Answer:
1.) 103.23 degree centigrade
2.) 126.96 degree centigrade
Explanation:
Given that a transparent film is to be bonded onto the top surface of a solid plate inside a heated chamber. For the bond to cure properly, a temperature of 70°C is to be maintained at the bond, between the film and the solid plate. The transparent film has a thickness of 1 mm and thermal conductivity of 0.05 W/m·K, while the solid plate is 13 mm thick and has a thermal conductivity of 1.2 W/m·K. Inside the heated chamber, the convection heat transfer coefficient is 70 W/m2·K. If the bottom surface of the solid plate is maintained at 52°C.
To determine the temperature inside the heated Chamber, let us first calculate the heat transfer rate per unit area through the plate by using the formula
Heat transfer rate R = k( Tb - T2)/L
Where
k = 1.2 W/mA.K
Tb = 70 degree
T2 = 52 degree
L = 13mm = 13/1000 = 0.013m
substitute all the parameters into the formula above.
R = 1.2 x ( 70 - 52 )/ 0.013
R = 1.2 x (18/0.013)
R = 1661. 5 W/m^2
the surface temperature of the transparent film will be
Ts = Tb + (RLf/Kf)
Ts = 70 + (1661.5 x 0.001)/0.05
Ts = 70 + (1.6615)/0.05
Ts = 70 + 33.23
Ts = 103.23 degree centigrade
the temperature inside the heated Chamber will be calculated by using the formula
Ti = Ts + (R/h)
Ti = 103.23 + (1661.5/70)
Ti = 103.23 + 23.74
Ti = 126.97 degree centigrade
what is the irregular past tense of hear?
Answer:
Past tense of hear = heard
what is the answer What will finally break the internet?
Which of the following is most characteristic of OSHA's relationship with professional trade organizations in residential construction?
O beneficial
confrontational
competitive
awkward
Answer:
its beneficial
Explanation:
took the test
Answer:
Beneficial would be the correct answer.
Explanation:
In this activity, you will conduct research on a famous work of architecture from the 1800s and write a report about the work and the architect who created it. You will then create an architectural drawing.
__________________________________________________________________________
Directions and Analysis
Task 1: Researching an Architectural Work of the 1800s
Use the Internet and other available resources to conduct basic research on architecture from the 1800s. Select a building designed by a significant (influential) architect from that period, and then write a paper about the building and architect. Make sure you include the following:
About the architect
basic biographical information
education
how the architect influenced society and design
About the building/work
main materials used
structural considerations
technology and tools used during construction
Answer:
can you explain how to do this
Explanation:
A hot brass plate is having its upper surface cooled by impinging jet of air at temperature of 15°C and convection heat transfer coefficient of 220 W/m^2•K. The 10-cm thick brass plate (rho = 8530 kg/m^3, cp = 380 J/kg•K, k = 110 W/m•K, and α = 33.9×10^–6 m^2/s) has a uniform initial temperature of 900°C, and the bottom surface of the plate is insulated.
Required:
Determine the temperature at the center plane of the brass plate after 3 minutes of cooling.
Answer:
809.98°C
Explanation:
STEP ONE: The first step to take in order to solve this particular Question or problem is to find or determine the Biot value.
Biot value = (heat transfer coefficient × length) ÷ thermal conductivity.
Biot value = (220 × 0.1)÷ 110 = 0.2.
Biot value = 0.2.
STEP TWO: Determine the Fourier number. Since the Biot value is greater than 0.1. Tis can be done by making use of the formula below;
Fourier number = thermal diffusivity × time ÷ (length)^2.
Fourier number = (3 × 60 × 33.9 × 10^-6)/( 0.1)^2 = 0.6102.
STEP THREE: This is the last step for the question, here we will be calculating the temperature of the center plane of the brass plate after 3 minutes.
Thus, the temperature of the center plane of the brass plane after 3 minutes = (1.00705) (0.89199) (900- 15) + 15.
= > the temperature of the center plane of the brass plane after 3 minutes = 809.98°C.
I need help with online lab
Answer:
I believe on the fist question it power rationg
Water that has evaporated returns to earth as
Answer:
rain
Explanation:
evaoration causes clouds
clouds condense and rain
Answer:
rain
Explanation:
Silicon carbide nanowires of diameter D = 15 nm can be grown onto a solid siliconcarbide surface by carefully depositing droplets of catalyst liquid onto a flat silicon carbide substrate. Siliconcarbide nanowires grow upward from the deposited drops, and if the drops are deposited in a pattern, an array ofnanowire fins can be grown, forming a silicon carbide nano-heat sink. Consider finned and unfinned electronicspackages in which an extremely small, 10 μm × 10 μm electronics device is sandwiched between two d = 100-nm-thick silicon carbide sheets. In both cases, the coolant is a dielectric liquid at 20°C. A heat transfer coefficient of h= 1 × 10^5 W/m^2 · K exists on the top and bottom of the unfinned package and on all surfaces of the exposed siliconcarbide fins, which are each L = 300 nm long. Each nano-heat sink includes a 200 × 200 array of nanofins.
Required:
Determine the maximum allowable heat rate that can be generated by the electronic device so that its temperatureis maintained at Tt < 85°C for the unfinned and finned packages.
Answer:
For unfined package = 1.30 × 10^-3 W.
Dor fined package = 8.64 × 10^-3 W.
Explanation:
STEP ONE: The first thing to do is to determine the cross sectional area and input the value into the convectional resistance formula/equation in order to determine the heat rate for the unfined package.
Thus, Area = (dimension of the electronic device) ^2. = (10 × 10^-6)^2 m.
Convectional Resistance = (100 × 10^-9)÷ (10 × 10^-6)^2 × 490 = 2.04 K/W.
Heat transfer coefficient = 1/ (10 × 10^-7)^2 × 10^5. = 1 × 10^5 K/W.
Thus, the heat rate for the unfined package = 2{(85 - 20) ÷ (10^5 + 2.04)} = 1.30 × 10^-3 W.
STEP TWO: Determine the surface area of each fin and the prime area.
Surface area of each fin = {300 + 15/4} × 10^-9 × 10^-9 × π × 15 = 1.43 × 10^-14 m^2.
Prime area = (10^-6 × 10)^2 - (200 × 200) × π × [ ( 15 × 10^-9)^2 ÷ 4] = 9.29 × 10^-11 m^2.
STEP THREE: Determine the efficiency
G ={ [300 + 15/4) × 10^9 } × [ 4 × 10^5÷ 490 × 15 × 10^-9] = 7.09 × 10^-2.
Efficiency= tanh ( 7.09 × 10^-2) ÷ 7.09 × 10^-2 = 0.998.
Thus, the coefficient = 1 - (1.43 × 10^-14 × (200 × 200) ÷ (6.65 × 10^-10) × (1 - 0.998) = 0.99.
STEP FOUR: Determine the heat rate for the finned package.
Thermal resistance = 1/ (10^5 × 0.999 × 10^-10 × 6.65 = 1.50 × 10^4 K/W.
The heat rate for the finned package = 2 { 85 - 20} ÷ { 1.50 × 10^4 + 2.04} = 8.64 × 10^-3 W
what is an undescribable outcome from cellphones?
Answer:
Cell phones emit low levels of non-ionizing radiation when in use. The type of radiation emitted by cell phones is also referred to as radio frequency (RF) energy. As stated by the National Cancer Institute, "there is currently no consistent evidence that non-ionizing radiation increases cancer risk in humans.
Explanation:
What is one advantage corporations have over other types of businesses
in software engineering how do you apply design for change?
Answer:
it is reducely very iloretable chance for a software engineer to give an end to this question