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.
To find the volume rate of flow, we can use Bernoulli's equation, which relates the pressure difference between two points in a fluid flow to the velocity difference between those points.
1. First, we need to calculate the velocity of the water at each section of the pipe. We can use the equation for velocity in terms of cross-sectional area and volumetric flow rate:
v = Q / A
where v is the velocity, Q is the volumetric flow rate, and A is the cross-sectional area.
2. To find the volumetric flow rate, we need to calculate the cross-sectional areas of the two sections of the pipe. The cross-sectional area of a circle is given by the formula:
A = π * (d/2)^2
where A is the cross-sectional area and d is the diameter.
So, the cross-sectional area of the 5.4 cm diameter section is:
A1 = π * (5.4 cm / 2)^2
And the cross-sectional area of the 3.7 cm diameter section is:
A2 = π * (3.7 cm / 2)^2
3. Now, we can calculate the velocities at each section of the pipe. We'll use the equation derived in step 1:
v1 = Q / A1
v2 = Q / A2
4. According to Bernoulli's equation, the pressure difference between the two sections is related to the velocity difference:
P1 - P2 = (1/2) * ρ * (v1^2 - v2^2)
where P1 and P2 are the gauge pressures at the two sections, ρ is the density of water, and v1 and v2 are the velocities at the two sections.
Also, recall that the density of water is approximately 1000 kg/m^3, and 1 Pa = 1 N/m^2.
We need to convert the pressure units from kPa to Pa:
P1 = 35.0 kPa = 35.0 * 1000 Pa
P2 = 21.0 kPa = 21.0 * 1000 Pa
5. Rearranging the equation from step 4, we can solve for the velocity difference:
v1^2 - v2^2 = 2 * (P1 - P2) / ρ
6. Substituting the values we have:
v1^2 - v2^2 = 2 * (35.0 * 1000 - 21.0 * 1000) / 1000
7. Simplifying the equation:
v1^2 - v2^2 = 2 * (35.0 - 21.0)
8. Solving for the velocity difference:
v1^2 - v2^2 = 28.0
(Q / A1)^2 - (Q / A2)^2 = 28.0
Q^2 / A1^2 - Q^2 / A2^2 = 28.0
Q^2 * (1 / A1^2 - 1 / A2^2) = 28.0
9. Now, we can solve for the volumetric flow rate, Q:
Q^2 = 28.0 / (1 / A1^2 - 1 / A2^2)
Q = √(28.0 / (1 / A1^2 - 1 / A2^2))
Finally, calculate the value of Q using the calculated values for A1 and A2.