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3 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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Any help is appreciated <3

Len [333]

Answer:

forwarder

Explanation:

8 0

2 years ago

The 30-kg gear is subjected to a force of P=(20t)N where t is in seconds. Determine the angular velocity of the gear at t=4s sta

tatyana61 [14]

Answer:

$\omega =\frac{24}{1.14375}=20.983\frac{rad}{s}$

Explanation:

Previous concepts

Angular momentum. If we consider a particle of mass m, with velocity v, moving under the influence of a force F. The angular momentum about point O is defined as the “moment” of the particle’s linear momentum, L, about O. And the correct formula is:

$H_o =r x mv=rxL$

Applying Newton’s second law to the right hand side of the above equation, we have that r ×ma = r ×F =

MO, where MO is the moment of the force F about point O. The equation expressing the rate of change of angular momentum is this one:

MO = H˙ O

Principle of Angular Impulse and Momentum

The equation MO = H˙ O gives us the instantaneous relation between the moment and the time rate of change of angular momentum. Imagine now that the force considered acts on a particle between time t1 and time t2. The equation MO = H˙ O can then be integrated in time to obtain this:

$\int_{t_1}^{t_2}M_O dt = \int_{t_1}^{t_2}H_O dt=H_0t2 -H_0t1$

Solution to the problem

For this case we can use the principle of angular impulse and momentum that states "The mass moment of inertia of a gear about its mass center is $I_o =mK^2_o =30kg(0.125m)^2 =0.46875 kgm^2$ ".

If we analyze the staritning point we see that the initial velocity can be founded like this:

$v_o =\omega r_{OIC}=\omega (0.15m)$

And if we look the figure attached we can use the point A as a reference to calculate the angular impulse and momentum equation, like this:

$H_Ai +\sum \int_{t_i}^{t_f} M_A dt =H_Af$

$0+\sum \int_{0}^{4} 20t (0.15m) dt =0.46875 \omega + 30kg[\omega(0.15m)](0.15m)$

And if we integrate the left part and we simplify the right part we have

$1.5(4^2)-1.5(0^2) = 0.46875\omega +0.675\omega=1.14375\omega$

And if we solve for $\omega$ we got:

$\omega =\frac{24}{1.14375}=20.983\frac{rad}{s}$

8 0

2 years ago

An alloy is evaluated for potential creep deformation in a short-term laboratory experiment. The creep rate (ϵ˙) is found to be

cupoosta [38]

Answer:

Activation energy for creep in this temperature range is Q = 252.2 kJ/mol

Explanation:

To calculate the creep rate at a particular temperature

creep rate, $\zeta_{\theta} = C \exp(\frac{-Q}{R \theta} )$

Creep rate at 800⁰C, $\zeta_{800} = C \exp(\frac{-Q}{R (800+273)} )$

$\zeta_{800} = C \exp(\frac{-Q}{1073R} )\\\zeta_{800} = 1 \% per hour =0.01\\$

$0.01 = C \exp(\frac{-Q}{1073R} )$ .........................(1)

Creep rate at 700⁰C

$\zeta_{700} = C \exp(\frac{-Q}{R (700+273)} )$

$\zeta_{800} = C \exp(\frac{-Q}{973R} )\\\zeta_{800} = 5.5 * 10^{-2} \% per hour =5.5 * 10^{-4}$

$5.5 * 10^{-4} = C \exp(\frac{-Q}{1073R} )$ .................(2)

Divide equation (1) by equation (2)

$\frac{0.01}{5.5 * 10^{-4} } = \exp[\frac{-Q}{1073R} -\frac{-Q}{973R} ]\\18.182= \exp[\frac{-Q}{1073R} +\frac{Q}{973R} ]\\R = 8.314\\18.182= \exp[\frac{-Q}{1073*8.314} +\frac{Q}{973*8.314} ]\\18.182= \exp[0.0000115 Q]\\$

Take the natural log of both sides

$ln 18.182= 0.0000115Q\\2.9004 = 0.0000115Q\\Q = 2.9004/0.0000115\\Q = 252211.49 J/mol\\Q = 252.2 kJ/mol$

3 0

2 years ago

A(n) is a detailed, structured diagram or drawing.

monitta

Answer:

Schematics

Explanation:

A schematic is a detailed structured diagram or drawing. It employs illustrations to help the viewer understand detailed information on the machine or object being described. Its main aim is not to help the observer know what the object looks like physically. It is rather aimed at helping the viewer know how the machine works. This is achieved by only including key and important details to the drawing.

It is most times used in the blueprint and user guides of machines and gadgets used in the home to help users know how these things work so that they can do little fixings should there be such needs.

6 0

2 years ago

Design a plan for ""your yard"" (no matter where you live) that will capture and utilize rainwater. You can prepare the plan in

Jlenok [28]

Answer:

Explanation:

<u><em>General Considerations</em></u>

The design of the yard will affect the natural surface and subsurface drainage pattern of a watershed or individual hillslope. Yard drainage design has as its basic objective the reduction or elimination of energy generated by flowing water. The destructive power of flowing water increases exponentially as its velocity increases. Therefore, water must not be allowed to develop sufficient volume or velocity so as to cause excessive wear along ditches, below culverts, or along exposed running surfaces, cuts, or fills.

A yard drainage system must satisfy two main criteria if it is to be effective throughout its design life:

1. It must allow for a minimum of disturbance of the natural drainage pattern.

2.It must drain surface and subsurface water away from the roadway and dissipate it in a way that prevents excessive collection of water in unstable areas and subsequent downstream erosion

The diagram below ilustrate diffrent sturcture of yard to be consider before planing to utiliza rainwater

4 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.