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Whitepunk [10]
3 years ago
7

True or False: Galvanized steel pipe can withstand moderate freezing.

Engineering
1 answer:
yaroslaw [1]3 years ago
5 0

Answer:

True

Explanation:

Older galvanized steel pipes, which have a tendency to freeze, are a bit more forgiving and will likely not burst. They can withstand extreme cold and warm temperatures.

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A Si sample contains 1016 cm-3 In acceptor atoms and a certain number of shallow donors, the In acceptor level is 0.16 eV above
creativ13 [48]

Answer:

6.5 × 10¹⁵/ cm³

Explanation:

Thinking process:

The relation N_{o} = N_{i} * \frac{E_{f}-E_{i}  }{KT}

With the expression Ef - Ei = 0.36 × 1.6 × 10⁻¹⁹

and ni = 1.5 × 10¹⁰

Temperature, T = 300 K

K = 1.38 × 10⁻²³

This generates N₀ = 1.654 × 10¹⁶ per cube

Now, there are 10¹⁶ per cubic centimeter

Hence, N_{d}  = 1.65*10^{16}  - 10^{16} \\           = 6.5 * 10^{15} per cm cube

5 0
3 years ago
Read 2 more answers
An experimentalist claims that, based on his measurements, a heat engine receives 300 Btu of heat from a source of 900 R, conver
lesya [120]

Answer:

Hook's law holds good up to. A elastic limit. B. plastic limit. C.yield point. D.Breaking point

3 0
3 years ago
Define a function pyramid_volume with parameters base_length, base_width, and pyramid_height, that returns the volume of a pyram
Maslowich

Hi, you haven't provided the programing language in which you need the code, I'll just explain how to do it using Python, and you can apply a similar method for any programming language.

Answer:

1. def pyramid_volume(base_length, base_width, pyramid_height):

2.     volume = base_length*base_width*pyramid_height/3

3.     return(volume)

Explanation step by step:

  1. In the first line of code, we define the function pyramid_volume and it's input parameters
  2. In the second line, we perform operations with the input values to get the volume of the pyramid with a rectangular base, the formula is V = l*w*h/3
  3. In the last line of code, we return the volume  

In the image below you can see the result of calling the function with input 4.5, 2.1, 3.0.

5 0
3 years ago
Consider a 2-shell-passes and 8-tube-passes shell-and-tube heat exchanger. What is the primary reason for using many tube passes
Maru [420]

Answer:

See explanation

Explanation:

Solution:-

- The shell and tube heat exchanger are designated by the order of tube and shell passes.

- A single tube pass: The fluid enters from inlet, exchange of heat, the fluid exits.

- A multiple tube pass: The fluid enters from inlet, exchange of heat, U bend of the fluid, exchange of heat, .... ( nth order of pass ), and then exits.

- By increasing the number of passes we have increased the "retention time" of a specific volume of tube fluid; hence, providing sufficient time for the fluid to exchange heat with the shell fluid.

- By making more U-turns we are allowing greater length for the fluid flow to develop with " constriction and turns " into turbulence. This turbulence usually at the final passes allows mixing of fluid and increases the heat transfer coefficient by:

                                U ∝ v^( 0.8 )    .... ( turbulence )

- The higher the velocity of the fluids the greater the heat transfer coefficient. The increase in the heat transfer coefficient will allow less heat energy carried by either of the fluids to be wasted ; hence, reduced losses.

Thereby, increases the thermal efficiency of the heat exchanger ( higher NTU units ).

5 0
3 years ago
What is the thermal efficiency of this regeneration cycle in terms of enthalpies and fractions of total flow?
irga5000 [103]

Answer:

\eta =\dfrac{(h_3-h_4)-(h_2-h_1)}{(h_3-h_5)}

Explanation:

generally regeneration of cycle is used in the case of gas turbine. due to regeneration efficiency of turbine is increased but there is no effect on the on the net work out put of turbine.Actually in regeneration net heta input is decreases that is why total efficiency  increase.

 Now from T-S diagram

    W_{net}=W_{out}-W_{in}

   W_{net}=(h_3-h_4)-(h_2-h_1)

  Q_{in}=h_3-h_5

  Due to generation (h_5-h_2) amount of energy has been saved.

  Q_{generation}=Q_{saved}

So efficiency of cycle \eta =\frac{W_{net}}{Q_{in}}

  \eta =\dfrac{(h_3-h_4)-(h_2-h_1)}{(h_3-h_5)}

Effectiveness of re-generator

  \varepsilon =\dfrac{(h_5-h_2)}{(h_4-h_2)}

So the efficiency of regenerative cycle

\eta =\dfrac{(h_3-h_4)-(h_2-h_1)}{(h_3-h_5)}

7 0
3 years ago
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