Answer:
Option C = Only 1 path for current to flow.
Your family has asked you to estimate the operating costs of your clothes dryer for the year. The clothes dryer in your home has a power rating of 2250 W. To dry one typical load of clothes the dryer will run for approximately 45 minutes. In Ontario, the cost of electricity is $0.11/kWh. Calculate the costs to run the dryer for your family for one year.
Answer:
The costs to run the dryer for one year are $ 9.03.
Explanation:
Given that the clothes dryer in my home has a power rating of 2250 Watts, and to dry one typical load of clothes the dryer will run for approximately 45 minutes, and in Ontario, the cost of electricity is $ 0.11 / kWh, to calculate the costs to run the dryer for one year the following calculation must be performed:
1 watt = 0.001 kilowatt
2250/45 = 50 watts per minute
45 x 365 = 16,425 / 60 = 273.75 hours of consumption
50 x 60 = 300 watt = 0.3 kw / h
0.3 x 273.75 = 82.125
82.125 x 0.11 = 9.03
Therefore, the costs to run the dryer for one year are $ 9.03.
Ideally speaking, bonds tend to form between two particles such that they are separated by a distance where force is exerted on them, and their overall energy is:________
a. a negative, minimized
b. a positive, minimized
c. zero, minimized
d. zero, maximized
e. a positive, maximized
f. a negative, maximized
Answer:
a g i
Explanation:
nnj
hãy trình bày sự hiểu biết của bạn về đo dòng điện
Answer:
-Khái niệm:
Đo dòng điện là sử dụng các dụng cụ như ôm kê, vôn kế, ampe kế, tần số kế… để xác định các đại lượng vật lý của dòng điện
-Đo lường điện để làm gì?
Phát hiện hư hỏng sự cố trong mạch điện và các thiết bị vi mạch
Xác định các giá trị cần đo
Đánh giá chất lượng của các thiết bị sau sản xuất
Xác định thông số kỹ thuật của thiết bị
-Phân loại dụng cụ đo điện: Hiên nay có 2 phương pháp phân loại chính
a. Theo nguyên lý làm việc
Dụng cụ đo kiểu điện từ
Dụng cụ đo kiểu điện động
Dụng cụ đo kiểu cảm ứng
Dụng cụ đo kiểu từ điện
b. Theo đai lượng, giá trị cần đo
+Đo điện năng: Ví dụ công tơ điện
+Đo điện áp: Ví dụ: Vôn kế
+Đo dòng điện: Ví dụ: Ampe kế
+Đo công suất: Oát kế
+Đo điện trở: Ôm kế
Sai số khi đo: Khi đo lường luôn xảy ra các sai số
Sai số tương đối: Là tỷ lệ % của sự chênh lệch giữa giá trị đo được và giá trị thự
A test bar of nonferrous material has a diameter of 0.253 inches. Upon applying a tensile load, the sample exhibited 0.002 plastic strain at 3400 lb and the maximum load during testing was 6200 lb and occurred at an engineering strain of 0.65; and breaking occurred at 4400 lb. The sample diameter at fracture was measured to be 0.15 inches.
Required:
a. The yield strength of the material is :________
b. The UTS of the material is:________
Answer:
a. 67607.9psi
b. 123278.33
Explanation:
to get the yield strength of the material
= load/ cross sectional area
cross sectional area = π * 0.253²/4
= 0.0502927
The yield strenght
= 3400/0.0502927
= 67609.9 psi
b. the uts of the material
= maximum load/cross sectional area
= 6200/0.0502927
= 123278.33
A balanced star-connected three-phase load is shown in Figure 4. Determine the value of the line currents IR,IY and IB using mesh-current analysis.
Answer:
Therefore the value of the line currents IR, IY, and IB are
[tex]I_{R}=I_{1}=83\angle 6.87^{o}A\\I_{B}=-I_{2}=-71.88\angle-23.13^{o}A\\I_{Y}=I_{2}-I_{1}\\I_{Y}=-I_{2}=41.50\angle113.3^{o}A[/tex]
Explanation:
Apply KVL for loop 1
[tex]415\angle 120^{o}=\left ( 3+j4+3+j4 \right )I_{1}-\left ( 3+j4 \right )I_{2}\\415\angle 120^{o}=\left ( 6+j8 \right )I_{1}-\left ( 3+j4 \right )I_{2}\\415\angle 120^{o}=\left ( 10\angle 53.13^{o} \right )I_{1}-\left ( 5\angle 53.13^{o}\right )I_{2} \rightarrow \left ( 1 \right )[/tex]
Apply KVL for loop 2
[tex]415\angle 0^{o}=\left ( 3+j4+3+j4 \right )I_{2}-\left ( 3+j4 \right )I_{1}\\415\angle 0^{o}=\left ( 6+j8 \right )I_{2}-\left ( 3+j4 \right )I_{1}\\415\angle 0^{o}=\left ( 10\angle 53.13^{o} \right )I_{2}-\left ( 5\angle 53.13^{o}\right )I_{1} \rightarrow \left ( 2 \right )[/tex]
Solving the above equations,
[tex]415\angle 120^{o}+830\angle 0^{o}=\left ( 10\angle 53.13^{o} \right )I_{1}-\left ( 5\angle 53.13^{o}\right )I_{2} +\left ( 20\angle 53.13^{o}\right )I_{2}- \left ( 10\angle 53.13^{o} \right )I_{1}\\415\angle 120^{o}+830\angle 0^{o}=\left ( 20\angle 53.13^{o}\right )I_{2}- \left ( 10\angle 53.13^{o} \right )I_{2}\\415\angle 120^{o}+830\angle 0^{o}=\left ( 10\angle 53.13^{o} \right )I_{2}\\I_{2}= 71.88\angle -23.13^{o}A[/tex]&
[tex]415\angle 0^{o}=\left ( 10\angle 53.13^{o} \right ) \times 71.88\angle -23.13^{o}-\left ( 5\angle 53.13^{o} \right )I_{1}\\415\angle 0^{o}=718.8\angle 30^{o}-\left ( 5\angle 53.13^{o} \right )I_{1}\\\left ( 5\angle 53.13^{o} \right )I_{1}=415\angle 60^{o}\\I_{1}= 83\angle 6.87^{o}A[/tex]
Hence,
[tex]I_{R}=I_{1}=83\angle 6.87^{o}A\\I_{B}=-I_{2}=-71.88\angle-23.13^{o}A\\I_{Y}=I_{2}-I_{1}\\I_{Y}=-I_{2}=41.50\angle113.3^{o}A[/tex]
From what year did 3.5G enter Vietnam?
Answer:
From what year did 3.5G enter Vietnam?
Explanation:
Vietnam does not have 3.5G network. MobiFone's first trial with 3.5G technology on this band was able to cover the whole waters of Vung Tau and Con Dao in August 2014.
In a certain pressing operation, the metallic powder fed into the open die has a packing factor of 0.5. The pressing operation reduces the powders to 70% of their starting volume. In the subsequent sintering operation, shrinkage amounts to 10% on a volume basis. Given that these are the only factors that affect the structure of the finished part, determine its final porosity.
Answer:
0.2063
Explanation:
Given data:
packing factor = 0.5
percentage of reduction of powders = 70%
Calculate the final porosity
after sintering Bulk specific volume = 0.9 * 0.7 = 0.63
assuming true specific volume = 1
packing factor = 0.5 , bulk specific volume = 2
packing factor after pressing and sintering
= 1 / ( 2 * 0.63 ) = 0.7937
hence : porosity = 1 - packing factor
= 1 - 0.7937 = 0.2063
A drum contains 3 black balls, 5 red balls and 6 green balls. If 4 balls are selected at random what is the probability that the 4 selected contain No red ball
A. 0.1258
B. 0.1587
C. 0.2356
D. 0.2289
Answer:
Probability of no red ball from 4 balls = 0.1258 (Approx.)
Explanation:
Given:
Number of black ball = 3
Number of red ball = 5
Number of green ball = 6
Find:
Probability of no red ball from 4 balls
Computation:
Probability of no red ball from 4 balls = 9c4 / 14c4
Probability of no red ball from 4 balls = 126 / 1001
Probability of no red ball from 4 balls = 0.12587
Probability of no red ball from 4 balls = 0.1258 (Approx.)
Una canica de masa 15g, con velocidad horizontal de 10 m/s impacta a otra de masa de 8g que se encuentra en el piso y en reposo. Suponga que el movimiento ocurre en la linea que une sus centros. Encuentra la velocidad de cada canica inmediatamente despues de la colision
Answer:
Una canica de masa 15g, con velocidad horizontal de 10 m/s impacta a otra de masa de 8g que se encuentra en el piso y en reposo. Suponga que el movimiento ocurre en la linea que une sus centros. Encuentra la velocidad de cada canica inmediatamente despues de la colisionExplanation:
làm giúp tôi hệ thống truyền lực trên xe toyota
Answer:
ay man ima be real, i just need the points yo
A heat pump heats the air in a rigid, insulated cuboid room of size 25m x 10m x 4m. The heat pump consumes 15 kW of power. The initial temperature and pressure in this room are 12°C and 1 bar, respectively. With an average coefficient of performance of COPHP= 3.0 over the range of air temperature in this room.
Requried:
How long will it take to raise the temperature in the room to 27 °C?
Answer:
Time required = 287.2 secs
Explanation:
Volume of room = 25 * 10 * 4 = 1000 m^3
power consumed by pump = 15 kW
T1 ( initial temperature ) = 12°C
P1 ( Initial pressure ) = 1 bar
COPhp = 3
Calculate time taken to raise room Temp to 27°C
average heat supplied ( ∅ ) = COPhp * power consumed by pump
= 3 * 15 = 45 kW
Time required can be calculated using the relation below
∅t = P*V*Cv ( T2 - T1 ) [ p = 1.2 kg/m^3 , Cv = 0.718 KJ/kg ( air properties ) ]
45 * 10^3 ( t ) = 1.2*1000* 718 ( 27 - 12 )
∴ solving for t
t = 287.2 secs ≈ 4.79 mins
Methane gas (CH4) at 25oC, 1 atm, and a volumetric flow rate of 27m3/h enters a furnace operating at steady-state. The methane burns completely with 140% of theoretical air that enters at 127oC, 1 atm. Products of combustion exit at 427oC, 1 atm. Determine:
(a) the volumetric flow rate of the air, in m3/h.
(b) the rate of heat transfer from the furnace, in kJ/h.
Answer:
a) [tex]r_a=37.8m^3/h[/tex]
b)[tex]Q=2.8Kw[/tex]
Explanation:
Temperature of CH_4[tex]t=25C[/tex]
CH_4Flow rate of [tex]r=27m3/h[/tex]
Air Percentage [tex]=140\%=1.4[/tex]
Temperature of air [tex]t_a=127=>400K[/tex]
Temperature at exit[tex]t_e=427C=>700k[/tex]
Generally the equation for Air's flow Rate is mathematically given by
[tex]r_a=air\%*r[/tex]
[tex]r_a=1.4*27[/tex]
[tex]r_a=37.8m^3/h[/tex]
Generally the equation for Ideal Gas is mathematically given by
[tex]PV=mRT[/tex]
[tex]m=\frac{PV}{RT}[/tex]
[tex]m=\frac{1.01*10^5*37.8}{0.287*10^3*400}[/tex]
[tex]m=33.35kg[/tex]
Therefore
The rate of heat transfer from the furnace, in kJ/h is
[tex]Q=mC_p(T_e-T_a)[/tex]
[tex]Q=33.35*1.005*(700-400)[/tex]
[tex]Q=2.8Kw[/tex]
Please choose a specific type of stability or control surface (e.g., a canard) and explain how it is used, what it is used for, and the pros and cons of the device or system.
Answer:
small forewing
pro : Can be used in place of tail plane configuration
con : Can be very complex and difficult to use
Explanation:
A canard is generally used to provide some form of stability to an unstable or semi stable system.
An example of a Canard is a small forewing placed in an aircraft that will help stabilize the aircraft when in motion( in air ). because an airplane is generally an unstable system on its own
pro : Can be used in place of tail plane configuration
con : Can be very complex and difficult to use
In the figure below, this “double” nozzle discharges water (at 10°C, density= 1000 kg/m3) into the atmosphere at a rate of 0.50 m3/s. The pressure at the inlet is to be 315612 Pa. If the nozzle is lying in a horizontal plane. Jet A is 10 cm in diameter, jet B is 12 cm in diameter, and the pipe (1) is 30 cm in diameter. The x-component of force (Rx) acting through the flange bolts is required to hold the nozzle in place is:
Solution :
Given data :
p = 315612 Pa
[tex]$V_1=7.07 \ m/sec$[/tex]
At exit of B,
p = [tex]$P_{atm}$[/tex]
[tex]$V_B = 26.1 \ m/sec$[/tex]
At exit of A,
[tex]p=P_{atm}[/tex]
[tex]$V_{A} = 26.1 \ m/s$[/tex]
We need to determine X component of force ([tex]$R_x$[/tex]) to hold in its place.
From figure,
[tex]$\sum F_x = m_0'V_{0x} - m_iV_{ix} $[/tex]
[tex]$=F_x+P_1A_1\sin 30=-mVA-mV_1 \sin 30$[/tex]
[tex]$=F_x=-pA_1\sin 30-m_AV_AA-m_B \sin30$[/tex]
Substitute all the values,
[tex]$=F_x=[-315612 \times \frac{\pi}{4}(0.3)^2 \sin 30]-[26.1 \times 1000 \times 26.1 \frac{\pi}{4}(0.1)^2]-[7.07 \times 1000\times 0.5 \sin 30]$[/tex][tex]$=F_x = -11154.64-5350.21-1767.28$[/tex]
[tex]$F_x = -18.2733 \ kN$[/tex]
Therefore, the force required to hold the nozzle in its place along horizontal direction.
[tex]$F_x = -18.2733 \ kN$[/tex]
What Is Soil Tunneling?
Answer:
A tunnel built in soft ground—such as clay, silt, sand, gravel or mud—requires specialized techniques compared to hard rock, to compensate for the shifting nature of the soil.
It's from web...
Soft ground tunneling describes the additional measures needed when Microtunneling through soil conditions that are vulnerable to collapse. ... This process ensures tunneling can happen effectively in soft grounds.
It's from me...
how Many years of college do you have To do in order To become a video game developer and designer
Answer:
Video game designers typically have a bachelor's degree in game design, computer engineering, or computer science which takes four to five years to complete. Courses in a game design degree program may include project management, integrated video design and technology, game prototyping and level design.
Here are the city gas mileages for 13 different midsized cars in 2008. 16, 15, 22, 21, 24, 19, 20, 20, 21, 27 , 18 , 21 , 48 What is the minimum ?
Answer:
Minimum city gas mileage is 15
Explanation:
Minimum city gas mileage among 13 different car sizes in 2008 is 15.
many manufacturers currently offer SAE level _________ equipped vehicles
A.2
B.3
C.4
D.5
find the volume of the pond with the following dimension length 40m breadth 10m height 1.2m depth 0.9m express in both meters and feet
Answer:
The volume for this is 29.7
Explanation:
Trust me on this I'm an expert
3. According to the drag equation the velocity of an object moving through a fluid can be modeled by the equation -- -ky- where k is a constant.
(a) Find the general solution to this equation.
(b) An object moving through the water has an initial velocity of 40 m/s. Two seconds later, the velocity has decreased to 30 m/s. What will the velocity be after ten seconds?
A find the ganeral solution to this equation
An object moving through the water has an initial velocity of 40 m/s. Two seconds later, the velocity has decreased to 30 m/s. The velocity after ten seconds is 0.
What is velocity?Velocity is defined as the speed at which an object's position changes in relation to time and a frame of reference. Speed is the rate at which an object travels along a path over time, whereas velocity is the speed and direction of an item's motion. In other words, speed is a scalar value, but velocity is a vector.
As given
Initial velocity = 40 m / sec
Velocity after 2 seconds = 30 m / sec
So the velocity after 10 seconds will be = 0 m / sec
That is the object will stop moving.
Thus. an object moving through the water has an initial velocity of 40 m/s. Two seconds later, the velocity has decreased to 30 m/s. The velocity after ten seconds is 0.
To learn more about velocity, refer to the link below:
https://brainly.com/question/28738284
#SPJ2
Your question is incomplete, but probably your complete question was
The pressure gage on a 2.5-m3 oxygen tank reads 500 kPa. Determine the amount of oxygen in the tank (mass in kg) if the temperature is 28°C and the atmospheric pressure is 97 kPa.
Answer:
[tex]n=5.36kg[/tex]
Explanation:
From the question we are told that:
Volume [tex]V=2.5m^3[/tex]
Pressure[tex]\rho=500Kpa[/tex]
Temperature [tex]T=28^o[/tex]
Atmospheric pressure [tex]\rho_{atm} =97 kPa.[/tex]
Generally the equation for an Ideal gas is mathematically given by
[tex]PV=nRT[/tex]
Therefore
[tex]n=\frac{500*2.5}{8.314*28}[/tex]
[tex]n=5.36kg[/tex]
A pressure transducer has the following specifications: Input rage: 0-100 psi and the corresponding Output Range: 0-5 Volts. Linearity Error: 0.10% of the Reading Hysteresis Error: 0.10% of the Reading Sensitivity Error: 0.15% of the Reading Zero Drift Error: 0.20% of the Reading Its output is read via a voltmeter with instrument error of 0.10% of the reading and resolution of 0.01 V. If the applied pressure on the transducer is 65 psi, what is the design stage uncertainty of this pressure measurement system?
Answer:
0.287
Explanation:
Design-stage uncertainty can be expressed as :
Ud = √ Uo^2 + Uc^2 ------ ( 1 )
where : Uo = 1/2( resolution value ) = 1/2 * 0.01 V = 0.005 V
Uc = √(0.10)^2 + (0.10)^2 + (0.15)^2 + (0.20)^2 = 0.287
back to equation 1
Ud = √ ( 0.005)^2 + ( 0.287 )^2 = 0.287
Answer:
0.287
Explanation:
Ethanol blended with gasoline can be used to power a "flex-fueled" car. One particular blend that is gaining in popularity is E85, which is 85% ethanol and 15% gasoline. E85 is 80% cleaner burning than gasoline alone, and it reduces our dependency on foreign oil. But a flex-fueled car costs $ more than a conventional gasoline-fueled car. Additionally, E85 fuel gets % less miles per gallon than a conventional automobile. Consider a 100% gasoline-fueled car that averages miles per gallon. The E85-fueled car will average about miles per gallon. If either car will be driven miles before being traded in, how much will the E85 fuel have to cost (per gallon) to make the flex-fueled car as economically attractive as a conventional gasoline-fueled car? Gasoline costs $ per gallon. Work this problem without considering the time value of money.
Answer: hello your question is poorly written below is the complete question
Ethanol blended with gasoline can be used to power a flex-fueled car. One particular blend that is gaining in popularity is E85 , which is 85% ethanol and 15% gasoline. E85 is 80% cleaner burning than gasoline alone, and it reduces our dependency on foreign oil. But a flex-fueled car costs $1,000 more than a conventional gasoline-fueled car. Additionally, E85 fuel gets 10% less miles per gallon than a conventional automobile. Consider a 100% gasoline -fueled car that averages 30 miles per gallon. The E85-fueled car will average about 27 miles per gallon. If either car will be driven 81000 miles before being traded in, how much will the E85 fuel have to cost (per gallon) to make the flex-fueled car as economically attractive as a conventional gasoline-fueled car? Gasoline costs $3.89 per gallon. Work this problem without considering the time value of money
answer :
$3.17
Explanation:
Determine how much E85 fuel have to cost
Fuel needed by 100% gasoline fueled car
= 81,000 miles / 30 = 2700 gallons
Fueled needed by E85 car
= 81,000 miles / 27 = 3000 gallons
next step : use the relation below
[ (Additional cost of flex fuel car) + (flex fuel consumption * cost of flex fuel per gallon( x ) ) ] = [ Gasoline consumption * cost of gasoline per gallon ]
= 1000 + 3000x = 2700 * 3.89
= 1000 + 3000x = 10503
∴ x = 9503 / 3000 = $3.17 per gallon
Consider a venturi with a throat-to-inlet area ratio of 0.75, mounted on the side of an airplane fuselage. The airplane is in flight at standard sea level. If the static pressure at the throat is 2050 lb/ft2 , calculate the velocity of the airplane.
This question is incomplete, the complete question is;
Consider a venturi with a throat-to-inlet area ratio of 0.75, mounted on the side of an airplane fuselage. The airplane is in flight at standard sea level. If the static pressure at the throat is 2050 lb/ft2 , calculate the velocity of the airplane.
Note that standard sea level density and pressure are 1.23 kg/m3 (0.002377 slug/ft3) and 1.01 x 105 N/m2 (2116lb/ft3), respectively.
Answer:
the velocity of the airplane is 267.2 ft/s
Explanation:
Given the data in the question;
throat-to-inlet area ratio A₂/A₁ = 0.75
density of air ρ = 0.002377 slug/ft³
the pressure at inlet p₁ = 2116 lb/ft³
the pressure at the throat p₂ = 2050 lb/ft³
Now, for a venturi duct, the velocity of the airplane V is given as;
V = √[ (2( p₁ - p₂ )) / (ρ( [A₁/A₂]² - 1 )) ]
so we substitute in our values
V = √[ (2( 2116 - 2050 )) / (0.002377 ( [1/0.75]² - 1 )) ]
V = √[ ( 2 × 66 ) / (0.002377 ( 1.7778 - 1 )) ]
V = √[ ( 2 × 66 ) / (0.002377 × 0.7778 ) ]
V = √[ 132 / 0.0018488 ]
V = √[ 71397.663349 ]
V = 267.2 ft/s
Therefore, the velocity of the airplane is 267.2 ft/s
?Why the efficiency of Class A amplifier is very poor
R-134a is throttled in a line flowing at 25oC, 750 kPa with negligible kinetic energy to a pressure of 165 kPa. Find the exit temperature and the ratio of the exit pipe diameter to that of the inlet pipe (Dex/Din) so that the velocity stays constant.
Solution :
For R-134a, we are given :
[tex]$T_i = 25^\circ C$[/tex]
[tex]$P_i=750 \ kPa$[/tex]
[tex]$P_e=165 \ kPa$[/tex]
Now we have one inlet and one exit flow, no work and no heat transfer. The energy equation is :
[tex]$h_e+\frac{1}{2}.v_e^2= h_i+\frac{1}{2}.v_i^2 $[/tex]
We also know that the gas is throttled and there is no change in the kinetic energy.
So, [tex]$v_e=v_i$[/tex]
Now from the energy equation above, we can see that the inlet and the exit enthalpies are also the same. Therefore,
[tex]$h_i=h_e$[/tex]
From the saturated R-134a table, corresponding to [tex]P_e = 165 \ kPa[/tex], we can find the exit saturation temperature.
[tex]$T_e=-15^\circ C$[/tex]
From the saturated R-134a table, corresponding to [tex]P_e = 165 \ kPa[/tex], we can find the specific enthalpies :
[tex]$h_f = 180.19 \ kJ/kg$[/tex]
[tex]$h_{fg} = 209 \ kJ/kg$[/tex]
Calculating the exit flow quality factor,
[tex]$x_e=\frac{h_e-h_f}{h_{fg}}$[/tex]
[tex]$=\frac{234.59-180.19}{209}$[/tex]
= 0.26
From the saturated R-134a table, corresponding to [tex]P_e = 165 \ kPa[/tex], we can find the specific volumes :
[tex]$v_f = 0.00746 \ m^3/kg$[/tex]
[tex]$v_{fg} = 0.11932 \ m^3/kg$[/tex]
Calculating the exit specific volume :
[tex]$v_e=v_f+x_e(v_{fg})$[/tex]
= 0.000746 + 0.26 (0.11932)
= 0.0318 [tex]m^3/kg[/tex]
The mass flow is equal to :
[tex]$\dot{m} = A_i . \frac{v}{v_i}$[/tex]
[tex]$=A_e . \frac{v}{v_e}$[/tex]
So, [tex]$\frac{A_e}{A_i}=\frac{v_e}{v_i}$[/tex]
Therefore, the ratio of the exit pipe and the inlet pipe diameter is equal to
[tex]$\frac{D_e}{D_i}=\sqrt{\frac{A_e}{A_i}}$[/tex]
[tex]$\frac{D_e}{D_i}=\sqrt{\frac{v_e}{v_i}}$[/tex]
[tex]$\frac{D_e}{D_i}=\sqrt{\frac{0.0318}{0.000829}}$[/tex]
[tex]$\frac{D_e}{D_i}=6.19$[/tex]
Request for proposal (RFP) is a type of document that contains the information and proposals mostly through the bidding process. This document is regarding the valuable assets, services, entity, commodity, etc.
Answer:
Answer to the following is as follows;
Explanation:
A request for proposal is a documentation that invites prospective contractors to submit business opportunities to an agency or corporation interested in procuring a commodities, product, or valuable resource through a bid procedure.
A request for proposal (RFP) is a commercial document that introduces a project, defines it, and invites eligible contractors to compete on its completion.
khái niệm về môi trường nhiệt nóng và môi trường nhiện lạnh ?
ảnh hưởng của môi trường và môi trường nhiệt lạnh đến con người như thế nào ?
Theo em môi trường nào gây nguy hiểm hơn đối với con người ? Vì sao ?
Explanation:
उह्ह्नमजज्ल्ह्ह्बनुतनकुहक्जो
Cite the phases that are present and the phase compositions for the following alloys: (a) 15 wt% Sn - 85 wt% Pb at 100 o C. (b) 1.25 kg of Sn and 14 kg Pb at 200 o C
Answer:
a) ∝ and β
The phase compositions are :
C[tex]_{\alpha }[/tex] = 5wt% Sn - 95 wt% Pb
C[tex]_{\beta }[/tex] = 98 wt% Sn - 2wt% Pb
b)
The phase is; ∝
The phase compositions is; 82 wt% Sn - 91.8 wt% Pb
Explanation:
a) 15 wt% Sn - 85 wt% Pb at 100⁰C.
The phases are ; ∝ and β
The phase compositions are :
C[tex]_{\alpha }[/tex] = 5wt% Sn - 95 wt% Pb
C[tex]_{\beta }[/tex] = 98 wt% Sn - 2wt% Pb
b) 1.25 kg of Sn and 14 kg Pb at 200⁰C
The phase is ; ∝
The phase compositions is; 82 wt% Sn - 91.8 wt% Pb
Csn = 1.25 * 100 / 1.25 + 14 = 8.2 wt%
Cpb = 14 * 100 / 1.25 + 14 = 91.8 wt%
The calculated value of the thermal conductivity of the carbon nano tube was found as: KCN = 3113 W/m-K, however, the theoretical value of the thermal conductivity of the wire is actually: K = 4500 W/m-K and the island separation is 5 μm (this is the actual spacing between the two islands). The difference between the measured and theoretical values is due to the contact resistance between the nano tube and the islands in the experiment.
Required:
a. Calculate the thermal contact resistance (Rtd) that exists between the carbon nano tube and the top surfaces of the heated and sensing islands.
b. Using the value of thermal contact resistance calculated in part A, calculate the fraction of the total resistance between the heated and sensing islands that is due to the thermal contact resistances for island separation distance of 5, 10, 15, and 20 μm.
Answer:
a) 1,607,973.9 K/W
b)
i) 0.3082 = 30.82%
ii) 0.1821 = 18.21%
iii) 0.1293 = 12.93%
iv) 0.1002 = 10.02%
Explanation:
Value of thermal conductivity ( calculated value ) KCN = 3113 W/m-k
Thermal conductivity ( theoretical value ) K = 4500 W/m-k
Island separation = 5 μm
a) Determine the thermal contact resistance
Resistance due to contact between carbon nano tube and top surfaces can be determined using the relation below
( I / A*K ) + 2Rc = ( l / A*KCN ) ------- ( 1 )
where ; I = 5 * 10^-6 m
A = π * ( 14 * 10^-9 )^2 m^2 = 153.93 * 10^-18 , K = 4500 , KCN = 3113
input values into equation 1 above
hence Rc = 1,607,973.9 K/W
b) Determine fraction of total resistance between heated and sensing
fraction of total resistance ; f1 = [tex]\frac{2 Rc}{I/KA + 2Rc}[/tex]
where : Rc = 1607973.9, K = 4500, A = 153.93 * 10^-18 ,
i) for I = 5 * 10^-6 m
fraction = 0.3082 = 30.82%
ii) for I = 10 * 10^-6 m
fraction = 0.1821 = 18.21%
iii) for I = 15 * 10^-6 m
fraction = 0.1293 = 12.93%
iv) for I = 20*10^-6
fraction = 0.1002 = 10.02%