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3 years ago

Can we modify the soil’s composition?

1 answer:

8 0

While changing a soil's basic texture is very difficult, you can improve its structure–making clay more porous, sand more water retentive–by adding amendments. The best amendment for soil of any texture is organic matter, the decaying remains of plants and animals.

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Underground water is to be pumped by a 78% efficient 5- kW submerged pump to a pool whose free surface is 30 m above the undergr

maksim [4K]

Answer:

a) The maximum flowrate of the pump is approximately 13,305.22 cm³/s

b) The pressure difference across the pump is approximately 293.118 kPa

Explanation:

The efficiency of the pump = 78%

The power of the pump = 5 -kW

The height of the pool above the underground water, h = 30 m

The diameter of the pipe on the intake side = 7 cm

The diameter of the pipe on the discharge side = 5 cm

a) The maximum flowrate of the pump is given as follows;

$P = \dfrac{Q \cdot \rho \cdot g\cdot h}{\eta_t}$

Where;

P = The power of the pump

Q = The flowrate of the pump

ρ = The density of the fluid = 997 kg/m³

h = The head of the pump = 30 m

g = The acceleration due to gravity ≈ 9.8 m/s²

$\eta_t$ = The efficiency of the pump = 78%

$\therefore Q_{max} = \dfrac{P \cdot \eta_t}{\rho \cdot g\cdot h}$

$Q_{max}$ = 5,000 × 0.78/(997 × 9.8 × 30) ≈ 0.0133 m³/s

The maximum flowrate of the pump $Q_{max}$ ≈ 0.013305 m³/s = 13,305.22 cm³/s

b) The pressure difference across the pump, ΔP = ρ·g·h

∴ ΔP = 997 kg/m³ × 9.8 m/s² × 30 m = 293.118 kPa

The pressure difference across the pump, ΔP ≈ 293.118 kPa

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3 years ago

PWM input and output signals are often converted to analog voltage signals using low-pass filters. Design and simulate the follo

Temka [501]

Answer:

Attached below

Explanation:

PWM signal source has 1 KHz base frequency

Analog filter : with time constant = 0.01 s

low pass transfer function = $\frac{1}{0.01s + 1 }$

PWM duty cycle is a constant block

Attached below is the design and simulation into Simulink at 25% , 50% and 100% respectively

5 0

3 years ago

As a human resources professional for a large company, Tommy is involved with the ___________ process. His goal is to find the r

zaharov [31]

Answer:

Recruitment

Explanation:

The process of recruitment involves several activities whose main objective is to get the right people with relevant qualifications at the appropriate time to meet the needs of an organisation whenever there’s need. The process involves screening a pool of candidates through several steps to obtain the best fit for the job.

4 0

3 years ago

Which two forms of energy are usually converted directly from solar radiation for houses in temperate climates. a. Thermal and E

katrin2010 [14]

Answer:

The correct answer is option 'a': Thermal and Electrical energy

Explanation:

We know that the solar radiation that we receive from sun provides us heat. Thus the solar radiation is a natural source of thermal energy which can be utilized in solar cooker's to cook food or to warm water to generate steam in Solar thermal power plants. The steam generated is used to drive a turbine and hence produce electricity.

As we know that solar radiation is converted into directly usable forms of energy by the solar panels. The solar panels work on the principle of photo-electric effect in which light energy is directly converted into electrical energy to run our electrical devices at home such as light bulbs, fans, e.t.c. An excellent application of this principle is the international space station that orbits the earth and is fully powered by solar energy.

3 0

3 years ago

A 10.2 mm diameter steel circular rod is subjected to a tensile load that reduces its cross- sectional area to 52.7 mm^2. Determ

VMariaS [17]

Answer:

The percentage ductility is 35.5%.

Explanation:

Ductility is the ability of being deform under applied load. Ductility can measure by percentage elongation and percentage reduction in area. Here, percentage reduction in area method is taken to measure the ductility.

Step1

Given:

Diameter of shaft is 10.2 mm.

Final area of the shaft is 52.7 mm².

Calculation:

Step2

Initial area is calculated as follows:

$A=\frac{\pi d^{2}}{4}$

$A=\frac{\pi\times(10.2)^{2}}{4}$

A = 81.713 mm².

Step3

Percentage ductility is calculated as follows:

$D=\frac{A_{i}-A_{f}}{A_{i}}\times100$

$D=\frac{81.713-52.7}{81.713}\times100$

D = 35.5%.

Thus, the percentage ductility is 35.5%.

5 0

3 years ago