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Svetradugi [14.3K]
3 years ago
5

An array of electronic chips is mounted within a sealedrectangular enclosure, and colling is implemented by attaching analuminum

heat sink (k=180 W/m*K). The base of the heat sinkhas dimensions of w1=w2=100mm, while the 6 fins are of thicknesst=10mm and pitch S=18mm. The fin length is Lf=50 mm, and thebase of the heat sink has a thickness of Lb=10mm.
If cooling is implemented by water flow through the heat sink,with u[infinity]=3 m/s and T[infinity]=17 C, what is the base temperatureTb of the heat sink when the power dissipation by the chips isPelec=1800W? The average convection coefficient for surfacesof the fins and the exposed base may be estimated by assumingparallel flow over a flat plate. Properties of the water maybe approximated as k=0.62 W/m*K, ν=7.73E-7 m2/s, andPr=5.2.

Engineering
1 answer:
Licemer1 [7]3 years ago
6 0

Answer:

Base temperature is 46.23 °C

Explanation:

I've attached explanations

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2. Other igneous rock forms from lava that cools quickly on Earth’s surface. Classify the rock as either intrusive or extrusive,
Darya [45]

Answer:

extrusive,

Explanation:

lava exlodes outwards, making it extrusive and not intrusive.

5 0
3 years ago
Comparison of copper and aluminium conductors looking at their properties
Alexeev081 [22]

Answer:

The density of the copper is higher than aluminium. Hence it is heavier compared to aluminium conductors it requires strong structures and hardware to bear the weight. More ductile and has high tensile strength.

...

Aluminium & Copper properties.

Property Copper (Cu) Aluminium (Al)

Density (g/cm3) 8.96 2.70

7 0
3 years ago
the oscillation of rod oa about o is defined by the relation θ= (3/pi)*(sin*pi*t), where theta and t are expressed in radians an
Mashcka [7]

Answer:

(a) The velocity of the collar = 1.624er-15.54eo in/s

(b) the acceleration of the collar=-49.94er-9.74eo in/s²

(c)the acceleration of collative rod =-3.284er in/s²

Explanation:

Check attachment for calculation

4 0
3 years ago
Design a filter that has infinite DC gain, a gain of one from 1Hz to 100 Hz and filters (1storder) any signals above 100 Hz.a) S
EastWind [94]

Answer:

Attached below are the  sketches

answer :

c) G(s) = 100 / ( s + 100 )

d) y'(t)  + 100Y(s) = 100 X(s)

e) g(t) = e^-100t  u(t)

Explanation:

a) Sketch the bode plot

The filter here is a low pass filter

b) Sketch the s-plane

attached below.     pole ( s ) is at 100

c) write the transfer function of the filter

Transfer function ; G(s) = 100 / ( s + 100 )

d) write the differential equation

Y(s) / X(s) = 100 / s + 100

Y(s) [ s + 100 ] = 100 X(s)

= sY(s) + 100Y = 100 X(s)

∴ differential equation = y'(t)  + 100Y(s) = 100 X(s)

e) write out the unforced transient response

g(t) = e^-100t  u(t)

f) write out the frequency response

attached below

4 0
3 years ago
Problem the pressure at a given point is 50 mmhg absolute
VLD [36.1K]

Answer:

6.65 kPa.

- 73.3 kPa.

Explanation:

Without much ado let's jump right into the solution to the problem given. It is given that the pressure = 50 mmHg and the solution to this question is to write out the value of the pressure in kpa and kPa gauge.

P(a) = 0.05m × 133 kN/m³ = 6.65 kPa.

The P(gauge) =( [ 0.05 × 13.6 × 9810] ÷ 1000 ) - 80 = - 73.3 kPa.

7 0
3 years ago
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