Homework
Assignment #14:
#1
Ch4 Cs7
(0.5pt)
(a)
Bernoulli's
equation can be rewritten to solve for the height (h) that the water can reach
by converting all the pressure potential energy into gravitational potential
energy. Bernoulli's equation tells us that P + pgh = constant. So with P1
= 500,000 Pa above atmospheric pressure and h1 = 0 (ground level) the
constant is just equal to the pressure P1 plus atmospheric pressure.
Up in the building (at h2) we know that P2 + pgh2
= P1. P2 is atmospheric pressure so we can solve for h2
to get h2 = (P1 + atmospheric pressure - P2)/(pg)
= P1/(pg) = 51m.
(0.5pt)
(b)
We again
use Bernoulli's equation but now we want to find the velocity of the water
leaving the hose. So initially we have P1 = 500,000 Pa plus atmospheric
pressure, v1 = 0 (the water movers slowly), and h1 = 0 (ground level). The
situation at the other end of the hose is P2 = atmospheric pressure, h2 = 0 (the
hose is still at ground level), and we want to solve for v2. Rewriting
Bernoulli's equation gives v = sqrt( 2*(P1-P2)/p) = 32 m/s.
(1pts)#2
Ch5 E14
The
behavior of the air that's responsible for the quiet (slow) and noise (fast) is
the type of air flow occurring around the stick. In the case of the slow and
quiet stick the air flow is laminar and the only forces acting on the stick are
viscous forces. When the stick is moving fast enough the air flow around it
becomes turbulent. There is a turbulent wake which results in the sound heard.
(1pts)#3
Ch5 E20
The
reason that the bicycle rider does not have to peddle as hard is due to the
pressure drag behind the truck resulting from turbulent air flow as well as a
reduction in air resistance. Without the truck in front of the bicycle the rider
must work against the force of air resistance. When behind the truck the rider
does not have to move the air (less air resistance force) and there is a region
behind the truck with a lower pressure, the pressure drag, which also pulls the
rider forward.
(1pts)#4
Ch5 E26
The
dry ball has many fuzzy projections that serve to break up the laminar flow
across its surface (the boundary layer). When the boundary layer is forced to
become turbulent by the fuzz it reduces the pressure drag behind the ball
(downwind). The combination of the air resistance and the pressure drag would
cause the ball to deflect. When we wet the ball we reduce the ability of the
fuzz to disrupt the boundary layer and the pressure drag is increased. This
increased pressure drag would cause slightly more deflection.
(1pts)#5
Ch5 E28
The
smooth taper of the rider’s helmet allows the air to flow around and past the
helmet without inducing turbulent flow and any associated aerodynamic forces
that might increase the drag forces acting on the rider.
(1pts)#6
Ch5 E30
The
lack of air means there would be no viscous and pressure drag forces, so the
ball would go further, but also would have no lift. Thus the golf ball would undergo simple projectile motion
(constant accelerated motion) under the relatively weak gravitational field of
the moon.
(1pts)#7
Ch5 E32
The
frisbee would also undergo simple projectile motion. There would be no air to
provide the lift that gives a frisbee the flight characteristics we are familiar
with.
(1pts)#8
Ch5 P2
A
5% narrowing of a blood vessel means that the narrower blood vessel has 0.95
times the diameter of its original. We use Poiseuille's law to solve for the
pressure difference. We can express the ratio (the ratio allows us to cancel all
of the quantities that are unknown and constant) of the pressure difference as
Dp2/Dp1
= (D1/D2)4 = (1/0.95)4 = 1.2
Thus
the final blood pressure is 120% of the original, or the blood pressure
increases by (1.2-1) x 100% = 20%. Please note that we were allowed to cancel
the volume flow rates because we have treated blood as an incompressible fluid.
(1pts)#9
Ch5 P6
We
can use Poiseuille's law to find the time to fill the cup with honey. First we
can write the equation (solved for t) for water as
tw = V*(128*L*h)/(p*Dp*D4) where tw
= 5 sec.
th = V*(128*L*h)/(p*Dp*D4)
The
difference in the two situations is the viscosity. For water this is 0.001 Pa*s
and for honey it’s 1000 Pa*s. We can write the same equation for the honey but
replacing the viscosity with that for honey. Taking the ratio of the two
equations (all the common factors cancel) gives us
th/
tw = hh/hw
= 1000000
So
that we find it would take 5*1000000 = 5x10^6 sec = 58 days!
(1pts)#10
Ch5 P8
In
order to maintain laminar airflow we know that the Reynolds number must be less
than about 2000. So we can calculate the flow speed from the Reynolds number as
Flow
speed = Reynolds number*h/r*obstacle length
We
use the viscosity and density of air and the given obstacle length to find that
the max speed of the blimp for laminar flow is v = 1.95x10^-3 m/s or about 2
mm/s.