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slavikrds [6]
3 years ago
9

A hollow pipe is submerged in a stream of water so that the length of the pipe is parallel to the velocity of the water. If the

water speed doubles and the cross-sectional area of the pipe triples, what happens to the volume flow rate of the water passing through it?
Engineering
1 answer:
Arlecino [84]3 years ago
8 0

Answer:

increases by a factor of 6.

Explanation:

Let us assume that the initial cross sectional area of the pipe is A m² while the initial velocity of the water is V m/s², hence the flow rate of the water is:

Initial flow rate = area * velocity = A * V = AV m³/s

The water speed doubles (2V m/s) and the cross-sectional area of the pipe triples (3A m²), hence the volume flow rate becomes:

Final flow rate = 2V * 3A = 6AV m³/s = 6 * initial flow rate

Hence, the volume flow rate of the water passing through it increases by a factor of 6.

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Write a program that asks the user to input a vector of integers of arbitrary length. Then, using a for-end loop the program exa
ELEN [110]

Answer:

%Program prompts user to input vector

v = input('Enter the input vector: ');

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fprintf('The input vector:\n ')

disp(v)

%Loop for checking all array elements

for i = 1 : length(v)

   %check if the element is a positive number

   if v(i) > 0

       %double the element

       v(i) = v(i) * 2;

   %else the element is negative number.

   else

       %triple the element

       v(i) = v(i) * 3;

   end

end

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fprintf('The modified vector:\n ')

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4 0
4 years ago
A piston-cylinder assembly has initially a volume of 0.3 m3 of air at 25 °C. Mass of the air is 1 kg. Weights are put on the pis
kogti [31]

Answer:

n=2.32

w= -213.9 KW

Explanation:

V_1=0.3m^3,T_1=298 K

V_2=0.1m^3,T_1=1273 K

Mass of air=1 kg

For polytropic process  pv^n=C ,n is the polytropic constant.

  Tv^{n-1}=C

  T_1v^{n-1}_1=T_2v^{n-1}_2

298\times .3^{n-1}_1=1273\times .1^{n-1}_2

n=2.32

Work in polytropic process given as

       w=\dfrac{P_1V_1-P_2V_2}{n-1}

      w=mR\dfrac{T_1-T_2}{n-1}

Now by putting the values

w=1\times 0.287\dfrac{289-1273}{2.32-1}

w= -213.9 KW

Negative sign indicates that work is given to the system or work is done on the system.

For T_V diagram

  We can easily observe that when piston cylinder reach on new position then volume reduces and temperature increases,so we can say that this is compression process.

5 0
3 years ago
Explain why the following scenario fails to meet the criteria for proper reverse engineering.
avanturin [10]

Answer:

he must document or remember the order he took it apart so he put it back together

Explanation:

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7 0
2 years ago
The theoretical maximum specific gravity of a mix at 5.0% binder content is 2.495. Using a binder specific gravity of 1.0, find
PSYCHO15rus [73]

Answer:

The theoretical maximum specific gravity at 6.5% binder content is 2.44.

Explanation:

Given the specific gravity at 5.0 %  binder content 2.495

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Where the  binder is S.G. = 1, Therefore

Per 100 mass unit we have (Mx + 5)/(Vx + 5) = 2.495

(95 +5)/(Vx +5) = 2.495

2.495 × (Vx + 5) = 100

Vx =35.08 to 95

Or density of mix = Mx/Vx = 95/35.08 = 2.7081

Therefore when we have 6.5 % binder content, we get

Per 100 mass unit

93.5 Mass unit of Mx has a volume of

Mass/Density = 93.5/2.7081 = 34.526 volume units

Therefore we have

At 6.5 % binder content.

(100 mass unit)/(34.526 + 6.5) = 2.44

The theoretical maximum specific gravity at 6.5% binder content = 2.44.

3 0
3 years ago
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