1969-08-30 — Page 14

報日僑筆

六期星日十三月八年九六九一层公年八十五圍民塞中育教僑華

小學畢業同學

暑期進修專欄

智慧出版社主編

英文算術

(1)政極分相式(Shaded-pole)多在小型 風扇应用,是将凸極分為两半,一半套 有短路銅圈,因感应関係,此短路銅 圈會產生相差大约90之電流其構造 如下面

頁二第張四第二日八十月七年己波

WAWAT PS

罗僑樂

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中學同學學科生活知識進修

堅道半文書院主编 應用科學

家庭電器浅释

缤三講電風扇

(三)高風扇旋轉的理論

重風翁雖是单相感应电动机,但其 旋轉的理論似和双相感应电动机更接 近,故現先說,双相感应震动机的旋 轉磁场理論

電源

假設有一双相田種的定部,若入電 源,则在足部中央會同時感受到磁场 互相垂直的作用力,如下面所示

Sin (ut+9%)

此两作用力My和Mx 不但互相垂直 一而且因電源相位的闺係其相传也相差

90。现再假設 My和M忪合力之大小為M, 南橫軸所成角度為日

MX

LESSON NINE

AREAS AND VOLUMES.

1. AREAS

短路銅圈

USEFUL FORMULAE

(2) 6 (Capacitor start 1 As of 極线绕外,还加上相差900空相」的輔助说 (Ausilary Winding)而此輔助绕号和一適当大

的電容器串联的如下湯

輔助统

主鏡

由於 電容器 関係,便经过辅绕的電流 超前了差不多90,故能產生旋轉磁场。阅

在上面所提告相,这名詞,是指磁極间 的相住角度,不論其实际角度如何,在 隣的不同磁極其空相都是180,故在单 相四極而言,空相900在实际上的角度是 450.

事实上,電容器串联的輔助绕只在 始初時有作用,在轉动我的電风角,实 在已变為其正的单相感应电动机;故 現在再談4单相的旋轉磁诊。

(A) AREA OF SQUARE = SIDE x SIDE A.wsxs:

(B) AREA OF RECTANGLE = LENGTH x BREADYTH A L x日

(C) AREA OF TRIANGLE = 专 x. BASE x HEIGHT AndxDxH

The PERIMETER of a figure bounded by straight line is

the sum of the lengths of the sides.

RECTANGLE:

TRIANGLE:

EXAMPLE 1

Perimeter (P) - 2 x (L + B) = 2 x L + 2^x Perimeter (P) = Sum of the three aides

A strip of carpet da.6 ft. uide, 7 yd. long. Find ita area and perimeter.

SOLUTION:

EXAMPLE ́2

The required area = 6 x 7 x 3 sq. ft. 126 sq.ft. or 14.

The required perimeter of the carpet 2 x (6 + 7x 3) £t. » 2 x 27 ft. -54 ft. or 18 yd.

The required area of the carpet is 14 sq, yd. The required perimeter of the carpet is. 18yd.

The area of a rectangular field is 9,600 sq.yd.

Find its breadth and its perimeter. 120 yd. long.

SOLUTION:

ANSWER:

EXAMPLE 3

The required breadth

(9,600 - 120) yd. = 80 yo.

The required perimeter = (120 + 80) yd.

5.2 x 200 yd.w400yd. The required breadth of the rectangular

field is 80 yd.

The.required perimeter of the rectangula

field is 40oyd...

The area of a triangle is 2 sq. 1 yd. 1 ft. Find its base. SOLUTION

ANSWER:

-EXAMPLE: 4

and dta height:

The required length of the base = 2 x 2 sq. yd. + 1 yd. 1. ft.

= (2 x 2 x 9 - 4) It. - 9.ft. or 3 ydu

The required length of the base is 3 yd.

A carpet 18 ft. Long, 15 ft. wide is laid in a room 19 ft square. Find the area of the floor left incovered ..

SOLUTION; Area of the floor = (19 x 19) sq

361 sq. ft.

則

My = a Im Sin (wt +90°

現假設有兩個方向相反的轉磁 吃,其大小均為M.

ANSWER:

EXAMPLE 50

Area of the carpet (18 x 15) sq.

- 270 sq. ft. Therefore, area of the part left uncovered =(361 - 270) sq. ft. = 91 sq. ft.

The required area left uncovered is

sq. ft.

Find the area of the unshaded part, Dimensions are in ft;

A Im Cos ut

Mx = a Im Sin wt

M=2MG+MP

A Imd Cos utt. Sin

a Im

are tan

A Im Carut

A Im Sin wt

arc tan ( Cot wt)

are Ton [ Tan(98+wt)] 90°rwt

由上面的简单計算,可知其合力大小永 遠不变,而其向度會因時间之坛加而坛 大,採句話說就是一個大小一定,向读 中心旋轉的向量,在電学上,这就是所称 旋轉磁场,其作用话像拿着一個永久磁 鉄,環繞着定部周團以一定方向和速度旋 轉一樣。

在轉部方面,因定部旋轉磁场的作 用,便其表面的金属條因感应而產生短路 高流,從而也引起旋轉磁场的磁伤互相 牽引的结果,轉部便旋轉起来。

不过,相信大家还記得,普通家庭用 電是单相電源,又何来相差90°的第二 相電源呢?一般風扇用的变动机有两 种方法

在上面中此西旋轉磁場分别以OB.oc 代表,在任何时间,此两磁吃在Y軸上之分量 方向相反,大小相导故剛好相谐,在工軸上 之合量為

-DA=2M Cor. B.

由上面的简单計,說明两相反方向的

不存在了軸上依

全法(或正弦波形变化的磁场,相白素 说,正弦波形的单相磁场,是可分解為两 方向相反的旋轉磁房。

当轉动後,和轉动方向相同的一個旋轉 磁场会断增强其对轉部的作用力而 的一個则漸,减低作用,在此消彼長下定 部对轉部的作用,实际上“剩一個单向 的旋轉磁场。

大家若有臭趣的話,不妨舌实验 是在电容器的一端串联一個闹闹學,當瓜 的轉动稳定後,電容器和輔助境实在可 以完全阕去。

SOLUTION:

ANSWER:

EXAMPLE 6

Area of ABHF - (12 x 15) sq. ft.

180 sq. ft. Area of DCHE = (2 x 3) q.ft.

aq.ft.

Area of the shaded part, AEF

x 10 x 15 sq. £t. = 75. sq. *** Therefore, arch of the unshaded part,

ABCDEA:

Area ABHF - Area DCHE Aren AFF ̈ - (180 - 75 - 6) sq. ft. - 99 sq. ft. Area of the unshaded part is 99 adi ft.

How many whole squares of side 9 inches can be cus out from a square sheet of metal of side 1 yard? SOLUTION: The required number of whole squares

(1 yd. x 1 yd.) – (9 in. x 9 in.) – 16 ANSWERF The required number of whole squares is 16,

VOLUMES.

Volume of a Cuboid = Length x Areadth x Height

V L x B x H Volume of a Cube

EXAMPLE.7.

Side x Side x Side

Find the volume of a cube whose edge is 11 in. SOLUTION: The required volume = 11 x 11 x 11 cu. in.

1331 ou, ini

ANSHER; The required volume is 1331 cu. in.

EXAMPLE 8

Two cubic feet of a liquid weighing 65 lb. per cu. ft. care mixed with 3 cu, ft. of another Liquid weighing. 90 lb. per cu. ft. If there is no change of volume, how much, does 1 cu, ft. of the mixture weigh?.

·SOLUTION $ The required weight of 1 cu. ft. of the

mixture

- (65 x 2 + 90 x 3) lb. + (2 + 3) (130270) lb. 5- 80 lb.

The required weight of the mixture is 80 lb. per cus Ite

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