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il63 [147K]
4 years ago
10

A wall in a house contains a single window. The window consists of a single pane of glass whose area is 0.15 m2 and whose thickn

ess is 5 mm. Treat the wall as a slab of the insulating material Styrofoam whose area and thickness are 10 m2 and 0.15 m, respectively. Heat is lost via conduction through the wall and the window. The temperature difference between the inside and outside is the same for the wall and the window. Of the total heat lost by the wall and the window, what is the percentage lost by the window
Engineering
1 answer:
Oxana [17]4 years ago
4 0

Answer:

The percentage loss of the window is  E  = 97.3%

Explanation:

From the question we are told that

      The area of pane of glass is  A = 0.15 m^2

      The thickness is d = 5mm = \frac{5}{1000} = 0.005m

       The thickness of the wall is  D = 0.15m

       The area of the wall is  a = 10m^2

Generally the heat lost as a result of conduction of the window is  

              Q_{window} = \frac{j_{glass} * A * (\Delta T) }{d}

Where j_{glass} is the thermal conductivity of glass which has a constant value of

          j_{glass} = 0.80 J/(s \cdot m \cdot  C^o)

 Substituting values

                 Q_{window} = \frac{ 0.80  * 0.15  * (\Delta T) }{0.005}

                 Q_{window} = 24 \Delta T

Generally the heat lost as a result of conduction of the wall is  

              Q_{wall} = \frac{j_{styrofoam} * A * (\Delta T) }{d}

j_{styrofoam} s the thermal conductivity of Styrofoam which has a constant value of  j_{styrofoam} = 0.010J / (s \cdot m \cdot C^o)

       Substituting values

                 Q_{wall} = \frac{ 0.010  * 10  * (\Delta T) }{0.15}

                 Q_{wall} = 0.667 \  \Delta T

Now the net loss of heat is

         Q_{net} = Q_{window} +  Q_{wall}

  Substituting values

         Q_{net} = 24 + 0.667

         Q_{net} =  24.667

Now the percentage loss by the window is  

            E  = \frac{Q_{window} }{Q_{net}}  * 100

  Substituting value  

           E  = \frac{24}{24 .667}  * 100

           E  = 97.3%

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When using bits to represent fractions of a number, can you create all possible fractions? Why or why not?
Strike441 [17]

Answer:

  no

Explanation:

There are an infinite number of fractions. You cannot create all possible fractions with a finite number of bits.

5 0
4 years ago
Energy is generated uniformly in a 6 cm thick wall. The steady-state temperature distribution
mezya [45]

Answer:

Heat generation per unit volume is 45 KW.

Explanation:

Given that

Thickness of wall = 6 cm

Temperature distribution

T(z)=145+3000z-1500z^2    -----1

K= 15 W/m.k

As we know that at steady state condition

\dfrac{d^2T}{dz^2}+\dfrac{q}{K}=0   -----2

Where q is the heat generation per unit volume.

So from equation 1

\dfrac{dT}{dz}=3000-3000z

\dfrac{d^2T}{dz^2}=-3000

Now from equation 2

\dfrac{d^2T}{dz^2}+\dfrac{q}{K}=0  

-3000+\dfrac{q}{15}=0  

So q= 45 KW

So heat generation per unit volume is 45 KW.

6 0
3 years ago
Calculate the energy in kJ added to 1 ft3 of water by heating it from 70° to 200°F.
Serga [27]

Answer:

The energy in kJ is 8558.16 kJ.

Explanation:

Data presented in the problem:

Water is heated from 70 (T1) to 200 °F (T2).

Volume (V) of the water is 1 ft3.

It is required for the specific heat of water(HW), which is 1 BTU/lb°F.

First, we need to calculate the mass of water (M) presented in the process. Water density (D) is 62. 4 lb/ft3.

M = V*D = (1ft3)*(62.4 lb/ft3) = 62.4 lb Water.

.After that, we can calculate the heat required (Q).

Q = M*HW*(T2 - T1) = (62.4 lb)*(1 BTU/lb°F)(200 °F - 70°F)

Q = 62.4 * 130 BTU = 8112 BTU.

Q is converted to kJ units using the conversion factor 1 BTU = 1.055 kJ

Q = 8112 BTU * (1.055 kJ/1BTU) = 8558.16 kJ.

Finally, the energy required is 8558.16 kJ.

3 0
3 years ago
Write the definition of a function printLarger, which has two int parameters and returns nothing. The function prints the larger
vredina [299]

Answer:

  1. #include <iostream>
  2. using namespace std;
  3. void printLarger(int a, int b){
  4.    
  5.    if(a > b){
  6.        cout<<a;
  7.    }else{
  8.        cout<<b;
  9.    }
  10. }
  11. int main()
  12. {
  13.    printLarger(4, 5);
  14.    return 0;
  15. }

Explanation:

The solution code is written in C++.

Firstly define a function printLarger that has two parameters, a and b with both of them are integer type (Line 5). In the function, create an if condition to check if a bigger than b, print a to terminal (Line 7-8). Otherwise print b (Line 9-10).

In the main program, test the function by passing 4 and 5 as arguments (Line 16) and we shall get 5 printed.  

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4 years ago
A) For Well A, provide a cross-section sketch that shows (i) ground elevation, (ii) casing height, (iii) depth to
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