Answer: Technician A is wrong , Technician B is correct.
Explanation:
The Air fuel ratio is used in the combustion process of engines whether internal or external, and defined as the mass ratio of air to fuel present in a combustion process. When the ratio of the mass of air and fuel is right, combustion is said to be efficient and balanced and this is refereed to as stoichiometrc mixture which is a standard that all engines are required to target.
when the air fuel Ratios is lower than the target stoichiometric, it is considered as a Rich mixture while higher ratios than stoichiometric are considered Lean mixtures. Lean mixtures are more efficient than rich mixtures although the main aim is to be at the stoichiometic mixture point.
Air fuel equivalence ratio, lambda and the amount of oxygen in an exhaust are two indicators for lean or rich mixtures in an air fuel ratio for combustion in engines
Now, Air–fuel equivalence ratio, λ (lambda), is defined as the ratio of actual AFR to stoichiometry for a particular mixture,
when Lambda λ = 1.0 , the mixture is at stoichiometry,
when λ < 1.0, the mixtrure is rich.
when λ > 1.0, the mixture is lean
Therefore Technician A is wrong
Also The amount of oxygen present in the exhaust determines if the air-fuel mixture supplied to the combustion chambers is at the target or best ratio. Too much oxygen indicates that the mixture is lean and too little indicates the mixture is rich. Therefore Technician B is right.
When the transportation of natural gas in a pipeline is not feasible for economic reasons, it is first liquefied using nonconventional refrigeration techniques and then transported in super-insulated tanks. In a natural gas liquefaction plant, the liquefied natural gas (LNG) enters a cryogenic turbine at 30 bar and –160°C at a rate of 20 kg/s and leaves at 3 bar. If 120 kW power is produced by the turbine, determine the efficiency of the turbine. Take the density of LNG to be 423.8 kg/m3.
Answer:
the isentropic efficiency of turbine is 99.65%
Explanation:
Given that:
Mass flow rate of LNG m = 20 kg/s
The pressure at the inlet [tex]P_1 =30 \ bar[/tex] = 3000 kPa
turbine temperature at the inlet [tex]T_1 = -160^0C[/tex] = ( -160+273)K = 113K
The pressure at the turbine exit [tex]P_2 = 3 bar[/tex] = 300 kPa
Power produced by the turbine W = 120 kW
Density of LNG [tex]\rho = 423.8 \ kg/m^3[/tex]
The formula for the workdone by an ideal turbine can be expressed by:
[tex]W_{ideal} = \int\limits^2_1 {V} \, dP[/tex]
[tex]W_{ideal} ={V} \int\limits^2_1 \, dP[/tex]
[tex]W_{ideal} ={V} [P]^2_{1}[/tex]
[tex]W_{ideal} ={V} [P_1-P_2][/tex]
We all know that density = mass * volume i.e [tex]\rho= m*V[/tex]
Then ;
[tex]V = \dfrac{m}{\rho}[/tex]
replacing it into the above previous derived formula; we have:
[tex]W_{ideal} ={ \dfrac{m}{\rho}} [P_1-P_2][/tex]
[tex]W_{ideal} ={ \dfrac{20}{423.8}} [3000-300][/tex]
[tex]W_{ideal} ={ \dfrac{20}{423.8}} [2700][/tex]
[tex]W_{ideal} =0.04719*[2700][/tex]
[tex]W_{ideal} =127.42 kW[/tex]
However ; the isentropic efficiency of turbine is given by the relation:
[tex]n_{isen} =\dfrac{W}{W_{ideal}}[/tex]
[tex]n_{isen} =\dfrac{120}{120.42}[/tex]
[tex]n_{isen} =0.9965[/tex]
[tex]n_{isen} =[/tex] 99.65%
Therefore, the isentropic efficiency of turbine is 99.65%
An Ideal gas is being heated in a circular duct as while flowing over an electric heater of 130 kW. The diameter of duct is 500 mm. The gas enters the heating section of the duct at 100 kPa and 27 deg C with a volume flow rate of 15 m3/s. If heat is lost from the gas in the duct to the surroundings at a rate of 80 kW, Calculate the exit temperature of the gas in deg C. (Assume constant pressure, ideal gas, negligible change in kinetic and potential energies and constant specific heat; Cp =1000 J/kg K; R = 500 J/kg K)
Answer:
Exit temperature = 32°C
Explanation:
We are given;
Initial Pressure;P1 = 100 KPa
Cp =1000 J/kg.K = 1 KJ/kg.k
R = 500 J/kg.K = 0.5 Kj/Kg.k
Initial temperature;T1 = 27°C = 273 + 27K = 300 K
volume flow rate;V' = 15 m³/s
W = 130 Kw
Q = 80 Kw
Using ideal gas equation,
PV' = m'RT
Where m' is mass flow rate.
Thus;making m' the subject, we have;
m' = PV'/RT
So at inlet,
m' = P1•V1'/(R•T1)
m' = (100 × 15)/(0.5 × 300)
m' = 10 kg/s
From steady flow energy equation, we know that;
m'•h1 + Q = m'h2 + W
Dividing through by m', we have;
h1 + Q/m' = h2 + W/m'
h = Cp•T
Thus,
Cp•T1 + Q/m' = Cp•T2 + W/m'
Plugging in the relevant values, we have;
(1*300) - (80/10) = (1*T2) - (130/10)
Q and M negative because heat is being lost.
300 - 8 + 13 = T2
T2 = 305 K = 305 - 273 °C = 32 °C
Time, budget, and safety are almost always considered to be
1. Efficiency
2. Constraints
3. Trade-offs
4. Criteria
Answer:
The answer is option # 2. (Constraints).
Eight switches are connected to PORTB and eight LEDs are connected to PORTA. We would like to monitor the first two least significant bits of PORTB (use masking technique). Whenever both of these bits are set, switch all LEDs of Port A on for one second. Assume that the name of the delay subroutine is DELAY. You do not need to write the code for the delay procedure.
Answer:
In this example, the delay procedure is given below in the explanation section
Explanation:
Solution
The delay procedure is given below:
LDS # $4000 // load initial memory
LDAA #$FF
STAA DDRA
LDAA #$00 //load address
STAA DDRB
THERE LDAA PORT B
ANDA #%00000011// port A and port B
CMPA #%0000011
BNE THERE
LDAA #$FF
STAA PORT A
JSR DELAY
LDAA #$00
STAA PORT A
BACK BRA BACK
A metal plate of 400 mm in length, 200mm in width and 30 mm in depth is to be machined by orthogonal cutting using a tool of width 5mm and a depth of cut of 0.5 mm. Estimate the minimum time required to reduce the depth of the plate by 20 mm
Complete Question:
A metal plate of 400 mm in length, 200mm in width and 30 mm in depth is to be machined by orthogonal cutting using a tool of width 5mm and a depth of cut of 0.5 mm. Estimate the minimum time required to reduce the depth of the plate by 20 mm if the tool moves at 400 mm per second.
Answer:
[tex]T_{min} =[/tex] 26 mins 40 secs
Explanation:
Reduction in depth, Δd = 20 mm
Depth of cut, [tex]d_c = 0.5 mm[/tex]
Number of passes necessary for this reduction, [tex]n = \frac{\triangle d}{d_c}[/tex]
n = 20/0.5
n = 40 passes
Tool width, w = 5 mm
Width of metal plate, W = 200 mm
For a reduction in the depth per pass, tool will travel W/w = 200/5 = 40 times
Speed of tool, v = 100 mm/s
[tex]Time/pass = \frac{40*400}{400} \\Time/pass = 40 sec[/tex]
minimum time required to reduce the depth of the plate by 20 mm:
[tex]T_{min} =[/tex] number of passes * Time/pass
[tex]T_{min} =[/tex] n * Time/pass
[tex]T_{min} =[/tex] 40 * 40
[tex]T_{min} =[/tex] 1600 = 26 mins 40 secs
Answer:
the minimum time required to reduce the depth of the plate by 20 mm is 26 minutes 40 seconds
Explanation:
From the given information;
Assuming the tool moves 100 mm/sec
The number of passes required to reduce the depth from 30 mm to 20 mm can be calculated as:
Number of passes = [tex]\dfrac{30-20}{0.5}[/tex]
Number of passes = 20
We know that the width of the tool is 5 mm; therefore, to reduce the depth per pass; the tool have to travel 20 times
However; the time per passes is;
Time/pass = [tex]\dfrac{20*L}{velocity \ of \ the \ feed}[/tex]
where;
length L = 400mm
velocity of the feed is assumed as 100
Time/pass [tex]=\dfrac{20*400}{100}[/tex]
Time/pass = 80 sec
Thus; the minimum time required to reduce the depth of the plate by 20 mm can be estimated as:
[tex]T_{min} = Time/pass *number of passes[/tex]
[tex]T_{min} = 20*80[/tex]
[tex]T_{min} = 1600 \ sec[/tex]
[tex]T_{min}[/tex] = 26 minutes 40 seconds
Using models helps scientists conduct research. How else can research using models save lives?
Answer: find the answer in the explanation.
Explanation:
Some of the methods of science are hypothesis, observations and experiments.
Scientific research and findings cut across many areas and field of life. Field like engineering, astronomy and medicine e.t.c
Using models in research is another great method in which researchers and scientists can master a process of a system and predict how it works.
Wilbur and Orville Wright wouldn't have died if the first airplane built by these two wonderful brothers was first researched by using models and simulations before embarking on a test.
Modelling in science involves calculations by using many mathematical methods and equations and also by using many laws of Physics. In this present age, many of these are computerised.
In space science, it will save lives, time and money to first model and simulate if any new thing is discovered in astronomy rather than people going to the space by risking their lives.
Also in medicine, drugs and many medical equipment cannot be tested by using people. This may lead to chaos and loss of lives.
So research using models in astronomy, medicine, engineering and many aspects of science and technology can indeed save lives.
A very large thin plate is centered in a gap of width 0.06 m with a different oils of unknown viscosities above and below; one viscosity is twice the other. When the plate is pulled at a velocity of 0.3 m/s, the resulting force on one square meter of plate due to the viscous shear on both sides is 29 N. Assuming viscous flow and neglecting all end effects calculate the viscosities of the oils.
Answer:
The viscosities of the oils are 0.967 Pa.s and 1.933 Pa.s
Explanation:
Assuming the two oils are Newtonian fluids.
From Newton's law of viscosity for Newtonian fluids, we know that the shear stress is proportional to the velocity gradient with the viscosity serving as the constant of proportionality.
τ = μ (∂v/∂y)
There are oils above and below the plate, so we can write this expression for the both cases.
τ₁ = μ₁ (∂v/∂y)
τ₂ = μ₂ (∂v/∂y)
dv = 0.3 m/s
dy = (0.06/2) = 0.03 m (the plate is centered in a gap of width 0.06 m)
τ₁ = μ₁ (0.3/0.03) = 10μ₁
τ₂ = μ₂ (0.3/0.03) = 10μ₂
But the shear stress on the plate is given as 29 N per square meter.
τ = 29 N/m²
But this stress is a sum of stress due to both shear stress above and below the plate
τ = τ₁ + τ₂ = 10μ₁ + 10μ₂ = 29
But it is also given that one viscosity is twice the other
μ₁ = 2μ₂
10μ₁ + 10μ₂ = 29
10(2μ₂) + 10μ₂ = 29
30μ₂ = 29
μ₂ = (29/30) = 0.967 Pa.s
μ₁ = 2μ₂ = 2 × 0.967 = 1.933 Pa.s
Hope this Helps!!!
Cathy works in a welding shop. While working one day, a pipe falls from scaffolding above and lands on her head, injuring her. Cathy complains to OSHA, but the company argues that because it has a "watch out for falling pipe" sign in the workplace that it gave fair warning. It also says that if Cathy wasn’t wearing a hardhat that she is responsible for her own injury. Which of the following is true?1. Common law rules could hold Cathy responsible for her own injury.2. Cathy’s employer may not be held liable for her injury if it fulfilled compliance and general duty requirements.3. OSHA rules can hold Cathy’s employer responsible for not maintaining a hazard-free workplace.4. More than one answer is correct.
Answer:1 common law
Explanation:
It also says that if Cathy wasn’t wearing a hardhat hat she is responsible for her own injury, more than one answer is correct.
What are OSHA rules?In this case, if Cathy's employer completes compliance and general duty requirements then the organization may not be held liable and again, the law can generally hold Cathy responsible for the injuries as she was not wearing the proper kits for such work.
According to OSHA, Cathy’s employer may not be held liable for her injury if it fulfilled compliance and general duty requirements.
You are entitled to a secure workplace. To stop workers from being murdered or suffering other types of harm at work, the Occupational Safety and Health Act of 1970 (OSH Act) was passed. According to the legislation, companies are required to give their workers safe working environments.
Therefore, more than one answer is correct.
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A nozzle receives an ideal gas flow with a velocity of 25 m/s, and the exit at 100 kPa, 300 K velocity is 250 m/s. Determine the inlet temperature if the gas is argon, helium, or nitrogen.
Given Information:
Inlet velocity = Vin = 25 m/s
Exit velocity = Vout = 250 m/s
Exit Temperature = Tout = 300K
Exit Pressure = Pout = 100 kPa
Required Information:
Inlet Temperature of argon = ?
Inlet Temperature of helium = ?
Inlet Temperature of nitrogen = ?
Answer:
Inlet Temperature of argon = 360K
Inlet Temperature of helium = 306K
Inlet Temperature of nitrogen = 330K
Explanation:
Recall that the energy equation is given by
[tex]$ C_p(T_{in} - T_{out}) = \frac{1}{2} \times (V_{out}^2 - V_{in}^2) $[/tex]
Where Cp is the specific heat constant of the gas.
Re-arranging the equation for inlet temperature
[tex]$ T_{in} = \frac{1}{2} \times \frac{(V_{out}^2 - V_{in}^2)}{C_p} + T_{out}$[/tex]
For Argon Gas:
The specific heat constant of argon is given by (from ideal gas properties table)
[tex]C_p = 520 \:\: J/kg.K[/tex]
So, the inlet temperature of argon is
[tex]$ T_{in} = \frac{1}{2} \times \frac{(250^2 - 25^2)}{520} + 300$[/tex]
[tex]$ T_{in} = \frac{1}{2} \times 119 + 300$[/tex]
[tex]$ T_{in} = 360K $[/tex]
For Helium Gas:
The specific heat constant of helium is given by (from ideal gas properties table)
[tex]C_p = 5193 \:\: J/kg.K[/tex]
So, the inlet temperature of helium is
[tex]$ T_{in} = \frac{1}{2} \times \frac{(250^2 - 25^2)}{5193} + 300$[/tex]
[tex]$ T_{in} = \frac{1}{2} \times 12 + 300$[/tex]
[tex]$ T_{in} = 306K $[/tex]
For Nitrogen Gas:
The specific heat constant of nitrogen is given by (from ideal gas properties table)
[tex]C_p = 1039 \:\: J/kg.K[/tex]
So, the inlet temperature of nitrogen is
[tex]$ T_{in} = \frac{1}{2} \times \frac{(250^2 - 25^2)}{1039} + 300$[/tex]
[tex]$ T_{in} = \frac{1}{2} \times 60 + 300$[/tex]
[tex]$ T_{in} = 330K $[/tex]
Note: Answers are rounded to the nearest whole numbers.
The benefit of using the generalized enthalpy departure chart prepared by using PR and TR as the parameters instead of P and T is that the single chart can be used for all gases instead of a single particular gas.
a. True
b. False
A tubular reactor has been sized to obtain 98% conversion and to process 0.03 m^3/s. The reaction is a first-order irreversible isomerization. The reactor is 3 m long, with a cross- sectional area of 25 dm^2. After being built, a pulse tracer test on the reactor gave the following data: tm = 10 s and σ2 = 65 s2. What conversion can be expected in the real reactor?
Answer:
The conversion in the real reactor is = 88%
Explanation:
conversion = 98% = 0.98
process rate = 0.03 m^3/s
length of reactor = 3 m
cross sectional area of reactor = 25 dm^2
pulse tracer test results on the reactor :
mean residence time ( tm) = 10 s and variance (∝2) = 65 s^2
note: space time (t) =
t = [tex]\frac{A*L}{Vo}[/tex] Vo = flow metric flow rate , L = length of reactor , A = cross sectional area of the reactor
therefore (t) = [tex]\frac{25*3*10^{-2} }{0.03}[/tex] = 25 s
since the reaction is in first order
X = 1 - [tex]e^{-kt}[/tex]
[tex]e^{-kt}[/tex] = 1 - X
kt = In [tex]\frac{1}{1-X}[/tex]
k = In [tex]\frac{1}{1-X}[/tex] / t
X = 98% = 0.98 (conversion in PFR ) insert the value into the above equation then
K = 0.156 [tex]s^{-1}[/tex]
Calculating Da for a closed vessel
; Da = tk
= 25 * 0.156 = 3.9
calculate Peclet number Per using this equation
0.65 = [tex]\frac{2}{Per} - \frac{2}{Per^2} ( 1 - e^{-per})[/tex]
therefore
[tex]\frac{2}{Per} - \frac{2}{Per^2} (1 - e^{-per}) - 0.65 = 0[/tex]
solving the Non-linear equation above( Per = 1.5 )
Attached is the Remaining part of the solution
You want to plate a steel part having a surface area of 160 with a 0.002--thick layer of lead. The atomic mass of lead is 207.19 . The density of lead is 11.36 . How many atoms of lead are required
Answer:
To answer this question we assumed that the area units and the thickness units are given in inches.
The number of atoms of lead required is 1.73x10²³.
Explanation:
To find the number of atoms of lead we need to find first the volume of the plate:
[tex] V = A*t [/tex]
Where:
A: is the surface area = 160
t: is the thickness = 0.002
Assuming that the units given above are in inches we proceed to calculate the volume:
[tex]V = A*t = 160 in^{2}*0.002 in = 0.32 in^{3}*(\frac{2.54 cm}{1 in})^{3} = 5.24 cm^{3}[/tex]
Now, using the density we can find the mass:
[tex] m = d*V = 11.36 g/cm^{3}*5.24 cm^{3} = 59.5 g [/tex]
Finally, with the Avogadros number ([tex]N_{A}[/tex]) and with the atomic mass (A) we can find the number of atoms (N):
[tex] N = \frac{m*N_{A}}{A} = \frac{59.5 g*6.022 \cdot 10^{23} atoms/mol}{207.19 g/mol} = 1.73 \cdot 10^{23} atoms [/tex]
Hence, the number of atoms of lead required is 1.73x10²³.
I hope it helps you!
An airplane flies from San Francisco to Washington DC at an air speed of 800 km/hr. Assume Washington is due east of San Francisco at a distance of 6000 km. Use a Cartesian system of coordinates centered at San Francisco with Washington in the positive x-direction. At cruising altitude, there is a cross wind blowing from north to south of 100 km/hr.
Required:
a. What must be the direction of flight for the plane to actually arrive in Washington?
b. What is the speed in the San Francisco to Washington direction?
c. How long does it take to cover this distance?
d. What is the time difference compared to no crosswind?
Answer:
A.) 7.13 degree north east
B.) 806.23 km/h
C.) 7.44 hours
D.) 0.06 hours
Explanation:
Assume Washington is due east of San Francisco and Francisco with Washington in the positive x-direction
Also, the cross wind is blowing from north to south of 100 km/hr in y coordinate direction.
A.) Using Cartesian system of coordinates, the direction of flight for the plane to actually arrive in Washington can be calculated by using the formula
Tan Ø = y/x
Substitute y = 100 km/h and x = 800km/h
Tan Ø = 100/800
Tan Ø = 0.125
Ø = Tan^-1(0. 125)
Ø = 7.13 degrees north east.
Therefore, the direction of flight for the plane to actually arrive in Washington is 7.13 degree north east
B.) The speed in the San Francisco to Washington direction can be achieved by using pythagorean theorem
Speed = sqrt ( 800^2 + 100^2)
Speed = sqrt (650000)
Speed = 806.23 km/h
C.) Let us use the speed formula
Speed = distance / time
Substitute the speed and distance into the formula
806.23 = 6000/ time
Make Time the subject of formula
Time = 6000/806.23
Time = 7.44 hours
D.) If there is no cross wind,
Time = 6000/800
Time = 7.5 hour
Time difference = 7.5 - 7.44
Time difference = 0.06 hours
An aggregate blend is composed of 65% coarse aggregate by weight (Sp. Cr. 2.635), 36% fine aggregate (Sp. Gr. 2.710), and 5% filler (Sp. Cr. 2.748). The compacted specimen contains 6% asphalt binder (Sp. Gr. 1.088) by weight of total mix, and has a bulk density of 143.9 lb/ft^3 Ignoring absorption, compute the percent voids in total mix, percent voids in mineral aggregate, and the percent voids filled with asphalt.
Answer:
Explanation:
From the given information:
Addition of all the materials = 65+ 36+ 5 +6 = 112 which is higher than 100 percentage; SO we need to find;
The actual percentage of each material which can be determined as follows:
Percentage of the coarse aggregate will be = 65 × 112/100
= 72.80%
Percentage of the Fine aggregate will be = 36 × 112/100
= 40.32%
Percentage of the filler will be = 5 × 112/100
= 5.6%
Percentage of the asphalt binder will be = 6 × 112/100
= 6.72 %
So; the theoretical specific gravity (Gt) of the mixture can be calculated as follows:
Gt = 100/( 72.80/2.635 + 40.32/2.710 + 5.6/2.748 + 6.72/1.088)
Gt = 100/( 27.628 + 14.878 + 2.039 + 6.177)
Gt = 100/ (50.722)
Gt =1.972
Also;Given that the bulk density = 143.9 lb/ft³
LIke-wsie ; as we know that unit weight of water is =62.43lb/cu.ft
Hence, the bulk specific gravity of the mix (Gm) = 143.9/62.43
=2.305
The percentage of air void = (Gt -Gm )× 100/ Gt
= (1.972 - 2.305) × 100/ 1.972
= -16.89%
The percentage of the asphalt binder is =(6.72/1.088*100)/(72.80+40.32+5.6+6.72)/2.305)
= 617.647/54.42
= 11.35%
Thus; the percentage voids in mineral aggregate = -16.89% + 11.35%
the percentage voids in mineral aggregate = -5.45%
The percent voids filled with asphalt. = 100 × 11.35/-5.45
The percent voids filled with asphalt = - 208.26 %
Which of the following is normally included in the criteria of a design?
1. Materials
2. Time
3. Budget
4. Efficiency
Answer:
i took the test and its budget
Explanation:
Budget, of the following is normally included in the criteria of a design. Thus, option (c) is correct.
What is design?The term design refers to the visual representation of image and shapes. The image is based on vector raster and bitmap. Who design the graphic is called graphic designer. The design main purpose to design of logo, word art, and advertisement design. The design mostly software as Coral Draw, Photoshop, Canva etc.
The criteria of a design are the budget. The budget was the financial as well as the monetary terminology. The criteria of the budget are the decided to how much they spend, and they save. The design on the country budget and the five-year plan.
As a result, the budget following is normally included in the criteria of a design. Therefore, option (c) is correct.
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An undersea research chamber is spherical with an external diameter of 3.50 mm . The mass of the chamber, when occupied, is 21700 kg. It is anchored to the sea bottom by a cable. Find the followings
Required:
a. The buoyant force on the chamber.
b. The tension in the cable?
Answer:
a. The buoyant force on the chamber is 220029.6 N
b. The tension in the cable is 7369.6 N
Explanation:
The diameter of the sphere cannot be in millimeter (mm), if the chamber must occupy a big mass as 21700kg
Given;
diameter of the sphere, d = 3.50 m
radius of the sphere, r = 1.75 mm = 1.75 m
mass of the chamber, m = 21700 kg
density of water, ρ = 1000 kg/m³
(a)
Buoyant force is the weight of water displaced, which is calculated as;
Fb = ρvg
where;
v is the volume of sphere, calculated as;
[tex]V = \frac{4}{3} \pi r^3\\\\V = \frac{4}{3} \pi (1.75)^3\\\\V = 22.452 \ m^3[/tex]
Fb = 1000 x 22.452 x 9.8
Fb = 220029.6 N
(b)
The tension in the cable will be calculated as;
T = Fb - mg
T = 220029.6 N - (21700 x 9.8)
T = 220029.6 N - 212660 N
T = 7369.6 N
for an electromotive force to be induced across a vertical loop from the field of an infinite length line of fixed current in the z axis the loop must be moving to?
Answer:
The correct answer to the following question will be "[tex]a_{x}[/tex] or [tex]a_{y}[/tex]".
Explanation:
Since along that same z-axis none electromagnetic field would be triggered as being in the same orientation loop movement of them across different line portions would allow some caused emf/voltage to be canceled. And the only logical choice seems to be either x or y-axes.The magnetic field of fluctuation should indeed be changed and changed across both X as well as Y directions.So that the above is the appropriate choice.
Engine oil (unused) flows at 1.81 x 10^-3 kg/s inside a 1-cm diameter tube that is heated electrically at a rate of 76 W/m. At a particular location where flow and heat transfer are fully developed, the wall temperature is 370K. Determine:
a. The oil mean temperature.
b. The centerline temperature.
c. The axial gradient of the mean temperature.
d. The heat transfer coefficient.
Answer:
(a)Tb = 330.12 K (b)Tc =304.73 K (c)19.81 K/m (d) h =60.65 W/m². K
Explanation:
Solution
Given that:
The mass flow rate of engine oil m = 1.81 x 10^-3 kg/s
Diameter of the tube, D = 1cm =0.01 m
Electrical heat rate, q =76 W/m
Wall Temperature, Ts = 370 K
Now,
From the properties table of engine oil we can deduce as follows:
thermal conductivity, k =0.139 W/m .K
Density, ρ = 854 kg/m³
Specific heat, cp = 2120 J/kg.K
(a) Thus
The wall heat flux is given as follows:
qs = q/πD
=76/π *0.01
= 2419.16 W/m²
Now
The oil mean temperature is given as follows:
Tb =Ts -11/24 (q.R/k) (R =D/2=0.01/2 = 0.005 m)
Tb =370 - 11/24 * (2419.16 * 0.005/0.139)
Tb = 330.12 K
(b) The center line temperature is given below:
Tc =Ts - 3/4 (qs.R/k)= 370 - 3/4 * ( 2419.16 * 0.005/0.139)
Tc =304.73 K
(c) The flow velocity is given as follows:
V = m/ρ (πR²)
Now,
The The axial gradient of the mean temperature is given below:
dTb/dx = 2 *qs/ρ *V*cp * R
=2 *qs/ρ*[m/ρ (πR²) *cp * R
=2 *qs/[m/(πR)*cp
dTb/dx = 2 * 2419.16/[1.81 x 10^-3/(π * 0.005)]* 2120
dTb/dx = 19.81 K/m
(d) The heat transfer coefficient is given below:
h =48/11 (k/D)
=48/11 (0.139/0.01)
h =60.65 W/m². K
7.D7 A cylindrical brass rod having a minimum tensile strength of 450 MPa (65,000 psi), a ductility of at least 13%EL, and a final diameter of 12.7 mm (0.50 in.) is desired. Some brass stock of diameter 19.0 mm (0.75 in.) that has been cold worked 35% is available. Describe the procedure you would follow to obtain this material. Assume that brass experiences cracking at 65%CW.
Answer: CW = 59%
Explanation:
Given Data:
Do = ?
D1 = 12.7mm
Average = 28% CW
%CW = π ([Do/2]^2 - π [ D1/2]^2 ) / π [ Do/2 ]^2 *100
Input data into the formula
28 = π ([Do/2]^2 - π [ 12.7/2]^2 ) / π [ Do/2 ]^2 *100
Do = 15.1mm
When CW = 35%
35 = π ([Do/2]^2 - π [ 35/2]^2 ) / π [ Do/2 ]^2 *100
= 23.6mm
Now we calculate the total CW needed.
CW= π ([Do/2]^2 - π [ D1/2]^2 ) / π [ Do/2 ]^2 *100
When Do = 23.6mm
D1 = 15.1mm
%CW = π ([23.6/2]^2 - π [ 15.1/2]^2 ) / π [ 23.6/2 ]^2 *100
CW = 59%
You are installing network cabling and require a cable solution that provides the best resistance to EMI.Which of the following will you choose for this installation?
Answer: b. STP
Explanation:
Twisted Pair Cables are best used for network cabling but are usually prone to EMI (Electromagnetic Interference) which affects the electrical circuit negatively.
The best way to negate this effect is to use Shielding which will help the cable continue to function normally. This is where the Shielded Twisted Pair (STP) cable comes in.
As the name implies, it comes with a shield and that shield is made out of metal which can enable it conduct the Electromagnetic Interference to the ground. The Shielding however makes it more expensive and in need of more care during installation.
Given the circuit at the right in which the following values are used: R1 = 20 kΩ, R2 = 12 kΩ, C = 10 µ F, and ε = 25 V. You close the switch at t = 0. Find (a) the current in R1 and R2 at t=0, (b) the voltage across R1 after a long time. (Careful with this one.)
Answer:
a.) I = 7.8 × 10^-4 A
b.) V(20) = 9.3 × 10^-43 V
Explanation:
Given that the
R1 = 20 kΩ,
R2 = 12 kΩ,
C = 10 µ F, and
ε = 25 V.
R1 and R2 are in series with each other.
Let us first find the equivalent resistance R
R = R1 + R2
R = 20 + 12 = 32 kΩ
At t = 0, V = 25v
From ohms law, V = IR
Make current I the subject of formula
I = V/R
I = 25/32 × 10^3
I = 7.8 × 10^-4 A
b.) The voltage across R1 after a long time can be achieved by using the formula
V(t) = Voe^- (t/RC)
V(t) = 25e^- t/20000 × 10×10^-6
V(t) = 25e^- t/0.2
After a very long time. Let assume t = 20s. Then
V(20) = 25e^- 20/0.2
V(20) = 25e^-100
V(20) = 25 × 3.72 × 10^-44
V(20) = 9.3 × 10^-43 V
If you've wondered about the flushing of toilets on the upper floors of city skyscrapers, how do you suppose the plumbing is designed so that there is not an enormous impact of sewage arriving at the basement level?
Answer:
The plumbing is designed to reduce the impact of pressure forces due to the height of skyscrapers. This is achieves by narrowing down the pipe down to the basement, using pipes with thicker walls down the basement, and allowing vents; to prevent clogging of the pipes.
Explanation:
Pressure increases with depth and density. In skyscrapers, a huge problem arises due to the very tall height of most skyscrapers. Also, sewage slug coming down has an increased density when compared to that of water, and these two factors can't be manipulated. The only option is to manipulate the pipe design. Pipes in skyscrapers are narrowed down with height, to reduce accumulation at the bottom basement before going to the sewage tank. Standard vents are provided along the pipes, to prevent clogging of the pipes, and pipes with thicker walls are used as you go down the basement of the skyscraper, to withstand the pressure of the sewage coming down the pipes.
Q: Draw shear and bending moment diagram for the beam shown in
the figure. EI= constant
Answer:
Explanation:
Please
Consider a steady developing laminar flow of water in a constant-diameter horizontal discharge pipe attached to a tank. The fluid enters the pipe with nearly uniform velocity V and pressure P1. The velocity profile becomes parabolic after a certain distance with a momentum correction factor of 2 while the pressure drops to P2. Identify the correct relation for the horizontal force acting on the bolts that hold the pipe attached to the tank.
Answer:
hello attached is the free body diagram of the missing figure
Fr = [tex]\frac{\pi }{4} D^2 [ ( P1 - P2) - pV^2 ][/tex]
Explanation:
Average velocity is constant i.e V1 = V2 = V
The momentum equation for the flow in the Z - direction can be expressed as
-Fr + P1 Ac - P2 Ac = mB2V2 - mB1V1 ------- equation 1
Fr = horizontal force on the bolts
P1 = pressure of fluid at entrance
V1 = velocity of fluid at entrance
Ac = cross section area of the pipe
P2 and V2 = pressure and velocity of fluid at some distance
m = mass flow rate of fluid
B1 = momentum flux at entrance , B2 = momentum flux correction factor
Note; average velocity is constant hence substitute V for V1 and V2
equation 1 becomes
Fr = ( P1 - P2 ) Ac + mV ( 1 - 2 )
Fr = ( P1 - P2 ) Ac - mV ---------------- equation 2
equation for mass flow rate
m = pAcV
p = density of the fluid
insert this into equation 2 EQUATION 2 BECOMES
Fr = ( P1 - P2) Ac - pAcV^2
= Ac [ (P1 - P2) - pV^2 ] ---------- equation 3
Note Ac = [tex]\frac{\pi }{4} D^2[/tex]
Equation 3 becomes
Fr = [tex]\frac{\pi }{4} D^2[/tex] [ (P1 -P2 ) - pV^2 ] ------- relation for the horizontal force acting on the bolts
what happens if your son makes a spark in a outlet and then the room goes dark
Answer:
He probably tripped the wiring, when metal hits the electricity it creates a reaction that burns the wiring.
Explanation:
A large tank is filled to capacity with 500 gallons of pure water. Brine containing 2 pounds of salt per gallon is pumped into the tank at a rate of 5 gal/minute. The well mixed solution is then pumped out at a rate of 10 gal/minute. Find the number A(t) of pounds of salt in the tank at time t. When is the tank empty?
Answer:
A) A(t) = 10(100 - t) + c(100 - t)²
B) Tank will be empty after 100 minutes.
Explanation:
A) The differential equation of this problem is;
dA/dt = R_in - R_out
Where;
R_in is the rate at which salt enters
R_out is the rate at which salt exits
R_in = (concentration of salt in inflow) × (input rate of brine)
We are given;
Concentration of salt in inflow = 2 lb/gal
Input rate of brine = 5 gal/min
Thus;
R_in = 2 × 5 = 10 lb/min
Due to the fact that the solution is pumped out at a faster rate, thus it is reducing at the rate of (5 - 10)gal/min = -5 gal/min
So, after t minutes, there will be (500 - 5t) gallons in the tank
Therefore;
R_out = (concentration of salt in outflow) × (output rate of brine)
R_out = [A(t)/(500 - 5t)]lb/gal × 10 gal/min
R_out = 10A(t)/(500 - 5t) lb/min
So, we substitute the values of R_in and R_out into the Differential equation to get;
dA/dt = 10 - 10A(t)/(500 - 5t)
This simplifies to;
dA/dt = 10 - 2A(t)/(100 - t)
Rearranging, we have;
dA/dt + 2A(t)/(100 - t) = 10
This is a linear differential equation in standard form.
Thus, the integrating factor is;
e^(∫2/(100 - t)) = e^(In(100 - t)^(-2)) = 1/(100 - t)²
Now, let's multiply the differential equation by the integrating factor 1/(100 - t)².
We have;
So, we ;
(1/(100 - t)²)(dA/dt) + 2A(t)/(100 - t)³ = 10/(100 - t)²
Integrating this, we now have;
A(t)/(100 - t)² = ∫10/(100 - t)²
This gives;
A(t)/(100 - t)² = (10/(100 - t)) + c
Multiplying through by (100 - t)²,we have;
A(t) = 10(100 - t) + c(100 - t)²
B) At initial condition, A(0) = 0.
So,0 = 10(100 - 0) + c(100 - 0)²
1000 + 10000c = 0
10000c = -1000
c = -1000/10000
c = -0.1
Thus;
A(t) = 10(100 - t) + -0.1(100 - t)²
A(t) = 1000 - 10t - 0.1(10000 - 200t + t²)
A(t) = 1000 - 10t - 1000 + 20t - 0.1t²
A(t) = 10t - 0.1t²
Tank will be empty when A(t) = 0
So, 0 = 10t - 0.1t²
0.1t² = 10t
Divide both sides by 0.1t to give;
t = 10/0.1
t = 100 minutes
what are the besl measures used for data variation or dispersion
Answer:
STANDARD DAVIATION OR SD
Explanation:
BECAUSE it is commonly in measuring of dispersion and it also the most roubst measure of variability
A horizontal turbine takes in steam with an enthalpy of h = 2.80 MJ/kg at 45 m/s. A steam-water mixture exits the turbine with an enthalpy of h = 1.55 MJ/kg at 20 m/s. If the heat loss to the surroundings from the turbine is 300 J/s, determine the power the fluid supplies to the turbine. The mass flow rate is 0.85 kg/s.
Answer:
The power that fluid supplies to the turbine is 1752.825 kilowatts.
Explanation:
A turbine is a device that works usually at steady state. Given that heat losses exists and changes in kinetic energy are not negligible, the following expression allows us to determine the power supplied by the fluid to the turbine by the First Law of Thermodynamics:
[tex]-\dot Q_{loss} - \dot W_{out} + \dot m \cdot \left[(h_{in}-h_{out}) + \frac{1}{2}\cdot (v_{in}^{2}-v_{out}^{2}) \right] = 0[/tex]
Output power is cleared:
[tex]\dot W_{out} = -\dot Q_{loss} + \dot m \cdot \left[(h_{in}-h_{out})+\frac{1}{2}\cdot (v_{in}^{2}-v_{out}^{2}) \right][/tex]
If [tex]\dot Q_{loss} = 0.3\,kW[/tex], [tex]\dot m = 0.85\,\frac{kg}{s}[/tex], [tex]h_{in} = 2800\,\frac{kJ}{kg}[/tex], [tex]h_{out} = 1550\,\frac{kJ}{kg}[/tex], [tex]v_{in} = 45\,\frac{m}{s}[/tex] and [tex]v_{out} = 20\,\frac{m}{s}[/tex], then:
[tex]\dot W_{out} = -0.3\,kW + \left(0.85\,\frac{kg}{s} \right)\cdot \left\{\left(2800\,\frac{kJ}{kg}-1550\,\frac{kJ}{kg} \right)+\frac{1}{2}\cdot \left[\left(45\,\frac{m}{s} \right)^{2}-\left(20\,\frac{m}{s} \right)^{2}\right] \right\}[/tex]
[tex]\dot W_{out} = 1752.825\,kW[/tex]
The power that fluid supplies to the turbine is 1752.825 kilowatts.
Steam is contained in a closed rigid container which has a volume of 2 initially the the pressure and the temperature is the remeraturedrops as a result of heat transfer to the surroundings. Determine
a) the temperature at which condensation first occurs, in °C,
b) the fraction of the total mass that has condensed when the pressure reaches 0.5 bar.
c) What is the volume, in m3, occupied by saturated liquid at the final state?
The given question is incomplete. The complete question is as follows.
Steam is contained in a closed rigid container with a volume of 1 m3. Initially, the pressure and temperature of the steam are 10 bar and 500°C, respectively. The temperature drops as a result of heat transfer to the surroundings. Determine
(a) the temperature at which condensation first occurs, in [tex]^{o}C[/tex],
(b) the fraction of the total mass that has condensed when the pressure reaches 0.5 bar.
(c) What is the volume, in [tex]m^{3}[/tex], occupied by saturated liquid at the final state?
Explanation:
Using the property tables
[tex]T_{1} = 500^{o}C[/tex], [tex]P_{1}[/tex] = 10 bar
[tex]v_{1} = 0.354 m^{3}/kg[/tex]
(a) During the process, specific volume remains constant.
[tex]v_{g} = v_{1} = 0.354 m^{3}/kg[/tex]
T = [tex](150 - 160)^{o}C[/tex]
Using inter-polation we get,
T = [tex]154.71^{o}C[/tex]
The temperature at which condensation first occurs is [tex]154.71^{o}C[/tex].
(b) When the system will reach at state 3 according to the table at 0.5 bar then
[tex]v_{f} = 1.030 \times 10^{-3} m^{3}/kg[/tex]
[tex]v_{g} = 3.24 m^{3} kg[/tex]
Let us assume "x" be the gravity if stream
[tex]v_{1} = v_{f} + x_{3}(v_{g} - v_{f})[/tex]
[tex]x_{3} = \frac{v_{1} - v_{f}}{v_{g} - v_{f}}[/tex]
= [tex]\frac{0.3540 - 0.00103}{3.240 - 0.00103}[/tex]
= 0.109
At state 3, the fraction of total mass condensed is as follows.
[tex](1 - x_{5})[/tex] = 1 - 0.109
= 0.891
The fraction of the total mass that has condensed when the pressure reaches 0.5 bar is 0.891.
(c) Hence, total mass of the system is calculated as follows.
m = [tex]\frac{v}{v_{1}}[/tex]
= [tex]\frac{1}{0.354}[/tex]
= 2.825 kg
Therefore, at final state the total volume occupied by saturated liquid is as follows.
[tex]v_{ws} = m \times v_{f}[/tex]
= [tex]2.825 \times 0.00103[/tex]
= [tex]2.9 \times 10^{-3} m^{3}[/tex]
The volume occupied by saturated liquid at the final state is [tex]2.9 \times 10^{-3} m^{3}[/tex].
A 2.75-kN tensile load is applied to a test coupon made from 1.6-mm flat steel plate (E = 200 GPa, ν = 0.30). Determine the resulting change in (a) the 50-mm gage length, (b) the width of portion AB of the test coupon, (c) the thickness of portion AB, (d) the cross- sectional area of portion AB.
Answer:
I have attached the diagram for this question below. Consult it for better understanding.
Find the cross sectional area AB:
A = (1.6mm)(12mm) = 19.2 mm² = 19.2 × 10⁻⁶m
Forces is given by:
F = 2.75 × 10³ N
Horizontal Stress can be found by:
σ (x) = F/A
σ (x) = 2.75 × 10³ / 19.2 × 10⁻⁶m
σ (x) = 143.23 × 10⁶ Pa
Horizontal Strain can be found by:
ε (x) = σ (x)/ E
ε (x) = 143.23 × 10⁶ / 200 × 10⁹
ε (x) = 716.15 × 10⁻⁶
Find Vertical Strain:
ε (y) = -v · ε (y)
ε (y) = -(0.3)(716.15 × 10⁻⁶)
ε (y) = -214.84 × 10⁻⁶
PART (a)For L = 0.05m
Change (x) = L · ε (x)
Change (x) = 35.808 × 10⁻⁶m
PART (b)
For W = 0.012m
Change (y) = W · ε (y)
Change (y) = -2.5781 × 10⁻⁶m
PART(c)
For t= 0.0016m
Change (z) = t · ε (z)
where
ε (z) = ε (y) ,so
Change (z) = t · ε (y)
Change (z) = -343.74 × 10⁻⁹m
PART (d)
A = A(final) - A(initial)
A = -8.25 × 10⁻⁹m²
(Consult second picture given below for understanding how to calculate area)
The resulting change in the 50-mm gauge length; the width of portion AB of the test coupon; the thickness of portion AB; the cross- sectional area of portion AB are respectively; Δx = 35.808 × 10⁻⁶ m; Δy = -2.5781 × 10⁻⁶m; Δ_z = -343.74 × 10⁻⁹m; A = -8.25 × 10⁻⁹m²
What is the stress and strain in the plate?Let us first find the cross sectional area of AB from the image attached;
A = (1.6mm)(12mm) = 19.2 mm² = 19.2 × 10⁻⁶m
We are given;
Tensile Load; F = 2.75 kN = 2.75 × 10³ N
Horizontal Stress is calculated from the formula;
σₓ = F/A
σₓ = (2.75 × 10³)/(19.2 × 10⁻⁶)m
σₓ = 143.23 × 10⁶ Pa
Horizontal Strain is calculated from;
εₓ = σₓ/E
We are given E = 200 GPa = 200 × 10⁹ Pa
Thus;
εₓ = (143.23 × 10⁶)/(200 × 10⁹)
εₓ = 716.15 × 10⁻⁶
Formula for Vertical Strain is;
ε_y = -ν * εₓ
We are given ν = 0.30. Thus;
ε_y = -(0.3) * (716.15 × 10⁻⁶)
ε_y = -214.84 × 10⁻⁶
A) We are given;
Gauge Length; L = 0.05m
Change in gauge length is gotten from;
Δx = L * εₓ
Δx = 0.05 × 716.15 × 10⁻⁶
Δx = 35.808 × 10⁻⁶ m
B) From the attached diagram, the width is;
W = 0.012m
Change in width is;
Δy = W * ε_y
Δy = 0.012 * -214.84 × 10⁻⁶
Δy = -2.5781 × 10⁻⁶m
C) We are given;
Thickness of plate; t = 1.6 mm = 0.0016m
Change in thickness;
Δ_z = t * ε_z
where;
ε_z = ε_y
Thus;
Δ_z = t * ε_y
Δ_z = 0.0016 * -214.84 × 10⁻⁶
Δ_z = -343.74 × 10⁻⁹m
D) The change in cross sectional area is gotten from;
ΔA = A_final - A_initial
From calculating the areas, we have;
A = -8.25 × 10⁻⁹ m²
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