Question

The questions concern water flowing in or from a hose. The cross section of the hose is circular. The circumference of a circle is 2πR and the area of a circle is πR², where R is the radius of the circle. 1 milliliter = 1 cm³.

A. If the inner diameter of the hose is 1.13 cm, and the water flows in the hose with a speed of 0.89 m/s, what is the volumetric current Q of the water flowing in the hose (in milliliters per second)?

B. If the inner diameter of the hose is 1.2 cm, but you partially cover the opening of the hose leaving an open area of 0.62 cm²at the end of the hose so that you can spray your friends. The water flows out at 3.3 m/s. What is the volumetric current in milliliters per second now?

Answer 1

Z at aj The volume flow rate throrga a cross section io givenas &=dr Au Va volume U=opeed of flow &= AU=TRU A = Area of froso section = TT (1:13)3789 cm R=113cm ; D=1:13.com ve 0.89 min. = 89000 =357cm3 To o Bj Again volumeric flow rate of rate of flow will be calculated by the same formula. &- dv S me = 89.26 ml Boros see یا A at

= 0-62 cm? = 330 cmp' Now Area A speed with the the which His coming outz 3-3 mis Now volume flow rates 0-62 x 33.0 cm3 S. = 204.6ml And

final answer A part 89.26 ml/s

B part 204.6 ml/s

Answer 2

SOLUTION 

A

Q = A * v = pi/4 (1.13)^2 * 0.89*100. cm^3/s (milliliteres per sec)

=> Q = 89.26 millilitres/sec (ANSWER).

B

Volumetric current remains unchanged.

So.

Volumetric current after partial blocking or before

= 0.62 * 3.3 * 100  

= 204.6  millilitres/sec (ANSWER)