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4 years ago

Define centrifugal pump. Give the construction and working of centrifugal pump.

1 answer:

7 0

Centrifugal pump is a hydraulic machine which converts mechanical energy into hydraulic energy by the use of centrifugal force acting on the fluid. These are the most popular and commonly used type of pumps for the transfer of fluids from low level to high level.

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A budding electronics hobbyist wants to make a simple 1.0-nF capacitor for tuning her crystal radio, using two sheets of aluminu

bazaltina [42]

Answer:

a. 8 sheets of paper is needed between her plates to get the proper capacitance

b. Area of Aluminum Foil needed = 0.45m²

c. To keep a 1.0-nF, a larger area of Teflon is required.

Explanation:

First, we need to calculate the distance between two plates.

This is given by

d = Kε0A/C

Where

K = 3

ε0 = Physical Constant = 8.854 * 10^-12 C²/Nm²

A = Area = 22 * 28 = 616cm² = 0.0616m²

C = 1.0-nF = 1 * 10^-12F

So, d = (3 * 8.854 * 10^-12 C²/Nm² * 0.0616) / (1 * 10^-12F)

d = 1.64 * 10^-3m

d = 1.64mm

Now, that the distance has been solved.

The Number of Sheets, N is given by

N = d/d,sheet where d, sheet =the sheet thickness = 0.2mm

N = 1.64/0.2

N = 8.2

N = 8 sheets --- Approximated

Here, she's changed the diameter of the sheets to 12mm

Well make use of the formula in (a) above

Using d = Kε0A/C

Where

d = 12 * 10^-3m

Other constraints remain unchanged

Make A the subject of formula

A = dC/Kε0

A = (12 * 10^-3m * 1 * 10^-12F)/(3 * 8.854 * 10^-12 C²/Nm²)

A= 0.45m²

c. From (b) above

A ∝ 1/K

As the dielectric constant increase, the area decreases

The dielectric constant of a Teflon is 2.1

This means that if she used a Teflon instead, the area will be larger.

So, to keep a 1.0-nF, a larger area of Teflon is required.

7 0

4 years ago

The moons light-colored highlights are

gulaghasi [49]

The light colored material in areas of the mood is the earliest crust on the moon.

3 0

3 years ago

A. Calculate the fraction of atom sites that are vacant for lead at its melting temperature of 327°C (600 K). Assume an energy f

NISA [10]

Answer:

a. Fraction of Atom = 2.41E-5 when T = 600K

b. Fraction of Atom = 5.03E-10 when T = 298K

Explanation:

Given

T = Temperature = 600K

Qv = Energy for formation = 0.55eV/atom

To calculate the fraction of atom sites, we make use of the following formula

Nv/N = exp(-Qv/kT)

Where k = Boltzmann Constant = 8.62E-5eV/K

Nv/N = exp(-0.55/(8.62E-5 * 600))

Nv/N = 0.000024078672493307

Nv/N = 2.41E-5

b. When T = 298K

Nv/N = exp(-0.55/(8.62E-5 * 298))

Nv/N = 5.026591237904E−10

Nv/N = 5.03E-10 ----- Approximated

6 0

4 years ago

The project's criteria.

Lena [83]

Answer: design 4

Explanation: I got it right

8 0

2 years ago

The top surface of an L = 5mmthick anodized aluminum plate is irradiated with G = 1000 W/m2 while being simultaneously exposed

puteri [66]

Answer:

$J=1963W/m^2$

$q_{rad}=963w/m^2$

$\triangle T= -0.378k/s$

Explanation:

From the question we are told that:

$L=5mm => 5*10^{-3}\\Irradiation G=1000W/m^2\\h=50W/m^2\\T_{infinity} = 25C.\\Plate\ temperature\ T_p= 400 K\\\alpha=0.14\\E=0.76$

at Temp=400K

$E=2702kg/m^2,c=949J/kgk$

Generally the equation for Radiosity is mathematically given by

$J=eG+\in E_p$

$J=(1-\alpha)G+\in \sigma T^4$

$J=(1-0.14)1000+0.76 (5.67*10^_{8}) (400)^4$

$J=1963W/m^2$

Generally the equation for net radiation heat flux $q_{rad}$ is mathematically given by

$q_{rad}=J-G\\q_{rad}=1963-1000$

$q_{rad}=963w/m^2$

Generally the equation for and the rate of plate temp $\triangleT$ is mathematically given by

$\triangle T= -\frac{q_{con} +q_{rad}}{Ecl}$

$\triangle T= \frac{45(400)-(30+273+963)}{(2702*949*0.005)}$

$\triangle T= -0.378k/s$

6 0

3 years ago

Define centrifugal pump. Give the construction and working of centrifugal pump. ​

Define centrifugal pump. Give the construction and working of centrifugal pump.