Answer:
The volume of sludge after filtration is 0.914 m
Explanation:
Solution
Given that:
We have to find the Volume of sludge after filtration
Now,
The Sludge concentration is = 32%,
The sludge volume = 100 m,
The sludge concentration after filtration = 35%
Then,
The mass balance equation is stated below
Cin∀in = Cout∀out
Now,
We Solve for ∀out
∀out =Cin∀in/Cout
By substituting the values
∀out = (0.32)(100 m)/(0.35) = 0.914 m
Transactional Vs Transformational Leadership. Using the Internet, each member of your team should read at least 2 articles each on Transactional Vs Transformational Leadership. Summarize the articles in 300 words or more. Provide appropriate reference. Combine each summarize in one paper but do not change the wording of the original summary. As a term, write a comprehensive summary of the articles. Present a discussion of what your team learned from this exercise?
Answer: Provided in the explanation section
Explanation:
Transactional Leadership - This leadership style is mainly focused on the transactions between the leader and employees. If the employees work hard, achieve the objectives and deliver the results, they are rewarded through bonuses, hikes, promotions etc. If the employees fail to achieve the desired results, they are punished by awarding lower ratings in the performance appraisal, denying opportunities etc.
In this style, leader lays emphasis on the relation with the followers.
It is a reactive style where the growth of the employee in the organization completely depends on the performance with respect to the activities and deliverables.
It is best suited for regular operations and for a settled environment by developing the existing organizational culture which is not too challenging.
It is a bureaucratic style of leadership where the leader concentrates on planning and execution rather than innovation and creation.
A transactional leader is short-term focused and result oriented. He/she doesn't consider long-term strategic objectives regarding the organization's future.
cheers i hope this helped !!
What's resistance in an electrical circuit? 1) Opposition to the flow of electricity 2) The ability of electricity to do work 3) The ability to make current flow
One piece of evidence that supports the Theory of Plate Tectonics is the discovery of what in both South America and Africa? The ancient atmosphere in both places was identical. The rates of weathering of rock are similar. Fossil remains of the same land-dwelling animal. Plants on both continents have similar flowers.
Answer: Fossil remains of the same land-dwelling animal.
Explanation: Fossil remains which were found to belong to same land dwelling animals, in South America and Africa was used as evidence to help support the theory of Tectonics plates, what this theory simply means is that the whole continents of earth were once fused together until a tectonic plate caused it’s division. Since same remains were found in Africa and South America this shows that both continents were once fused together.
Answer:
Fossil remains of the same land-dwelling animal
Explanation:
Fossil remains tell us where the animals once lived and how by the movement of plate spearated their remaind that was burried thousands of years ago.
A 60-m-long steel wire is subjected to a 6-kN tensile load. Knowing that E = 200 GPa and that the length of the rod increases by 48 mm, determine a) the smallest diameter that can be selected for the wire.b) the corresponding normal stress.
Answer:
(a) 6.91 mm (b) 160 MPa
Explanation:
Solution
Given that:
E = 200 GPa
The rod length = 48 mm
P =P¹ = 6 kN
Recall that,
1 kN = 10^3 N
1 m =10^3 mm
I GPa = 10^9 N/m²
Thus
The rod deformation is stated as follows:
δ = PL/AE-------(1)
σ = P/A----------(2)
Now,
(a) We substitute the values in equation and obtain the following:
48 * 10 ^⁻3 m = (6 * 10³ N) (60 m)/A[ 200 * 10^9 N/m^2]
Thus, we simplify
A = (6 * 10³) (60)/ ( 200 * 10^9) (48 * 10 ^⁻3)m²
A =0.0375 * 10 ^⁻3 m²
A =37.5 mm²
A = π/4 d²
Thus,
d² = 4A /π
After inserting the values we have,
d = √37.5 * 4/3.14 mm
= 6.9116 mm
or d = 6.91 mm
Therefore, the smallest that should be used is 6.91 mm
(B) To determine the corresponding normal stress that is caused by the tensile force, we input the values in equation (2)
Thus,
σ = P/A
σ= 6 * 10 ^ 3 N/ 37. 5 * 10 ^⁻6 m²
σ= 160 MPa
Note: I MPa = 10^6 N/m²
Hence the the corresponding normal stress is σ= 160 MPa
A .02 kg of R-134a fills a 0.14-m3 weighted piston–cylinder device at a temperature of –26.4°C. The container is now heated until the temperature is 100°C. Determine the final volume of R-134a. Use data from the refrigerant tables.
Answer:
The answer is 0.06027 m³
Explanation:
Solution
Given that:
The first step to take is to determine the initial state of the volume for R-134a refrigerant
Now,
v₁ =V/m
V = the volume of weighted piston cylinder device at the normal state
m = the mass of the R-134a refrigerant
Thus,
We substitute the values 0.14 for V and 0.2 kg for m
Which results in
v₁ = 0.14/0.2
v₁ = 0.7 m₃/kg
The next step is to find the saturated pressure of the R-134a refrigerant from the temperature table of saturated refrigerant R-134a which is equivalent to the normal temperature of 26.4°C.
Thus, by applying the method of interpolation we have the following
P₁ =101.73 - ((101.73-92.76) * (-26-(-26.4)/-26- (-28))
P₁ = 99.936 kPa
So,
The refrigerant in the weighted piston–cylinder device is then heated until the temperature gets to a 100°C
Hence, the temperature and pressure at a state of two becomes
P₂ = 99.936 kPa
T₂ = 100°C
The next step is to determine the specific volume of the refrigerant R-134a at a final state from the super heated refrigerant R-134a which is equivalent to the pressure of 99.936 kPa
v₂ =0.30138 m³ /kg
Now,
we now calculate the final state of the weighted piston cylinder device
V₂ = mv₂
V₂ = 0.2 * 0.30138
V₂ = 0.06027 m³
Hence ,the final volume of the weighted cylinder piston device is 0.06027 m³
The wheel and the attached reel have a combined weight of 50lb and a radius of gyration about their center of 6 A k in = . If pulley B that is attached to the motor is subjected to a torque of M ft lb = − 50 , determine the velocity of the 200lb crate after the pulley has turned 5 revolutions. Neglect the mass of the pulley
The complete question is;
The wheel and the attached reel have a combined weight of 50 lb and a radius of gyration about their center of ka = 6 in. If pulley B that is attached to the motor is subjected to a torque of M = 50 lb.ft, determine the velocity of the 200lb crate after the pulley has turned 5 revolutions. Neglect the mass of the pulley.
The image of this system is attached.
Answer:
Velocity = 11.8 ft/s
Explanation:
Since the wheel at A rotates about a fixed axis, then;
v_c = ω•r_c
r_c is 4.5 in. Let's convert it to ft.
So, r_c = 4.5/12 ft = 0.375 ft
Thus;
v_c = 0.375ω
Now the mass moment of inertia about of wheel A about it's mass centre is given as;
I_a = m•(k_a)²
The mass in in lb, so let's convert to slug. So, m = 50/32.2 slug = 1.5528 slug
Also, let's convert ka from inches to ft.
So, ka = 6/12 = 0.5
So,I_a = 1.5528 × 0.5²
I_a = 0.388 slug.ft²
The kinetic energy of the system would be;
T = Ta + Tc
Where; Ta = ½•I_a•ω²
And Tc = ½•m_c•(v_c)²
So, T = ½•I_a•ω² + ½•m_c•(v_c)²
Now, m_c is given as 200 lb.
Converting to slug, we have;
m_c = (200/32.2) slugs
Plugging in the relevant values, we have;
T = (½•0.388•ω²) + (½•(200/32.2)•(0.375ω)²)
This now gives;
T = 0.6307 ω²
The system is initially at rest at T1 = 0.
Resolving forces at A, we have; Ax, Ay and Wa. These 3 forces do no work.
Whereas at B, M does positive work and at C, W_c does negative work.
When pulley B rotates, it has an angle of; θ_b = 5 revs × 2π rad/revs = 10π
While the wheel rotates through an angle of;θ_a = (rb/ra) • θ_b
Where, rb = 3 in = 3/12 ft = 0.25 ft
ra = 7.5 in = 7.5/12 ft = 0.625 ft
So, θ_a = (0.25/0.625) × 10π
θ_a = 4π
Thus, we can say that the crate will have am upward displacement through a distance;
s_c = r_c × θ_a = 0.375 × 4π
s_c = 1.5π ft
So, the work done by M is;
U_m = M × θ_b
U_m = 50lb × 10π
U_m = 500π
Also,the work done by W_c is;
U_Wc = -W_c × s_c = -200lb × 1.5π
U_Wc = -300π
From principle of work and energy;
T1 + (U_m + U_Wc) = T
Since T1 is zero as stated earlier,
Thus ;
0 + 500π - 300π = 0.6307 ω²
0.6307ω² = 200π
ω² = 200π/0.6307
ω² = 996.224
ω = √996.224
ω = 31.56 rad/s
We earlier derived that;v_c = 0.375ω
Thus; v_c = 0.375 × 31.56
v_c = 11.8 ft/s
The temperature distribution across a wall 0.3 m thick at a certain instant of time is T(x) a bx cx2 , where T is in degrees Celsius and x is in meters, a 200 C, b 200 C/m, and c 30 C/m2 . The wall has a thermal conductivity of 1 W/mK. (a) On a unit surface area basis, determine the rate of heat transfer into and out of the wall and the rate of change of energy stored by the wall. (b) If the cold surface is exposed to a fluid at 100 C, what is the convection coefficient
Answer:
the rate of heat transfer into the wall is [tex]\mathbf{q__{in}} \mathbf{ = 200 W/m^2}[/tex]
the rate of heat output is [tex]\mathbf{q_{out} =182 \ W/m^2}[/tex]
the rate of change of energy stored by the wall is [tex]\mathbf{ \Delta E_{stored} = 18 \ W/m^2 }[/tex]
the convection coefficient is h = 4.26 W/m².K
Explanation:
From the question:
The temperature distribution across the wall is given by :
[tex]T(x) = ax+bx+cx^2[/tex]
where;
T = temperature in ° C
and a, b, & c are constants.
replacing 200° C for a, - 200° C/m for b and 30° C/m² for c ; we have :
[tex]T(x) = 200x-200x+30x^2[/tex]
According to the application of Fourier's Law of heat conduction.
[tex]q_x = -k \dfrac{dT}{dx}[/tex]
where the rate of heat input [tex]q_{in} = q_k[/tex] ; Then x= 0
So:
[tex]q_{in}= -k (\dfrac{d( 200x-200x+30x^2)}{dx})_{x=0}[/tex]
[tex]q_{in}= -1 (-200+60x)_{x=0}[/tex]
[tex]\mathbf{q__{in}} \mathbf{ = 200 W/m^2}[/tex]
Thus , the rate of heat transfer into the wall is [tex]\mathbf{q__{in}} \mathbf{ = 200 W/m^2}[/tex]
The rate of heat output is:
[tex]q_{out} = q_{x=L}[/tex]; where x = 0.3
[tex]q_{out} = -k (\dfrac{dT}{dx})_{x=0.3}[/tex]
replacing T with [tex]200x-200x+30x^2[/tex] and k with 1 W/m.K
[tex]q_{out} = -1 (\dfrac{d(200x-200x+30x^2)}{dx})_{x=0.3}[/tex]
[tex]q_{out} = -1 (-200+60x)_{x=0.3}[/tex]
[tex]q_{out} = 200-60*0.3[/tex]
[tex]\mathbf{q_{out} =182 \ W/m^2}[/tex]
Therefore , the rate of heat output is [tex]\mathbf{q_{out} =182 \ W/m^2}[/tex]
Using energy balance to determine the change of energy(internal energy) stored by the wall.
[tex]\Delta E_{stored} = E_{in}-E_{out} \\ \\ \Delta E_{stored} = q_{in}- q_{out} \\ \\ \Delta E_{stored} = (200 - 182 ) W/m^2 \\ \\[/tex]
[tex]\mathbf{ \Delta E_{stored} = 18 \ W/m^2 }[/tex]
Thus; the rate of change of energy stored by the wall is [tex]\mathbf{ \Delta E_{stored} = 18 \ W/m^2 }[/tex]
We all know that for a steady state, the heat conducted to the end of the plate must be convected to the surrounding fluid.
So:
[tex]q_{x=L} = q_{convected}[/tex]
[tex]q_{x=L} = h(T(L) - T _ \infty)[/tex]
where;
h is the convective heat transfer coefficient.
Then:
[tex]Replacing \ 182 W/m^2 \ for \ q_{x=L} , (200-200x +30x \ for \ T(x) \ , 0.3 m \ for \ x \ and \ 100^0 C for \ T[/tex] We have:
182 = h(200-200×0.3 + 30 ×0.3² - 100 )
182 = h (42.7)
h = 4.26 W/m².K
Thus, the convection coefficient is h = 4.26 W/m².K
The thrust F of a screw propeller is known to depend upon the diameter d,speed of advance \nu ,fluid density p, revolution per second N, and the coefficient of viscosity μ of the fluid. Determine the dimensions of each of the variables in terms of L,M,T,and find an expression for F in terms of these quantities
Answer:
thrust = ML[tex]T^{-2}[/tex]
Explanation:
T = p[tex]V^{2}[/tex][tex]D^{2}[/tex] x [tex]\frac{ND}{V}[/tex]Re
where D is diameter
p is the density
N is the revolution per second
Re is the Reynolds number which is equal to puD/μ
where p is the fluid density
u is the fluid velocity
μ is the fluid viscosity = kg/m.s = M[tex]L^{-1}[/tex][tex]s^{-1}[/tex]
Reynolds number is dimensionless so it cancels out
diameter is m = L
speed is in m/s = L[tex]T^{-1}[/tex]
fluid density is in kg/[tex]m^{3}[/tex] = M[tex]L^{-3}[/tex]
N is in rad/s = L[tex]L^{-1}[/tex][tex]T^{-1}[/tex] = [tex]T^{-1}[/tex]
combining these dimensions into the equation, we have
thrust = M[tex]L^{-3}[/tex][tex]( LT^{-1}) ^{2}[/tex][tex]L^{2}[/tex][tex]\frac{T^{-1}L }{LT^{-1} }[/tex]
= M[tex]L^{-3}[/tex][tex]L^{2}[/tex][tex]T^{-2}[/tex][tex]L^{2}[/tex]
thrust = ML[tex]T^{-2}[/tex] which is the dimension for a force which indicates that thrust is a type of force
difference between a pillar drill and radial arm pillar
drill
Answer:
Pillar drill are small and used for woodworking while Radial arm pillar is mounted on a very large column.
21.Why are throttling devices commonly used in refrigeration and air-conditioning
applications?
Answer is given below
Explanation:
we know that some common types of throttling devices are
Hard -throttling devices Capillary valve Constant pressure throttling devicesThermostatic expansion valve Float expansion valveso here throttling devices commonly used in refrigeration and air-conditioning because
To reduce the coolant pressure, the high pressure of the refrigerant from the condenser is necessary to reduce the evaporation to obtain evaporation at the right temperature To meet the refrigerated load, the throttling valve flows through the coolant to cool the load at high temperatures.A process is in control with mean 50 and standard deviation of 2. The upper specification limit for the product being produced is 60 and the lower specification limit is 42. What is the value of Cpk?
Answer:
Cpk = 1.33
Explanation:
Given:
Mean = 50
Sd = 2
USL = 60
LSL = 42
The Cpk means process capability index. it helps decide the specification limit when the nominal value is not the central value of upoer specification limint (USL) and lower specification limit (LSL).
The Cpk can be derived using the formula:
[tex]Cpk = min [\frac{(usl - mean)}{3 * \sigma}, \frac{(mean - lsl)}{3*\sigma}] [/tex]
[tex]Cpk = min [\frac{(60 - 50)}{3*2} , \frac{(50 - 42)}{3*2}] [/tex]
Solving further,
[tex]Cpk = min [\frac{10}{6} , \frac{8}{6}] [/tex]
Cpk = min ( 1.67 , 1.33)
Cpk = 1.33
Cpk = 1.33
Pressure in U.S. customary units is measured in psi (pound per square inch). In SI metric units pressure is measured in Pa (N/m2). Write a user-defined MATLAB function that converts pressure given in units of psi to pressure in units of Pa. For the function name and arguments, use [Pa] = Psi-ToPa(psi). The input argument psi is the pressure in units of psi to be converted, and the output argument Pa is the converted pressure in units of Pa (rounded to the nearest integer). Use the function in the Command Window to: (a) Convert 120 psi to units of Pa. (b) Convert 3,000 psi to units of Pa.
Answer:
Check below for answers
Explanation:
Matlab code:
function[Pa] = Psi-ToPa(psi)
Pa = psi * 6894.75728;
end
a) To convert 120 psi to units of Pa, just call the function Pa using the command:
Psi-ToPa(120)
ans =
8.2737e+05
b) To convert 3000 psi to units of Pa, just call the function Pa using the command:
Psi-ToPa(3000)
ans =
2.0684e+07
A thin-walled pressure vessel is constructed by rolling a 6 mm thick steel sheet into a cylindrical shape, welding the seam along line A-B, and capping the ends. The vessel is subjected to an internal pressure of 1.25 MPa. What is the normal stress normal to line A-B?
Answer:
41.015
Explanation:
Solution
Given that:
Th first step to take is to find the longitudinal stress in the cylinder
σl = PD/4t
P = the pressure
D = the diameter
t = the thickness
Thus,
σl = 1.25 * ^ 6 * 0.45 / 4 * 6 * 10 ^ ⁻3
=23.475Mpa
Now. we find the hoop stress in the cylinder∠
σh = PD/2t
σh = 1.25 * ^ 6 * 0.45/ 2 * 6 * 10 ^ ⁻3
σh =46.875 Mpa
Then
we find the normal stress in the line of the 30° angle with the longitudinal axis stated below:
σab = σh + σl/2 + ( σh - σl/2) cos 2θ + t sin 2θ
So,
σab =46.875 + 23.4375/2 + ( 46.875 - 23.4375/2) cos 2(30°) + 0
σab= 41.015
Therefore the normal stress to line A-B is 41.015
what are the non-functional requirements of a washing machine
Answer:
a laudry automaat has no nonfunctional parts
whywould they put something in there that has no function ?
You built an android that has a subcircuit containing a power supply, a tactile sensor, and a fuse where safe operation should keep current below 250 mA. You measured that your sensor is dissipating 12 W, the power supply is pro- viding 12.2 W, and the voltage drop across the fuse is 500 mV. Is your circuit properly protected?
Answer:
a co mam zroic!!
Explanation:
Explain the advantages of continuous insulation in envelope assemblies and why it is better to have the continuous insulation on the exterior face of the envelope. Where is inside placement of continuous insulation in the envelope desirable
Answer:
Continuous insulation helps in eliminating the necessity of applying extra materials to achieve moisture barrier demands, reducing labor and material cost.
Continuous insulation helps building last much longer over a period of time,without the need to upgrade and repair.
Inside the placement of continuous insulation in the envelope needed includes the following; the foundation wall or slab insulation, balcony interface, bond joist insulation, insulation against sub floor, on the interior of masonry wall.
Explanation:
Solution:
The analytical path to success contains a well planned and designed building exterior, which avoids the loss of energy, control cost and the maximization of technology advancement in materials.
Property installed continuous insulation on the exterior can also execute as an air barrier. the flashing of wall penetrations can form into a drainage plane. this plane can stop potentially damaging moisture from entering into the wall assembly.
By making use of continuous insulation it helps in removing necessity of applying extra materials to achieve moisture barrier demands, reducing labor and material cost.
Continuous insulation helps building to withstand the test of time without the need to upgrade and repair
Now inside the placement of continuous insulation in the envelope is needed as follows:
Foundation wall or slab insulationInsulation against sub floorBalcony interfacesOn the interior of masonry wallBond joist insulationQUESTIONS
DOK S STANDARD RST.1
MS-PS2-2
A car driving at a constant speed of 20 m/s(meters per second)
turns right, down a street.
1. Is the car accelerating during the turn? Why or why not?
2. Would you feel any of the movement during this event? Why
or why not?
ANSWER:
1) The car is not accelerating, because for the the car to turn while accelerating down the street, it has to deaccelerate (reduce speed) due to CENTRIPETAL FORCE that is acting on the car during the turn. This force acts to bring the car in the direction where it has turned to.
2) You will feel the movement as everyone in the car will tend to fall to the left, while the car makes a right turn. This is because of CENTRIFUGAL FORCE, which is acting to take the car to a left direction, while the car is turning to the right direction.
(TCO 4) In a CAN bus, there are three computers: Computer A, Computer B, and Computer C. The messages being sent are 2 milliseconds long. Computer A starts sending a message on the bus; 250 microseconds later, Computer B starts to send a message, and 250 microseconds after that, Computer C starts to send a message. When does Computer A retransmit its message?
A) After 2.5 milliseconds
B) After 1 millisecond
C) After 1.5 milliseconds
D) Never; Computer A will not retransmit.
Answer:
A) After 2.5 milliseconds
Explanation:
Given that :
In a CAN bus, there are three computers: Computer A, Computer B, and Computer C.
Length in time of the message sent are 2 milliseconds long
As the computer start sending the message; the message are being sent at a message duration rate of 2 milliseconds
250 microseconds later, Computer B starts to send a message; There is a delay of 250 microseconds = 0.25 milliseconds here at Computer B
and 250 microseconds after that, Computer C starts to send a message
Similarly; delay at Computer C = 0.25 milliseconds
Assuming [tex]T_{RT}[/tex] is the retransmit time for Computer A to retransmit its message, Then :
[tex]T_{RT}[/tex] = [tex]T_A + T_B + T_C[/tex]
[tex]T_{RT}[/tex] = 2 milliseconds + 0.25 milliseconds + 0.25 milliseconds
[tex]T_{RT}[/tex] = 2.5 milliseconds
Thus; the correct option is A) After 2.5 milliseconds
Air flows steadily and isentropically from standard atmospheric conditions to a receiver pipe through a converging duct. The cross-sectional area of the throat of the converging duct is 0.05 ft2. Determine the mass flowrate in lbm/s through the duct if the receiver pressure is 10 psia.
Answer:
The answer is "0.0728"
Explanation:
Given value:
[tex]P_0= 14.696\ ps\\\\\ p _{0}= 0.00238 \frac{slue}{ft^{3}}\\\\\ A= 0.05 ft^2\\\\\ T_0= 59^{\circ}f = 518.67R\\\\\ air \ k= 1\\\\ \ cirtical \ pressure ( P^*)=P_0\times \frac{2}{k+1}^{\frac{k}{k-1}}\\[/tex]
[tex]= 14.696\times (\frac{2}{1.4+1})^{\frac{1.4}{1.4-1}}\\\\=7.763 Psia\\\\[/tex]
if [tex]P<P^{*} \to[/tex] flow is chocked
if [tex]P>P^{*} \to[/tex] flow is not chocked
When P= 10 psia < [tex]P^{*}[/tex] [tex]\to[/tex] not chocked
match number:
[tex]\ for \ P= \ 10\ G= \sqrt{\frac{2}{k-1}[(\frac{\ p_{0}}{p})^{\frac{k-1}{k}}-1]}[/tex]
[tex]= \sqrt{\frac{2}{1.4-1}[(\frac{14.696}{10})^{\frac{1.4-1}{1.4}}-1]}[/tex]
[tex]M_0=7.625[/tex]
[tex]p=p_0(1+\frac{k-1}{2} M_0 r)^\frac{1}{1-k}[/tex]
[tex]=0.00238(1+\frac{1.4-1}{2}0.7625`)^{\frac{1}{1-1.4}}\\\\\ p=0.001808 \frac{slug}{ft^3}[/tex]
[tex]\ T= T_0(1+\frac{k-1}{2} Ma^r)^{-1}\\\\\ T=518.67(1+\frac{1.4-1}{2} 0.7625^2)^{-1}\\\\\ T=464.6R\\\\[/tex]
[tex]\ velocity \ of \ sound \ (C)=\sqrt{KRT}\\\\[/tex]
[tex]=\sqrt{1.4\times1716\times464.6}\\\\=1057 ft^3\\\\[/tex]
R= gas constant=1716
[tex]m=PAV\\\\[/tex]
[tex]=0.001808\times0.05\times(Ma.C)\\\\=0.001808\times0.05\times0.7625\times 1057\\\\=0.0728\frac{slug}{s}[/tex]
A three-phase, 415 V load takes a line current of 800 A from a 3300/415V delta – star transformer. The 3300V system is supplied from an 11000/3300V star – star transformer. Draw the circuit diagram and, assuming no losses, find both line and phase values of voltages and currents in each part of the circuit. What will be the turns ratios of both transformers?
Answer:
Delta star transformer
Secondary side
[tex]V_L[/tex] = 415 V
[tex]V_P[/tex] = 239.6 V
[tex]I_P[/tex] = Line current, [tex]I_L[/tex] = 800 A
Primary side
[tex]V_L[/tex] = Phase voltage, [tex]V_P[/tex] = 3300 V
Primary phase current = 58.08 A
[tex]I_L[/tex] = 100.6 A
Turns ratio = 83/(660·√3)
Star-star transformer
Secondary side
[tex]V_L[/tex] = 3300 V
[tex]V_P[/tex] = 1905.26 V
[tex]I_P[/tex] = Line current, [tex]I_L[/tex] = 100.6 A
Primary side
[tex]V_L[/tex] = 11000 V
[tex]V_P[/tex] = 6350.9 V
Line current, [tex]I_L[/tex] = Phase current, [tex]I_P[/tex] = 3.018 A
Turns ratio = 3/100.
Explanation:
Delta star transformer
For the star Connection to the 3-Phase load,
Secondary line voltage, [tex]V_L[/tex] = 415 V
Secondary phase voltage, [tex]V_P[/tex] = [tex]V_L[/tex] ÷ √3 = 415 ÷ √3 = 239.6 V
Secondary phase current, [tex]I_P[/tex] = Line current, [tex]I_L[/tex] = 800 A
For the primary side of the delta star transformer
Line voltage, [tex]V_L[/tex] = Phase voltage, [tex]V_P[/tex] = 3300 V
K = 415/(3300×√3) = 83/(660·√3)
Primary phase current = 83/(660·√3)× 800 A = 58.08 A
Line current, [tex]I_L[/tex] = Phase current, [tex]I_P[/tex] × √3 = 100.6 A
The turns ratio = 83/(660·√3)
For the star-star transformer, we have;
Secondary line voltage, [tex]V_L[/tex] = 3300 V
Secondary phase voltage, [tex]V_P[/tex] = [tex]V_L[/tex] ÷ √3 = 3300 ÷ √3 = 1905.26 V
Secondary phase current, [tex]I_P[/tex] = Line current, [tex]I_L[/tex] = 100.6 A
For the primary side of the star-star transformer
Line voltage, [tex]V_L[/tex] = 11000 V
Phase voltage, [tex]V_P[/tex] = 11000 V ÷ √3 = 6350.9 V
K = 3300/11000 = 3/100
Primary phase current = 3/100× 100.6 = 3.018 A
Line current, [tex]I_L[/tex] = Phase current, [tex]I_P[/tex] = 3.018 A
The turns ratio = k = 3/100.
I. The time till failure of an electronic component has an Exponential distribution and it is known that 10% of components have failed by 1000 hours. (a) Find the mean and standard deviation of the time till failure. (b) What is the probability that a component is still working after 5000 hours? (c) What is the probability that three components fail in one hour? (d) What is the probability that at least two components fail in one half hour?
Answer:
(a) The mean time to fail is 9491.22 hours
The standard deviation time to fail is 9491.22 hours
(b) 0.5905
(c) 3.915 × 10⁻¹²
(d) 2.63 × 10⁻⁵
Explanation:
(a) We put time to fail = t
∴ For an exponential distribution, we have f(t) = [tex]\lambda e^{-\lambda t}[/tex]
Where we have a failure rate = 10% for 1000 hours, we have(based on online resource);
[tex]P(t \leq 1000) = \int\limits^{1000}_0 {\lambda e^{-\lambda t}} \, dt = \dfrac{e^{1000\lambda}-1}{e^{1000\lambda}} = 0.1[/tex]
e^(1000·λ) - 0.1·e^(1000·λ) = 1
0.9·e^(1000·λ) = 1
1000·λ = ㏑(1/0.9)
λ = 1.054 × 10⁻⁴
Hence the mean time to fail, E = 1/λ = 1/(1.054 × 10⁻⁴) = 9491.22 hours
The standard deviation = √(1/λ)² = √(1/(1.054 × 10⁻⁴)²)) = 9491.22 hours
b) Here we have to integrate from 5000 to ∞ as follows;
[tex]p(t>5000) = \int\limits^{\infty}_{5000} {\lambda e^{-\lambda t}} \, dt =\left [ -e^{\lambda t}\right ]_{5000}^{\infty} = e^{5000 \lambda} = 0.5905[/tex]
(c) The Poisson distribution is presented as follows;
[tex]P(x = 3) = \dfrac{\lambda ^x e^{-x}}{x!} = \frac{(1.0532 \times 10^{-4})^3 e^{-3} }{3!} = 3.915\times 10^{-12}[/tex]
p(x = 3) = 3.915 × 10⁻¹²
d) Where at least 2 components fail in one half hour, then 1 component is expected to fail in 15 minutes or 1/4 hours
The Cumulative Distribution Function is given as follows;
p( t ≤ 1/4) [tex]CDF = 1 - e^{-\lambda \times t} = 1 - e^{-1.054 \times 10 ^{-4} \times 1/4} = 2.63 \times 10 ^{-5}[/tex].
A. The mean time to fail is:
For an exponential distribution, we have f(t) = e^(1000·λ) - 0.1·e^(1000·λ)= 1
0.9·e^(1000·λ) = 1
1000·λ = ㏑(1/0.9)
λ = 1.054 × 10⁻⁴
Hence the mean time to fail, E = 1/λ = 1/(1.054 × 10⁻⁴) = 9491.22 hours
The standard deviation = √(1/λ)² = √(1/(1.054 × 10⁻⁴)²)) = 9491.22 hours
Exponential distribution
Exponential distribution is the probability the distribution of the time between events in a Poisson point process.
B. Here we have to integrate from 5000 to ∞ as follows: 0.5905
C. The Poisson distribution is presented as follows;
p(x = 3) = 3.915 × 10⁻¹²
D. Where at least 2 components fail in one half hour, then 1 component is expected to fail in 15 minutes or 1/4 hours.
The Cumulative Distribution Function is given as follows; p( t ≤ 1/4) = 2.63 × 10⁻⁵
Therefore, the correct answers are:
(a) The mean time to fail is 9491.22 hours
The standard deviation time to fail is 9491.22 hours
(b) 0.5905
(c) 3.915 × 10⁻¹²
(d) 2.63 × 10⁻⁵
Read more about exponential here:
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With reference to the vector loop of Section 4.6, the cross and the open configurations of the circuit will produce:_______1. only different angles for theta 3 andTheta 4 2. should have the same values in both cases 3. different angles for theta 3 and for theta 4 sames angles for theta 3 and for theta 4 4. only different angles for theta Theta 3 should have the same values in both cases
Answer:
Option 4 => only different angles for theta 4, Theta 3 should have the same values in both cases.
Explanation:
The vector loop of Section 4.6 that was made as the reference diagram has the vector loop closing itself and the vectors summing around loop zero. Also, it has one DOF mechanism and the vector length is equal to the vector link.
NOTE: Kindly check the attached file or picture for the diagram of Section 4.6 diagram.
We can see from the diagram below that the V3 and V4 can be determined or Calculated with parameters such as theta 2, all links length and length d.
Therefore, the cross and the open configurations of the circuit will produce from the diagram will be " only different angles for theta 4, Theta 3 should have the same values in both cases."( That is option 4)
What is the transfer function of a filter? What is the transfer function of a filter? The transfer function shows how a filter affects the phase of input components as a function of frequency. It is defined as the output phasor divided by the input phasor as a function of frequency. The transfer function shows how a filter affects the amplitude and phase of input components as a function of frequency. It is defined as the input phasor divided by the output phasor as a function of frequency. The transfer function shows how a filter affects the amplitude and phase of input components as a function of frequency. It is defined as the output phasor divided by the input phasor as a function of frequency. The transfer function shows how a filter affects the phase of input components as a function of frequency. It is defined as the input phasor divided by the output phasor as a function of frequency.
Answer: Provided in the explanation section
Explanation:
A precise explanation is provided here to make it easy for understanding
(a). The transfer option shows how a filter affects the amplitude and phase of input components as a function of frequency. It is defined as the output phasor divided by the input phasor as a functoon of frequency.
H(jw) = V₀ (jw) / Vin(jw)
(b). To determine the transfunction of a filter, we connect a sinusoidal source to be the input part, measure the amplitude and phases of both the input signal and resulting output signal using voltmeters.
Oscilloscope or other instruments, and divide the Output phasor by the input phasor. This is is repeated for all frequencies.
cheers i hope this helped !!
With a reservoir pressure of 1.0 MPa and temperature of 750 K, air enters a converging-diverging nozzle, in a steady fashion. Flow is isentropic and k=1.4. If exit Mach number is 2 and throat area is 20 cm2 , find (a) the throat conditions (static pressure, temperature, density, and mach number), (b) the exit plane conditions i
Answer:
a) P* = 0.5283 MPa , T* = 624.75 K , ρ* = 2.945 kg/m^3 , V* = 501.023 m/s
b) Pe = 0.1278 MPa , Te = 416.7 K , ρe = 1.069 kg/m^3 , Ve = 818.36 m/s, Ae = 33.75 cm^2
c) m' = 2.915 kg/s
Explanation:
Given:-
- The inlet pressure, Pi = 1.0 MPa
- The inlet temperature, Ti = 750 K
- Inlet velocity is negligible
- Steady, Isentropic Flow
- The specific heat ratio of air, k = 1.4
- Exit Mach number, Mae = 2
- The throat area, Ath = 20 cm^2
- Gas constant of air, R = 0.287 KJ / kg.K
Find:-
(a) the throat conditions (static pressure, temperature, density, and mach number)
b) the exit plane conditions
c) the mass flow rate
Solution:-
- For this problem we will assume air to behave like an ideal gas with constant specific heat at RTP. Also the flow of air through the nozzle is assumed to be steady, one dimensional, and Isentropic with constant specific heat ratio ( k ).
- First we will scrutinize on the exit conditions. We have a Mach number of 2 at the exit. The flow at the exit of converging-diverging nozzle is in super-sonic region this is only possible only if sonic ( Ma = 1 ) conditions are achieved by the flow at the throat area ( minimum cross-sectional area ).
- Moreover, the flow is almost still at the inlet. Hence, we can assume that the flow has negligible velocity ( vi = 0 m/s ) at the inlet and the reservoir temperature and pressure can be assumed to be stagnation temperature and pressures as follows:
[tex]P_o = 1.0 MPa\\\\T_o = 750 K[/tex]
- Using the ideal gas law we can determine the stagnation density ( ρo ) as follows:
[tex]p_o = \frac{P_o}{RT_o} = \frac{1000}{0.287*750} = 4.64576\frac{kg}{m^3}[/tex]
- We will use the already developed results for flow which has reached sonic velocity ( Ma = 1 ) at the throat region. Use Table A - 13, to determine the critical static values at the throat region:
[tex]\frac{P^*}{P_o} = 0.5283\\\\P^* = 0.5283*1 = 0.5283 MPa\\\\\frac{T^*}{T_o} = 0.8333\\\\T^* = 0.8333*750 = 624.75 K\\\\ \frac{p^*}{p_o} = 0.6339\\\\p^* = 0.6339*4.64576 = 2.945 \frac{kg}{m^3} \\\\[/tex]
[tex]V^* = \sqrt{kRT^*} =\sqrt{1.4*287*624.75} = 501.023 \frac{m}{s}[/tex]
- Similarly, we will again employ the table A - 13 to determine the exit plane conditions for ( Ma = 2 ) as follows:
[tex]\frac{P_e}{P_o} = 0.1278 \\\\P_e = 0.1278*1.0 = 0.1278 MPa\\\\\frac{T_e}{T_o} = 0.5556 \\\\T_e = 0.5556*750 = 416.7 K\\\\\frac{p_e}{p_o} = 0.23 \\\\p_e = 0.23*4.64576 = 1.069 \frac{kg}{m^3} \\\\\frac{A_e}{A_t_h} = 1.6875 \\\\A_e =1.6875*20 = 33.75 cm^2\\[/tex]
- The velocity at the exit plane ( Ve ) can be determined from the exit conditions as follows:
[tex]V_e = Ma_e*\sqrt{kRT_e} = 2*\sqrt{1.4*287*416.7} = 818.36 \frac{m}{s}[/tex]
- For steady flows the mass flow rate ( m' ) is constant at any section of the nozzle. We will use the properties at the throat section to determine the mass flow rate as follows:
[tex]m' = p^* A_t_h V^*\\\\m' = 2.945*20*10^-^4*501.023\\\\m' = 2.951 \frac{kg}{s}[/tex]
19.11 The atmosphere is stable with a lapse rate of –0.2 °C/100 m. The surface air temperature is 15 °C. A parcel of air is released at the ground with a temperature of 25 °C. Calculate the maximum mixing height, in m.
Answer:
The maximum mixing height in meters is 5,000 m
Explanation:
In this question, we are expected to calculate the maximum mixing height in meters.
We identify the following parameters;
Lapse rate(L) = 0.2 °C/100 m = 0.002 °C/m (divide 0.2 by 100 to get this)
Mathematically;
Lapse rate(L) = Temperature difference/altitude (h)
Where the temperature difference is (25 °C -15 °C) = 10°C
Substituting this in the formula, we have;
0.002 = 10/h
h * 0.002 = 10
h = 10/0.002
Altitude(h) = 5,000 m
An engineering firm just lost one of their larger customers. The firm president says that the solution to this problem is to fire three engineers to balance the firm’s workforce with the current level of business. He says it is a simple problem with a simple solution.a) The three engineers disagree. Why?b) What are the ethical factors form the perspective of the firm president and the engineers?
Answer:
The engineers disagreed because their jobs were on the line
The ethical factors are:
The reason for the customer dumping the business is yet to be figured out
The need to keep cost to the lowest ebb in order to keep maintain profitability at the expense of employees' welfare
The are several ways of growing customer base which are yet to be exploited.
Explanation:
The engineers disagree because there is no direct connection between the company's loss of the customer and their proposed layoff of the engineers,at least no one strong evidence has been given by the president.
The ethical factors inherent in this case are as follows:
The reason for the customer dumping the business is yet to be figure out
The need to keep cost to the lowest ebb in order to keep maintain profitability.
The are several ways of growing customer base which are yet to be exploited.
There is a need for a fact-finding exercise to establish the main motive behind losing such customer,without which the company can run into more troubles in future,otherwise the company would keep firing its good hands each time a customer dumps it.
Also,the president had resulted into such decision in order to maintain company's margins,where then lies ethics of welfare economics?Welfare economics is about looking beyond margins and looking at issues from a wider perspective of fulfilling the needs of employees in order for them to put in their best performance,at least by granting them job safety.
The company could have also grown business by investing in new technology that sets it apart from competitors instead of just jumping into the conclusions of sacking employees in a business where the company's strength lies in quality of engineers that it has.
Determine the flow velocities at the inlet and exit sections of an
inclined tapering pipe using fluid flow theory and given pressure
readings and flow rates.
There is a sloping pipeline that has one end 1.35 m higher than the
other. The pipe section tapers from 0.95 m diameter at the top end to
0.44m diameter at the lower end. The difference in pressure between
the two sections is 12.35kPa, with pressure being greater at higher
level.
Your task is to determine the inlet and exit velocities and the
volume
flow rate of the inclined pipe.
Answer:
The inlet velocity v₁ = 6.66 m/s, the exit velocity v₂ = 7.32 m/s and the volume flow rate Q = 4.72 m³/s
Explanation:
Using Bernoulli's equation
P₁ + ρgh₁ + 1/2ρv₁² = P₂ + ρgh₂ + 1/2ρv₂²
P₁ - P₂ + ρgh₁ - ρgh₂ = 1/2ρv₂² - 1/2ρv₁²
ΔP + ρgΔh = 1/2ρ(v₂² - v₁²) (1)
where ΔP = pressure difference = 12.35 kPa = 12350 Pa
Δh = height difference = 1.35 m
From the flow rate equation Q = A₁v₁ = A₂v₂ and v₁ = A₂v₂/A₁ = d₂²v₂/d₁² where v₁ and v₂ represent the inlet and exit velocities from the pipe and d₁ = 0.95 m and d₂ = 0.44 m represent the diameters at the top end and lower end of the pipe respectively.
Substituting v₁ into (1), we have
ΔP + ρgΔh = 1/2ρ(v₂² - (d₂²v₂/d₁² )²)
ΔP + ρgΔh = 1/2ρ(v₂² - (d₂/d₁)⁴v₂²)
v₂ = √[2(ΔP + ρgΔh)/ρ(1 - (d₂/d₁)⁴)}
substituting the values of the variables, we have
v₂ = √[2(12350 Pa + 1000 kg/m³ × 9.8 m/s² × 1.35 m)/(1000 kg/m³ (1 - (0.44 m/0.95 m)⁴))}
= √[2(12350 Pa + 13230 Pa)/(1000 kg/m³ × 0.954)]
= √[2(25580 Pa)/954 kg/m³]
= √[51160 Pa/954 kg/m³]
= √53.627
= 7.32 m/s
v₁ = d₂²v₂/d₁²
= (0.44 m/0.95 m)² × 7.32 m/s
= (0.954)² × 7.32 m/s
= 6.66 m/s
The volume flow rate Q = A₁v₁
= πd₁²v₁/4
= π(0.95 m)² × 6.66 m/s ÷ 4
= 18.883 m³/s ÷ 4
= 4.72 m³/s
So, the inlet velocity v₁ = 6.66 m/s, the exit velocity v₂ = 7.32 m/s and the volume flow rate Q = 4.72 m³/s
How is the minimum circuit ampacity for an air-conditioning condensing unit calculated?
1) compressor amps + fan amps
2) (compressor amps x 1.25) + fan amps
3) compressor amps x fan amps
4) (compressor amps + fan amps) x 1.25
The minimum circuit ampacity for an air-conditioning condensing unit calculated by compressor amps + fan amps. The correct option is 1.
What is ampacity?Ampacity is the term used for the greatest current conveying limit, in amperes, of a specific electrical gadget. The current carrying capacity is generally depended upon the electrical cable and is calculated as the maximum amount of current a cable can withstand before it warms past the most extreme working temperature.
So, in air conditioning units, the greatest current carrying capacity is equal to the addition of compressor and fan current capacity.
Thus, the correct option is 1.
Learn more about ampacity.
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1. The 1992 Ford Crown Victoria came with a 190 horsepower engine. It has a drag coefficient of 0.33 and a frontal projected area of 26.36 ft2. Assuming that all of the available power from the engine is used to overcome the drag force, then what would the maximum speed (in miles per hour) be for this car
Answer:
1.50 miles/hour is the correct answer to the given question
Explanation:
As mention in the question engine power = 190 horsepower engine.......eq(1)
we have to converted into the watt as we know that 1 horse power=746 watt
putting this value in eq(1) we get
[tex]Engine \ power\ =190 *746\\\ Engine \ power =146216\ NM/S[/tex]
Drag coefficient = 0.33
Projected area = 26.36 [tex]ft^{2}[/tex]............eq(2)
we have to converted into the [tex]m^{2}[/tex] as we know that [tex]1m^{2}\ =\ 10.7639\ ft^{2}[/tex]
putting this value in eq(2) we get
Projected area = [tex]2.448921\ m^{2}[/tex]
Now we to calculated the Drag Force
[tex]F\ =\frac{\ Drag \ coefficient\ *P*\ V^{2}\ *A }{2}[/tex]....................eq(3)
Putting the above value in eq(3) we get
[tex]F=\ \frac{0.33\ * 1.2 * \ V^{2 \ * 2.44821\ *N} }{2} \\F=\ 0.4848V^{2}[/tex]
As we know that
[tex]F=0.4848\ V^{2}\ *V \\F= \frac{\ 0.4848V^{3} NM}{S}[/tex]
As mention in the question
Power of Engine =Drag free force
[tex]146216\ =\ 0.4848\ V^{3}\\ V^{3} =301550.889\\V=67.05M/S[/tex]
We have to converted this value into the miles/hour
1 M/S =2.22369 M/H
Putting this value into the V we get
V=67.05 * 2.2234 M/H
V=150 M/H
. Air at 200 C blows over a 50 cm x 75 cm plain carbon steel (AISI 1010) hot plate with a constant surface temperature of 2500 C. The convection heat transfer coefficient is 25 W/m2K. Assume that 300 W is lost from the surface by radiation. Calculate the inside plate temperature. Assume that the plate is 2 cm thick. State all assumptions
Answer:
The inside temperature, [tex]T_{in}[/tex] is approximately 248 °C.
Explanation:
The parameters given are;
Temperature of the air = 20°C
Carbon steel surface temperature 250°C
Area of surface = 50 cm × 75 cm = 0.5 × 0.75 = 0.375 m²
Convection heat transfer coefficient = 25 W/(m²·K)
Heat lost by radiation = 300 W
Assumption,
Air temperature = 20 °C
Hot plate temperature = 250 °C
Thermal conductivity K = 65.2 W/(m·K)
Steady state heat transfer process
One dimensional heat conduction
We have;
Newton's law of cooling;
q = h×A×([tex]T_s[/tex] - [tex]T_{\infty[/tex]) + Heat loss by radiation
= 25×0.325×(250 - 20) + 300
= 2456.25 W
The rate of energy transfer per second is given by the following relation;
[tex]P = \dfrac{K \times A \times \Delta T}{L}[/tex]
Thermal conductivity K = 65.2 W/(m·K)
Therefore;
[tex]2456.25 = \dfrac{65.2 \times 0.375 \times (250 - T_{in})}{0.02}[/tex]
[tex]T_{in} = 250 - \dfrac{2456.25 \times 0.02}{65.2 \times 0.375} = 247.99 ^{\circ}C[/tex]
The inside temperature, [tex]T_{in}[/tex] = 247.99 °C ≈ 248 °C.