physics
posted by Cara on .
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?

(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 ! 
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 
Stephanie's answer for m/s will also work for m^3/s.