Question 10.3: A steady jet of water is redirected by a deflector, as shown...
A steady jet of water is redirected by a deflector, as shown in Figure 10.12. The jet has mass flow rate of 32 kg/s, cross-sectional area 2 cm×40 cm, and speed V1 when it encounters the deflector. What force per unit width of the deflector (into the page) is needed to hold the deflector in place?
Given: Geometry of flow deflector.
Find: Reaction forces from deflector support on fluid in CV.
Assume: Jet has constant cross-sectional area, even after being deflected.
Flow is steady and incompressible. Density of water is constant, uniformly 1000 kg/m³. Gravity and viscous effects may be neglected.
The flow of water imparts a force to the deflector. Reaction forces from the deflector balance these forces and act on the fluid in the CV drawn. We are asked for these reactions, R_x and R_y , if the deflector has width of 1 m into the page. We begin by finding the inlet velocity V_1 . We are given the mass flow rate of the jet, \dot{m} , which is ρV_1 A_1 . As the cross-sectional area A_1 is also given, we use this to find V_1 :
V_1=\frac{\dot{m}}{ρ A_1}=\frac{32 ^{kg}/_s}{(1000 \frac{\textrm{kg}}{m^3})(0.02 \textrm{ m})(0.40 \textrm{ m})}=4 ^{m}/_s .
If the flow is steady, we must have a constant mass flow rate (what flows into our CV must flow back out again), or ρV_1A_1 = ρV_2A_2 . Since we have an incompressible flow and since the jet’s cross-sectional area does not change, we must have V_2 = V_1 = 4 m/s.
To find the requested reaction forces, we must apply the conservation of
linear momentum, or \underline{F}=m \underline{a}, in both x and y directions. These equations are
and
F_y= (F_{visc})_{_y}+(F_{external})_{_y}+\int\limits_{CV}{ρ g_y dV}- \int\limits_{CS}{p\hat{\underline{j}}\cdot \underline{n}dA }=\frac{∂}{∂t}\int\limits_{CV}{ρ v dV}+\int\limits_{CS}{ρ v \underline{V}\cdot \underline{n} dA} .
Since our jet is steady, with negligible contributions from gravity, viscosity, and pressure gradients, these equations simplify greatly:
-R_x= \int\limits_{CS}{ρ u \underline{V}\cdot \underline{n} dA} .
R_y = \int\limits_{CS}{ρ v \underline{V}\cdot \underline{n} dA} .
The x momentum flux therefore balances the reaction force R_x, negative as it is in the negative x direction. Writing out the flux at each of the two control surfaces, we have
-R_x= \int\limits_{CS}{ρ u \underline{V}\cdot \underline{n} dA}=-ρ V_1 V_1 A_1+ρ (V_2 \cos \theta) V_2A_2 .
Note that V_2 cos θ is the x component of velocity at surface A_2 and that V_2 \textrm{ is } \underline{V}\cdot \underline{n} \textrm{ at } A_2:
R_x= \dot{m}(V_1-V_2 \cos \theta)= (32 ^{kg}/_s)(4 ^{m}/_s) (1- cos(30°)).
= 17.2 N
in the y direction we have
R_y= \int\limits_{CS}{ρ u \underline{V}\cdot \underline{n} dA}=0+ρ (V_2 \sin \theta) V_2 A_2
R_y = \dot{m} V_2 sinθ
= (32 ^{kg}/_s)(4 ^{m}/_s) sin(30°).
=64 N