Question 6.SP.8: A total of 20,000 cfs of air (0.072 lb/ft³) flows through tw......
A total of 20,000 cfs of air (0.072 lb/ft³) flows through two 6.5-ft-diameter propellers that are attached to an airplane moving at 150 mph through still air. Find (a) the total thrust and (b) the efficiency of the propellers.
Also find (c) the pressure rise across the propellers and (d) the horsepower input
to each propeller. Neglect eddy losses.
(a) The velocity of air relative to the airplane is
V_1=150 \mathrm{mph}=150\left(\frac{44}{30}\right)=220 \mathrm{fps}
The velocity of air through the actuating disk is
V=V_1+\frac{\Delta V}{2}=\frac{Q}{A}=\frac{20,000 / 2}{(\pi / 4)(6.5)^2}=301 \mathrm{fps}
Thus : \Delta V=2(301-220)=162.7 \mathrm{fps}
Eq. (6.38): F_T=\rho Q \Delta V=\frac{0.072}{32.2}(20,000) 162.7
=7280 \mathrm{lb} \text { (total thrust of both propellers) }
\text { (b) Eq. (6.42): } \eta=\frac{1}{1+\Delta V / 2 V_1}=\frac{1}{1+162 / 440}=0.730=73 \%(c) F_T on one propeller =7280 / 2=3640 \mathrm{lb}. But from Eq. (6.37) F_T=\Delta p A, thus 3640=\Delta p(\pi / 4)(6.5)^2,
F_T= (p_3-p_2)\frac{πD^2}{4}=\Delta p A (6.37)
\Delta p=109.6 \mathrm{psf}=0.761 \mathrm{psi}
\text { (d) } \quad P_{\mathrm{in} / \text { propeller }}=\frac{\gamma Q(\Delta p / \gamma)}{550}=\frac{Q \Delta p}{550}=\frac{10,000(109.6)}{550}=1994 \mathrm{hp} \text { Check: } \quad P_{\text {in/propeller }}=\frac{F_T V}{550}=\frac{3640(301)}{550}=1994 \mathrm{hp}
F_T=\rho Q(\Delta V)=\rho A V\left(V_4-V_1\right) (6.38)
\eta=\frac{P_{\text {out }}}{P_{\text {in }}}=\frac{(\rho Q \Delta V) V_1}{(\rho Q V) \Delta V}=\frac{V_1}{V}=\frac{V_1}{V_1+\frac{1}{2} \Delta V}=\frac{1}{1+\frac{1}{2}\left(\Delta V / V_1\right)} (6.42)