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konstantin123 [22]
3 years ago
8

Consider insulation on a circular pipe For the same thickness and type of insulation, the thermal resistance of the insulation i

s (a) Higher for larger diameter pipes (b) The same for all pipes independent of the diameter (c) Lower for larger diameter pipe
Engineering
1 answer:
leonid [27]3 years ago
5 0

Answer:

b). The same for all pipes independent of the diameter

Explanation:

We know,

R_{conduction}=\frac{ln(\frac{r_{2}}{r_{1}})}{2\pi LK}

R_{convection}=\frac{1}{h(2\pi r_{2}L)}

From the above formulas we can conclude that the thermal resistance of a substance mainly depends upon heat transfer coefficient,whereas radius has negligible effects on heat transfer coefficient.

We also know,

Factors on which thermal resistance of insulation depends are :

1. Thickness of the insulation

2. Thermal conductivity of the insulating material.

Therefore from above observation we can conclude that the thermal resistance of the insulation is same for all pipes independent of diameter.

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Write a complete C++ program that is made of functions main() and rShift(). The rShift() function must be a function without ret
erma4kov [3.2K]

Answer:

Explanation:

attached is an uploaded picture to support the answer.

the program is written thus;

#include<iostream>

using namespace std;

// function declaration

void rShift(int&, int&, int&, int&, int&, double&);

int main()

{

   // declare the variables

   int a1, a2, a3, a4;

   int maximum = 0;

   double average = 0;

   // inputting the numbers

   cout << "Enter all the 4 integers seperated by space -> ";

   cin >> a1 >> a2 >> a3 >> a4;

   cout << endl << "Value before shift." << endl;

   cout << "a1 = " << a1 << ", a2 = " << a2 << ", a3 = "  

        << a3 << ", a4 = " << a4 << endl << endl;

   // calling rSift()

   // passing the actual parameters

   rShift(a1,a2,a3,a4,maximum,average);

   // printing the values

   cout << "Value after shift." << endl;

   cout << "a1 = " << a1 << ", a2 = " << a2 << ", a3 = "  

           << a3 << ", a4 = " << a4 << endl << endl;

   cout << "Maximum value is: " << maximum << endl;

   cout << "Average is: " << average << endl;

}

// function to right shift the parameters circularly

// and calculate max, average of the numbers

// and return it to main using pass by reference

void rShift(int &n1, int &n2, int &n3, int &n4, int &max, double &avg)

{

   // calculating the max

   max = n1;

   if(n2 > max)

     max = n2;

   if(n3 > max)

     max = n3;

   if(n4 > max)

     max = n4;

   // calculating the average

   avg = (double)(n1+n2+n3+n4)/4;

   // right shifting the numbers circulary

   int temp = n2;

   n2 = n1;

   n1 = n4;

   n4 = n3;

   n3 = temp;

}

8 0
4 years ago
Briefly explain why small-angle grain boundaries are not as effective in interfering with the slip process as are high-angle gra
Vlada [557]

Answer:

Explanation:

Small-angle grain boundaries are not as effective in interfering with the slip process as are high-angle grain boundaries because there is not as much crystallographic misalignment in the grain boundary region for small-angle, and therefore not as much change in slip direction.

Low angle grain boundaries (quasi-coherent) are formed by the dislocation network positioned along the geometric plane with small tilt angle differences between successive peers that is tilt boundary made up edge dislocations therefore it may only divert the slip direction of the incoming gliding dislocation with very little frictional stresses. And on the other hand, a high angle grain boundary region because of their disordered almost liquid like structure which acts as a strong barrier against dislocation slip motion and causes actually formation of dislocations file-up against it by arresting their motion unless that the stress concentration at the leading dislocation becomes high enough to go though the barrier.

5 0
4 years ago
You want to determine whether the race of the defendant has an impact on jury verdicts. You assign participants to watch a trial
Andru [333]

Answer:

The confidence scale represents an ordinal scale of measurement

Explanation:

An ordinal scale or level of measurement is used to measure attributes that can be ranked or ordered, but the interval between the attributes do not have quantitative significance. In this case, the measurement was done on a scale of 1 - 7, with a "1" being; not all that race of defendant has an impact on jury verdicts and a "7" being "very" meaning that race indeed has impact on jury verdicts. Another example can be a survey carried out on the level of customer satisfaction on a particular product, with "1" most dissatisfied and "10 " representing most satisfied. In the first example, it is wrong to say that the difference between 1 being "not at all" and maybe 3 is the same as the difference between 5 and 7 which have different connotations, because the numbers are merely for tagging and not to quantify.

Other levels of measurement include:

1. Nominal: this is the simplest level of measurement and it is simply used to categorize the attributes. Example is taking a survey on gender in the categories of male, female and transgender.

2. Interval: the interval scale is used when the distance between two attributes have meanings but there is no true zero value associated with the scale.

3. Ratio: this combines all the other three levels of measurement and is used to categorize, used to show ranking, has meaningful distances between the attributes and the scale has a true zero point. Example is the measurement of temperature using the celcius scale thermometer, where there is a true zero point at 0°C and the distance between 5°C and 10°C is the same as the distance between 10°C and 15°C.

6 0
3 years ago
Chapter 15 – Fasteners: Determine the tensile load capacity of a 5/16 – 18 UNC thread and a 5/16 – 24 UNF thread made of the sam
Roman55 [17]

Answer:

The 5/16 – 24 UNF is stronger because it has more tensile load capacity.

Tensile load capacity for M8 -1.25 = 5670 lb

Tensile load capacity for M8 -1 = 6067 lb

Explanation:

For 5/16 - 18 UNC thread:

D = 0.3125

n = 18

Therefore the tensile load capacity is = 100000 X (0.7854 X (0.3125 - 0.9743/ 18) ^2

= 5243 lb.

Similarly for 5/16 - 24 UNF , only the n value changes to 24

we get the tensile load capacity = 5806.6 lb

Hence the 5/16 – 24 UNF is stronger because it has more tensile load capacity.

For metric Bolts:

We have to consider all values in SI units

Strength = 689 MPa

We get for M8 -1.25:

Tensile load capacity as = 689 X 36.6 = 25223 N = 5670 lb

For M8 -1:

Tensile load capacity as = 689 X 39.167 = 26986 N = 6067lb

7 0
3 years ago
Read 2 more answers
These tadpoles are confined to a limited environment. What are they all competing for in that environment
Dovator [93]

Answer: to the earth air

Explanation:

5 0
3 years ago
Read 2 more answers
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