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serg [7]
3 years ago
15

What is the function of deaerator in thermal power plant? ​

Engineering
1 answer:
avanturin [10]3 years ago
3 0

Answer:

The function of the Deaerator is to remove dissolved non-condensable gases and to heat boiler feed water.

Explanation:

You might be interested in
Going back the b beginning of the process is common in engineering true or false?
Yuliya22 [10]

Answer:

true !!

Explanation:

hope i helped :)! brainliest ?

6 0
3 years ago
Using the following data, determine the percentage retained, cumulative percentage retained, and percent passing for each sieve.
vekshin1

Solution :

<u>Sieve Size</u> (in)                   <u>Weight retain</u><u>(g)</u>

3                                         1.62

2                                         2.17

$1\frac{1}{2}$                                       3.62

$\frac{3}{4}$                                        2.27

$\frac{3}{8}$                                        1.38

PAN                                    0.21

Given :

Sieve       weight       % wt. retain    % cumulative       % finer

size        retained                               wt. retain

No. 4        59.5            10.225%          10.225%            89.775%

No. 8        86.5            14.865%          25.090%           74.91%

No. 16       138              23.7154%        48.8054%         51.2%

No. 30      127.8           21.91%              70.7154%          29.2850%

No. 50      97               16.6695%         87.3849%         12.62%

No. 100     66.8            11.4796%         98.92%              1.08%

Pan          <u>  6.3    </u>           1.08%              100%                   0%

                581.9 gram

Effective size = percentage finer 10% ($$D_{20})

0.149 mm, N 100, % finer 1.08

0.297, N 50 , % finer 12.62%

x  ,   10%

$y-1.08 = \frac{12.62 - 1.08}{0.297 - 0.149}(x-0.149)$

$(10-1.08) \times \frac{0.297 - 0.149}{12.62 - 1.08}+ 0.149=x$

x = 0.2634 mm

Effective size, $D_{10} = 0.2643 \ mm$

Now, N 16 (1.19 mm)  ,  51.2%

N 8 (2.38 mm)  ,  74.91%

x,  60%

$60-51.2 = \frac{74.91-51.2}{2.38-1.19}(x-1.19)$

x = 1.6317 mm

$\therefore D_{60} = 1.6317 \ mm$

Uniformity co-efficient = $\frac{D_{60}}{D_{10}}$

   $Cu= \frac{1.6317}{0.2643}$

Cu = 6.17

Now, fineness modulus = $\frac{\Sigma \text{\ cumulative retain on all sieve }}{100}$

$=\frac{\Sigma (10.225+25.09+48.8054+70.7165+87.39+98.92+100)}{100}$

= 4.41

which lies between No. 4  and No. 5 sieve [4.76 to 4.00]

So, fineness modulus = 4.38 mm

7 0
2 years ago
A school is playing $0.XY per kWh for electric power. To reduce its power bill, the school installs a wind turbine with a rated
Semmy [17]

Answer: Your question has some missing figures so kindly plug in the values into the solution provided to get the exact amount of money saved

answer : Electric power generated = 216 * 10^6 kJ

             money saved = $0.XY * 60000 kwh

Explanation:

<u>Calculating  the amount of electric power generated by wind turbine</u>

power generated = ( 30 * 2000 ) kWh  = 60000 kWh

Electric energy generated = 60000 kWh * 3600 kJ = 216 * 10^6 kJ

<u>Calculate money saved by school per year </u>

$0.XY * 60000 kwh

5 0
2 years ago
In contouring, it is necessary to measure position and not velocity for feedback.
EastWind [94]

Answer:

(1). False, (2). True, (3). False, (4). False, (5). True.

Explanation:

The term ''contouring'' in this question does not have to do with makeup but it has to deal with the measurement of all surfaces in planes. It is a measurement in which the rough and the contours are being measured. So, let us check each questions again.

(1). In contouring, it is necessary to measure position and not velocity for feedback.

ANSWER : b =>False. IT IS NECESSARY TO MEASURE BOTH FOR FEEDBACK.

(2). In contouring during 2-axis NC machining, the two axes are moved at the same speed to achieve the desired contour.

ANSWER: a=> True.

(3). Job shop is another term for process layout.

ANSWER: b => False

JOB SHOP IS A FLEXIBLE PROCESS THAT IS BEING USED during manufacturing process and are meant for job Production. PROCESS LAYOUT is used in increasing Efficiency.

(4). Airplanes are normally produced using group technology or cellular layout.

ANSWER: b => False.

(5). In manufacturing, value-creating time is greater than takt time.

ANSWER: a => True.

8 0
2 years ago
Determine the hydraulic radius for the following rectangular open channel width =23m water depth =3m
Romashka-Z-Leto [24]

Answer:

2.379m

Explanation:

The width = 23m

The depth = 3m

The radius is denoted as R

The wetted area is = A

The perimeter perimeter = P

Hydraulic radius

R = A/P

The area of a rectangular channel

= Width multiplied by Depth

A = 23x3

A = 69m²

Perimeter = (2x3)+23

P = 6+23

P= 29

Hydraulic radius R = 69/29

= 2.379m

This answers the question

Thank you!

8 0
2 years ago
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