As a project manager of Permagam Construction you want to plan renting a fleet of 25 cu yd tractor-scrapers and have them hauling between the pit and a road construction job. The haul road is a rutted dirt road that deflects slightly under the load of the scraper. There is a slight grade of 5% from the pit at the fill location. The return road is level. The haul distance to the dump location is 0.90 miles and the return distance is 0.75 miles. The scrapers are push loaded in the pit. The cycle time for the pusher is 1.5 minutes and the cycle time for the scrapers is 8 minutes. Assume that the weight on the wheels is 75 tons (full) and 50 tons
(empty). (Use Tables 14.1 and 14.2)
What are the rolling resistances and grade resistance?
What are the effective grades?
How many scrapers you recommend to be rented? Explain.
What is the production of the system in case 4 scrapers would be rented?
Answers
Answer:
See explaination
Explanation:
Rolling resistance which in some occassions can be called rolling friction or rolling drag, is the force resisting the motion when a body rolls on a surface.
In order to calculate our rolling resistance, there should be a force.
Please kindly check attachment for the step by step solution of the given problem.
Related Questions
Estimate the theoretical fracture strength of a brittle material if it is known that fracture occurs by the propagation of an elliptically shaped surface crack of length 0.28 mm and that has a tip radius of curvature of 0.002 mm when a stress of 1430 MPa is applied.
Answers
Answer:
theoretical fracture strength = 16919.98 MPa
Explanation:
given data
Length (L) = 0.28 mm = 0.28 × 10⁻³ m
radius of curvature (r) = 0.002 mm = 0.002 × 10⁻³ m
Stress (s₀) = 1430 MPa = 1430 × 10⁶ Pa
solution
we get here theoretical fracture strength s that is express as
theoretical fracture strength = [tex]s_{0} \times \sqrt{\frac{L}{r} }[/tex] .............................1
put here value and we get
theoretical fracture strength = [tex]1430 \times 10^6\times \sqrt{\frac{0.28\times 10^{-3}}{0.002\times 10^{-3}} }[/tex]
theoretical fracture strength = [tex]16919.98 \times 10^6[/tex]
theoretical fracture strength = 16919.98 MPa
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?
Answers
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]
Using a set of values from 0 to 5, perform the following unions using union-by-size. Show the result of each union. When sizes are the same, make the second tree a child of the first tree. Notice the finds return roots, and the union will union the roots. union(find(0),find(1)) union(find(3),find(4)) union(find(5),find(1)) union(find(2),find(5)) union(find(3),find(2)) 10 points
Answers
Answer:
Explanation:
Please kindly go through the attached file for a step by step approach to the solution of this problem
When will stemuless checks come I was told not to file my taxes because I get a pension check every month direct deposit money into my bank account is this true I don't have to do my taxes because of this
Answers
The IRS and the U.S Department of the treasury declared that social security recipients are not required to file a simple tax return to receive stimulus payments under the CARES Act.
Explanation:
Due to the impact of corona virus problem, the CARES Act calls for stimulus payment to be sent to Americans based on their gross income.
The Social security recipients are not required to file a tax return and do not take action and they will receive the payments directly to their bank accounts.
The Automatic payments will begin by next week. The eligible taxpayers who filed tax returns for 2019 or 2018 and chose direct deposit for their refund will automatically receive a stimulus payment of $1,200 for individuals or $2,400 for married couples and $500 for each qualifying child.
What is the frequency response of the stable, causal LTI system defined by the differential equation:fraction numerator d squared y (t )over denominator d t squared end fraction plus 6 fraction numerator d y (t )over denominator d t end fraction plus 2 y (t )equals fraction numerator d x (t )over denominator d t end fraction plus 4 x (t )Use Matlab syntax for your response, assuming w is the frequency vector. Make sure you use parentheses correctly (try plotting your code in Matlab)
Answers
Answer:
Explanation:
first convert difference equation to transfer function form,
apply laplase transform to difference equation
s2Y(s) + 6 * s * y(s) + 2 * y(s) = s * X(s) + 4 * X(s)
(s2 + 6s+2) * y(s) = (s+4)*X(s)Y s)
Lets write below code in matlab command prompt
lets use lodspace to create values from 10^-1 to 10^5 and use freqs to plot frequency response of above system with frequency w
>> n=[1 4];
>> d=[1 6 2];
>> w = logspace(-1,5);
>> freqs(n,d,w)
Attached is the written solution and the MATLAB diagram
2. Similar to problem 1, assume your computer system has a 32-bit byte-addressable architecture where addresses and data are each 32 bits. It has a 16K-byte (16,384 bytes) direct-mapped cache, but now the block size is 32 bytes. Answer the following question to observe how the design change impacts the cache size. [10pts]
Answers
Question:
The question is not complete. The question to answer was not added. See below the possible question and the answer.
a. How many blocks are in the cache with this new arrangement?
b. Calculate the number of bits in each of the Tag, Index, and Offset fields of the memory address.
C. Using the values calculated in part b, what is the actual total size of the cache including data, tags, and valid bits?
Answer:
(a) Number of blocks = 512 blocks
(b) Tag is 18
(c) Total size of the cache = 8388608 bytes
Explanation:
a .
block size = 32 bytes
cache size = 16384 bytes
No.of blocks = 16384 / 32
No.pf blocks = 512 blocks
b.
Total address size = 32 bits
Address bits = Tag + Line index +block offset
Block Size = 32 bytes.
So block size = 25 bytes.
Hence Offset is 5
No . of Cache blocks = 512 blocks = 29 blocks
Hence line offset is 9
We know that Address bits = Tag + Line index +block offset
So , 32 =tag+9+5
tag = 32-(9+5)
So Tag is 18
c.
Data bits = 32 bits
Tag=18 bits
Valid bit is 1 bit
so Total cache size = 25+218+20
= 223
=8388608 bytes
3) A mixture of nitrogen and oxygen (xN2=0.7) behaves as an ideal gas mixture. 50 moles of this mixture at 1 bar and 25 °C are fed into an initially-empty, rigid, diathermal vessel causing the pressure in the vessel to reach 1 bar. Assuming the surroundings are also at 25 °C, calculate the heat transfer needed for the gas mixture in the vessel to be at 25 °C. At this temperature, Cp for nitrogen = 1.040 J/(g K) and Cp for oxygen = 0.918 J/(g K).
Answers
Answer:
435.032 kj
Explanation:
We can describe Heat transfer as a discipline of thermal engineering that concerns the generation, use, conversion, and exchange of thermal energy between physical systems.
Please refer to the attached file for the detailed step by step solution of the given problem
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.
Answers
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
Read more about Aquifers at; https://brainly.com/question/1052965
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.
Answers
Answer:
Check the explanation
Explanation:
Kindly check the attached image below for the full step by step explanation to your question.
This is a multi-part question. Once an answer is submitted, you will be unable to return to this part As steam is slowly injected into a turbine, the angular acceleration of the rotor is observed to increase linearly with the time t Know that the rotor starts from rest at t = 0 and that after 10 s the rotor has completed 20 revolutions.
Determine the angular velocity at t20 s. (You must provide an answer before moving on to the next part)
The angular velocity is [ ] rpm.
Answers
Answer:
60 rpm
Explanation:
At t = 0,
Angular speed = 0
At t = 10 sec
Angular speed = 20/10 = 2 rev/s
Average speed = (2 - 0)/2 = 1 rev/s
= 1 x 60 = 60 rpm
A square power screw has a mean diameter of 30 mm and a pitch of 4 mm with single thread. The collar diameter can be assumed to be 35 mm. The screw is to be used to lift and lower a load of 7 kN. A coefficient of friction of 0.05 is to be used for friction at the thread and at the collar. Determine the following: (i) Torque required to raise the load, TR, (Equation 8.1) (ii) Torque required to lower the load, TL, (Equation 8.2) (iii) A conservative estimate of self locking condition is to set TL in equation 8.2 to zero and calculate the minimum coefficient of friction to ensure self locking. What is the minimum coefficient of friction to ensure self locking
Answers
Answer:
i) The torque required to raise the load is 15.85 N*m
ii) The torque required to lower the load is 6.91 N*m
iii) The minimum coefficient of friction is -0.016
Explanation:
Given:
dm = mean diameter = 0.03 m
p = pitch = 0.004 m
n = number of starts = 1
The lead is:
L = n * p = 1 * 0.004 = 0.004 m
F = load = 7000 N
dc = collar diameter = 0.035 m
u = 0.05
i) The helix angle is:
[tex]tan\alpha =\frac{L}{\pi *d_{m} } =\frac{0.004}{\pi *0.03} \\\alpha =2.43[/tex]
The torque is:
[tex]T=F\frac{d_{m} }{2} (\frac{\pi *u*d_{m}+L }{\pi *d_{m}-uL } )+(u_{c} F+\frac{d_{2} }{2} )=7000*\frac{0.03}{2} (\frac{\pi *0.05*0.03+0.004}{\pi *0.03-0.05*0.004} )+(0.05*7000*\frac{0.035}{2} )=15.85Nm[/tex]
ii) The torque to lowering the load is:
[tex]T=7000*\frac{0.03}{2} (\frac{\pi *0.05*0.03-0.004}{\pi *0.03+0.05*0.004} )+(0.05*7000*\frac{0.035}{2} )=6.91Nm[/tex]
iii)
[tex]T=F\frac{d_{m} }{2} (\frac{u*\pi *d_{m}-L}{\pi *d_{m}+uL} )+u_{c} *F*\frac{d_{c}}{2}\\ 0=F\frac{d_{m} }{2} (\frac{u*\pi *d_{m}-L}{\pi *d_{m}+uL} )+u_{c} *F*\frac{d_{c}}{2}\\F\frac{d_{m} }{2} (\frac{u*\pi *d_{m}-L}{\pi *d_{m}+uL} )=-u_{c} *F*\frac{d_{c}}{2}\\\frac{d_{m} }{2} (\frac{u*\pi *d_{m}-L}{\pi *d_{m}+uL} )=-u_{c} *\frac{d_{c}}{2}\\\\\frac{0.03}{2} (\frac{u*\pi *0.03-0.004}{\pi *0.03+u*0.004} )=-0.05*\frac{0.035}{2}[/tex]
Clearing u:
u = -0.016
Write a program named CheckZips that is used by a package delivery service to check delivery areas. The program contains an array that holds the 10 zip codes of areas to which the company makes deliveries. Prompt a user to enter a zip code, and display a message indicating whether the zip code is in the company’s delivery area.
Answers
Answer:
# list of 10 zip codes assigned to zip
zips = ["12789", "54012", "54481", "54982", "60007", "60103", "60187", "60188", "71244", "90210"]
# user is prompt to enter zip code and assigned to user_zip
user_zip = input("Enter your zip code: ")
# if else statement to check if user input is available for delivery
# if statement check if user zip is in zip, if it is, it display
# delivery is okay to specified zip
if user_zip in zips:
print("Delivery to {} ok.".format(user_zip))
# else it display no delivery to such zip code
else:
print("Sorry - no delivery to {}.".format(user_zip))
Explanation:
The question doesn't specify programming language to use. Since no programming language was stated, the problem was solved using Python3. List structure is the equivalent of array in Python.
Assumption was also made on the array holding 10 zip codes of areas to which the company make deliveries.
The program first initialized a list of 10 zip codes and assigned it to zips. Then it prompt the user to enter a zip code which is assigned to user_zip.
Then if-else statement is used to check if user inputted zip is available with the zips variable.
If it is available, "Delivery ok" is displayed to the user else "no delivery" is displayed to the user.
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.
Answers
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
Water flows inside a smooth circular thin-walled tube of diameter D = 25 mm at a mass flow rate of 50 g/s. Outside of the tube, air moves in cross flow over the tube at a velocity of V = 20 m/s and a temperature of T[infinity] = 10°C. If the mean temperature of the water is Tm = 50°C, determine (a) The Darcy friction factor for the water flow inside the tube
Answers
Answer:
See explaination
Explanation:
We can say that that the The Darcy Friction factor or Equation is a theoretical equation that predicts the frictional energy loss in a pipe based on the velocity of the fluid and the resistance due to friction. It is used almost exclusively to calculate head loss due to friction in turbulent flow.
Please kindly check attachment for the step by step solution of the given problem.
4. A 25 km2 watershed has a time of concentration of 1.6 hr. Calculate the NRCS triangular UH for a 10-minute rainfall event and plot it. Determine the runoff hydrograph for a 30-minute storm where there is 4 cm of runoff in the first 10 min, 2.5 cm of runoff in the second 10 min, and 2 cm of runoff in the third 10 min. Plot the runoff hydrograph for each 10-minute rainfall excess along with the aggregated (total) runoff hydrograph on the same axes. Also report the peak flow of the aggregated runoff hydrograph. Also report the total volume of runoff. You do not need to report any tables of data.
Answers
Answer:
The NRCS triangular UH for a 10-minute rainfall is Qp = 49.84 m³/s
Peak flow of the aggregated runoff hydrograph is 420.58 m³/s
The total volume of runoff is 2125000 m³/s
Explanation:
We have
A = 25 km²
tr = 10 min = 1/6 hr
tc = 1.6 hr
lag time = 0.6 tc = 0.96 hr
Tp = tr/2 + 0.6 tc = 1/12 + 0.96 = 1.043 hr
Qp = 2.08×25/1.043 = 49.84 m³/s
Tb = 8/3×Tp = 8/3×1.043 = 2.782 hr
Since the area is
Time (min) Runoff (cm) Volume of runoff m³
0 0 0
10 4 1000000 m³
20 2.5 625000 m³
30 2 500000 m³
Total volume of runoff = 1000000 + 625000 + 500000 = 2125000 m³/s
For the 1st 10 minutes, we have
A = 25 km²
tr = 30 min = 1/2 hr
tc = 1.6 hr
lag time = 0.6 tc = 0.96 hr
Tp = tr/2 + 0.6 tc = 1/4 + 0.96 = 1.21 hr
Qp = 2.08×25×4/1.043 = 197.92 m³/s
Tb = 8/3×Tp = 8/3×1.21 = 3.227 hr
For the 2nd 10 minutes, we have
A = 25 km²
tr = 30 min = 1/2 hr
tc = 1.6 hr
lag time = 0.6 tc = 0.96 hr
Tp = tr/2 + 0.6 tc = 1/4 + 0.96 = 1.21 hr
Qp = 2.08×25×2.5/1.043 = 123.7 m³/s
Tb = 8/3×Tp = 8/3×1.21 = 3.227 hr
For the 3rd 10 minutes, we have
A = 25 km²
tr = 30 min = 1/2 hr
tc = 1.6 hr
lag time = 0.6 tc = 0.96 hr
Tp = tr/2 + 0.6 tc = 1/4 + 0.96 = 1.21 hr
Qp = 2.08×25×2.5/1.043 = 98.96 m³/s
Tb = 8/3×Tp = 8/3×1.21 = 3.227 hr
Peak flow of aggregate runoff is given by
Qp (total) = 98.96 + 123.7 +197.92 = 420.58 m³/s
Total volume of runoff is given by
Total volume of runoff = 1000000 + 625000 + 500000 = 2125000 m³/s
A cylinder fitted with a movable piston contains water at 3 MPa with 50% quality, at which point the volume is 20 L. The water now expands to 1.2 MPa as a result of receiving 600 kJ of heat from a large source at 300◦C. It is claimed that the water does 124 kJ of work during this process. Is this possible?
Answers
Answer:
The process is possible:
Explanation:
We are going to find out if the entropy generated is greater than 0, if it is greater than 0, then the process is feasible. If it is not, the process is not feasible.
[tex]P_{1} = 3 MPa[/tex]
[tex]x_{1} = 50 % = 0.5[/tex]
[tex]V_{1} = 20 L = 0.02 m^{3}[/tex]
[tex]P_{2} = 1.2 MPa[/tex]
[tex]T_{H} = 300^{0} C = 573 K[/tex]
Received heat energy, [tex]Q_{12} = 600 kJ[/tex]
Work done, [tex]W_{12} = 124 kJ[/tex]
At state 1, using the steam table:
[tex]T_{1} = T_{s} = 233.9^{0} C\\v_{f1} = 0.001216 m^{3} /kg\\v_{fg1} = 0.06546m^{3} /kg\\u_{f1} = 1004.76 kJ/kg\\u_{fg1} = 1599.34 kJ/kg\\s_{f1} = 2.6456 kJ/kg-K\\s_{fg1} = 3.5412kJ/kg-K[/tex]
[tex]v_{1} = v_{f1} + x_{1} * v_{fg1}[/tex]
[tex]v_{1} = 0.001216 + 0.5*(0.06546)\\v_{1} = 0.03395 m^{3} /kg[/tex]
[tex]M = \frac{V_{1} }{v_{1} } \\M = 0.02/0.03395\\M = 0.5892 kg[/tex]
[tex]u_{1} = u_{f1} + x_{1} * u_{fg1}\\u_{1} = 1004.76 + 0.5*1599.34\\u_{1} = 1804.43 kJ/kg[/tex]
[tex]s_{1} = s_{f1} + x_{1} * s_{fg1}\\s_{1} = 2.6456 + 0.5*3.5412\\s_{1} = 4.4162 kJ/kg[/tex]
[tex]Q_{12} = m(u_{2} - u_{1} ) + W_{12} \\600 = 0.5892(u_{2} -1804.43) + 124\\[/tex]
Solving for u₂
[tex]u_{2} = 2612.3 kJ/kg[/tex]
Since P₂ = 1.2 MPa, u₂ = 2612.2 kJ/kg,
then from steam table, T₂ = 200°C, S₂ = 6.5898 kJ/kg-K
The entropy generated will be:
[tex]\triangle S = m(S_{2} -S_{1} ) - \frac{Q_{12} }{T_{H} }\\ \triangle S= 0.5892(6.5898 - 4.4162) - \frac{600 }{573 }\\ \triangle S =0.233 kJ/K[/tex]
Since ΔS > 0, this process is possible
Answer:
Yes it is possible
Explanation:
Attached is the solution
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
Answers
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.
A 500 turn coil is wound on an iron core. When a 120Vrms 60Hz voltage is applied to the coil, the current is 1A rms. Neglect the resistance of the coil. Determine the reluctance of the core. Given that the cross-sectional area of the core is 5cm2 and the length is 20cm, determine the relative permeability of the core material.
Answers
Answer:
R = 7.854 x 10⁵ anpert turns / Wb
Relative permeability = 405.3
Explanation:
Detailed explanation is given in the attached document.
Answer: 85398.16, 405.28473.
Explanation:
We are given that the number of turns on the core is 500 2 , the cross sectional area is :
A=5cm^{2}({1meter}/{100cm})^{2} =0.0005meters^{2}
and the length of the core is l=20cm ({1meter}/{100cm})= 0.2meters .
In this solution, we are meant to neglect the resistance of the coil , and the current through the coil is I=1amPrms when the voltage applied across it is:
V=120voltsrms at f=60Hz. From this, We can calculate the inductance(L) whiof the coil (a coil have an inductance value of one Henry when an electromotive force of one volt is induced in the coil were the current flowing through the said coil changes at a rate of one ampere/second).
The natural frequency of the applied voltage is:
ω =2π f=2π(60)=120π{radians}/{second} .
The inductive reactance of the coil is equal to X=ω L=120π L . We then know that current is :
I=V/X
=I =20/120πL
L=120/120π
=1/π henries .
For reluctance (R) (which is a unit measuring the opposition to the flow of magnetic flux within magnetic materials and is analogous to resistance in electrical circuits). Looking at the relationship between inductance and reluctance . You will note that it is :
L=n^2/R .
We can use this relationship to find reluctance for our closed iron core coil :
L=1/π = 500^2/R
R=500^2π =785398.16{amps}/{volt-seconds}}
We can therefore use the other equation for reluctance.
R=1/μ A= 1/μA or μRA
To calculate the relative permeability of the core :
R=500^2π
=0.2/(4π ×10^-7)/ μR(0.0005)
μR=0.2/(4π ×10^-7)/ 500^2π(0.0005)
= 405.28473
An activated sludge plant receive 5.0 MGD of wastewater with a BOD of 220 mg/L. The primary clarifier removes 35% of the BOD. The sludge is aerated for 6 hr. The food-to-microorganism ratio is 0.30. The recirculation ratio is 0.2. The surface loading rate of the secondary clarifier is 800 gal/day-ft2. The final effluent has a BOD of 15 mg/L. What are the (a) BOD removal efficiency of the activated sludge treatment processes (secondary BOD removal), (b) aeration tank volume, (c) MLSS, and (d) secondary clarifier surface area? If 0.5 pound of oxygen is required for each pound of BOD entering the aeration tank and the density of air is approximately 0.075 lb/ft3, and the air is 20.9% oxygen by volume, calculate air requirements per day.
Answers
Answer:
(a) BOD removal efficiency = 89.51%
(b) Aeration tank volume = 4732m³
(c) MLSS = 1706.669 mg/L
(d) Secondary clarifier surface area 6250ft²
(e) Air requirement = 12930.284 lb
Explanation:
See the attached file for explanation. This is continuation from page 3 of the attached file.
Air required = 172403.792 ft3
since, density of air = 0.075lb/ft3
air required = 0.075*172403.792 lb
= 12930.284 lb
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.
Answers
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
Calculate the length of a metal cylinder while it is subjected to a tensile stress of 10,000 psi. You are given the following data: Original length = 1 in Original cross-sectional area = 0.1 in2 Yield strength, σy = 9 ksi Young’s modulus, E = 1000 ksi
Answers
Answer:
length of cylinder can not calculated
Explanation:
given data
tensile stress = 10,000 psi
Original length = 1
Original cross-sectional area = 0.1 in²
Yield strength, σy = 9 ksi
Young’s modulus, E = 1000 ksi
solution
we can see that here that applied stress is greater than yield stress of material that is express
1000 ksi > 9 ksi
so here hooks law and strain relation is not working
so length of cylinder can not calculated
as stress applied 10000 psi
Under 10,000 psi tensile stress, the metal cylinder elongates by 0.01 inches, resulting in a final length of 1.01 inches.
To calculate the elongation (change in length) of the metal cylinder under tensile stress, we can use Hooke's Law, which states that the elongation [tex](\( \Delta L \))[/tex] is directly proportional to the applied tensile stress [tex](\( \sigma \))[/tex] and the original length [tex](\( L_0 \))[/tex], and inversely proportional to the Young's modulus [tex](\( E \))[/tex]:
[tex]\[ \Delta L = \frac{\sigma \cdot L_0}{E} \][/tex]
Given:
- Original length [tex](\( L_0 \))[/tex] = 1 in
- Applied tensile stress [tex](\( \sigma \))[/tex] = 10,000 psi
- Young's modulus [tex](\( E \))[/tex] = 1000 ksi = 1,000,000 psi
Substitute the values into the formula:
[tex]\[ \Delta L = \frac{10,000 \times 1}{1,000,000} \][/tex]
[tex]\[ \Delta L = \frac{10,000}{1,000,000} \][/tex]
[tex]\[ \Delta L = 0.01 \, \text{in} \][/tex]
So, the elongation of the metal cylinder under the given tensile stress is 0.01 inches.
To find the final length, we add the elongation to the original length:
[tex]\[ \text{Final length} = \text{Original length} + \Delta L \][/tex]
[tex]\[ \text{Final length} = 1 + 0.01 \][/tex]
[tex]\[ \text{Final length} = 1.01 \, \text{in} \][/tex]
Therefore, the final length of the metal cylinder under the given tensile stress is 1.01 inches.
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?
Answers
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
A system consists of N very weakly interacting particles at a temperature T sufficiently high so that classical statistical mechanics is applicable. Each particle has a mass m and is free to perform one-dimensional oscillations about its equilibrium position. Calculate the heat capacity of this system of particles at this temperature in each of the following cases:
Answers
Answer:
the restoring force is = 3/4NKT
Explanation:
check the attached files for answer.
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?
Answers
Answer:
Horse power = 167.84 hp
Explanation:
Horsepower is calculated using the formula;
P = T * w
See the attached file for the calculation
Calculate the headloss through a filter bed consisting of 30.0 in. of stratified sand with the gradation given below. Assume a filtration rate of 5.0 gpm/ft2, a clean bed porosity of 0.42 and a temperature of 40i F.
U.S. Sieve Number 50
40 30 20 18 16
Sieve Opening (mm) 0.297
0.420 0.590 0.840 1.000 1.190
Percent Passing (by wt.) 0.10
6.50 22.00 76.00 90.00 98.00
Answers
Answer:
1.5258m
Explanation:
Please see attachment
(A) Develop the activity sequence model and determine the normal time for the following work activity:
An assembly worker on a production line obtains an Allen key within reach, positions it 15 cm (6 in) onto a bolt head, cranks it 7 times to seat the bolt, and then sets the key aside.
(B) Express the MTM-1 motion elements in (a) as one or more MOST activity sequence models with index numbers.
(i) Determine the normal times in TMUs for these sequence activity models.
(ii) What is the total time for this (these) sequence activity model(s) in secs?
Answers
Answer:
Activity sequence model = A1B0G1A0B0P3F16A1B0P1A0
Tn= 10(1 + 1 + 3 + 16 + 1 + 1) = 10(23) = 230 TMU (8.3 sec)
The purification of hydrogen gas by diffusion through a palladium sheet. Compute the number of kilograms of hydrogen that pass per hour through a 5-mm-thick sheet of palladium having an area of 0.20 m2 at 500°C. Assume a diffusion coefficient of 1.0 × 10-8 m2/s and that the concentrations at the high and low pressure sides of the plate are 2.4 and 0.6 kg of hydrogen per m3 of palladium. Assume steady state conditions.
Answers
Answer:
The mass of the hydrogen for one hour would be 0.0039168 kg/hr
Explanation:
The concentration and the distance at concentration point is calculated before calculating the mass of hydrogen. The attached images show a clear explanation;
The mass of the hydrogen for one hour would be 0.0039168 kg/hr. The derivation of the diffusion coefficient has been attached in the image below:
According to Fick's law, the mass of a substance dM is proportional to the concentration gradient grad c of this substance as it diffuses in time dt over a surface dF normal to the diffusion direction: dM = D grad c dF dt.
Physically, the diffusion coefficient, therefore, suggests that for a concentration gradient of unity, the mass of the substance diffuses through a unit surface in a unit of time. A square meter per second corresponds to D in the SI system. Physical constants such as temperature, pressure, and the size of the dispersing substance's molecules all affect the diffusion coefficient.
Learn more about diffusion coefficient here:
https://brainly.com/question/31430680
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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?
Answers
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
Steam enters an insulated pipe at 200 kPa and 200°C and leaves at 150 kPa and 150°C. The inlet-to-outlet diameter ratio for the pipe is D1/D2 = 1.80. Determine the inlet and exit velocities of the steam.
Answers
Answer:
Inlet and exit velocities are 143.71 m/sec and 465.697 m/sec
Explanation:
At inlet of the pipe
[tex]P_1=150kPa[/tex] and [tex]T_1=150^{\circ}C[/tex]
At this pressure and temperature from steam table.
[tex]h_1=2870kj/kg[/tex] and [tex]s_1=7.508kj/kgK[/tex]
At pressure [tex]P_2=200kPa[/tex] and [tex]T_2=200^{\circ}C[/tex]
By steam table from interpolation method.
[tex]h_2=2776.38+(2768.80-2776.38)(\frac{150-100}{200-100})[/tex]
[tex]h_2=2772.59kj/kg[/tex]
[tex]Q=A_1V_1=A_2V_2[/tex]
[tex]D_1^2V_1=D_2^2V_2[/tex]
We have given [tex]\frac{D_1}{D_2}=1.80[/tex]
[tex]\frac{V_1}{V_2}=\frac{D_2^2}{D_1^2}[/tex]
[tex]\frac{V_1}{V_2}=(\frac{1}{1.80})^2[/tex]
[tex]V_1=0.3086V_2[/tex]
Now energy equation in the pipe
[tex]h_1+\frac{V_1^2}{2}=h_2+\frac{V_2^2}{2}[/tex]
[tex]2870.7\times 10^3+\frac{0.3086V_2^2}{2}=2772.59\times 10^3+\frac{V^2}{2}[/tex]
[tex]V_2=465.697m/sec[/tex]
[tex]V_1=0.3086\times 465.697=143.714m/sec[/tex]
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.
Answers
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.