Chapter 13 Homework Solutions
Problems: 13: 103, 12, 13, 14, 15, 16, 17, 18, 19, 20, 32, 38, 48, 55
Force = ρQ(Vx2-Vx1)
V1 = V, V2 = -½ V
Stationary plate: Force = ρQ(Vx2-Vx1) = ρQ(V- 0) = ρQ 1.5 V = m-dot V
Moving toward you plate: Force = ρQ(Vx2-Vx1) = ρQ(V- - 1/2V) = ρQ 1.5 V = m-dot 1.5 V
Note: one can also put this in terms of the volumetric flow rate (Q in hydraulics textbooks, but V-dot in this textbook) by
noting the volumetric flow rate is velocity of the jet times area of the jet, however since the jet mixes with air the area of
it changes along the path, making the area ambiguous & the textbook answer below somewhat ambiguous; for it to
work you would take the area of the jet. Better to leave it at the simple answer above.
Talk about kinetic energy factor
Force momentum x direction =1.03 ρ Q(Vx2-Vx1) = 1.03 m-dot (Vx2-Vx1) = 1.03 m-dot (Vx2-0) = 1.03*30 kg/s * 3.82 m/s =
118 N
Force of momentum y direction = 1.03*30 kg/s *(0 - 3.82 m/s) = -118 N so this force is in the negative y direction and
the reactive force must be in the (+) y direction
We first need the velocity = Q/A = m-dot/rho/area = (see below) 3.82 m/s
Assumptions: water and pipe weight are small and will be ignored; head losses are small and will be ignored
Pressure at outlet is assumed to be zero, Solve energy equation for pressures, assuming zero head loss then velocities
will cancel and change in pressure comes out as rho g delta z = 3.92 kPa
Pressure force at outlet is zero so reactive force is 118 N
Pressure force at inlet is P*A = (see below) 30.86 N
∑ Forces = 0, both the pressure and the momentum forces are in the positive x direction so the balancing reactive force
must be in the negative x direction or:
Total reaction (anchor) = -(118 + 30.86) = -148.9
We can use square root of each component squared to get the total reactive force and inverse tangent of Fy/Fx to get
direction; it points upwards and left
The reactive force is thus in the upper left quadrant of the unit circle.
Looking at this again I see that the control volume should be taken around the cart (moving control volume). That gives
the answer below. Note that we also must conserve mass, so if we reduce the mass flow rate where does the extra
water go? If the mass flow rate reduces from 30 kg/s at the nozzle to 22.5 kg/s at the cart then we must have done
something with the lost 7.5 kg/s. The answer is that the missing water (30-22.5 = 7.5 kg/s) is in the longer and longer jet
that chases the cart.
Note that the textbook solution ignores the change from pressure*area and momentum coming into the nozzle; since
this information was not given the simplification was necessary; but it overestimates the reaction force!
A more complete presentation, as illustrated in the schematic below, would be more educational. The momentum
coming in and the P*A term coming in would cancel some of the nozzle force.
Note: don’t assign 13-38 in the future, too long and tedious