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Vera_Pavlovna [14]
3 years ago
8

Consider the thermocouple and convection conditions of Example 1, but now allow for radiation exchange with the walls of a duct

that encloses the gas stream. If the duct walls are at 400℃ and the emissivity of the thermocouple bead is 0.9, calculate the steady-state temperature of the junction

Engineering
1 answer:
alex41 [277]3 years ago
6 0

Answer:

hello your question has some missing part attached below is the complete question

answer : steady state temperature = 419.713k ≈ 218.7⁰c

              Time required to reach a junction ≈ 5 secs

Explanation:

The detailed solution of the given problem is attached below but the solution to the subsequent problem from which the question you asked is referenced to( problem 1 ), is not attached because it was not part of the question you asked

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Number pattern Write a recursive method called print Pattern() to output the following number pattern. Given a positive integer
lilavasa [31]

Answer:

See explaination

Explanation:

Code;

import java.util.Scanner;

public class NumberPattern {

public static int x, count;

public static void printNumPattern(int num1, int num2) {

if (num1 > 0 && x == 0) {

System.out.print(num1 + " ");

count++;

printNumPattern(num1 - num2, num2);

} else {

x = 1;

if (count >= 0) {

System.out.print(num1 + " ");

count--;

if (count < 0) {

System.exit(0);

}

printNumPattern(num1 + num2, num2);

}

}

}

public static void main(String[] args) {

Scanner scnr = new Scanner(System.in);

int num1;

int num2;

num1 = scnr.nextInt();

num2 = scnr.nextInt();

printNumPattern(num1, num2);

}

}

See attachment for sample output

3 0
3 years ago
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The future goal of health information is appropriate access to ubiquitous, personalized, specific information relevant to the si
erastovalidia [21]
I thinks it’s a A.True
3 0
3 years ago
I don’t get this it’s hella hard
qwelly [4]

Answer:

V₂ = 20 V

Vt = 20 V

V₁ = 20 V

V₃ = 20 V

I₁ = 10 mA

I₃ = 3.33 mA

It = 18.33 mA

Rt = 1090.91 Ω

Pt = 0.367 W

P₁ = 0.2 W

P₂ = 0.1 W

P₃ = 0.067 W

Explanation:

Part of the picture is cut off.  I assume there is a voltage source Vt there?

First, use Ohm's law to find V₂.

V = IR

V₂ = (0.005 A) (4000 Ω)

V₂ = 20 V

R₁ and R₃ are in parallel with R₂ and the voltage source Vt.  That means V₁ = V₂ = V₃ = Vt.

V₁ = 20 V

V₃ = 20 V

Vt = 20 V

Now we can use Ohm's law again to find I₁ and I₃.

V = IR

I = V/R

I₁ = (20 V) / (2000 Ω)

I₁ = 0.01 A = 10 mA

I₃ = (20 V) / (6000 Ω)

I₃ = 0.00333 A = 3.33 mA

The current It passing through Vt is the sum of the currents in each branch.

It = I₁ + I₂ + I₃

It = 10 mA + 5 mA + 3.33 mA

It = 18.33 mA

The total resistance is the resistance of the parallel resistors:

1/Rt = 1/R₁ + 1/R₂ + 1/R₃

1/Rt = 1/2000 + 1/4000 + 1/6000

Rt = 1090.91 Ω

Finally, the power is simply each voltage times the corresponding current.

P = IV

Pt = (0.01833 A) (20 V)

Pt = 0.367 W

P₁ = (0.010 A) (20 V)

P₁ = 0.2 W

P₂ = (0.005 A) (20 V)

P₂ = 0.1 W

P₃ = (0.00333 A) (20 V)

P₃ = 0.067 W

7 0
3 years ago
2. A trapezoidal channel has a bottom width of 4 m and side slopes of 2:1 (H:V). If the flow rate is 50 m3/s at a depth of 3.6 m
Daniel [21]

Answer:

Explanation:

Sum of the side slope = 2 + 1 = 3

Length of first slope = 2/3 X 3.6 = 2 X 1.2 = 2.4m

Lenght of second slope = 1/3 X 3.6 = 1.2m

Area of the trapezoidal channel = (2.4 + 1.2)/2 X 3.6 = 1.8 X 3.6 = 6.48m²

Alternate dept = 50m³/6.48m²= 7.716m

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3 years ago
In python/Spyder
lianna [129]

Answer:

Answer to both the parts A and B are provided in the word document attached.

Explanation:

The explanation of the programs is provided in the attached file along with the screenshot of ATM application output.

Download docx
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