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ÈûÇ»£¨ÎÞ¸ËÇ»£©Ãæ»ý A?

A_1 = \frac{\pi D^2}{4}

¸ËÇ»£¨ÓиËÇ»£©Ãæ»ý A?

A_2 = \frac{\pi (D^2 - d^2)}{4}

Èý¡¢ÍÆÁ¦¹«Ê½£¨Á¦ = ѹÁ¦ ¡Á Ãæ»ý£©

Éì³öÍÆÁ¦£¨ÈûÇ»½øÓÍ£©

F_1 = P \times A_1

¡ú ÍÆÁ¦´ó

ÊÕ»ØÍÆÁ¦£¨¸ËÇ»½øÓÍ£©

F_2 = P \times A_2

¡ú ÍÆÁ¦Ð¡

ËÄ¡¢Ëٶȹ«Ê½£¨ËÙ¶È = Á÷Á¿ ¡Â Ãæ»ý£©

Éì³öËÙ¶È

v_1 = \frac{Q}{A_1}

¡ú Ãæ»ý´ó ¡ú ËÙ¶ÈÂý

ÊÕ»ØËÙ¶È

v_2 = \frac{Q}{A_2}

¡ú Ãæ»ýС ¡ú Ëٶȿì

Îå¡¢Ò»¾ä»°×ܽá¼ÍÂÉ

Ãæ»ýÔ½´ó ¡ú ÍÆÁ¦Ô½´ó¡¢ËÙ¶ÈÔ½Âý

Ãæ»ýԽС ¡ú ÍÆÁ¦Ô½Ð¡¡¢ËÙ¶ÈÔ½¿ì

ÈûÇ»Ãæ»ý > ¸ËÇ»Ãæ»ý

¡ú Éì³ö£º´óÁ¦ÂýËÙ£»ÊջأºÐ¡Á¦¿ìËÙ

¶¨²ÎÊý£¨×î³£¼ûµÄÓ͸ף©

¸×¾¶£ºD = 50 mm

¸Ë¾¶£ºd = 28 mm

ϵͳѹÁ¦£ºP = 16 MPa

Á÷Á¿£ºQ = 20 L/min

1. ËãÃæ»ý

ÈûÇ»Ãæ»ý£¨ÎÞ¸ËÇ»£©

A_1 = \frac{\pi D^2}{4}

= \frac{\pi \times 50^2}{4}

\approx 1963.5\ \mathrm{mm^2}

¸ËÇ»Ãæ»ý£¨ÓиËÇ»£©

A_2 = \frac{\pi (D^2-d^2)}{4}

= \frac{\pi (50^2-28^2)}{4}

\approx 1347.7\ \mathrm{mm^2}

2. ËãÍÆÁ¦£¨Á¦ = ѹÁ¦ ¡Á Ãæ»ý£©

ѹÁ¦µ¥Î»£º16\ \mathrm{MPa} = 16\ \mathrm{N/mm^2}

Éì³öÍÆÁ¦£¨ÈûÇ»½øÓÍ£©

F_1 = P \times A_1

= 16 \times 1963.5

\approx 31416\ \mathrm{N} \approx \mathbf{3.14\ ¶Ö}

ÊÕ»ØÍÆÁ¦£¨¸ËÇ»½øÓÍ£©

F_2 = P \times A_2

= 16 \times 1347.7

\approx 21563\ \mathrm{N} \approx \mathbf{2.16\ ¶Ö}

3. ËãËٶȣ¨ËÙ¶È = Á÷Á¿ ¡Â Ãæ»ý£©

Á÷Á¿»»Ë㣺20\ \mathrm{L/min} = \frac{20}{1000}\ \mathrm{m^3/min}

Éì³öËÙ¶È

v_1 = \frac{Q}{A_1}

\approx \mathbf{0.17\ \mathrm{m/s}}

ÊÕ»ØËÙ¶È

v_2 = \frac{Q}{A_2}

\approx \mathbf{0.25\ \mathrm{m/s}}

½áÂÛ£¨¼Ç×ÅÕâ¸ö£©

Éì³ö£ºÍÆÁ¦´ó£¨¡Ö3.14 ¶Ö£©£¬ËÙ¶ÈÂý£¨0.17 m/s£©

ÊջأºÍÆÁ¦Ð¡£¨¡Ö2.16 ¶Ö£©£¬Ëٶȿ죨0.25 m/s£©