Forward Error Correction (FEC) is a popular method of error control in networks where bandwidth utilization is a secondary issue. (FEC transmits information using a redundant coding that allows full recovery of the data even in cases where some parts of the transmission were not delivered). Can FEC fully replace the standard timeout/acknowledgment-based error control mechanisms? Explain your answer.
Answer: NO
Explanation:
FEC transmits information using a redundant coding that allows full recovery of the data even in cases where some parts of the transmission were not delivered.FEC cannot be always the alternative for standard timeout/acknowledgment-based error control mechanisms. One of the reasons is that it can recover full data only when some part of transmission is not delivered.
2. * Give a scatterplot of this data and comment on the direction, form and strength of this relationship. a. Determine the least-squares estimate equation for this data set. b. Give the r2, comment on what that means. c. Give the residual plot based on the least-squares estimate equation. d. Test if this least-squares estimate equation specify a useful relationship between commuting distance and commuting time.
Answer
The answer and procedures of the exercise are attached in the following archives.
Step-by-step explanation:
You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.
Explain the difference, on the basis of the test results, between the ultimate strength and the "true" stress at fracture.
Answer / Explanation:
On the basis of the test result, Ultimate strength which is mostly known as the ultimate tensile strength is the strength attached to the ability or capacity of a structural element or material used in the test to withstand elongation forces or pull force applied to it.
WHILE,
True stress at fracture can be classified as stress or load associated to the point where yielding or fracture occurred divided by the cross-sectional area at the yield point.
A series RC circuit contains a 0.5 microfarad capacitor and a 500 ohm resistor. The circuit is operating at a frequency of 2,000 Hz. What is the impedance of the circuit?
(A) 500 ohms
(B) 809 ohms
(C) 2,006 ohms
(D) 525 ohms
Answer:
[tex]\left | Z_{in}} \right | = 524.72 \approx 525[/tex]ohm
Explanation:
Given data:
capacitor = 0.5 micro farad
resistor = 500 ohm
Frequency = 2000 Hz
Impedance of any circuit is calculated by using following equation
[tex]Z_{in} = Resistor - \frac{j}{\omega_c}[/tex]
[tex] = 500 - \frac{j}{2\pi*2000*0.5 \mu}[/tex]
=500 - j 159.155
[tex]\left | Z_{in}} \right | = \sqrt{(500^2 + 159.155^2)}[/tex]
[tex]\left | Z_{in}} \right | = 524.72 \approx 525[/tex]ohm
2.73 A fan within an insulated duct delivers moist air at the duct exit at 358C, 50% relative humidity, and a volumetric flow rate of 0.4 m3 /s. At steady state, the power input to the fan is 1.7 kW. The pressure in the duct is nearly 1 atm throughout. Using the psychrometric chart, determine the temperature, in 8C, and relative humidity at the duct inlet.
Answer:
Temperature at 8C = 32 °C × 8 = 256
Relative humidity is = 60% × 8 of 256°C accordingly.
Explanation:
Considering the volumetric flow rate = 0.4 m³/s
Moist air delivered (Temperature T1 = 358 C)
With Relative humidity at duct outlet = 50%
Power input at steady rate = 1.7 KW
Pressure in the duct = 1 atmosphere
Temperature at 8C = ?
Relative humidity at the duct inlet = ?
Recalling that the value for specific enthalpy and specific volume is
Specific Enthalpy = h1 = 81 kj/kilogram of dry air
and Specific Volume = v1 = 0.9 m³/kg
Now, recalling the formula for mass flow rate,
We have, m = Volumetric flow rate / Specific volume
Therefore, 0.4 m³/s ÷ 0.9 m³/kg
= 0.44 kilogram / second
Recalling the enthalpy at inlet,
we have, h2 = h1 - p/m
Where h1 = Specific enthalpy
p = power input at steady rate
m = calculated mass flow rate
Now, if we substitute the values into the equation,
we have h2 = 81 kj/kilogram of dry air - 1.7 KW / 0.44 kilogram / second
h2 = 77.175 kj / kilogram of dry air.
Therefore, the properties of air at constant absolute humidity and specific enthalpy is 77.175 kj/kg.
Temperature at 8C = 32 °C × 8 = 256
Relative humidity is = 60% × 8 of 256°C accordingly.
A room in a building has the following characteristics: Sensible heat gain: 90,000 Btu/hr, Latent heat gain: 40,000 Btu/hr, Supply air: 3,600 cfm @ 55 deg. F db
Air ventilation: 700 cfm, The room is kept at 78 deg. F db and 45% rh. Outdoor air is 92 deg. F db 76 deg. F wb.
What is the wet bulb wb air temperature entering EAT and wb temperature of air leaving the cooling coils LAT in the room?
[LAT = 55 deg. F db, given, need to calculate EAT db, LAT wb. The EAT db of the air entering the cooling coils is the mixed temperature = [(room temp db)(2,900) cfm + (outside air temp db)(700)]/3,600 cfm. Find EAT wb and h of EAT from chart. Find h of outside air from chart. Then Q for outside air = (4.5)(?h)(cfm) where cfm is outside air ventilation = about 26,800 Btu/hr (show this), then total Q = (sensible heat gain) + (latent heat gain) + (outside air heat gain) = about 157,000 Btu/hr (show this), then total Q = (4.5)(?h)(total cfm), then find (?h) from total Q, then find h of leaving air from ?h, then find wb temp of leaving air LAT wb from chart. We get EAT = 80.7 deg. F db/66.2 deg. F wb, LAT = 51.7 deg. F wb.]
Answer
The answer and procedures of the exercise are attached in the following archives.
Step-by-step explanation:
You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.
Air enters the compressor of an air-standard Brayton cycle with a volumetric flow rate of 60 m3/s at 0.8 bar, 280 K. The compressor pressure ratio is 20, and the maximum cycle temperature is 2100 K. For the compressor, the isentropic efficiency is 92% and for the turbine the isentropic efficiency is 95%. Determine
(a) the net power developed, in MW
(b) the rate of heat addition in the combustor, in MW
(c) the thermal efficiency of the cycle
To determine the net power developed, the rate of heat addition, and the thermal efficiency for an air-standard Brayton cycle with specified inlet conditions, compressor pressure ratio, maximum cycle temperature, and efficiencies, detailed thermodynamic calculations considering the given parameters are required.
Explanation:The Brayton cycle is a thermodynamic cycle that describes the workings of a constant-pressure heat engine, commonly used in jet engines and gas turbine engines. The cycle involves air entering a compressor, being compressed, then heated in a combustor before expanding through a turbine, and finally being released to the environment. This question asks to calculate the net power developed, the rate of heat addition in the combustor, and the thermal efficiency of the cycle given specific conditions and efficiencies of the compressor and turbine.
Unfortunately, without the necessary thermodynamic equations and properties of air provided, along with the specific steps and intermediate states of the cycle, it is impossible to perform the detailed calculations needed to answer this question precisely. Typically, solving this problem would involve using the thermodynamic relations for an ideal gas, the isentropic process equations for the compressor and turbine, and the definitions of isentropic efficiency. The conservation of energy principle would be applied to find the net work output and heat added, and subsequently, the thermal efficiency would be calculated.
As such, this question requires a detailed understanding of thermodynamics, specifically the principles governing the Brayton cycle and the equations for calculating work done, heat transfer, and efficiency in thermodynamic cycles.
Lot ABCD between two parallel street lines is 350.00 ft deep and has a 220.00-ft frontage (AB) on one street and a 260.00-ft frontage (CD) on the other. Interior angles at A and B are equal, as are those at C and D. What distances AE and BF should be laid off by a surveyor to divide the lot into two equal areas by means of a line EF parallel to AB?
Answer:
240.83 ft
Explanation:
The distances AE and BF will be equal = 182.58 ft
The area of the lot will be the product of the depth and the average of the two frontages
= ( 350 * (220 + 260)/2) = 84000 ft
Half of the area becomes 42000 ft
<A = arctan (20/350) = 3. 3°
Hence 42000=h/2*(220 + 220 + 2h*tan3.3)
Solving, we obtain h= 182.28ft
EF = 220 + (2*182.28*tan 3.3)
= 240.83ft
The nozzle bends the flow from vertically upward to 30 degrees with the horizontal as it discharges the water (at 20 degrees C) to the atmosphere at V = 125 ft/s.
The volume of water within the nozzle itself (above the flange) is 100 lb.
Find the horizontal and vertical forces that must be applied to the flange (by the pipe below it) to hold it in place.Area of flange = 1.0 ft^2Area of nozzle = 0.50 ft^2Volume of area above flange = 1.8 ft^3Vertical height from flange to nozzle = 2 ft
Answer
The answer and procedures of the exercise are attached in the following archives.
Step-by-step explanation:
You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.
Answer:
Fy = -11267.294 lbf
Explanation:
given data
nozzle flow = 30 degrees
discharges the water = 20 degrees C
volume of water = 100 lb
Area of flange = 1.0 ft²
Area of nozzle = 0.50 ft²
Volume of area flange = 1.8 ft³
Vertical height flange to nozzle = 2 ft
solution
we will apply here continuity equation that is
A1 × V1 = A2 × V2 .............1
put here value and we get volume V1 that is
V1 = [tex]\frac{0.5\times 125}{1}[/tex]
V1 = 62.5 ft/s
and
now we will apply here Bernoulli equation that is
[tex]\frac{p1}{\gamma 1} + \frac{V1^2}{2g} + z1 = \frac{p2}{\gamma 2} + \frac{V2^2}{2g} + z2[/tex] .............................2
put here value and we will get
p1 = 0 + [tex]\frac{62.4}{2\times 32.2}(125^2 - 62.5^2) + 62.4 (2)[/tex]
p1 = 11479.614 psf
so here moment in y will be
∑ Fy = m [ (Vo)y - (Vi)x ]
so here we get
p1 ×A1 + Fy - Wn - Ww = [tex]\rho[/tex] Q [ V2 × sin30 - V1 ]
put here value and we get Fy
1147.614 × 1 + Fy - 100 - (62.4 × 1.8) = (1.94) × (0.5 ×130) × (125sin30 - 62.5)
solve it we get
Fy = -11267.294 lbf
Depending on the environmental demands, there are different types of organizational structures, including __________.
Answer:
Functional (Centralized) Organization Divisional Organization Team-Based Organization Product-Based Organization Modular OrganizationMatrix OrganizationExplanation:
Organization structure:
refers to the idea of how people are supposed to work and coordinate in an organization to maintain a healthy and effective work environment.
There are various types of organizational structures which depends on several factors. There is no single best organization structure. Each structure has its own advantages and disadvantages. In order to select a structure the organization's vision, mission, culture values, goals are to be identified first.
Consider a Carnot refrigeration cycle executed in a closed system in the saturated liquid-vapor mixture region using 0.96 kg of refrigerant-134a as the working fluid. Is it known that the maximum absolute temperature in the cycle is 1.2 times the minimum absolute temperature, and the net work input to the cycle I s 22 kJ. If the refrigerant changes from saturated vapor to saturated liquid during the heat rejection process, determine he minimum pressure in the cycle.
Answer:
[tex]P_m_i_n= 356.9 KPa[/tex]
Explanation:
Coefficient of performance (COP) of refrigeration cycle is given by:
[tex]COP=\frac{1}{\frac{T_H}{T_L}-1 }[/tex]
We are given:
[tex]T_H=1.2T_L[/tex]
[tex]COP=\frac{1}{1.2-1 }[/tex]
COP= 5
We can also write Coefficient of performance (COP) of refrigeration cycle as:
[tex]COP_R=\frac{Q_L}{W_i_n}[/tex]
Amount of heat absorbed by low temperature reservoir can be found as:
[tex]Q_L=COP_R * W_i_n[/tex]
[tex]Q_L=5 * 22 KJ[/tex]
[tex]Q_L=110 KJ[/tex]
According to first law of thermodynamics amount of heat rejected by hot reservoir is given by:
[tex]Q_H=Q_L + W_i_n[/tex]
[tex]Q_H=110 KJ + 22 KJ[/tex]
[tex]Q_H=132 KJ[/tex]
We are given the mass of 0.96 kg. So,
[tex]q_H=\frac{Q_H}{m}[/tex]
[tex]q_H=\frac{132 KJ}{0.96Kg}[/tex]
[tex]q_H=137.5 KJ/Kg[/tex]
Since it is a saturated liquid-vapour mixture [tex]q_H=h_f_g[/tex].
[tex]q_H=h_f_g=137.5 KJ/Kg[/tex]
From Refrigerant 134-a tables [tex]T_H[/tex] at [tex]h_f_g=137.5 KJ/Kg[/tex] is 61.3 C. (We calculated this by interpolation)
Converting [tex]T_H[/tex] from Celsius to Kelvin:
[tex]61.3^{o} C+273 = 334.3^{o} K[/tex]
[tex]T_H= 334.3^{o} K[/tex]
We are given:
[tex]T_H=1.2T_L[/tex]
[tex]T_L=\frac{T_H}{1.2}[/tex]
[tex]T_L=\frac{334.3}{1.2}[/tex]
[tex]T_L=278.58^{o} K[/tex]
Converting [tex]T_L[/tex] from Kelvin to Celsius:
[tex]278.58^{o} K-273 = 5.58^{o} C [/tex]
[tex]T_L= 5.58^{o} C [/tex]
From Refrigerant 134-a tables [tex]P_m_i_n[/tex] at [tex]T_L=5.58^{o} C[/tex] is 356.9 KPa. (We calculated this by interpolation).
[tex]P_m_i_n= 356.9 KPa[/tex]
The minimum pressure in the Carnot refrigeration cycle is 356.9 kPa.
Using,
Coefficient of performance (COP) of the refrigeration cycle. The COP is a measure of how efficient the refrigeration cycle is.
It is given by the formula:
COP = 1 / ([tex]T_H[/tex] /[tex]T_L[/tex] - 1),
where
[tex]T_H[/tex] is the maximum absolute temperature and [tex]T_L[/tex] is the minimum absolute temperature.
We are told that [tex]T_H[/tex] is 1.2 times [tex]T_L[/tex].
COP = 1 / (1.2 - 1).
Simplifying
COP = 5.
We can also express the COP as the ratio of the heat absorbed by the low-temperature reservoir to the net work input.
Using this relationship,
The heat absorbed ([tex]Q_L[/tex]) is equal to 5 times the net work input, which is 5 * 22 kJ
= 110 kJ.
According to the first law of thermodynamics, the heat rejected by the hot reservoir ([tex]Q_H[/tex]) is equal to the sum of the heat absorbed ([tex]Q_L[/tex]) and the net work input (22 kJ).
So,
[tex]Q_H[/tex] = 110 kJ + 22 kJ
= 132 kJ.
Since we are given the mass of the refrigerant (0.96 kg), we can find the heat transfer per unit mass ([tex]q_H[/tex]) using QH/m.
[tex]q_{H}[/tex]= 132 kJ / 0.96 kg
= 137.5 kJ/kg.
Since the refrigerant is in a saturated liquid-vapor mixture state, [tex]q_H[/tex] represents the enthalpy change from liquid to vapor.
By looking up the tables for refrigerant-134a, we can find the corresponding temperature ([tex]T_H[/tex]) at [tex]q_H[/tex] = 137.5 kJ/kg.
After conversion to Kelvin, [tex]T_H[/tex] is found to be 334.3 K.
Using the relationship [tex]T_H[/tex] = 1.2[tex]T_L[/tex],
Dividing [tex]T_H[/tex] by 1.2,
[tex]T_L[/tex] = 334.3 K / 1.2 =
278.58 K.
Converting this back to Celsius, we get [tex]T_L[/tex] = 5.58 °C.
Finally, by looking up the tables for refrigerant-134a, we can find the minimum pressure ([tex]P_{min}[/tex]) at [tex]T_L[/tex] = 5.58 °C.
The value is determined to be 356.9 kPa.
So, the minimum pressure in the cycle is 356.9 kPa.
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Find vC(t) for t ≥ 0 (in V), assuming the switch was open for a long time when closed at t = 0.
Assume the capacitor and inductor are initially uncharged.
(1) 3 – 4e–2t + e–8t
(2) 2 – e–2t + 2e–8t
(3) 6 – 6e–4t + 2e–6t
(4) 6 – 6(1+ t)e–8t
(5) 3 + (1+ 4t)e–2t
(6) 3 + [–cos(4t) + 4sin(2t)]e–2t
(7) 3 + [–4cos(4t) + sin(t)]e–t
(8) 3 + [–2cos(4t) + 4sin(2t)]e–4t
(9) None of the above
Answer:
So, we have to analyze the whole circuit system provided, then we will get to know that its an RL-Circuit along with that we can easily pick the best option from the given, which is mentioned below:
Explanation:
Option (1) 3- 4e-2t+e-8t is the best option to chose from the given options.
Letm(t) = sinc2(t) andmh(t) be its Hilbert transform. Denote the upper side band modulatedwaveform asxUSB(t) =Am(t) cos(2000πt)−Amh(t) sin(2000πt)
(a) Sketch the Fourier transforms ofm(t) andmh(t).
(b) Sketch the Fourier transformX(f) of the complex baseband ofxUSB(t) with respectto carrier frequency 1000.
Answer:
The Hilbert transform we use in mathematics it usually apply on a function consisting of a real variable and produces another function of a real variable.
Explanation:
Water vapor at 6 MPa, 500°C enters a turbine operating at steady state and expands to 20 kPa. The mass flow rate is 3 kg/s, and the power developed is 2626 kW. Stray heat transfer and kinetic and potential energy effects are negligible.
Determine:
(a) the isentropic turbine efficiency and
(b) the rate of entropy production within the turbine, in kW/K.
Answer:
a)75.8%
b)2.517KW/K
Explanation:
Hello!
To solve this problem follow the steps below
Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)
through prior knowledge of two other properties such as pressure and temperature.
1. use thermodynamic tables to find the following variables.
a.enthalpy and entropy at the turbine entrance
h1=Enthalpy(Water;T=500;P=6000)
=3422KJ/kg
s1=Entropy(Water;T=500;P=6000)
=6.881KJ/kgK
b. enthalpy and ideal entropy at the turbine outlet
h2i=Enthalpy(Water;s=6.881;P=20)
=2267KJ/kg
s2i=s1=6.881KJ/kgK
2. uses the output power and the first law of thermodynamics to find the real enthalpy at the turbine's output
W=m(h1-h2)
h2=h1-W/m
h2r=3422-2626/3=2546.6KJ/kg
3.
find efficiency with the following equation
[tex]eficiency=\frac{h1-h2r}{h1-h2i}[/tex]
[tex]\frac{h1-h2r}{h1-h2i}=\frac{3422-2546.6}{3422-2267} =0.758=75.8%[/tex]
4.
find the real entropy at the turbine exit
s2=Entropy(Water;h=2546,6;P=20)=7.72KJ/kgK
5.Finally find the entropy generated, using the following equation
ΔS=m(s2-s1)=(3kg/s)(7.72 KJ/kgK-6.881 KJ/kgK)=2.517KW/K
One kilogram of ammonia initially at 8.0 bar and 50°C undergoes a process to 4.5 bar, 20°C while being rapidly expanded in a piston–cylinder assembly. Heat transfer between the ammonia and its surroundings occurs at an average temperature of 40°C. The work done by the ammonia is 40 kJ. Kinetic and potential energy effects can be ignored. Determine the heat transfer, in kJ, and the entropy production, in kJ/K.
The heat transfer for the process is -20.9 kJ, and the entropy production is 1 kJ/K.
Explanation:The heat transfer for the given ammonia process is -20.9 kJ. To calculate the entropy production, we can use the equation: ΔS = Q/T, where temperature T is in Kelvin. Given Q = 40 kJ, and the average temperature is 40°C, the entropy production is 1 kJ/K.
An engineer is considering time of convergence in a new Layer 3 environment design. Which two attributes must be considered? (Choose two)
A.Addition of a valid forwarding path
B.Loss of a valid forwarding path
C.SPT timers update
D.OSPF database updates
E.Forwarding table updates
Answer:
Options B and E
Explanation:
To give sustainable environmental design when considering time of convergence in layer 3, an engineer must consider the loss of a valid forwarding path and the table updates since these will determine whether the design becomes fit or not.
Technician A says that load-carrying ball joints should always have some play in them. Technician B says that follower ball joints should never have some play in them. Who is correct?
a. Technician A
b. Technician B
c. Both A and B
d. Neither A nor B
Answer:
The right answer is d.
Explanation:
The amount of play that a ball joint can have (if any) depends on the application of the joint itself and the suspension system. For each vehicle, the manufacturer must warn wich is the maximum tolerance of play. For some load-carrying ball joints, the amount of play should be minimum or null (like in some heavy-duty vehicles). Therefore technician A is wrong. In most cases in urban vehicles, it is acceptable some level of play for follower ball joints (these joints are normally unloaded). Therefore the technician B is also wrong. That means neither technicians A nor B statements are correct.
There are different functions of Technician. The two Technician are wrong so Neither A nor B is the right answer.
In Ball Joint Inspection , using the old rule of thumb which state that ball joints if more than .050 inches of play are worn and it does will not hold true for all vehicles.
There are some ball joints which does not have visible play while others can hold up to . 250 inch or more of play and still be work well.
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A steel rotating-beam test specimen has an ultimate strength of 120 kpsi. Estimate the life of the specimen if it is tested at
a completely reversed stress amplitude of 70 kpsi.
Answer:
104,576 cycles
Explanation:
Step 1: identify given parameters
Ultimate strength of steel ([tex]S_{ut}[/tex])= 120 Kpsi
stress amplitude ([tex]\alpha_{a}[/tex])= 70 kpsi
life of the specimen (N) = ?
[tex]N = (\frac{\alpha_{a}}{a})^\frac{1}{b}[/tex]
where a and b are coefficient of fatigue cycle
Step 2: calculate the the endurance limit of specimen
[tex]S_{e} = 0.5*S_{ut}[/tex]
[tex]S_{e}[/tex] = 0.5*120 = 60 kpsi
Step 3: calculate coefficient 'a'
[tex]a=\frac {(0.8XS_{ut})^2}{S_{e}}[/tex]
[tex]a=\frac {(0.8X120)^2}{60}[/tex]
[tex]a= 153.6 kpsi
Step 4: calculate the coefficient 'b'
[tex]b =-\frac{1}{3}log(\frac{f*S_{ut} }{S_{e}})[/tex]
[tex]b =-\frac{1}{3}log(\frac{0.8*120}{60})[/tex]
[tex]b =-0.0680
Step 5: calculate the life of the specimen
[tex]N=(\frac{\alpha_{a}}{a})^\frac{1}{b}[/tex]
[tex]N=(\frac{70}{153.6})^\frac{1}{-0.068}[/tex]
[tex]N=104,576 cycles [/tex]
∴ the life (N) of the steel specimen is 104,576 cycles
Consider each statement below and determine which are correct concerning dietary fiber. Select all that apply. View Available Hint(s) Select all that apply. Increasing the amount of fiber in your diet can aid in achieving and maintaining a healthy weight. Consuming a high-fiber diet most likely promotes the health of the digestive system. Fiber and other carbohydrates like starch and sugar are digested and absorbed in the same manner. Consuming a diet high in dietary fiber increases LDL "the bad" cholesterol. Most American women consume more than 20 g of fiber per day, and most American men consume more than 30 g per day.
Answer:
Increasing the amount of fiber in your diet can aid in achieving and maintaining a healthy weight. This is correct if you are consuming less than 25-30gms of fiber per day. exceding this limit won't be beneficial.
Consuming a high-fiber diet most likely promotes the health of the digestive system. This is correct. Fibers are important for the digestive system´s health, especially for intestines and colon.
Fiber and other carbohydrates like starch and sugar are digested and absorbed in the same manner. This is Incorrect. Fiber is absorbed and digested at a much slower rate than sugar or starch.
Consuming a diet high in dietary fiber increases LDL "the bad" cholesterol. This is incorrect. Consuming a diet high in dietary fiber would decrease the LDL.
Most American women consume more than 20 g of fiber per day, and most American men consume more than 30 g per day. This is incorrect. The data obtained by the University of California San Francisco said that currently the amount of fiber intake by Americans adults is about 15g a day, which is half the recommended amount.
Answer:
Consuming a high-fiber diet most likely promotes the health of the digestive system.
Increasing the amount of fiber in your diet can aid in achieving and maintaining a healthy weight.
Explanation:
A horizontal 2-m-diameter conduit is half filled with a liquid (SG=1.6 ) and is capped at both ends with plane vertical surfaces. The air pressure in the conduit above the liquid surface is 200kPa. Determine the resultant force of the fluid acting on one of the ends caps, and locate this force relative to the bottom of the conduit.
Answer:
Resultant force = 639 kN and it acts at 0.99m from the bottom of the conduit
Explanation:
The pressure is given as 200 KPa and the specific gravity of the liquid is 1.6.
The resultant force acting on the vertical plate, Ft, is equivalent to the sum of the resultant force as a result of pressurized air and resultant force due to oil, which will be taken as F1 and F2 respectively.
Therefore,
Ft = F1 + F2
According to Pascal's law which states that a change in pressure at any point in a confined incompressible fluid is transmitted throughout the fluid such that the same change occurs everywhere, the air pressure will act on the whole cap surface.
To get F1,
F1 = p x A
= p x (πr²)
Substituting values,
F1 = 200 x π x 1²
F1 = 628.32 kN
This resultant force acts at the center of the plate.
To get F2,
F2 = Π x hc x A
F2 = Π x (4r/3π) x (πr²/2)
Π - weight density of oil,
A - area on which oil pressure is acting,
hc - the distance between the axis of the conduit and the centroid of the semicircular area
Π = Specific gravity x 9.81 x 1000
Therefore
F2 = 1.6 x 9.81 x 1000 x (4(1)/3π) x (π(1)²/2)
F2 = 10.464 kN
Ft = F1 + F2
Ft = 628.32 + 10.464
Ft = 638.784 kN
The resultant force on the surface is 639 kN
Taking moments of the forces F1 and F2 about the centre,
Mo = Ft x y
Ft x y = (F1 x r) + F2(1 - 4r/3π)
Making y the subject,
y = (628.32 + 10.464(1 - 4/3π)/ 638.784
y = 0.993m
A 0.5m diameter sphere containing pollution monitoring equipment is dragged through the Charles River at a relative velocity of 10m/s. The sphere has a specific gravity of 0.5, is fully submerged, and tethered to the towing device by a 2m cable. What is the angle the towing cable makes with the horizontal? Assume the water is at 10°C.
Answer:
[tex]\phi = 155.57[/tex]
Explanation:
from figure
taking summation of force in x direction be zero
[tex]\sum x = 0 [/tex]
[tex]F_D = Tsin \theta[/tex] .....1
[tex]\frac{c_d \rho v^2 A}{2} =Tsin \theta[/tex]
taking summation of force in Y direction be zero
[tex]F_B - W- Tcos \theta[/tex]
[tex]T = \frac{F_B -W}{cos \theta}[/tex] .........2
putting T value in equation 1
[tex]F_D - \frac{F_B -W}{cos \theta} sin\theta[/tex]
[tex]F_D = \rho g V ( 1 -Sg) tan \theta [/tex].........3
[tex]F_D = \rho g [\frac{\pi d^3}{6}] ( 1 -Sg) tan \theta [/tex]
[tex]tan \theta = \frac{6 c_D \rho v&2 A}{ 2 \rho g V \pi D^3 (1- Sg)}[/tex]
Water at 10 degree C has kinetic viscosity v = 1.3 \times 10^{-6} m^2/s
Reynold number
[tex]Re = \frac{ VD}{\nu} = \frac{10\times 0.5}{1.3 \times 10^{-6}} = 3.84 \times 10^6[/tex]
so for Re =[tex] 3.84 \times 10^6 [/tex] cd is 0.072
[tex]tan \theta = \frac{3\times 0.072 \times 10^2 \times \pi \times 0.5^2}{ 2\times 9.81 \pi 0.5^3(1- 1.5)}[/tex]
[tex]\theta = tan^{-1} [\frac{3\times 0.072 \times 10^2 \times \pi \times 0.5^2}{ 2\times 9.81 \pi 0.5^3(1- 1.5)}][/tex]
[tex]\theta = - 65.57 degree[/tex]
[tex]\phi = 90 - (-65.57) = 1557.57[/tex] degree
Water at 20 bar, 400°C enters a turbine operating at steady state and exits at 1.5 bar. Stray heat transfer and kinetic and potential energy effects are negligible. A hard-to-read data sheet indicates that the quality at the turbine exit is 98%. Can this quality value be correct? If no, explain. If yes, determine the power developed by the turbine, in kJ per kg of water flowing.
Answer:
quality value is not 98%
Explanation:
given data
inlet pressure p1 = 20 bar
outlet pressure p2 = 1.5 bar
temperature t1 = 400°C
solution
as assume here assume isentropic process so equation that states stray heat transfer is negligible so Q = 0 and S1 = S2
S1 = Sf2 + x Sfg2 ........................1
here x is quantity of steam and we get all other value by steam table
so at pressure 20 bar and 400°C and at pressure 1.5 bar
S1 = 7.127 kJ/kg K and Sf2 = 1.4336 kJ/kg K
and Sfg2 = 5.7898 kJ/kg K
so put all value in equation 1 we get x that is
x = [tex]\frac{7.127-1.4336}{5.7898}[/tex]
x = 0.9833
x = 98.33 %
so here we can say quality value is not 98%
There is no charge at the upper terminal of the ele- ment in Fig. 1.5 for t 6 0. At t = 0 a current of 125e-2500t mA enters the upper terminal. a) Derive the expression for the charge that accu- mulates at the upper terminal for t 7 0. b) Find the total charge that accumulates at the upper terminal. c) If the current is stopped at t = 0.5 ms, how much charge has accumulated at the upper terminal?
Answer:
A) The charge accumulated on upper plate for t>0 is
[tex]q(t)=50[1-e^{-2500t}]\mu C[/tex]
B) The total charge that accumulates at the upper terminal is 50μC
C) If the current is stopped at t = 0.5 ms then total charge stored on upper terminal is 35.67μC
Explanation:
Given that:
[tex]I(t)=0 \quad \quad \quad \quad \quad t<0\\\\I(t)= 125e^{-2500t} \quad t\geq 0[/tex]
A) The charge that accumulates at the upper terminal for t > 0:
As we know
[tex]q(t)=\int {I(t)} \, dt[/tex]
for t > 0
[tex]q(t)=\int\limits^t_0 {I(t)} \, dt\\q(t)=\int\limits^t_0 {125e^{-2500t} mA} \, dt\\q(t)=(125\times 10^{-3})[\frac{e^{-2500t}}{-2500}]^{t}_{0}\\\\q(t)=( 50\times 10^{-6})[-e^{-2500t}]^{t}_{0}\\\\q(t)=( 50\times 10^{-6})[-e^{-2500t}+1]\\\\[/tex]
The charge accumulated on upper plate for t>0 is
[tex]q(t)=50[1-e^{-2500t}]\mu C---(1)[/tex]
B) The total charge that accumulates at the upper terminal can be found by substituting t → ∞ in equation (1)
[tex]q(t)=( 50\times 10^{-6})[1-e^{-2500(\infty)}]\\q(t)=(50\times 10^{-6})[1-0]\\q(t) =50\mu C[/tex]
C) If the current is stopped at t = 0.5 ms then
[tex]q(t)=( 50\times 10^{-6})[1-e^{-2500(0.5\times10^{-3})}]\\q(0.5ms)=35.67\mu C[/tex]
Neon gas enters an insulated mixing chamber at 300 K, 1 bar with a mass flow rate of 1 kg/s. A second steam of carbon monoxide enters at 575 K, 1 bar at a mass flow rate of 0.5 kg/s. Assuming the mixture exits at 1 bar and the specific heat ratios for Neon and CO are constant, determine:
(a) The molar composition of the exiting mixture.
(b) The temperature of the exiting mixture, in K.
(c) The rate of entropy production, in kW/K.
Answer:
a) the molar fraction of neon at the exit is
xₙ= 0.735
and carbon monoxide
xₓ = 0.265
b) the final temperature is
T = 410.55 K
c) the rate of entropy production is
ΔS = 1.83 KW/K
Explanation:
denoting n for neon and x for carbon monoxide:
a) from a mass balance, the molar fraction of neon at the exit is:
outflow mass neon=inflow mass neon
xₙ = outflow mass neon/ (total outflow of mass) = inflow mass neon/ (total outflow of mass) = (1 kg/seg / 20.18 kg/kmol) / (1 kg/seg / 20.18 kg/kmol + 0.5 kg/seg / 28.01 kg/kmol) = 0.735
and the one of carbon monoxide is
xₓ = 1-xₙ = 1-0.735 = 0.265
b) from the first law of thermodynamics applied to an open system, then
Q - Wo = ΔH + ΔK + ΔV
where
Q= heat flow to the chamber = 0 ( insulated)
Wo= external work to the chamber = 0 ( there is no propeller to mix)
ΔH = variation of enthalpy
ΔK = variation of kinetic energy ≈ 0 ( the changes are small with respect to the one of enthalpy)
ΔV = variation of kinetic energy ≈ 0 ( the changes are small with respect to the one of enthalpy)
therefore
ΔH = 0 → H₂ - H₁ = 0 → H₂=H₁
if we assume ideal has behaviour of neon and carbon monoxide, then
H₁ = H ₙ₁ + H ₓ₁ = mₙ₁*cpₙ*Tn + mₓ₁*cpₓ*Tc
H₂ = (m ₙ+mₓ)*cp*T
for an ideal gas mixture
cp = ∑ cpi xi
therefore
mₙ*cpₙ*Tₙ + mₓ*cpₓ*Tₙ₁ = (m ₙ+mₓ)*∑ cpi xi*T
mₙ/(m ₙ+mₓ)*cpₙ*Tₙ + mₓ/(m ₙ+mₓ)*cpₓ*Tₓ = T ∑ cpi xi
xₙ* cpₙ*Tₙ +xₓ*cpₓ*Tₓ = T*( xₙ* cpₙ+xₓ*cpₓ)
T= [xₙ* cpₙ/( xₙ* cpₙ+xₓ*cpₓ)]*Tₙ₁ +[xₓ*cpₓ/( xₙ* cpₙ+xₓ*cpₓ)]*Tₓ₁
denoting
rₙ = xₙ* cpₙ/( xₙ* cpₙ+xₓ*cpₓ)
and
rₓ= xₓ*cpₓ/( xₙ* cpₙ+xₓ*cpₓ)
T= rₙ *Tₙ +rₓ*Tₓ
for an neon , we can approximate its cv through the cv for an monoatomic ideal gas
cvₙ= 3/2 R , R= ideal gas constant=8.314 J/mol K=
since also for an ideal gas: cpₙ - cvₙ = R → cpₙ = 5/2 R
for the carbon monoxide , we can approximate its cv through the cv for an diatomic ideal gas
cvₓ= 7/2 R → cpₓ = 9/2 R
replacing values
rₙ = xₙ* cpₙ/( xₙ* cpₙ+xₓ*cpₓ) = xₙ₁*5/2 R/ ( xₙ₁*5/2 R+xₓ*9/2 R) =
xₙ₁*5/(xₙ₁*5 + xₓ*9) = 5xₙ₁/(5 + 4*xₓ) = 5*0.735/(5+ 4*0.265) =0.598
since
rₙ + rₓ =1 → rₓ = 1-rₙ = 1- 0.598 = 0.402
then
T = rₙ *Tₙ +rₓ*Tₓ = 0.598 * 300 K + 0.402 * 575 K = 410.55 K
c) since there is no entropy changes due to heat transfer , the only change in entropy is due to the mixing process
since for a pure gas mixing process
ΔS = n*Cp* ln T₂/T₁ -n*R ln (P₂/P₁)
but P₂=P₁ (P=pressure)
ΔS = n*Cp* ln T₂/T₁ = n*Cp*ln T₂ - n*Cp*ln T₁ = S₂-S₁
for a gas mixture as end product
ΔS = (nₓ+nₙ)*Cp*ln T - (nₓ*Cpₓ*ln Tₓ + nₙ*Cpₙ*ln Tₙ)
ΔS = nₙ*Cpₙ ( (nₓ+nₙ)*Cp/[nₙ*Cpₙ]* ln T - ( nₓ*Cpₓ/ (nₙ*Cpₙ) *ln Tₓ + ln Tₙ)
ΔS = nₙ*Cpₙ [ 1/rₙ * ln T -( (rₓ/rₙ)*ln Tₓ + ln Tₙ)]
replacing values ,
ΔS = nₙ*Cpₙ [ 1/rₙ * ln T -( (rₓ/rₙ)*ln Tₓ + ln Tₙ)] = (1 kg/s/ 20.18*10 kg/kmol)* 5/2* 8.314 kJ/kmol K *[ 1/0.598 * ln 410.55 K-( 0.402/0.598 *ln 575 K + ln 300K)]
= 1.83 KW/K
What is the stress concentration factor of a shaft in torsion, where D=1.25 in. and d=1 in. and the fillet radius is, r=0.2 in.a. K=1.4b. K=1.5c. K=1.2d. K=1.3
Answer:
Concentration factor will be 1.2
So option (C) will be correct answer
Explanation:
We have given outer diameter D = 1.25 in
And inner diameter d = 1 in and fillet ratio r = 0.2 in
So [tex]\frac{r}{d}[/tex] ratio will be [tex]=\frac{0.2}{1}=0.2[/tex]
And [tex]\frac{D}{d}[/tex] ratio will be [tex]=\frac{1.25}{1}=1.25[/tex]
Now from the graph in shaft vs torsion the value of concentration factor will be 1.2
So concentration factor will be 1.2
So option (C) will be correct answer.
Oil with a density of 850 kg/m3 and kinematic viscosity of 0.00062 m2 /s is being discharged by a 8-mm-diameter, 40-m-long horizontal pipe from a storage tank open to the atmosphere. The height of the liquid level above the center of the pipe is 4 m. Disregarding the minor losses, determine the flow rate of oil through the pipe.
Answer:
Q = 5.06 x 10⁻⁸ m³/s
Explanation:
Given:
v=0.00062 m² /s and ρ= 850 kg/m³
diameter = 8 mm
length of horizontal pipe = 40 m
Dynamic viscosity =
μ = ρv
=850 x 0.00062
= 0.527 kg/m·s
The pressure at the bottom of the tank is:
P₁,gauge = ρ g h = 850 x 9.8 x 4 = 33.32 kN/m²
The laminar flow rate through a horizontal pipe is:
[tex]Q = \dfrac{\Delta P \pi D^4}{128 \mu L}[/tex]
[tex]Q= \dfrac{33.32 \times 1000 \pi\times 0.008^4}{128 \times 0.527 \times 40}[/tex]
Q = 5.06 x 10⁻⁸ m³/s
Which one of the following activities is not exempt from licensure pursuant to Chapter 471, F.S.? A person practicing engineering on property owned by him or her. A full time electrical engineer of Progress Energy Corp. A civil engineer employed full time by the U.S. Army Corps of Engineers. An Independent consultant working on the design of an electrical distribution system project for Progress Energy Corp. Applicants for licensure with degrees from foreign institutions are required to document "substantial equivalency to ABET criteria to the FBPE. They can do this by: Providing a transcript from their institution to the Board. Providing a notarized certification that they have completed the requisite college credit hours set forth in Rules 61615-20.007(2)(a) thru (2)(d), F.A.C. Getting the evaluation of substantial equivalency from a provider of the service that is approved by the FBPE Passing the Principles & Practice examination In order to verify an applicant's experience, the FBPE: Follows guidelines set forth in Rule 61615 20.002, F.A.C. Relies on information obtained from personal references Requires evidence of employment from employers or supervisors who are employed in the engineering profession. • All of the above.
An Independent consultant working on the design of an electrical distribution system for a corporation is not exempt from licensure. Applicants with foreign degrees must prove ABET equivalency, and the FBPE verifies experience through multiple documentation and procedures.
Explanation:The activity that is not exempt from licensure pursuant to Chapter 471, F.S. is 'An Independent consultant working on the design of an electrical distribution system project for Progress Energy Corp.' This scenario indicates an individual offering engineering services to the public, which requires a license. On the other hand, individuals practicing engineering on their own properties, full-time employees of a utility company, and civil engineers employed by the U.S. Army Corps of Engineers typically fall under exemptions to licensure requirements in many states, including Florida.
For applicants with degrees from foreign institutions to document "substantial equivalency to ABET criteria to the FBPE," they can achieve this by providing a transcript from their institution, providing a notarized certification of completed credit hours as per the specified rules, getting the evaluation from a service provider approved by the FBPE, or passing the Principles & Practice examination. To verify an applicant's experience, the FBPE requires various forms of documentation and follows several procedures, including adherence to specific rules, consideration of personal references, and requiring evidence of employment in the engineering profession – essentially, they consider all of the above methods.
Chairs and bolsters are used to:
a. keep formwork from moving during concrete placement.
b. maintain proper spacing between members during installation of precast concrete members.
c. help during posttensioning of concrete slabs.
d. splice reinforcing bars that require greater length.
e. support reinforcing bars in beams and slabs, prior to concrete placement.
Answer:e. support reinforcing bars in beams and slabs, prior to concrete placement.
Explanation: Chairs are used for construction of foundations,large steel supports,deck constructions and for underground works. Chairs can be for rebar support ( rebar support chairs),post tension chairs.
Bolsters are usually long made of metallic materials mainly used as support for construction of different infrastructures like roads it helps to ensure the concretes and other construction materials stay firmly connected. Image 1 is a metal be chair
Image 2 is a metal bolster
Consider a 4.45-m × 4.45-m × 4.45-m cubical furnace whose floor and ceiling are black and whose side surfaces are reradiating. The floor and the ceiling of the furnace are maintained at temperatures of 547 K and 1100 K, respectively. Determine the net rate of radiation heat transfer between the floor and the ceiling of the furnace. (Given: The view factor from the ceiling to the floor of the furnace F12 = 0.2, σ = 5.67 × 10-8.)
Answer: -1543371.65837 W
= - 1543.372 kW.
Explanation:
Using the equation;
Q= EσA -------------------------------------------------------------------------------------------(1).
Where Q= net rate of radiation heat transfer between the floor and the ceiling of the furnace, σ = Boltzmann's constant, A= area of the cube, E = emissitivity.
Recall that the emissitivity of a black body is equals to one(1).
From the question, the parameters given are; The view factor from the ceiling to the floor of the furnace,F12 = 0.2, σ = 5.67 × 10-8., A= (4.45×4.45) m.
Slotting in the parameters into the equation;
Q= EσA[T(2)^4 - (T(1)^4] ---------------------------------------------------------------------(2).
Therefore, Q= (1)× (5.67×10^-8) × (4.45×4.45) m × [(547)^4 - (1100)^4]
= 0.0000011228 × (89526025681 - 1.4641×10^12).
= 0.0000011228×(-1.374574×10^12)
= -1543371.65837 W
= -1543.372 kW.
In an aligned and continuous carbon fiber-reinforced nylon 6,6 composite, the fibers are to carry 97% of a load applied in the longitudinal direction. (a) Using the data provided, determine the volume fraction of fibers required.
Answer:
a) 0.26
b) 1077 MPa
Explanation:
a) The following equation can be used to determine the volume fraction:
[tex]\frac{F_f}{F_m} =\frac{E_fV_f}{E_m(1-V_f)}[/tex]
[tex]\frac{0.97}{1-0.97} =\frac{260V_f}{2.8(1-V_f)}[/tex]
[tex]32.3 = \frac{260V_f}{2.8-2.8V_f}[/tex]
[tex]V_f = 0.26[/tex]
b) Tensile strength can be found by using the following equation:
[tex]\sigma_{cl} = \sigma_m(1-V_f)+\sigma_fV_f = 50*(1-0.26)+4000*0.26 = 1077[/tex] MPa