Posted by **Cara** on Monday, May 9, 2011 at 9:20pm.

A 5.4 cm diameter horizontal pipe gradually narrows to 3.7 cm. When water flows through this pipe at a certain rate, the gauge pressure in these two sections is 35.0 kPa and 21.0 kPa, respectively. What is the volume rate of flow?

- physics -
**Damon**, Monday, May 9, 2011 at 9:39pm
(1/2) rho v^2 + p = constant (Bernoulli)

Q = area* speed = constant

area of big pipe = pi d^2/4 = pi(.054)^2/4 = .00229 m^2

area of small pipe = pi (.037)^2/4 = .00108 m^2

Vsmall = Vbig *.00229/.00108

so

Vsmall = 2.12 Vbig

Pbig = 35*10^3

Psmall = 21*10^3

rho = 1000 kg/m^3

so

35*10^3 + (1/2) (10)^3 Vbig^2 = 21*10^3 + (1/2)(10)^3 (2.12 Vbig)^2

35 - 21 = (1/2)(4.5 - 1) Vbig^2

28 = 3.5 Vbig^2

Vbig = .894 m/s

Q = Area big* Vbig = .00229 m^2 * .894 m/s

= .00205 m/s = 2.05 cm/s

check my arithmetic !

- physics -
**Stephanie**, Tuesday, January 17, 2012 at 11:36pm
when i was doing it ... everything was alright until the last part...

(28 = 3.5 Vbig^2

--> Vbig = 2.828 m/s <--

Q = Area big* Vbig = .00229 m^2 * 2.828 m/s

= .00647 m/s = 6.047 cm/s

- physics -
**Wendi**, Saturday, April 2, 2016 at 8:42pm
Stephanie's answer for m/s will also work for m^3/s.

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