Physics volume & mass flow rate

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Fluid Problems
# h2phy1 - Pressure converted inside Aorta
# h2phy2 - Pressure inside vein
# h2phy3 - Pressure on the floor
# h2phy4 - Pressure under the sea	# h2phym1- Hydraulic Lift - 1
# h2phy5 - Pressure between oil and water	# h2phym2- Hydraulic Lift - 2
# h2phy6 - Buoyant Force	# h2phym3- Hydraulic Lift - 3
# h2phy7 - Flow rate & Velocity
# h2phy8 - Surface Tension Force(1)
# h2phy9 - Volume & Mass Flow rate ↰
# h2phy10 - Surface Tension Force(2)

(a) A water line necks down from a pipe with a 12.5 mm.
       radius to a pipe with a 9 mm radius. If the speed of the
       water in the 12.5 mm pipe is 1.8 m.s^-1, what is the speed
       in the smaller pipe?
(b) What is the volume flow rate?
(c) What is the mass flow rate?

Strategy - the rule "Must have" of Mr.Zhang ®

ΔM	=	ρΔV	=	ρAV&#916t
ΔM &#916t	=	ρAV	→	(1) The rate of flow of mass
ρV1A1 = ρV2A2			→	The equation of continuty.
V1A1 = V2A2			→	(2) Density is constant.

Solution

(a) Compute the speed in the smaller pipe by solving the equation (2)

V2

=

V1A1 A2

V2

=

V1 π(r1)2 π(r2)2

V2	=	1.8 m.s-1(12.5 x 10-3)2 (9 x 10-3)2
Ans. The speed in the smaller pipe is 3.47 m.s-1,therefore, it make sense.
How doest it make sense?

(b) Compute the volume flow rate is;
V1A1	=	V2A2
	=	π(12.5x10-3m)2(1.8 m.s-1)
	=	8.8 x 10-4 m3.s-1

Ans. The volume flow rate is 8.8 x 10^-4m³s^-1.

(c) The mass flow rate is;
Compute it from eqaution(2);
ρV1A1	=	ρV2A2
	=	(1.0x103 kg.m-3)(8.8x10-4 m3.s-1)

Ans. The mass flow rate is 0.88 kg.s^-1.

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