Question 6.SP.4: The 2-in-diameter water jet with a velocity of 100 fps shown......

(a) The velocity vector diagrams at entrance and exit to the vane are given in Fig. S6.4. Since v_{2}=0.9(40)=36 \mathrm{fps}

Eq. (6.17):    V_{2 u}=V_{2} \cos \alpha_{2}=u+v_{2} \cos \beta_{2}=60+36 \cos 150^{\circ}=28.8 \mathrm{fps}          (1)

Eq. (6.18):   \quad V_{2 n}=V_{2} \sin \alpha_{2}=v_{2} \sin \beta_{2}=-36 \sin 150^{\circ}=-18 \mathrm{fps}                      (2)

Q^{\prime}=A_{1} v_{i}=A_{i}\left(V_{1}-u\right)=\frac{\pi}{4}\left(\frac{2}{12}\right)^{2}(100-60)=0.873 \mathrm{cfs}

Eq. (6.22):

 -F_{x}=\rho Q^{\prime}\left(V_{2 u}-V_{1 u}\right)=1.938(0.873)(28.8-100)=-120.4 \mathrm{lb}

So   F_{x}=120.4 \mathrm{lb}. The force of the vane on the water is to the left, as assumed; hence the force of water on the vane is 120.4 \mathrm{lb} to the right.

-F_{y}=\rho Q^{\prime}\left(V_{2 n}-V_{1 n}\right)=1.938(0.873)(-18-0)=-30.4 \mathrm{lb}

Thus F_{y}=30.4 lb in the direction shown. The force of water on the vane is equal and opposite and thus 30.4 lb upward.

Therefore the net

F_{W / B}=124.2 \mathrm{lb} \text { at } 14.19^{\circ} \measuredangle

If needed, we may be solve (1) and (2) simultaneously to yield V_{2}=34.0 \mathrm{fps}, \alpha_{2}=32.0^{\circ}.

(b) If the blade were one of a series of blades,

Q=A_{1} V_{1}=\frac{\pi}{4}\left(\frac{2}{12}\right)^{2}(100)=2.18 \mathrm{cfs}

Eq. (6.24):

-F_{x}=\rho Q\left(V_{2} \cos \alpha_{2}-V_{1}\right)=1.938(2.18)(28.8-100)=-301 \mathrm{lb}

The horsepower transferred (delivered) to the blades (i.e., out of the fluid) is Sec. 5.9: \quad P_{\text {transfer }}=\frac{F u}{550}=\frac{(301)(60)}{550}=32.8 \mathrm{hp}

Also, for a series of blades, we can use energy considerations for the solution. The horsepower of the original jet is

Sec. 5.9:       \quad P_{\text {in }}=\frac{\gamma Q\left(V_{1}^{2} / 2 g\right)}{550}=\frac{62.4(2.18)(100)^{2}}{550(2) 32.2}=38.4 \mathrm{hp}

(c) The horsepower of the water as it leaves the system is

P_{\text {out }}=\frac{\gamma Q\left(V_{2}^{2} / 2 g\right)}{550}=\frac{62.4(2.18)(34.0)^{2}}{550(2) 32.2}=4.44 \mathrm{hp}

(d) An equation for conservation of energy expressed in terms of power is

P_{\text {in }}-P_{\text {out }}-P_{\text {transfer }}-P_{\text {friction loss }}=0

Thus                    38.4-4.44-32.8=P_{\text {friction loss }}

Therefore         P_{\text {friction loss }}=1.168 \mathrm{hp}

We may verify this by computing

\frac{\gamma Q\left(v_{1}^{2} / 2 g\right)-\gamma Q\left(v_{2}^{2} / 2 g\right)}{550}=\frac{62.4(2.18)\left[(40)^{2}-(36)^{2}\right]}{550(2) 32.2}=1.168 \mathrm{hp}

Note: The horsepower loss due to friction is a small percentage of the power of the original jet. Therefore, in problems of this type with free jets, the common assumption that v_{1}=v_{2} in magnitude gives reasonably good results.

F_u=\dot{m} \Delta V_u=\rho Q \Delta V_u=\frac{\gamma A_1 V_1}{g} \Delta V_u=\frac{\gamma A_1 V_1}{g} \Delta v_u          (6.24)

\sum F_u=0-F_u=\rho Q^{\prime}\left(\Delta V_u\right)=\rho Q^{\prime}\left(\Delta v_u\right)       (6.22)

V_n=v_n=V \sin \alpha=v \sin \beta          (6.18)

V_u=u+v_u=V \cos \alpha=u+v \cos \beta      (6.17)