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Naya [18.7K]
2 years ago
7

One-dimensional plane wall of thickness 2L=80 mm experiences uniform thermal generation of q dot =1000 W/m^3 and is convectively

cooled at x=±40m by an ambient fluid characterized by T [infinity] = 30 degrees C. If the steady-state temperature distribution within the wall is T(x) = a(L2-x2)+b where a = 15o C/m^2 and b=40oC, what is the thermal conductivityof the wall? What is the value of the convection heat transfer coefficient?
Engineering
1 answer:
Black_prince [1.1K]2 years ago
7 0

Answer:

Thermal Conductivity (K) = 33.33 W/m. ° C

The value of the convection heat transfer coefficient = 3 W/m².° C

Explanation:

The attached document file gives a detailed and clear explanation about the question.

Download docx
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Prompt the user to enter five numbers, being five people's weights. Store the numbers in an array of doubles. Output the array's
pochemuha

Answer:

import java.util.Scanner;

  public class PeopleWeights {

    public static void main(String[] args) {

    Scanner reader = new Scanner(System.in);  

    double weightOne = reader.nextDouble();

    System.out.println("Enter 1st weight:");

    double weightTwo = reader.nextDouble();

    System.out.println("Enter 2nd weight :");

    double weightThree = reader.nextDouble();

    System.out.println("Enter 3rd weight :");

    double weightFour = reader.nextDouble();

    System.out.println("Enter 4th weight :");

    double weightFive = reader.nextDouble();

    System.out.println("Enter 5th weight :");

     double sum = weightOne + weightTwo + weightThree + weightFour + weightFive;

     double[] MyArr = new double[5];

     MyArr[0] = weightOne;

     MyArr[1] = weightTwo;

     MyArr[2] = weightThree;

     MyArr[3] = weightFour;

     MyArr[4] = weightFive;

     System.out.printf("You entered: " + "%.1f %.1f %.1f %.1f %.1f ", weightOne, weightTwo, weightThree, weightFour, weightFive);

     double average = sum / 5;

     System.out.println();

     System.out.println();

     System.out.println("Total weight: " + sum);

     System.out.println("Average weight: " + average);

     double max = MyArr[0];

     for (int counter = 1; counter < MyArr.length; counter++){

        if (MyArr[counter] > max){

           max = MyArr[counter];

        }

     }

     System.out.println("Max weight: " + max);

  }

import java.util.Scanner;

  public class PeopleWeights {

    public static void main(String[] args) {

    Scanner reader = new Scanner(System.in);  

    double weightOne = reader.nextDouble();

    System.out.println("Enter 1st weight:");

    double weightTwo = reader.nextDouble();

    System.out.println("Enter 2nd weight :");

    double weightThree = reader.nextDouble();

    System.out.println("Enter 3rd weight :");

    double weightFour = reader.nextDouble();

    System.out.println("Enter 4th weight :");

    double weightFive = reader.nextDouble();

    System.out.println("Enter 5th weight :");

     double sum = weightOne + weightTwo + weightThree + weightFour + weightFive;

     double[] MyArr = new double[5];

     MyArr[0] = weightOne;

     MyArr[1] = weightTwo;

     MyArr[2] = weightThree;

     MyArr[3] = weightFour;

     MyArr[4] = weightFive;

     System.out.printf("You entered: " + "%.1f %.1f %.1f %.1f %.1f ", weightOne, weightTwo, weightThree, weightFour, weightFive);

     double average = sum / 5;

     System.out.println();

     System.out.println();

     System.out.println("Total weight: " + sum);

     System.out.println("Average weight: " + average);

     double max = MyArr[0];

     for (int counter = 1; counter < MyArr.length; counter++){

        if (MyArr[counter] > max){

           max = MyArr[counter];

        }

     }

     System.out.println("Max weight: " + max);

  }

8 0
2 years ago
Read 2 more answers
Calculate the mean piston speed, bmep, and specific power of: (a) the spark-ignition engines in Figs. 1.11 and 1.13 at their max
Kisachek [45]

Answer:

Explanation:

See the attached picture for detailed answer.

5 0
2 years ago
Q4. What happen when a steady potential is connected across the end points of the [3] conductor? Illustrate with an example.
den301095 [7]

Answer: Flow of current and the resistance to the flow of the current.

Explanation:

When a steady potential is connected across the end points of the conductor, there will be a of current through the conductor and the internal resistance in the conductor will tend to resist the flow of the current

A good example is a simple circuit comprises of a battery, wires and a touch bulb.

When the battery is connected to the two ends of the torch bulb, current flow through and the bulb is lit up.

4 0
2 years ago
generally compound curves are not filtered recommended for A. Road B. water way C. underground road D. rail way​
vfiekz [6]

Answer:

C. underground road

Explanation:

Generally compound curves are not filtered and recommended for use in an underground road. However, they are best used in the road, water way, and rail way.

3 0
2 years ago
Air enters a horizontal, constant-diameter heating duct operating at steady state at 290 K, 1 bar, with a volumetric flow rate o
Taya2010 [7]

Answer:

a) m = 0.3003 kg/s

b) Vel1 = 6.25 m/s, Vel2 = 7.3725 m/s

c) 12.845 KW

Explanation:

a)

using ideal gas law:

PV = nRT

since, n = no. of moles = m/M

therefore,

PV = (m/M)RT

P1 v1 = RT1/M

where,

P1 = inlet pressure = 1 bar = 100000 Pa

v1 = specific volume at inlet = ?

R = universal gas constant = 8.314 KJ/Kmol.k

T1 = inlet temperature = 290 K

M = Molecular mass of air = 28.9628 Kg/kmol

Therefore,

v1 = (8.314 KJ/Kmol.k)(290 k)/(28.9628 kg/kmol)(100 KPa)

v1 = 0.83246 m³/kg

Now, the mass flow rate can be given as:

Mass Flow Rate = (Volume Flow Rate)/(v1)

Mass Flow Rate = (0.25 m³/s)/(0.83246 m³/kg)

<u>Mass Flow Rate = 0.3003 kg/s</u>

b)

using general gas equation to find the specific volume at the exit first, we get:

P1v1/T1 = P2v2/T2

v2 = P1 v1 T2/T1 P2

where,

P1 = inlet pressure = 1 bar

P2 = exit pressure = 0.95 bar

T1 = inlet temperature = 290 k

T2 = exit temperature = 325 k

v1 = specific volume at inlet = 0.83246 m³/kg

v2 = specific volume at exit = ?

Therefore,

v2 = (1 bar)(0.83246 m³/kg)(325 k)/(290 k)(0.95 bar)

v2 = 0.98203 m³/kg

Now, for velocity, we use formula:

Vel = v/A

where,

A = Area = 0.04 m²

For inlet:

Vel1 = Inlet Volume flow Rate/A = (0.25 m³/s)/(0.04 m²)

<u>Vel1 = 6.25 m/s</u>

For exit:

Vel2 = Exit Volume flow Rate/A = (Mass Flow Rate)(v2)/(0.04 m²)

Vel2 = (0.98203 m³/kg)(0.3003 kg/s)/0.04 m²

<u>Vel2 = 7.3725 m/s</u>

c)

using first law of thermodynamics, with no work done, we can derive the formula given below:

Q = m(h2 - h1) + (m/2)(Vel2² - Vel1²)

where,

Q = rate of heat transfer

m = mass flow rate = 0.3003 kg/s

m(h2 - h1) = change in enthalpy = mCpΔT   , for ideal gas

Vel1 = 6.25 m/s

Vel2 = 7.3725 m/s

Also,

Cp in this case will be function of temperature and given as:

Cp = KR/(K-1)

where,

K = 1.4

R = Gas constant per molecular mass of air = 0.28699 KJ/kg.k

Therefore,

Cp = 1.0045 KJ/kg.k

Now, using the values in the expression of first law of thermodynamics, we get:

Q = (0.3003 kg/s)(1.0045 KJ/kg.k)(35 K) + [(0.3003 kg/s)/2][(7.3725 m/s)² - (6.25 m/s)²]

Q = 10.55 KW + 2.29 KW

<u>Q = 12.845 KW</u>

6 0
2 years ago
Read 2 more answers
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