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Xelga [282]
3 years ago
11

A pipe is insulated such that the outer radius of the insulation is less than the critical radius. Now the insulation is taken o

ff. Will the rate of heat transfer from the pipe increase or decrease for the same pipe surface temperature?
Engineering
1 answer:
irina [24]3 years ago
7 0

Answer:

the heat transfer from the pipe will decrease when the insulation is taken off for r₂< r_{cr}

where;

r₂ = outer radius

r_{cr} = critical radius

Explanation:

Note that the critical radius of insulation depends on the thermal conductivity of the insulation k and the external convection heat transfer coefficient h .

r_{cr} =\frac{k}{h}

The rate of heat transfer from the cylinder increases with the addition of insulation for outer radius less than  critical radius (r₂< r_{cr}) 0,  and reaches a maximum when r₂ = r_{cr}, and starts to decrease for r₂< r_{cr}. Thus, insulating the pipe may actually increase the rate of heat transfer from the pipe instead of decreasing it when r₂< r_{cr} .

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Which option identifies why Ethan’s skills are valuable to his team in the following scenario?
larisa [96]

Answer:

Explanation:

The options are:

- In an isometric drawing, multiple angles and axes can be shown in one sketch.

- There is no room for detail in an isometric drawing, so the detail is shown in the orthographic projection.

- Only one sketch will be needed since all other previous designs will no longer be necessary.

- Computer programs will not be necessary to create the exact dimensions of the design.

Orthographic projections are in either the First or Third Angles but the angles are fixed and do not provide perspective view.  Isometric drawings are perspective views from different angles.

So Ethan's skill is valuable because "In an isometric drawing, multiple angles and axes can be shown in one sketch."

4 0
2 years ago
The wall of drying oven is constructed by sandwiching insulation material of thermal conductivity k = 0.05 W/m°K between thin me
masha68 [24]

Answer:

86 mm

Explanation:

From the attached thermal circuit diagram, equation for i-nodes will be

\frac {T_ \infty, i-T_{i}}{ R^{"}_{cv, i}} + \frac {T_{o}-T_{i}}{ R^{"}_{cd}} + q_{rad} = 0 Equation 1

Similarly, the equation for outer node “o” will be

\frac {T_{ i}-T_{o}}{ R^{"}_{cd}} + \frac {T_{\infty, o} -T_{o}}{ R^{"}_{cv, o}} = 0 Equation 2

The conventive thermal resistance in i-node will be

R^{"}_{cv, i}= \frac {1}{h_{i}}= \frac {1}{30}= 0.033 m^{2}K/w Equation 3

The conventive hermal resistance per unit area is

R^{"}_{cv, o}= \frac {1}{h_{o}}= \frac {1}{10}= 0.100 m^{2}K/w Equation 4

The conductive thermal resistance per unit area is

R^{"}_{cd}= \frac {L}{K}= \frac {L}{0.05} m^{2}K/w Equation 5

Since q_{rad}  is given as 100, T_{o}  is 40 T_ \infty  is 300 T_{\infty, o}  is 25  

Substituting the values in equations 3,4 and 5 into equations 1 and 2 we obtain

\frac {300-T_{i}}{0.033} +\frac {40-T_{i}}{L/0.05} +100=0  Equation 6

\frac {T_{ i}-40}{L/0.05}+ \frac {25-40}{0.100}=0

T_{i}-40= \frac {L}{0.05}*150

T_{i}-40=3000L

T_{i}=3000L+40 Equation 7

From equation 6 we can substitute wherever there’s T_{i} with 3000L+40 as seen in equation 7 hence we obtain

\frac {300- (3000L+40)}{0.033} + \frac {40- (3000L+40)}{L/0.05}+100=0

The above can be simplified to be

\frac {260-3000L}{0.033}+ \frac {(-3000L)}{L/0.05}+100=0

\frac {260-3000L}{0.033}=50

-3000L=1.665-260

L= \frac {-258.33}{-3000}=0.086*10^{-3}m= 86mm

Therefore, insulation thickness is 86mm

8 0
3 years ago
13. Which stroke of the four-stroke cycle is shown in the above figure?
lianna [129]

Answer:

the cycle is on the power just before the exhaust as both the valves are closed

7 0
2 years ago
100 kg of R-134a at 200 kPa are contained in a piston–cylinder device whose volume is 12.322 m3. The piston is now moved until t
LekaFEV [45]

Answer:

T=151 K, U=-1.848*10^6J

Explanation:

The given process occurs when the pressure is constant. Given gas follows the Ideal Gas Law:

 pV=nRT

For the given scenario, we operate with the amount of the gas- n- calculated in moles. To find n, we use molar mass: M=102 g/mol.  

Using the given mass m, molar mass M, we can get the following equation:  

 pV=mRT/M

To calculate change in the internal energy, we need to know initial and final temperatures. We can calculate both temperatures as:

T=pVM/(Rm); so initial T=302.61K and final T=151.289K

 

Now we can calculate change of U:

U=3/2 mRT/M using T- difference in temperatures

 U=-1.848*10^6 J

Note, that the energy was taken away from the system.  

5 0
3 years ago
In part A you are asked to write the pseudocode for the program. In part B you are asked to write the syntax of the code for the
Naya [18.7K]

Answer:

C++.

Explanation:

#include <iostream>

#include <string>

using namespace std;

///////////////////////////////////////////////////////////////

int main() {

   string quote, book;

   int page;

   

   cout<<"What is your favorite quote from a book?"<<endl;

   getline(cin, quote);

   cout<<endl;

   /////////////////////////////////////////////

   cout<<"What book was that quote from?"<<endl;

   getline(cin, book);

   cout<<endl;

   /////////////////////////////////////////////

   cout<<"What page was that quote from?"<<endl;

   cin>>page;

   cout<<endl;

   /////////////////////////////////////////////

   int no_of_upper_characters = 0;

   for (int i=0; i<quote.length(); i++) {

       if (isupper(quote[i]))

          no_of_upper_characters++;

   }

   

   cout<<"No. of upper case characters: "<<no_of_upper_characters<<endl;

   /////////////////////////////////////////////

   int no_of_characters = quote.length();

   cout<<"No. of characters: "<<no_of_characters<<endl;

   /////////////////////////////////////////////

   bool isDog = false;

   for (int i=0; i<quote.length(); i++) {

       if (isDog == true)

           break;

       else if (quote[i] == 'd') {

           for (int j=i+1; j<quote.length(); j++) {

               if (isDog == true)

                   break;

               else if (quote[j] == 'o') {

                   for (int z=j+1; z<quote.length(); z++) {

                       if (quote[z] == 'g') {

                           isDog = true;

                           break;

                       }

                   }

               }

           }

       }

   }

   

   if (isDog == true)

       cout<<"This includes 'd' 'o' 'g' in the quote";

   //////////////////////////////////////////////

   return 0;

}

3 0
3 years ago
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