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Based on the results, which statement is most likely a criterion for the gaming system design? Choose the correct answer.
a. width of about 8 inches
b. ability to connect wifi
c. battery life of at least 24hrs
d. production cost of less than $40
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Answer:
B
Explanation:
You need to connect to WiFi without it no multiplayer
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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.
Answers
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.
What is one advantage corporations have over other types of businesses
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what is the answer What will finally break the internet?
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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.
Answers
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
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