Answer:
Each of the metal specimens HAS an indentation near the center to ensure that the fracture point would occur in this region. Tension tests WERE CONDUCTED as follows. Two pieces of reflective tape WERE PLACED approximately 1 inch apart in the center of the specimen where the indentation 4 WAS LOCATED. The width and the thickness of the specimen at this location WAS MEASURED using a Vernier caliper. Then the specimen WAS SECURED in the MTS Load Frame. A laser extensometer WAS PLACED into position to measure the deformation of the specimen. The laser extensometer WAS USED to measure the original distance between the pieces of reflective tape. The MTS WAS SET to elongate the specimen one tenth of an inch every minute.
how does load transfer of space needle
Answer:
The Space Needle is a cut away with minimal residual deflection due to load transfer.
how many types of lavatory there is?
Answer:
there are generally two types of toilet bowl types- round and elongated.
Activity 1. Fill the blank with the correct answer. Write your answer on the blank. 1. ___________________ is a regular pattern of dots displayed on the screen which acts as a visual aid and also used to define the extent of your drawing. 2. Ortho is short for ___________________, which means either vertical or horizontal. 3. Tangent is a point where two _______________________ meet at just a single point. 4. If you want to create a new drawing, simply press ___________________ for the short cut key. 5. There are _______________Osnap that can help you performs your task easier.
Answer:
1. Drawing grid.
2. Orthogonal.
3. Geometries.
4. CTRL+N.
5. Thirteen (13).
Explanation:
CAD is an acronym for computer aided design and it is typically used for designing the graphical representation of a building plan. An example of a computer aided design (CAD) software is AutoCAD.
Some of the features of an AutoCAD software are;
1. Drawing grid: is a regular pattern of dots displayed on the screen of an AutoCAD software, which acts as a visual aid and it's also used to define the extent of a drawing.
2. Ortho is short or an abbreviation for orthogonal, which means either vertical or horizontal.
3. Tangent is a point where two geometries meet at just a single point.
4. If you want to create a new drawing, simply press CTRL+N for the short cut key.
5. There are thirteen object snaps (Osnap) that can help you perform your task on AutoCAD easily. The 13 object snaps (Osnap) are; Endpoint, Midpoint, Apparent intersect, Intersection, Quadrant, Extension, Tangent, Center, Insert, Perpendicular, Node, Parallel, and Nearest.
The temperature gradient in a spherical (or cylindrical) wall at steady state will always decrease (in magnitude) with increasing distance from the center (line), i.e. radial distance.
A. True
B. False
Answer:
True
Explanation:
Yes it is true that the Temperature gradient would also decrease with magnitude just as the distances rise from the centre line.
We have this cylinder equation as
[T1-T2 / ln(r1-r2)]2πKL
The radial distance is r2-r1
The gradient of temperature is T1-T2
From the equation,
The temperature gradient has a direct and proportional relationship to radial distance
T1-T2 ∝ ln(r2-r1)
1/T1-T2 = k(r2-r1)
This inverse relationship above confirms that the statement is true
A well insulated rigid tank contains 4 kg of argon gas at 450 kPa and 30 C. A valve is opened, allowing the argon to escape until the tank pressure drops to 200 kPa. Assuming that the argon remaining in the tank experiences a reversible adiabatic process, find the final mass of argon in the tank. Since you don't have argon gas tables, assume cp, cv, k as needed at some appropriate temperature(s).
Answer:
Final mass of Argon= 2.46 kg
Explanation:
Initial mass of Argon gas ( M1 ) = 4 kg
P1 = 450 kPa
T1 = 30°C = 303 K
P2 = 200 kPa
k ( specific heat ratio of Argon ) = 1.667
assuming a reversible adiabatic process
Calculate the value of the M2
Applying ideal gas equation ( PV = mRT )
P₁V / P₂V = m₁ RT₁ / m₂ RT₂
hence : m2 = P₂T₁ / P₁T₂ * m₁
= (200 * 303 ) / (450 * 219 ) * 4
= 2.46 kg
Note: Calculation for T2 is attached below
Define chart name the different types of charts explain any three types of charts
Answer:
There are several different types of charts and graphs. The four most common are probably line graphs, bar graphs and histograms, pie charts, and Cartesian graphs. They are generally used for, and are best for, quite different things. ... Pie charts to show you how a whole is divided into different parts.
#carryonlearning
Use pseudocode. 1) Prompt for and input a saleswoman's sales for the month (in dollars) and her commission rate (percentage). Output her commission for that month. Note that you will need the following Variables: SalesAmount CommissionRate CommissionEarned
You will need the following formula: CommissionEarned= Sales Amount * (commissionrate/100)
Answer:
The pseudocode is as follows:
Input SalesAmount
Input CommissionRate
CommissionEarned= SalesAmount * (CommissionRate/100)
Print CommissionEarned
Explanation:
This gets input for SalesAmount
Input SalesAmount
This gets input for CommissionRate
Input CommissionRate
This calculates the CommissionEarned
CommissionEarned= SalesAmount * (CommissionRate/100)
This prints the calculated CommissionEarned
Print CommissionEarned
A container can be made from steel [β = 36 × 10-6 (C°)-1] or lead [β = 87 × 10-6 (C°)-1]. A liquid is poured into the container, filling it to the brim. The liquid is either water [β = 207 × 10-6 (C°)-1] or ethyl alcohol [β = 1120 × 10-6 (C°)-1]. When the full container is heated, some liquid spills out. To keep the overflow to a minimum, from what should the container be made and what should the liquid be
Answer:
The container should be made of lead and the liquid should be water.
Explanation:
Since the volume of the container of liquid after expansion is V = V₀(1 + βΔθ) where V = initial volume, β = coefficient of volume expansion, Δθ = temperature change.
So, the volume change V₂ - V₁ where V₁ = volume of liquid and V₂ = volume of container
For steel, V₂ = V₀(1 + β₂Δθ) and V₁ = V₀(1 + β₁Δθ)
So, ΔV = V₀(1 + β₂Δθ) - V₀(1 + β₁Δθ) = V₀[1 + β₂Δθ - 1 - β₁Δθ] = V₀[β₂Δθ - β₁Δθ]
Since we want a minimum value for ΔV and V₀ and Δθ are the same, we need ΔV/V₀Δθ = β₂ - β₁ to be a minimum
where β₂ = coefficient of volume expansion of liquid and β₁ = coefficient of volume expansion of container.
So, trying each combination, with β₂ = 207 × 10⁻⁶ (C°)⁻¹] and β₁ = 36 × 10⁻⁶ (C°)⁻¹
β₂ - β₁ = 207 × 10⁻⁶ (C°)⁻¹ - 36 × 10⁻⁶ (C°)⁻¹ = 171 × 10⁻⁶ (C°)⁻¹
With β₂ = 207 × 10⁻⁶ (C°)⁻¹] and β₁ = 87 × 10⁻⁶ (C°)⁻¹
β₂ - β₁ = 207 × 10⁻⁶ (C°)⁻¹ - 87 × 10⁻⁶ (C°)⁻¹ = 120 × 10⁻⁶ (C°)⁻¹
With β₂ = 1120 × 10⁻⁶ (C°)⁻¹] and β₁ = 36 × 10⁻⁶ (C°)⁻¹
β₂ - β₁ = 1120 × 10⁻⁶ (C°)⁻¹ - 36 × 10⁻⁶ (C°)⁻¹ = 1084 × 10⁻⁶ (C°)⁻¹
With β₂ = 1120 × 10⁻⁶ (C°)⁻¹] and β₁ = 87 × 10⁻⁶ (C°)⁻¹
β₂ - β₁ = 207 × 10⁻⁶ (C°)⁻¹ - 87 × 10⁻⁶ (C°)⁻¹ = 1033 × 10⁻⁶ (C°)⁻¹
The combination that gives the lowest value for β₂ - β₁ is β₂ = 207 × 10⁻⁶ (C°)⁻¹] and β₁ = 87 × 10⁻⁶ (C°)⁻¹
Since β₁ = 87 × 10⁻⁶ (C°)⁻¹ = coefficient of expansion for lead β₂ = 207 × 10⁻⁶ (C°)⁻¹] = coefficient of expansion for water, the container should be made of lead and the liquid should be water.
The following is a correlation for the average Nusselt number for natural convection over spherical surface. As can be seen in the above, the Nusselt number approaches 2 as Rayleigh number approaches zero. Prove that this situation corresponds to conduction heat transfer and in conduction heat transfer over sphere, the Nusselt number becomes 2. Hint: First step: Write an expression for heat transfer between two spherical shells that share the same center. Second step: Assume the outer spherical shell is infinitely large.
Answer:
Explanation:
[tex]r_2=[/tex]∞
[tex]q=4\pi kT_1(T_2-T_1)\\[/tex]
[tex]q=2\pi kD.[/tex]ΔT--------(1)
[tex]q=hA[/tex] ΔT[tex]=4\pi r_1^2(T_2_s-T_1_s)\\[/tex]
[tex]N_u=\frac{hD}{k} = 2+\frac{0.589 R_a^\frac{1}{4} }{[1+(\frac{0.046}{p_r}\frac{9}{16} )^\frac{4}{9} } ------(3)[/tex]
By equation (1) and (2)
[tex]2\pi kD.[/tex]ΔT=h.4[tex]\pi r_1^2[/tex]ΔT
[tex]2kD=hD^2\\\frac{hD}{k} =2\\N_u=\frac{hD}{k}=2\\[/tex]-------(4)
From equation (3) and (4)
So for sphere [tex]R_a[/tex]→0
Bainite has finer grains because the transformation takes place at a lower temperature where the nucleation rate is high relative to the growth rate. True or False
Answer:
False
Explanation:
Bainite is a type of steel. It is formed by the decomposition of austenite. This happens at a temperature above MS but given that MS temperature is below the one where fine pearlite is formed.
In other words, when iron goes through a metastable crystallization phase Bainite is formed. It takes the rapid cooling, or quenching, of austenite for that to happen.
Metastability is the phenomenon where matter remains in an apparent state of equilibrium or stability even though it can achieve a more stable state. It can also be referred to as the condition of remaining in that pseudo unstable state for a very protracted period of time.
Bainite steel is used for the construction of components that require frequent usage and where plasticity, or tendency for deformity, as well as wear, cannot be tolerated.
Cheers
Determine the resolution of a manometer required to measure the velocity of air at 50 m/s using a pitot-static tube and a manometer fluid of mercury (density: 13,600 kg/m3) to achieve uncertainty of 5% (i.e., 2.5 m/s) and 1 % (0.5 m/s).
Answer:
a) Δh = 2 cm, b) Δh = 0.4 cm
Explanation:
Let's start by using Bernoulli's equation for the Pitot tube, we define two points 1 for the small entry point and point 2 for the larger diameter entry point.
P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂
Point 1 is called the stagnation point where the fluid velocity is reduced to zero (v₁ = 0), in general pitot tubes are used in such a way that the height of point 2 of is the same of point 1
y₁ = y₂
subtitute
P₁ = P₂ + ½ ρ v₂²
P₁ -P₂ = ½ ρ v²
where ρ is the density of fluid
now we measure the pressure on the included beforehand as a pair of communicating tubes filled with mercury, we set our reference system at the point of the mercury bottom surface
ΔP =ρ_{Hg} g h - ρ g h
ΔP = (ρ_{Hg} - ρ) g h
as the static pressure we can equalize the equations
ΔP = P₁ - P₂
(ρ_{Hg} - ρ) g h = ½ ρ v²
v = [tex]\sqrt{\frac{2 (\rho_{Hg} - \rho) g}{\rho } } \ \sqrt{h}[/tex]
in this expression the densities are constant
v = A √h
A =[tex]\sqrt{\frac{2(\rho_{Hg} - \rho ) g}{\rho } }[/tex]
They indicate the density of mercury rhohg = 13600 kg / m³, the density of dry air at 20ºC is rho air = 1.29 kg/m³
we look for the constant
A = [tex]\sqrt{\frac{2( 13600 - 1.29) \ 9.8}{1.29} }[/tex]
A = 454.55
we substitute
v = 454.55 √h
to calculate the uncertainty or error of the velocity
h = [tex]\frac{1}{454.55^2} \ v^2[/tex]
Δh = [tex]\frac{dh}{dv}[/tex] Δv
[tex]\frac{\Delta h}{h } = 2 \ \frac{\Delta v}{v}[/tex]
Suppose we have a height reading of h = 20 cm = 0.20 m
a) uncertainty 2.5 m / s ( 0.05)
[tex]\frac{\delta v}{v} = 0.05[/tex]
[tex]\frac{\Delta h}{h}[/tex] = 2 0.05
Δh = 0.1 h
Δh = 0.1 20 cm
Δh = 2 cm
b) uncertainty 0.5 m / s ( Δv/v= 0.01)
[tex]\frac{\Delta h}{h}[/tex] = 2 0.01
Δh = 0.02 h
Δh = 0.02 20
Δh = 0.1 20 cm
Δh = 0.4 cm = 4 mm
James the Pilot James is a pilot. He is wearing a flight suit. He flies to Paris. He loves flying. 1. James is a a) teacher b) doctor c) pilot. whatisthe 2. He is wearing a a) shirt b) t-shirt c) flight suit. 3. Where does he fly to? a) Italy b) Luxembourg c) Paris http https://whatistheurl.com Please visit our site for worksheets and charts
Answer:
1.c
2.c
3.c
Explanation:
James is a pilot, whistle. He is wearing a flight suit. Paris is the palace where does he fly to. Hence, option C, C, and C are correct.
What is the point of a flight suit?When flying an aircraft, such as a military aircraft, a glider, or a helicopter, one must wear a full-body suit called a flight suit. These outfits are typically meant to keep the user warm and are also functional (they have many of pockets) (including fire ). In most cases, it looks like a jumpsuit.
The G suit, sometimes known as a "anti-G suit," is a one-piece jumpsuit that shields a pilot from the pressure of G forces pressing down on him and causing discomfort or unconsciousness.
The traditional attire for pilots of military and commercial aircraft, helicopters, and even gliders is flight suits or flyers coveralls. In areas where there is a risk of fire, ground personnel—including aircrews—often wear flight suits as well.
Thus, option C, C, and C are correct.
For more information about point of a flight suit, click here:
https://brainly.com/question/12302183
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Technician A says that a graphing multi-meter may be used to verify signals going to and from electrical and electronic components. Technician B says that digital storage oscilloscope may be used to verify signals going to and from electrical and electronic components. Who is correct
Answer:
Both are correct.
Explanation:
Graphing multi meter is used to verify signals that move from electrical components. Digital oscilloscope is an equipment which stores and analyzes input signals with digital technique.
Consider the equation y = 10^(4x). Which of the following statements is true?
A plot of log(y) vs. x would be linear with a slope of 4.
A plot of log(y) vs. log (x) would be linear with a slope of 10.
A plot of log(y) vs. x would be linear with a slope of 10.
A plot of y vs. log(x) would be linear with a slope of 4.
A plot of log(y) vs. log (x) would be linear with a slope of 4.
A plot of y vs. log(x) would be linear with a slope of 10.
Answer: Plot of [tex]\log y[/tex] vs [tex]x[/tex] would be linear with a slope of 4.
Explanation:
Given
Equation is [tex]y=10^{4x}[/tex]
Taking log both sides
[tex]\Rightarrow \log y=4x\log (10)\\\Rightarrow \log y=4x[/tex]
It resembles with linear equation [tex]y=mx+c[/tex]
Here, slope of [tex]\log y[/tex] vs [tex]x[/tex] is 4.
A circular rod with a gage length of 3.1 m and a diameter of 3 cm is subjected to an axial load of 68 kN . If the modulus of elasticity is 200 GPa , what is the change in length
Answer:
1.49 mm
Explanation:
The modulus of elasticity, Y = stress/strain = σ/ε
σ = F/A where F = load = 68 kN = 68 × 10³ N and A = cross-sectional area of rod = πd²/4 where d = diameter of rod = 3 cm = 3 × 10⁻² m.
ε = ΔL/L where ΔL = change in length of the circular rod and L = length of circular rod = 3.1 ,
So, Y = σ/ε
Y = F/A ÷ ΔL/L
Y = FL/AΔL
making the change in length ΔL subject of the formula, we have
ΔL = FL/AY
substituting the value of A into the equation, we have
So, ΔL = FL/(πd²/4)Y
ΔL = 4FL/πd²Y
Since Y = 200 GPa = 200 × 10⁹ Pa
Substituting the values of the variables into the equation, we have
ΔL = 4FL/πd²Y
ΔL = 4 × 68 × 10³ N × ×3.1 m/[π(3 × 10⁻²m)² × 200 × 10⁹ Pa]
ΔL = 843.2 × 10³ Nm/[9π × 10⁻⁴m² × 200 × 10⁹ Pa]
ΔL = 843.2 × 10³ Nm/[1800π × 10⁵ N]
ΔL = 843.2 × 10³ Nm/5654.87 × 10⁵ N
ΔL = 0.149 × 10⁻² m
ΔL = 1.49 × 10⁻³ m
ΔL = 1.49 mm
The change in length of the circular rod is 1.49 mm
All of the following are instruments involved in changing a tire EXCEPT:
Answer:
can you please give us the option for this question
it is a small sharp and printed item for fine worker in trimming scallops clipping threads and cutting large eyelets
Answer:
embroidery scissor
Explanation:
is small, sharp and pointed good for fine work use trimming scallops,clipping threads,and cutting large eyelets.
hope this helps
plsssssss help me here
Steam enters an adiabatic turbine at 6 MPa, 600°C, and 80 m/s and leaves at 50 kPa, 100°C, and 140 m/s. If the power output of the turbine is 5 MW, determine (a) the reversible power output and (b) the second-law efficiency of the turbine. Assume the surroundings to be at 25°C.
Answer:
(a) the reversible power output of turbine is 5810 kw
(b) The second-law efficiency of he turbine = 86.05%
Explanation:
In state 1: the steam has a pressure of 6 MPa and 600°C. Obtain the enthalpy and entropy at this state.
h1 = 3658 kJ/kg s1=7.167 kJ/kgK
In state 2: the steam has a pressure of 50 kPa and 100°C. Obtain the enthalpy and entropy at this state
h2 = 2682kl/kg S2= 7.694 kJ/kg
Assuming that the energy balance equation given
Wout=m [h1-h2+(v1²-v2²) /2]
Let
W =5 MW
V1= 80 m/s V2= 140 m/s
h1 = 3658kJ/kg h2 = 2682 kJ/kg
∴5 MW x1000 kW/ 1 MW =m [(3658-2682)+ ((80m/s)²-(140m/s)²)/2](1N /1kg m/ s²) *(1KJ/1000 Nm)
m = 5.158kg/s
Consider the energy balance equation given
Wrev,out =Wout-mT0(s1-s2)
Substitute Wout =5 MW m = 5.158kg/s 7
s1= 7.167 kJ/kg-K s2= 7.694kJ/kg-K and 25°C .
Wrev,out=(5 MW x 1000 kW /1 MW) -5.158x(273+25) Kx(7.167-7.694)
= 5810 kW
(a) Therefore, the reversible power output of turbine is 5810 kw.
The given values of quantities were substituted and the reversible power output are calculated.
(b) Calculating the second law efficiency of the turbine:
η=Wout/W rev,out
Let Wout = 5 MW and Wrev,out = 5810 kW
η=(5 MW x 1000 kW)/(1 MW *5810)
η= 86.05%
A stream of oxygen enters a compressor at a rate of 200 SCMH. The oxygen exits at 360 K and 500 bar. Determine the volumetric flowrate exiting the compressor using the compressibility factor equation of state.
Answer:
≈ 0.516 m^3/hr
Explanation:
Inlet of compressor = 200 SCMH
sheer standard conditions = 1 atm and 288.5 K
For oxygen :
critical pressure(Pc) = 49.8 atm
critical temperature Tc = 154.6 K
hence at compressor inlet
Tr = T / Tc = 288.5/154.6 = 1.866
Pr = P / Pc = 1 / 49.8 = 0.0204
Z1 ( from compressibility chart ) = 0.98
at compressor outlet
P2 = 500 bar = 500*0.9869 = 493.45 atm , T2 = 360 k
hence : Pr = P / Pc = 493.45 / 49.8 = 9.91
Tr = T / Tc = 360 / 154.6 = 2.33
Z2 ( from compressibility chart ) ≈ 1
V2( volumetric flow rate ) = V1*(P₁Z₂T₂) / (P₂Z₁T₁)
= 200 ( 1 * 1* 360) / (493.45 *0.98*288.5)
= 0.516 m^3/hr
Air is compressed in a well insulated compressor from 95 kPa and 27 C to 600 kPa and 277 C. Use the air tables; assume negligible changes in kinetic and potential energy. Find the isentropic efficiency of the compressor. Find the exit temperature of the air if the compressor was reversible.
Answer:
a) 1.9%
b) T2s = 505.5 k = 232.5°C
Explanation:
P1 = 95 kPa
T1 = 27°C = 300 k
P2 = 600 kPa
T1 = 277°c = 550 k
Table used : Table ( A - 17 ) Ideal gas properties of air
a) determining the isentropic efficiency of the compressor
Л = ( h2s - h1 ) / ( h2a - h1 ) ---- ( 1 )
where ; h1 = 300.19 kJ/kg , T1 = 300 K , h2a = 554.74 kJ/kg , T2 = 550 k
To get h2s we have to calculate the the value of Pr2 using Pr1(relative pressure)
Pr2 = P2/P1 * Pr = ( 600 / 95 ) * 1.306 hence; h2s = 500.72 kJ/kg
back to equation1
Л = 0.019 = 1.9%
b) Calculate the exit temperature of the air if compressor is reversible
if compressor is reversible the corresponding exit temperature
T2s = 505.5 k = 232.5°C
given that h2s = 500.72 kJ/kg
Given resistance 30ohms Inductance 200mH is connected to a 230v,50hZ supply. Impedance 69.6ohms Calculate current consumed?
Answer:
the current consumed is 3.3 A
Explanation:
Given;
resistance, R = 30 ohms
inductance, L = 200 mH
Voltage supply, V = 230 V
frequency of the coil, f = 50 Hz
impedance, Z = 69.6 Ohms
The current consumed is calculated as;
[tex]I = \frac{V}{Z} \\\\I = \frac{230}{69.6} \\\\I = 3.3 \ A[/tex]
Therefore, the current consumed is 3.3 A
Determine the pressure difference in N/m2,between two points 800m apart in horizontal pipe-line,150 mm diameter, discharging water at the rate of 12.5litres per second. Take the frictional coefficient ,f, as being 0.008
Answer: [tex]10.631\times 10^3\ N/m^2[/tex]
Explanation:
Given
Discharge is [tex]Q=12.5\ L[/tex]
Diameter of pipe [tex]d=150\ mm[/tex]
Distance between two ends of pipe [tex]L=800\ m[/tex]
friction factor [tex]f=0.008[/tex]
Average velocity is given by
[tex]\Rightarrow v_{avg}=\dfrac{12.5\times 10^{-3}}{\frac{\pi }{4}(0.15)^2}\\\\\Rightarrow v_{avg}=\dfrac{15.9134\times 10^{-3}}{2.25\times 10^{-2}}\\\\\Rightarrow v_{avg}=7.07\times 10^{-1}\\\Rightarrow v_{avg}=0.707\ m/s[/tex]
Pressure difference is given by
[tex]\Rightarrow \Delta P=f\ \dfrac{L}{d}\dfrac{\rho v_{avg}^2}{2}\\\\\Rightarrow \Delta P=0.008\times \dfrac{800}{0.15}\times \dfrac{997\times (0.707)^2}{2}\\\\\Rightarrow \Delta P=10,631.45\ N/m^2\\\Rightarrow \Delta P=10.631\ kPa[/tex]
Alice and Bob both have RSA Public-Private key pairs: (PUA, PRA) and (PUB, PRB). They also have cryptographic functions E_AES / D_AES to encrypt / decrypt using AES; and E_RSA and D_RSA to encrypt / decrypt using RSA. Alice wants to sent a high resolution video of a large secret facility to Bob.
A. Show how Alice can securely and efficiently send the video to Bob. You are required to use the cryptographic functions above to get full credit;
B. Does your solutions assure confidentiality? How / Why not?
C. Does your solutions assure non-repudiation? How / Why not?
D. Does your solutions assure integrity? How / Why not?
E. Does your solutions assure replay attacks? How / Why not?
Solution :
B. yes, the given solution assures confidentiality. The sender Alice encrypting his messages with its own private key PRA which provides authentication. Sender Alice further encrypts his messages with the receiver's public key PUB provides confidentiality.
C. So the given solution provides non repudiation. Alice and Bob who are exchanging messages. In one case, Alice denies sending a messages to Bob that he claims to have received being able to counter Alice's denial is caused non repudiation of origin.
D. The given solution provides integrity. Because it provides authentication and have not been changed.
E. It does not provide replay attacks because it does not captures the traffic. The client does not receive the messages twice.
A signalized intersection has a sum of critical flow ratios of 0.72 and a total cycle lost time of 12 seconds. Assuming a critical intersection v/c ration of 0.9, calculate the minimum necessary cycle length.
Answer:
[tex]T_o=82.1sec[/tex]
Explanation:
From the question we are told that:
Lost Time [tex]t=12secs[/tex]
Sum of critical flow ratios [tex]X=0.72[/tex]
Generally the Webster Method's equation for Optimum cycle time is is mathematically given by
[tex]T_o=\frac{1.5t+5}{1-x}[/tex]
[tex]T_o=\frac{1.5*12+5}{1-0.72}[/tex]
[tex]T_o=82.1sec[/tex]
bending stress distribution is a.rectangle b.parabolic c.curve d.i section
An industrial load with an operating voltage of 480/0° V is connected to the power system. The load absorbs 120 kW with a lagging power factor of 0.77. Determine the size of the capacitor in vars that is necessary to raise the power factor to 0.9 lagging.
Answer:
[tex]Q=41.33 KVAR\ \\at\\\ 480 Vrms[/tex]
Explanation:
From the question we are told that:
Voltage [tex]V=480/0 \textdegree V[/tex]
Power [tex]P=120kW[/tex]
Initial Power factor [tex]p.f_1=0.77 lagging[/tex]
Final Power factor [tex]p.f_2=0.9 lagging[/tex]
Generally the equation for Reactive Power is mathematically given by
Q=P(tan \theta_2-tan \theta_1)
Since
[tex]p.f_1=0.77[/tex]
[tex]cos \theta_1 =0.77[/tex]
[tex]\theta_1=cos^{-1}0.77[/tex]
[tex]\theta_1=39.65 \textdegree[/tex]
And
[tex]p.f_2=0.9[/tex]
[tex]cos \theta_2 =0.9[/tex]
[tex]\theta_2=cos^{-1}0.9[/tex]
[tex]\theta_2=25.84 \textdegree[/tex]
Therefore
[tex]Q=P(tan 25.84 \textdegree-tan 39.65 \textdegree)[/tex]
[tex]Q=120*10^3(tan 25.84 \textdegree-tan 39.65 \textdegree)[/tex]
[tex]Q=-41.33VAR[/tex]
Therefore
The size of the capacitor in vars that is necessary to raise the power factor to 0.9 lagging is
[tex]Q=41.33 KVAR\ \\at\\\ 480 Vrms[/tex]
A venturimeter of 400 mm × 200 mm is provided in a vertical pipeline carrying oil of specific gravity 0.82, flow being upward. The difference in elevation of the throat section and entrance section of the venturimeter is 300 mm. The differential U-tube mercury manometer shows a gauge deflection of 300 mm. Calculate: (i) The discharge of oil, and (ii) The pressure difference between the entrance section and the throat section.Take the coefficient of meter as 0.98 and specific gravity of mercury as 13.6
Answer:
the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s
Explanation:
Given:
Diameter of the pipe = 100mm = 0.1m
Contraction ratio = 0.5
thus, diameter at the throat of venturimeter = 0.5×0.1m = 0.05m
The formula for discharge through a venturimeter is given as:
Where,
is the coefficient of discharge = 0.97 (given)
A₁ = Area of the pipe
A₁ =
A₂ = Area at the throat
A₂ =
g = acceleration due to gravity = 9.8m/s²
Now,
The gauge pressure at throat = Absolute pressure - The atmospheric pressure
⇒The gauge pressure at throat = 2 - 10.3 = -8.3 m (Atmosphric pressure = 10.3 m of water)
Thus, the pressure difference at the throat and the pipe = 3- (-8.3) = 11.3m
Substituting the values in the discharge formula we get
or
or
Q = 29.28 ×10⁻³ m³/s
Hence, the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s
Hope This Helps :D
CO2 enters an adiabatic nozzle, operating at steady state, at 200 kPa, 1500 K, 5 m/s and exits at 100 kPa, 1400 K. The exit area of the nozzle is 10 cm2. Using the PG model, determine the exit velocity
Answer:
[tex]v_2=549.2 m/s\\[/tex]
Explanation:
Given:
[tex]P_1=2500kPa\\T_1=1500 k\\V_1=5 m/s\\P_2=100 kPa\\T_2=1400 k\\A_2=10 cm^2[/tex]
Solution:
For [tex]Co_2[/tex] y=1.4
Since Nozzle is adiababic
So,
[tex]h_1+\frac{V_1^2}{2}=h_2+\frac{V_2^2}{2}\\\frac{v_2^2}{2}=(h_2-h_2)+\frac{r^2}{2}\\v_2^2=2(h_1-h_2)+v_1^2\\v_2=\sqrt{2(h_1-h_2)+v_1^2}[/tex]
Now,
[tex]h_1-h_2=Cp_1T_1-CP_2T_2\\h_1-h_2=(1989-1838.2)*10^3\\ =150.8 * 10^3\\Cp for co_2\\C_{p1}=1.326 kj/kg\\C_{p2}=1.313 kj/kg\\v_2=\sqrt{301600+25}\\ =549.2 m/s[/tex]
1. A manufacturing cell with two workers is responsible for producing a small frying pan with a required takt time of 496 seconds. The material passes through two processes: a deep drawing process and a trimming process. The average cycle time for the deep drawing process is 450 seconds and average cycle time for trimming is 430 seconds. (2 pts.)
a. Does the work cell have adequate capacity to meet demand? Explain.
b. What is the required daily production capacity of the work cell (in number of frying pans per day)? Assume 480 minutes/workday of available time.
2. What is the total daily idle time for both workers in Problem 1? Report your answer in (a) seconds of idle time and (b) as a percentage of total working time for the cell. (2 pts.)
Answer:
Explanation:
[tex]496=\frac{480\times 60}{demand}[/tex]
demand per day = 58 pans
Due to availability of two workers we can have parallel we can have deep drawing and trimming operations simultaneously.
Hence the cycle time would be the greater time of the two operations.
cycle time = 450 seconds
[tex]\text{capacity of work cell}=\frac{\text{available working time}}{\text{cycle time}}[/tex]
[tex]\text{capacity of work cell}=\frac{480\times 60}{450}[/tex]
[tex]\text{capacity of work cell}=64 ~pans[/tex] (which is greater than the demand of 58 pans)
Therefore the work cell has sufficient capacity and time (496 sec.>cycle time 450 sec) to meet the demand.
b)
Required daily production is 58 pans
