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katrin [286]
4 years ago
10

A 15 cm x 15 cm circuit board dissipating 20 W of power uniformly is cooled by air, which approaches the circuit board at 20°C w

ith a velocity of 6 m/s. Disregarding any heat transfer from the back surface of the board, determine the surface temperature of the electronic components (a) at the leading edge and (b) at the end of the board. Assume the flow to be turbulent since the electronic components are expected to act as turbulators.

Engineering
1 answer:
Illusion [34]4 years ago
7 0

Answer:

(a) Surface temperature at the leading edge = 49.9°C

(b)  Surface temperature at the end of the board = 51.1°C

Explanation:

See the attached files for the calculation

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A beam of span L meters simply supported by the ends, carries a central load W. The beam section is shown in figure. If the maxi
saw5 [17]

Answer:

W = 11,416.6879 N

L ≈ 64.417 cm

Explanation:

The maximum shear stress, \tau_{max}, is given by the following formula;

\tau_{max} = \dfrac{W}{8 \cdot I_c \cdot t_w} \times \left (b\cdot h^2 - b\cdot h_w^2 + t_w \cdot h^2_w \right )

t_w = 1 cm = 0.01

h = 29 cm = 0.29 m

h_w = 25 cm = 0.25 m

b = 15 cm = 0.15 m

I_c = The centroidal moment of inertia

I_c = \dfrac{1}{12} \cdot \left (b \cdot h^3 - b \cdot h_w^3 + t_w \cdot h_w^3 \right )

I_c = 1/12*(0.15*0.29^3 - 0.15*0.25^3 + 0.01*0.25^3) = 1.2257083 × 10⁻⁴ m⁴

Substituting the known values gives;

I_c = \dfrac{1}{12} \cdot \left (0.15 \times 0.29^3 - 0.15 \times 0.25^3 + 0.01 \times 0.25^3 \right )  = 1.2257083\bar 3 \times 10^{-4}

I_c = 1.2257083\bar 3 × 10⁻⁴ m⁴

From which we have;

4,500,000 = \dfrac{W}{8 \times 1.225708\bar 3 \times 10 ^{-4}\times 0.01} \times \left (0.15 \times 0.29^2 - 0.15 \times 0.25^2 + 0.01 \times 0.25^2 \right )

Which gives;

W = 11,416.6879 N

\sigma _{b.max} = \dfrac{M_c}{I_c}

\sigma _{b.max} = 1500 N/cm² = 15,000,000 N/m²

M_c = 15,000,000 × 1.2257083 × 10⁻⁴ ≈ 1838.56245 N·m²

From Which we have;

M_{max} = \dfrac{W \cdot L}{4}

L = \dfrac{4 \cdot M_{max}}{W} = \dfrac{4 \times 1838.5625}{11,416.6879} \approx 0.64417

L ≈ 0.64417 m ≈ 64.417 cm.

4 0
3 years ago
Wqqwfqwfqwfqfqfqffqwffqwqfqqfqfqffqqfqfwccc
almond37 [142]

Answer:

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Explanation:

4 0
3 years ago
Read 2 more answers
What is an air mass?​
kotegsom [21]

Answer:

An air mass is a body of air with horizontally uniform temperature, humidity, and pressure.

Explanation:

Because it is

8 0
3 years ago
Read 2 more answers
B1) 20 pts. The thickness of each of the two sheets to be resistance spot welded is 3.5 mm. It is desired to form a weld nugget
kap26 [50]

Answer:

minimum current level required =  8975.95 amperes

Explanation:

Given data:

diameter = 5.5 mm

length = 5.0 mm

T = 0.3

unit melting energy = 9.5 j/mm^3

electrical resistance = 140 micro ohms

thickness of each of the two sheets = 3.5mm

Determine the minimum current level required

first we calculate the volume of the weld nugget

v = \frac{\pi }{4} * D^2 * l = \frac{\pi }{4} * 5.5^2 * 5 = 118.73 mm^3

next calculate the required melting energy

= volume of weld nugget * unit melting energy

= 118.73 * 9.5 = 1127.94 joules

next find the actual required electric energy

= required melting energy / efficiency

= 1127 .94 / ( 1/3 )  = 3383.84 J

TO DETERMINE THE CURRENT LEVEL REQUIRED  use the relation below

electrical energy =  I^2 * R * T

3383.84 / R*T = I^2

3383.84 / (( 140 * 10^-6 ) * 0.3 ) = I^2

therefore  8975.95 = I ( current )

4 0
3 years ago
Fix the code so the program will run correctly for MAXCHEESE values of 0 to 20 (inclusive). Note that the value of MAXCHEESE is
GarryVolchara [31]

Answer:

Code fixed below using Java

Explanation:

<u>Error.java </u>

import java.util.Random;

public class Error {

   public static void main(String[] args) {

       final int MAXCHEESE = 10;

       String[] names = new String[MAXCHEESE];

       double[] prices = new double[MAXCHEESE];

       double[] amounts = new double[MAXCHEESE];

       // Three Special Cheeses

       names[0] = "Humboldt Fog";

       prices[0] = 25.00;

       names[1] = "Red Hawk";

       prices[1] = 40.50;

       names[2] = "Teleme";

       prices[2] = 17.25;

       System.out.println("We sell " + MAXCHEESE + " kind of Cheese:");

       System.out.println(names[0] + ": $" + prices[0] + " per pound");

       System.out.println(names[1] + ": $" + prices[1] + " per pound");

       System.out.println(names[2] + ": $" + prices[2] + " per pound");

       Random ranGen = new Random(100);

       // error at initialising i

       // i should be from 0 to MAXCHEESE value

       for (int i = 0; i < MAXCHEESE; i++) {

           names[i] = "Cheese Type " + (char) ('A' + i);

           prices[i] = ranGen.nextInt(1000) / 100.0;

           amounts[i] = 0;

           System.out.println(names[i] + ": $" + prices[i] + " per pound");

       }        

   }

}

7 0
4 years ago
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