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MaRussiya [10]
3 years ago
5

Discuss four (4) advantages of direct and indirect water supply system.

Engineering
2 answers:
Reil [10]3 years ago
7 0

Answer and Explanation:

Some of the advantages of direct water supply system are:

  • Smaller size of the cistern
  • No risk of polluted water
  • Ease of installation
  • Lower pipe work

Some of the advantages of indirect water supply system are:

  • It can be pumped completely
  • Lower demand on the main source
  • Minimization of wear and tear on taps
  • Larger size of the cistern.
poizon [28]3 years ago
7 0

Answer:

Direct water supply system:

1. Smaller cold water cistern.

2.Less costly system

3.Smaller storage cistern and required less pipe work

4.Drinking water can be available at the wash basin

Indirect water supply system:

1.First water stored in the stored tank then come out to from the stored tank.That is why it produce less noise.

2.Large quantity of water can stored.

3.Pressure is low in the pipe because water did not come out directly from the main line.

4.it prevent water from become dirty.

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10. What does a profile of a river from its headwaters to its mouth typically show?
irina1246 [14]

Answer:

c. an abrupt increase followed by a gradual decrease

Explanation:

At the headwater, the flow gradient starts high but then slowly decreases as the river moves downstream to its mouth.

3 0
2 years ago
Select the correct answer.
juin [17]
Orthographic projection, common method of representing three-dimensional objects, usually by three two-dimensional drawings in each of which the object is viewed along parallel lines that are perpendicular to the plane of the drawling.
4 0
2 years ago
For some transformation having kinetics that obey the Avrami equation, the parameter n is known to have a value of 2. If, after
kotegsom [21]

This question is incomplete, the complete question is;

For some transformation having kinetics that obey the Avrami equation, the parameter n is known to have a value of 2. If, after 100 s, the reaction is 40% complete, how long (total time in seconds) will it take the transformation to go to 95% completion

y = 1 - exp( -ktⁿ )

Answer: the time required for 95% transformation is 242.17 s

Explanation:

First, we calculate the value of k which is the dependent variable in Avrami equation

y = 1 - exp( -ktⁿ )

exp( -ktⁿ ) = 1 - y

-ktⁿ = In( 1 - y )

k = - In( 1 - y ) / tⁿ

now given that; n = 2, y = 40% = 0.40, and t = 100 s

we substitute

k = - In( 1 - 0.40 ) / 100²

k = - In(0.60) / 10000

k = 0.5108 / 10000

k = 0.00005108 ≈ 5.108 × 10⁻⁵

Now calculate the time required for 95% transformation

tⁿ = - In( 1 - y ) / k

t = [- In( 1 - y ) / k ]^1/n

n = 2, y = 95% = 0.95 and k = 5.108 × 10⁻⁵

we substitute our values

t = [- In( 1 - 0.95 ) / 5.108 × 10⁻⁵ ]^1/2

t = [2.9957 / 5.108 × 10⁻⁵]^1/2

t = [ 58647.22 ]^1/2

t = 242.17 s

Therefore the time required for 95% transformation is 242.17 s

8 0
3 years ago
A 60-cm-high, 40-cm-diameter cylindrical water tank is being transported on a level road. The highest acceleration anticipated i
dlinn [17]

Answer:

h_{max} = 51.8 cm

Explanation:

given data:

height of tank = 60cm

diameter of tank =40cm

accelration = 4 m/s2

suppose x- axis - direction of motion

z -axis - vertical direction

\theta = water surface angle with horizontal surface

a_x =accelration in x direction

a_z =accelration in z direction

slope in xz plane is

tan\theta = \frac{a_x}{g +a_z}

tan\theta = \frac{4}{9.81+0}

tan\theta =0.4077

the maximum height of water surface at mid of inclination is

\Delta h = \frac{d}{2} tan\theta

            =\frac{0.4}{2}0.4077

\Delta h  0.082 cm

the maximu height of wwater to avoid spilling is

h_{max} = h_{tank} -\Delta h

            = 60 - 8.2

h_{max} = 51.8 cm

the height requird if no spill water is h_{max} = 51.8 cm

3 0
3 years ago
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