Bernoulli and Friends - It's a Family Program!

Q 18,19 P37, 39, 41, 43, 72 (pressure head is the source of the water), 74

We will have a quiz Tuesday on fluid mechanics, so tomorrow (Monday) will be a day of practicing problems and deciding which method to use to solve parts of problems.


Questions:
18. When the streamlines of the wind curve over the roof, this creates a low pressure area above the roof, while the pressure below the roof remains high. This difference in pressure creates an upward force that pushes the roof off the house.

19. The increase in speed of the air passing between the sheets causes a lower static pressure between them. Since the air pressure on the outside of the sheets is the same, the outside pressure pushes into the space between where the pressure is lower.

Problems:

37. Using Bernoulli, with 1 at the ground and 2 at the top of the water flow, we can assume that at the MINIMUM pressure, the fluid is not moving.

P1 + 0 + 0 = Patm + 0 + rho*g*h

P1 - Patm = 1.176 x 10⁵ Pa

39. The pressure is atmospheric underneath the roof. We can take 1 to be at a point far from the roof (where it is still), and 2 to be in the wind above the roof:

Patm + 0 + 0 = P2 + 1/2*rho*v^2 + 0

The net pressure difference will be the gauge pressure of P2, or P2 - Patm

P2 - Patm= - 1/2*rho*v^2 = 580.5 Pa


Since this is pressure, the force is this pressure times the area of the roof:
F = 580.5 Pa * 240 M^2 = 1.39 x 10⁵ N

41.
Consider a streamline that goes from the outside of the hurricane (where P = Patm and the air is still) to the center. Remember to convert 300 km/h to 83.3 m/s.

Patm + 0 + 0 = P2 + .5*rho*v^2 + 0

P2 = 9.68 x 10⁴ Pa

43. We have to use continuity to find the speed of the water in the pipe when the pipes have smaller diameter. Since it is a liquid, we can assume the flow is incompressible.

A1v1 = A2v2, so v2 = A1/A2*v1 = 2.22 m/s

The gauge pressure at the pump is 3.8 atm gauge, which means it is 4.8 atm = 4.86 x 10⁵ Pa. The question asks for the static pressure at the top of the building.

Bernoulli:

P_pump + .5*rho*v1^2 + 0 = P_top + .5*rho*v2^2 + rho*g*h

P_top = 2.88 x 10⁵ Pa absolute, or 1.87 x 10⁵ Pa gauge.

72. The pressure head is the source of the water coming out of the shower, a tank on the roof perhaps. We can assume that the water has no velocity initially in the tank, and that the initial pressure is atmospheric.

Patm + 0 + rho*g*H = Patm + .5*rho*v2^2 + 0
H = 2.64 m

74. If the raft holds the maximum number of people, the logs are completely submerged in the water, though not yet sinking. This means the logs and people are in equilibrium. Note the factor of 10 since there are ten logs.

Fb - M_totalg = 0
10*rho_water*g*pi*R^2*L = (10*rho_wood*pi*R^2*L + M_people)*g

Since we know the specific gravity of wood, we can solve for the mass of the people.

M_people = 2087 kg = N*70 kg
N = 29.8 people

Thus there can be no more than 29 people on the raft.