influence of the pumping well; (c) the expected drawdown in the pumping well if the radius of the well is 0.4 m.
Answers
Answer:
Part (a)
K = 0.00406 m / s
T = 0.14 m2 / s
Part (b)
R = Radius of influence of Pumping well = 237.94 m
Part (c)
S = Drawdown at well = 3.68 m
Explanation:
General formula for wells at confined aquifer is
Q = [tex]{2*3.1416*K*B*(S1-S2)} / {2.303 log (R/r)}[/tex] ............ eq (A)Where ,
Q = Discharge
K = Hydraulic conductivity
B = Thickness of aquifer
S1 = Draw-down at point 1
S2 = Draw-down at point 2
R = Radius of influence of well
r = Radius of well
Part (a)
We will use
Q = [tex]{2*3.1416*K*B*(S1-S2)} / {2.303 log (r2/r1)}[/tex] ......... eq (1)Where,
r 1 = Distance of first observation well from main well
r2 = Distance of 2nd observation well from main well
Given data: Q = 0.5 m3/s B = 34 m r 1 = 50 m
r2 = 100 m S1 = 0.9 m S2 = 0.4 m
Put all these values in equation 1
0.5 = 2*3.1416*K*34*(0.9 - 0.5) / {2.303 log(100 / 50)}
Write Equation in terms of K = hydraulic conductivity
K = {0.5*2.303 log (100 / 50)} / {2*3.1416*34*(0.9 - 0.5)}
K = 0.00406 m / s
Now Transmissivity
T = K*B
T = 0.14 m2 / s
Part (b)
To calculate radius of influence, use equation (A)
Q = [tex]{2*3.1416*K*B*(S1-S2)} / {2.303 log (R/r)}[/tex]
At distance R from main well drawdown (S2) = 0
use r = r1 = 50 m and S1 = 0.9
use ln ( R / r) = 2.303 log (R / r)
0.5 = 2* 3.1416 * 0.00406*34* (0.9 - 0) / ln(R / 50)
Write equation in terms of R
ln( R / 50) = 2* 3.1416 * 0.00406*34* (0.9 - 0) / 0.5
ln( R / 50) = 1.56
Take anti log (e) of above equation
R / 50 = 4.76
R = Radius of influence of Pumping well = 237.94 m
Part (c)
Use equation A
Q = [tex]{2*3.1416*K*B*(S1-S2)} / {2.303 log (R/r)}[/tex]
S1 = ?
Put S2 = 0 R = 237.94 r = 0.4
0.5 = 2* 3.1416* 0.00406*34*(S1 - 0) / 2.303 log(237.94 / 0.4)
Write equation in terms of S1
S1 = 0.5* 2.303 log(237.94 / 0.4) / 2*3.1416*0.00406*34
S1 = Drawdown at well = 3.68 m
The hydraulic conductivity and transmissivity of the aquifer are respectively; 0.0032 m/s and 0.11 m³/s
What is the hydraulic conductivity?
A) We are given;
Pump rate; Q = 0.5 m³/s
thickness of quifer; b = 34 m
depth 1; r₁ = 50 m
depth 2; r₂ = 100 m
distance 1; S₁ = 0.9 m
distance 2; S₂ = 0.4 m
Formula for the pump rate is;
Q = 2π × b × k × (S₁ - S₂)/(In r₂/r₁)
making k the subject gives;
k = Q(In r₂/r₁)/(2π × b × (S₁ - S₂))
k = 0.5(In 100/50)/(2π × 34 × (0.9 - 0.4))
Solving for K gives;
Hydraulic conductivity is; k = 0.0032 m/s
Transmissivity is;
T = K * b
T = 0.0032 * 34
T = 0.11 m³/s
B) Formula for radius of incfluence is;
S_w = S₁ - [(Q/2π × b × k) In (r_w/r₁)]
Plugging in the relevant values gives;
S_w = 4.338 m
C) Formula for expected drawdown is;
R = r₁ e^(2πbk(S_w - S₁)/Q)
R = 100 * e^(2π*34*0.0032(-78.9)/0.5)
R = 147.7 m
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Air at 27oC and 50% relative humidity is cooled in a sensible cooling process to 18oC. The air is then heated to 45oC in a sensible heating process. Finally, the air experiences an adiabatic saturation process that increases the relative humidity back to 50%. Find the specific energy that is removed when the air is cooled to 18°C.
Answers
Answer:
[tex]q_{out} = 9.25\,\frac{kJ}{kg}[/tex]
Explanation:
First, it is required to find the absolute humidity of air at initial state:
[tex]\omega = \frac{0.622\cdot \phi \cdot P_{g}}{P-\phi \cdot P_{g}}[/tex]
The saturation pressure at [tex]T = 27^{\textdegree}C[/tex] is:
[tex]P_{g} = 3.601\,kPa[/tex]
Then,
[tex]\omega = \frac{0.622\cdot (0.5)\cdot (3.601\,kPa)}{101.325\,kPa-(0.5)\cdot (3.601\,kPa)}[/tex]
[tex]\omega = 0.0113\,\frac{kg\,H_{2}O}{kg\,air}[/tex]
A simple cooling process implies a cooling process with constant absolute humidity. The specific entalphies for humid air are:
Initial state:
[tex]h_{1} = (1.005\,\frac{kJ}{kg\cdot ^{\textdegree}C})\cdot (27^{\textdegree}C)+(0.0113)\cdot (2551.96\,\frac{kJ}{kg} )[/tex]
[tex]h_{1} = 55.972\,\frac{kJ}{kg}[/tex]
Final state:
[tex]h_{2} = (1.005\,\frac{kJ}{kg\cdot ^{\textdegree}C})\cdot (18^{\textdegree}C)+(0.0113)\cdot (2533.76\,\frac{kJ}{kg} )[/tex]
[tex]h_{2} = 46.722\,\frac{kJ}{kg}[/tex]
The specific energy that is removed is:
[tex]q_{out}= h_{1} - h_{2}[/tex]
[tex]q_{out} = 9.25\,\frac{kJ}{kg}[/tex]
A pipe of inner radius R1, outer radius R2 and constant thermal conductivity K is maintained at an inner temperature T1 and outer temperature T2. For a length of pipe L find the rate at which the heat is lost and the temperature inside the pipe in the steady state.
Answers
Answer:
Heat lost from the cylindrical pipe is given by the formula,
[tex]d Q= \frac{2 \pi K L (T_{1} - T_{2} )}{log_{e}(\frac{R_{2} }{R_{1} } ) }[/tex]
Temperature distribution inside the cylinder is given by,
[tex]\frac{T - T_{1} }{T_{2} - T_{1} } = \frac{log_{e} \frac{R}{R_{1} }}{log_{e} \frac{R_{2}}{R_{1} }}[/tex]
Explanation:
Inner radius = [tex]R_{1}[/tex]
Outer radius = [tex]R_{2}[/tex]
Constant thermal conductivity = K
Inner temperature = [tex]T_{1}[/tex]
Outer temperature = [tex]T_{2}[/tex]
Length of the pipe = L
Heat lost from the cylindrical pipe is given by the formula,
[tex]d Q= \frac{2 \pi K L (T_{1} - T_{2} )}{log_{e}(\frac{R_{2} }{R_{1} } ) }[/tex]------------ (1)
Temperature distribution inside the cylinder is given by,
[tex]\frac{T - T_{1} }{T_{2} - T_{1} } = \frac{log_{e} \frac{R}{R_{1} }}{log_{e} \frac{R_{2}}{R_{1} }}[/tex] ------------ (2)
An insulated piston-cylinder device contains 5 L of saturated liquid water at a constant pressure of 175 kPa. Water is stirred by a paddle wheel while a current of 8 A flows for 45 min through a resistor placed in the water. If one-half of the liquid is evaporated during this constant-pressure process and the paddle-wheel work amounts to 400 kJ, determine the voltage of the source. Also, show the process on a
Answers
Answer:
note:
solution is attached in word form due to error in mathematical equation. furthermore i also attach Screenshot of solution in word due to different version of MS Office please find the attachment
The voltage of the source can be determined by dividing the total energy transferred (sum of the paddle-wheel work and heat required to evaporate water) by the product of current and time. The process will be seen as a horizontal line on a piston-cylinder diagram
Explanation:The heat transferred to the water can be calculated from the energy supplied by the electrical source and the work done by the paddle wheel. Since power is the energy transferred per unit time, we can use the equation Power = Voltage x Current. We know that the current is 8 A and the energy transferred is the sum of the work done by the paddle wheel and the heat required to evaporate half of the water, which is 45 minutes (converted to seconds) times the power. Solving for voltage we get:
Voltage = (Energy transferred) / (Current x Time)
In terms of a piston-cylinder diagram, the process will appear as a horizontal line since the pressure is constant. The line will move upwards as heat is added and some of the water is turned into steam.
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Checkpoint 4.72 Write a statement that uses a conditional expression that determines if the credits variable is less than 0. If the condition is true, assign the value 0 to the credits variable, otherwise the value of the credits variable should remain unchanged.
Answers
Answer:
See step to step explanations for answer.
Explanation:
This is the expression to check if credits is less than 0 assuming variable name is credits:
if(credits<0)
{
credits = 0;
}
Q 4.69:
this is the expression checking the name is Swordfish or not
if(name=="Swordfish")
{
cout<<"We have a match";
}
Q 4.67
This is the expression to check the character assuming that character is ch
if(ch>='1' && ch<='9')
{
cout<<"Digit detected";
}
Q 4.59:
This is the statement to check if the number is between 0 and 500 inclusive . Assuming the variable is n
if(n>=0 && n<=500)
{
cout<<"The number is valid";
}
To address the programming question about conditional expressions, a ternary operator can be used to set the variable 'credits' to 0 if its value is less than 0, and otherwise keep the current value.
Explanation:The student's question pertains to the use of a conditional expression in a programming context. A conditional expression evaluates a condition and based on its truth value, executes one of two expressions. The statement required should check if the credits variable is less than 0 and if so, assign the value 0 to credits; otherwise, it should leave the value of credits as is.
An example in a generic programming language would be:
credits = 0 if credits < 0 else creditsThis line of code is known as a ternary operator or conditional assignment. It reads as 'Set credits to 0 if credits is less than 0; otherwise, keep the value of credits.'
A platinum resistance temperature sensor has a resistance of 120 Ω at 0℃ and forms one arm of a Wheatstone bridge. At this temperature the bridge is balanced with each of the other arms being 120 Ω. The temperature coefficient of resistance of the platinum is 0.0039/K. What will be the output voltage from the bridge for a change in temperature of 20℃? The loading across the output is effectively open circuit and the supply voltage to the bridge is from a source of 6.0 V with negligible internal resistance.
Answers
The output voltage from a Wheatstone bridge with a change in temperature of 20℃ in one of its platinum resistance temperature sensor arms is calculated to be approximately 0.233 V, taking into account the specific temperature coefficient of resistance for platinum.
Explanation:The student's question involves calculating the output voltage from a Wheatstone bridge when a platinum resistance temperature sensor, which forms one arm of the bridge, changes its resistance due to a temperature change. With a temperature coefficient of resistance for platinum of 0.0039/K and an initial balance condition at 0℃ with each arm having a resistance of 120 Ω, the temperature change of 20℃ will lead to a change in resistance in the platinum arm, affecting the bridge's balance and generating an output voltage.
To calculate the change in resistance (ΔR) for the platinum sensor due to the temperature change: ΔR = Ro·α·ΔT, where Ro is the initial resistance (120 Ω), α is the temperature coefficient of resistance (0.0039/K), and ΔT is the temperature change (20℃). Therefore, ΔR = 120·0.0039·20 = 9.36 Ω. The new resistance of the platinum sensor at 20℃ is 120 Ω + 9.36 Ω = 129.36 Ω.
Given the supply voltage (Vs) is 6.0 V, and considering the bridge was initially balanced, the output voltage (Vo) from the bridge can be calculated using the formula derived from the Wheatstone bridge principles: Vo = Vs · (ΔR / (2Ro + ΔR)). Substituting the values gives Vo = 6.0 · (9.36 / (240 + 9.36)) = 0.233 V. Thus, the output voltage from the bridge for a change in temperature of 20℃ is approximately 0.233 V.
The water behind Hoover Dam is 206m higher than the Colorado river below it. At what rate must water pass through the hydraulic turbines of this dam to produce 100 MW of power if the turbines are 100 percent efficient?
Answers
Answer:
m' = 4948.38 kg/s
Explanation:
For a case of 100% efficiency, the power produced must be equal to the rate of potential energy conversion
GIVEN THAT
Power = 100 MW
rate of Potential energy = (m')*g*h
100*10^6 = (m')*9.81*206
m' = 4948.38 kg/s
Answer:
49.484 m³ / s
Explanation:
Volume flow rate = Power in W / (efficiency × density × height × acceleration due to gravity)
Volume flow rate = 100 × 10⁶ / ( 1 × 1000 kg/m³ × 206 m × 9.81 m/s²)
V = 49.484 m³ / s
g Create an array of five animals. Use a for loop to display the values stored in that array. Add two more animals to the end of that array. Sort the array and display the sorted array on the screen.
Answers
Answer:
The most common approach to accessing an array is to use a for loop:
var mammals = new Array("cat","dog","human","whale","seal");
var animalString = "";
for (var i = 0; i < mammals. length; i++) {
animalString += mammals[i] + " ";
}
alert(animalString);
Discussion
A for loop can be used to access every element of an array. The array begins at zero, and the array property length is used to set the loop end.
Though support for both indexOf and lastIndexOf has existed in browsers for some time, it’s only been formalized with the release of ECMAScript 5. Both methods take a search value, which is then compared to every element in the array. If the value is found, both return an index representing the array element. If the value is not found, –1 is returned. The indexOf method returns the first one found, the lastIndexOf returns the last one found:
var animals = new Array("dog","cat","seal","walrus","lion", "cat");
alert(animals.indexOf("cat")); // prints 1
alert(animals.lastIndexOf("cat")); // prints 5
Both methods can take a starting index, setting where the search is going to start:
var animals = new Array("dog","cat","seal","walrus","lion", "cat");
alert(animals.indexOf("cat",2)); // prints 5
alert(animals.lastIndexOf("cat",4)); /
Answer:
animals = ["Dog", "Lion", "Goat", "Zebra", "Cat"]
for animal in animals:
print(animal)
x = animals.insert(5, "Lizard")
y = animals.insert(6, "Bat")
z = sorted(animals)
print(z)
Explanation:
The question can be solved using various back-end coding language like python, java, JavaScript etc. But I will be writing the code with python.
The first question said we should create an array or list of five animals.
animals = ["Dog", "Lion", "Goat", "Zebra", "Cat"] → I created the five animals in a list and stored them in the variable animals.
for animal in animals: I used a for loop to iterate through the list print(animal) I used print statement to display the values stored in the array or list.
x = animals.insert(5, "Lizard") I added an animal, lizard to the end of the array
y = animals.insert(6, "Bat") I added another animal , Bat to the end of the array.
z = sorted(animals) I sorted the array according to alphabetical number.
print(z) I displayed the sorted array .