Question 7.18: It is proposed to use high pressure water in the design of a...

We will choose a CV enclosing the power unit and rotating with it as shown in Figure 7.30, and assume that the unit has achieved a constant angular velocity. Assuming equal flowrates out of each jet, we can use a mass balance to write \dot M = ρQ = N\dot M_{jet}, where \dot M_{jet} = ρA_{jet} \bar V_{jet} is the jet mass flowrate, A_{jet} is the jet area, and \bar V_{jet} is the jet exit velocity relative to the device. To apply a steady process angular momentum balance we will employ Eq. 7.41:

\sum{}± \dot M_{port} [r_{port} ×u_{port} ]=r_G ×(−W_{CV} k)+\int_{CV}r×ρ[−(2\pmb{\Omega} ×u)−(\pmb{\Omega} ×(\pmb{\Omega} ×x)]dV

+ T_{shaft}+T_{solid}+\sum{}[r_{port} ×(−p_{port}A_{port} n)]+T_{exterior}

and choose the origin on the rotation axis as shown in Figure 7.30. To account for the angular momentum transport at each port of the CV, we write

\sum{}± \dot M_{port} [r_{port} ×u_{port}]=−\dot M[r_B ×u_B]+ N\dot M_{jet} [r_{jet}×u_{jet}]

The angular momentum transport at the inlet port can be neglected. To understand why, recall that the term −\dot M[r_B ×u_B] representing the transport at the inlet is actually given by the integral \int_{inlet} (r × ρu)(u • n)dS. The velocity vector in this integral is relative to the CV, and since the CV is rotating, the velocity entering the inlet is actually not unidirectional and uniform but swirling relative to the CV. The angular momentum of this incoming flow is very small, however, since the moment arm is always less than the radius of the inlet. Since this radius is small, we can neglect this term. The total angular momentum transport is thus N\dot M_{jet} [r_{jet}×u_{jet}]. To evaluate this term, consider the jet leaving the tube aligned with the +y axis. The moment arm to this jet is r_jet = Hi + Lj, the jet velocity vector is u_jet = \bar V_{jet}i, and we obtain

\dot M_{jet} [r_{jet}×u_{jet}]=\dot M_{jet}[(H\mathrm{i} +Lj)\times \bar V_{jet}]=-\dot M_{jet}L\bar V_{jet}k

The angular momentum transport for each of the remaining jets is the same, thus the total angular momentum transport by the jets is

\sum{}± \dot M_{port} [r_{port} ×u_{port}]=N\dot M_{jet} [r_{jet}×u_{jet}]=-N\dot M_{jet}L\bar V_{jet}k=-\dot ML\bar V_{jet}k                    (A)

Next consider the body force terms. Symmetry causes the term r_G × (−W_CVk) representing the torque due to gravity to be zero. The term \int_{CV}r × ρ{−(2\pmb{\Omega} ×u)− [\pmb{\Omega}×(\pmb{\Omega}×x)]}dV is evaluated as follows. Consider the portion of the CV that includes the entire jet tube aligned with the y axis. We must account for the torque exerted by the Coriolis and centrifugal forces acting on both the solid and fluid inside this part of the CV. The centrifugal force does not create a torque about the rotation axis. We can show this by observing that since the jet tube is of small diameter, the moment arm to any point in the control volume is r = yj. The location of the point is also x = yj, and \pmb{\Omega} = ωk. Thus the centrifugal force \pmb{\Omega} × (\pmb{\Omega}× x) = ω²yj acts along the tube in the y direction, and the cross product r × ρ[−(\pmb{\Omega} × (\pmb{\Omega} × x)] is zero.

To evaluate the torque created by the Coriolis force on this portion of the CV, we need only consider the fluid, since the solid parts inside the CV are not moving relative to the rotating reference frame. Let the uniform velocity of the fluid inside the portion of the tube of length L lying along the y axis be given by u_L=\bar V_{tube}j and that in the remaining section of the tube of length H be given by u_H=\bar V_{tube}\mathrm{i} . For the tube section of length L, we have −2\pmb{\Omega} ×u=− 2ωk× \bar V_{tube}j=2ω \bar V_{tube}\mathrm{i} . The moment arm to a point in the fluid inside is r = yj, andr×ρ[−(2\pmb{\Omega} ×u)]= ρ[yj×2ω \bar V_{tube}i]=−2ωρy\bar V_{tube}k. The integral over this portion of the CV for the N jet tubes can be evaluated for a tube of area A_{tube} as

N\int_{L−tube}r×ρ[−(2\pmb{\Omega} ×u)]dV=NA_{tube}\int_{0}^{L}−2ωρy\bar V_{tube} k  dy                                    (B)

=−ωL^2N(ρA_{tube}\bar V_{tube} )k=−ωL^2\dot Mk

where we used a mass balance to show that ρA_{tube}\bar V_{tube}=ρA_{jet}\bar V_{jet}=\bar M_{jet},  and  \dot M_{jet}=N\dot M_{jet}.

For the section of the tube of length H, we have −2\pmb{\Omega} × u = − 2ωk × V_tubei = −2ωV_tubej. The moment arm to a point in this section of the CV can be written as r = xi + yj,  and  thus r × ρ[−(2\pmb{\Omega} × u)] = ρ[(xi + yj) × (−2ωV_tubej)] = −2ωρV_tubexk. The integral for all N tubes is

N\int_{H−tube}r×ρ[−(2\pmb{\Omega} ×u)]dV=NA_{tube}\int_{0}^{H}−2ωρx\bar V_{tube} k  dx                                    (C)

=−ωH^2N(ρA_{tube}\bar V_{tube} )k=−ωH^2\dot Mk

We are told to neglect friction in the inlet bearing, so we can set T_solid to zero. Since T_exterior includes the effect of atmospheric pressure on the exterior (which is always present), we will neglect the unknown aerodynamic drag but account for atmospheric pressure by using gage pressures at the ports. Since each jet leaves at atmospheric pressure, the gage pressures are all zero, and the term ­\sum{} [r_port × ( − p_portA_portn)] needs be evaluated only at the inlet. Because of the symmetry and alignment of the rotation axis, the pressure on the inlet does not contribute a torque.

The term T_shaft represents the torque applied to the CV by the shaft on which the brush mounts. By the principle of action–reaction, we can write the torque T_brushk supplied to the brush as

−T_brushk = T_shaft                                                      (D)

Inserting (A) through (D) into the angular momentum balance, we find

−\dot ML\bar V_{jet} k=−ωL^2\dot Mk−ωH^2\dot Mk−T_{brush}k

Solving for the torque available to turn the brush, we have

T_{brush}=\dot ML\bar V_{jet}−\dot M ω(L^2 + H^2)                                        (E)

From this result we see that there is no apparent advantage to employing multiple jets, since the torque delivered to the brush is independent of N. Notice also that the maximum torque is delivered to the brush when it is not turning. As the rpm of the brush increases, the torque delivered decreases. The result also shows that it is advantageous to keep the length H as short as possible.