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Stells [14]
3 years ago
13

If the head loss in a 30 m of length of a 75-mm-diameter pipe is 7.6 m for a given flow rate of water, what is the total drag fo

rce exerted by the water on this length of pipe?
Engineering
1 answer:
Stolb23 [73]3 years ago
3 0

Answer:

526.5 KN

Explanation:

The total head loss in a pipe is a sum of pressure head, kinetic energy head and potential energy head.

But the pipe is assumed to be horizontal and the velocity through the pipe is constant, Hence the head loss is just pressure head.

h = (P₁/ρg) - (P₂/ρg) = (P₁ - P₂)/ρg

where ρ = density of the fluid and g = acceleration due to gravity

h = ΔP/ρg

ΔP = ρgh = 1000 × 9.8 × 7.6 = 74480 Pa

Drag force over the length of the pipe = Dynamic pressure drop over the length of the pipe × Area of the pipe that the fluid is in contact with

Dynamic pressure drop over the length of the pipe = ΔP = 74480 Pa

Area of the pipe that the fluid is in contact with = 2πrL = 2π × (0.075/2) × 30 = 7.069 m²

Drag Force = 74480 × 7.069 = 526468.1 N = 526.5 KN

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Wastewater flows into a _________ once it is released into a floor drain.
rodikova [14]

Answer:

A) Sump pit

Explanation:

A wastewater typically refers to a body of water that has contaminated through human use in homes, offices, schools, businesses etc. Wastewater are meant to be disposed in accordance with the local regulations and standards because they are unhygienic for human consumption or use.

Generally, many homes use a floor drain in their bathrooms and toilets to remove wastewater in order to mitigate stagnation and to improve hygiene. A floor drain can be defined as a material installed on floors for the continuous removal of any stagnant wastewater in buildings. Wastewater flows into a sump pit once it is released into a floor drain through the use of a pipe such as a polyvinyl chloride (PVC) pipe, which directly connects the floor drain to the sump pit. The wastewater can the be removed from the sump pit when it is filled up through the use of a pump.

6 0
2 years ago
How would an engineer know if a product design were feasible?
Masteriza [31]
- the last one ‘design meets all the criteria...’
8 0
3 years ago
Read 2 more answers
HW6P2 (20 points) The recorded daily temperature (°F) in New York City and in Denver, Colorado during the month of January 2014
Maurinko [17]

Answer & Explanation:

function Temprature

NYC=[33 33 18 29 40 55 19 22 32 37 58 54 51 52 45 41 45 39 36 45 33 18 19 19 28 34 44 21 23 30 39];

DEN=[39 48 61 39 14 37 43 38 46 39 55 46 46 39 54 45 52 52 62 45 62 40 25 57 60 57 20 32 50 48 28];

%AVERAGE CALCULATION AND ROUND TO NEAREST INT

avgNYC=round(mean(NYC));

avgDEN=round(mean(DEN));

fprintf('\nThe average temperature for the month of January in New York city is %g (F)',avgNYC);

fprintf('\nThe average temperature for the month of January in Denvar is %g (F)',avgDEN);

%part B

count=1;

NNYC=0;

NDEN=0;

while count<=length(NYC)

   if NYC(count)>avgNYC

       NNYC=NNYC+1;

   end

   if DEN(count)>avgDEN

        NDEN=NDEN+1;

   end

   count=count+1;

end

fprintf('\nDuring %g days, the temprature in New York city was above the average',NNYC);

fprintf('\nDuring %g days, the temprature in Denvar was above the average',NDEN);

%part C

count=1;

highDen=0;

while count<=length(NYC)

   if NYC(count)>DEN(count)

       highDen=highDen+1;

   end

   count=count+1;

end

fprintf('\nDuring %g days, the temprature in Denver was higher than the temprature in New York city.\n',highDen);

end

%output

check the attachment for additional Information

8 0
3 years ago
Turn on your____
storchak [24]

Answer:

b

Explanation:

5 0
2 years ago
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A 12-ft circular steel rod with a diameter of 1.5-in is in tension due to a pulling force of 70-lb. Calculate the stress in the
padilas [110]

Answer:

The stress in the rod is 39.11 psi.

Explanation:

The stress due to a pulling force is obtained dividing the pulling force by the the area of the cross section of the rod. The respective area for a cylinder is:

A=\pi*D^2/4

Replacing the diameter the area results:

A= 17.76 in^2

Therefore the the stress results:

σ = 70/17.76 lb/in^2 = 39.11 lb/in^2= 39.11 psi

5 0
2 years ago
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