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Lelu [443]
3 years ago
9

A composite wall is made of two layers of 0.3 m and 0.15 m thickness with surfaces held at 600°C and 20°C respectively. If the c

onductivities are 20 and 50 W/mK, determine the heat conducted. In order to restrict the heat loss to 5 kW/m2 another layer of 0.15 m thickness is proposed. Determine the thermal conductivity of the material required

Engineering
1 answer:
kogti [31]3 years ago
7 0

Answer:

Q=32.22\dfrac{KW}{m^2}

K_3=1.5\ \frac{W}{m.K}

Explanation:

At initial condition

As we know that  thermal resistance

R_{th}=\dfrac{L}{KA}

So the total thermal resistance

R_{th}=\dfrac{L_1}{K_1A}+\dfrac{L_2}{K_2A}

Now by putting the values

R_{th}=\dfrac{L_1}{K_1A}+\dfrac{L_2}{K-2A}\

R_{th}=\dfrac{0.3}{20A}+\dfrac{0.15}{50A}

R_{th}=\dfrac{0.018}{A}\ \frac{K}{W}

So the heat conduction

Q=\dfrac{\Delta T}{R_{th}}

Q=\dfrac{600-20}{0.018}

Q=32.22\dfrac{KW}{m^2}

At final condition another layer is added

Given\ that\ heat\ flux\ is\ 5\ \frac{KW}{m^2}

So the total thermal resistance

R_{th}=\dfrac{L_1}{K_1A}+\dfrac{L_2}{K_2A}+\dfrac{L_3}{K_3A}

R_{th}=\dfrac{0.018}{A}+\dfrac{0.15}{K_3A}\ \frac{K}{W}

\dfrac{0.018}{A}+\dfrac{0.15}{K_3}=\dfrac{\Delta T}{q}

\dfrac{0.018}{A}+\dfrac{0.15}{K_3A}=\dfrac{580}{5000A}

K_3=1.5\ \frac{W}{m.K}

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Nutka1998 [239]

Answer:

β = \frac{c}{\sqrt{km} } =  0.7071 ≈ 1 ( damping condition )

closed-form expression for the response is attached below

Explanation:

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computing a solution :

M = 1,

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next we determine the damping condition using the damping formula

β = \frac{c}{\sqrt{km} } =  0.7071 ≈ 1

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How can I solve 23.5 million Nona meters to millimeters using no calculator because I have to show my work
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Define an ADT for a two-dimensional array of integers. Specify precisely the basic operations that can be performed on such arra
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Answer:

Explanation:

ADT for an 2-D array:

struct array{

int arr[10];

}arrmain[10];

An application that stores an array with 1000 rows and 1000 columns, where less than 10,000 of the array values are non-zero. The two different implementations for such arrays that would be more space efficient than a standard two-dimensional array implementation requiring one million positions are :

1) struct array{

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2) struct array{

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3 years ago
Radioactive wastes generating heat at a rate of 3 x 106 W/m3 are contained in a spherical shell of inner radius 0.25 m and outsi
MariettaO [177]

Answer:

Inner surface temperature= 783K.

Outer surface temperature= 873K

Explanation:

Parameters:

Heat,e= 3×10^6 W/m^3

Inner radius = 0.25 m

Outside radius= 0.30 m

Temperature at infinity, T(¶)= 10°c = 273. + 10 = 283K.

Convection coefficient,h = 500 W/m^2 . K

Temperature of the surface= T(s) = ?

Temperature of the inner= T(I) =?

STEP 1: Calculate for heat flux at the outer sphere.

q= r × e/3

This equation satisfy energy balance.

q= 1/3 ×3000000(W/m^3) × 0.30 m

= 3× 10^5 W/m^2.

STEP 2: calculus the temperature for the surface.

T(s) = T(¶) + q/h

T(s) = 283 + 300000( W/m^2)/500(W/m^2.K)

T(s) = 283+600

T(s)= 873K.

TEMPERATURE FOR THE OUTER SURFACE is 873 kelvin.

The same TWO STEPS are use for the calculation of inner temperature, T(I).

STEP 1: calculate for the heat flux.

q= r × e/3

q= 1/3 × 3000000(W/m^3) × 0.25 m

q= 250,000 W/m^2

STEP 2:

calculate the inner temperature

T(I) = T(¶) + q/h

T(I) = 283K + 250,000(W/m^2)/500(W/m^2)

T(I) = 283K + 500

T(I) = 783K

INNER TEMPERATURE IS 783 KELVIN

5 0
3 years ago
Write a naive implementation (i.e. non-vectorized) of matrix multiplication, and then write an efficient implementation that uti
erik [133]

Answer:

import numpy as np  

import time  

def matrixMul(m1,m2):      

   if m1.shape[1] == m2.shape[0]:  

       

       t1 = time.time()

       r1 = np.zeros((m1.shape[0],m2.shape[1]))

       for i in range(m1.shape[0]):

           for j in range(m2.shape[1]):

               r1[i,j] = (m1[i]*m2.transpose()[j]).sum()

       t2 = time.time()

       print("Native implementation: ",r1)

       print("Time: ",t2-t1)

       

       t1 = time.time()

       r2 = m1.dot(m2)

       t2 = time.time()

       print("\nEfficient implementation: ",r2)

       print("Time: ",t2-t1)

       

   else:

       print("Wrong dimensions!")

Explanation:

We define a function (matrixMul) that receive two arrays representing the two matrices to be multiplied, then we verify is the dimensions are appropriated for matrix multiplication if so we proceed with the native implementation consisting of two for-loops and prints the result of the operation and the execution time, then we proceed with the efficient implementation using .dot method then we return the result with the operation time. As you can see from the image the execution time is appreciable just for large matrices, in such a case the execution time of the efficient implementation can be 1000 times faster than the native implementation.

7 0
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