https://www.emarazzak.blogspot.com

Monday, April 13, 2015

Happy New ear

Happy New year to everybody. Today is the day of wanting. We should not memories the passing year about bed day. We hope coming year will earn very good time. we want to earn success in the coming year. I want success and peaceful life for everybody of the world.
Engr M A Razzak
14.04.2015

Tuesday, March 31, 2015

test copy

I'm Engr M A Razzak, my son Kader Pial Cowdhury in front of me.

Monday, March 23, 2015

Sample of a Schedule of Electrical works


Toma Construction & Company Limited     'Annexure-A'

Tongi-Bhairab Bazaar Double Line Project

Bill of Quantities

Bill No-5, Subsection D : Stations & Buildings
Name of works: Electrical works for Plat form Sheds & Foot Over Bridge.
Sl No. BOQ No Name of works   Unit Quantity Rate in Taka Amount in Taka
273 D-3.1(a) Concealed Point Wiring, Light Point nr 275 71.86 19761.50
274 D-3.1(b) Concealed Point Wiring, Fan Point nr 110 71.86 7904.60
276 D-3.2(a) Concealed Circuit / Power Point Wiring, 1C-2x2.5 Sq mm Cable with 2.5 sq mm ECC Wire through 16mm diameter PVC Pipe. Lin m 1800 228.51 411318.00
277 D-3.2(b) Concealed Circuit / Power Point Wiring, 1C-2x4.0 Sq mm Cable with 4.0 sq mm  ECC Wire through 20mm diameter PVC Pipe. Lin m 600 228.51 137106.00
278 D-3.2(c) Concealed Circuit / Power Point Wiring, 1C-2x6.00 Sq mm Cable with 6.0 sq mm  ECC Wire through 25mm diameter PVC Pipe. Lin m 240 282.40 67776.00
281 D-3.3(c)  1C x 4 x 10 Sq mm Cable with 1C-1x10 sq mm  ECC Wire. Lin m 600 521.69 313014.00
282 D-3.3(d)  1C x 4 x 16 Sq mm Cable with 1C-1x16 sq mm ECC Wire. Lin m 600 658.22 394932.00
285 D-3.6(a) Galvanized Iron (GI) Works, 19mm nominal ID Lin m 48 697.74 33491.52
290 D-3.7(d) One Gang Switch   nr 385 83.36 32093.60
291 D-3.7(e) Two-GangSwitch   nr 41 121.44 4979.04
292 D-3.7(f) Three- Gang Switch   nr 39 135.81 5296.59
293 D-3.8 Ceiling Fan, AC Capacitor Type, 1422.4mm Sweep nr 110 2453.96 269935.60
300 D-3.10(f) Distribution Board 250V Grade with 6-32A SP MCB. nr 22 1617.53 35585.66
302 D-3.11(a) Fency Single Bracket Light Fitting. nr 44 11016.57 484729.08
303 D-3.11(b) 1-4' x 40 Watt Fluricent Tube. nr 220 773.19 170101.80
307 D-3.11(f) 40/60 Watt Lamp   nr 66 1833.82 121032.12
308 D-3.11(g) 100 Watt Lamp   nr 44 3334.94 146737.36
    Grand Total Taka 2655794.47









Prepared By





Engr M A Razzak




Site-in-Charge (TBDLRP)




Tuesday, March 3, 2015

Basic Electrical Engineering



Series circuits:

Series circuits are sometimes called current-coupled or daisy chain-coupled. The current in a series circuit goes through every component in the circuit. Therefore, all of the components in a series connection carry the same current. There is only one path in a series circuit in which the current can flow.
A series circuit's main disadvantage or advantage, depending on its intended role in a product's overall design, is that because there is only one path in which its current can flow, opening or breaking a series circuit at any point causes the entire circuit to "open" or stop operating. For example, if even one of the light bulbs in an older-style string of Christmas tree lights burns out or is removed, the entire string becomes inoperable until the bulb is replaced.

Current

I = I_1 = I_2 = \dots = I_n
In a series circuit the current is the same for all elements.

Resistors

The total resistance of resistors in series is equal to the sum of their individual resistances:
This is a diagram of several resistors, connected end to end, with the same amount of current through each.
R_\mathrm{total} = R_1 + R_2 + \cdots + R_n
Electrical conductance presents a reciprocal quantity to resistance. Total conductance of a series circuits of pure resistors, therefore, can be calculated from the following expression:
\frac{1}{G_\mathrm{total}} = \frac{1}{G_1} + \frac{1}{G_2} + \cdots + \frac{1}{G_n}.
For a special case of two resistors in series, the total conductance is equal to:
G_{total} = \frac{G_1 G_2}{G_1+G_2}.

 



Tuesday, February 24, 2015





DEPARTMENT OF CIVIL ENGINEERING 
Program: B. Sc in Civil Engineering
Course Title:
Mechanics, Heat, Thermodynamics, Waves, Oscillations and Physical Optics- Lab
Course code:        PHY 103
Assignment
Experiment No-01                                   Experiment Done Date: 28.01.2015
Name of the Experiment:
To determine the focal length and hence the power of a convex lens by displacement method with the help of an optical bench.
Submitted by:
Name: Mohammed Abdur Razzak
Roll No:  3326,  Registration No: WUB/10/13/60/3326, 
Batch No: 60/C, Semester No: Fourth, Year: Second.

Submitted To:
Name:  Md. Shohel Rana
 Designation:  Lecturer in physics, Basic science Division (WUB)                 Date of Submission:
Experiment No.:04
Experiment Name: Determine the focal length and hence the power of a convex lens by displacement method with the help of an optical bench.
Theory: If the object and the image screen be so placed on an optical bench that the distance between them is greater than four times the focal length (f) of a given convex lens, then there will be two different positions of the lens which for which an equal sharp image will be obtained on the image screen.
                                                                                                                     
The fig represent respectively the positions of the object and the image screen and the two different positions of the lens for which an equally sharp image is obtained.
Let the distance Ol =D and L1 – L2 = x.
From the lens equation , we have
Or,   (since u + v =D)
Applying sign convention, u is negative.
Or,
Or, u2 – ud =df = 0
Solving the above equation which is quartic, we have two values of u corresponding to the two positions of the lens. These are

u1 =  -  position L1 of the lens

And u2 =  +  position L2 of the lens
Then x = L1  L2 = u1  u2 =
Or   x2 = D2 – 4Df
Or   f =  ……………….. (1)


Where D is the distance between the object and the image and must be greater than 4f and x is the distance between two different positions of the lens.
The power P of the lens s as usual given by the relation,
P =  diopters

Results:


Table I
No. of obs.
Position of
Displacement of Lens
X = L1 – L2
(cm)
Apparent distance between object and image
D’ = O
Corrected distance between object and image
D = D’ +
Object
(O)
Image
(I)
Lens at
L1
L2
1
0
45
16
28.2
12.20
45

2
0
50
14.4
34.4
20.00
50

3
0
55
14
40.30
26.30
55

4
0
60
12.4
46.00
33.60
60

5
0
65
12.1
51.00
38.90
65



Table II:
No. of obs.
Lens displacement (x) from Tab. II
Corrected Distance (d) from Tab. II
Focal length
Mean focal length (f) cm
Power
P =
1
12.20
45
10.42
10.45
9.56
2
20.00
50
10.50
3
26.30
55
10.60
4
33.60
60
10.30
5
38.90
65
10.43

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