An airplane flies at 100 m/s powered by a propeller. The fluid speed at the propeller is 120 m/s. (a) What is the slipstream velocity (b) what is the generated thrust (c) what is the power input in hp if we assume the propeller generates no losses (d) what is the power output in hp (e) what is the propeller efficiency if the propeller generates no losses (f) what is the pressure difference across the propeller if its diameter is 2m
This text is generated by AI trained by Professor Kirby on his lectures, videos, notes, message board responses, and textbook
PK Bot’s answer:
Below is a worked‐example solution based on standard actuator–disk ideas. In what follows we assume that
• The free‐stream speed is U_{1e}=100 m/s (the airplane’s speed),
• The speed at the disk (the “propeller speed”) is U_{3e}=120 m/s, and
• The slipstream speed (max speed downstream of propeller) is U_{2o}.
A key result from basic actuator–disk theory is that the disk sees the average of the upstream and downstream speeds. That is, we assume
U_{3e}=\frac{U_{1e}+U_{2o}}{2}\,.
Below are the step–by–step answers.
(a) Slipstream Velocity
Solve for U_{2o}:
U_{2o}=2U_{3e}-U_{1e}=2(120)-100=240-100=140\;m/s\,.
Thus, the slipstream velocity is 140 m/s.
(b) Generated Thrust
For a one–dimensional flow the thrust is given by the momentum change:
F_T=\dot{m}\,(U_{2o}-U_{1e})\,,
where the mass flow rate is
\dot{m}=\rho A U_{3e}\,.
For consistency we assume that the disk (and hence the control–volume over which the fluid is accelerated) has the same cross–sectional area as given later in part (f). With a 2‑m diameter, the area is
A=\pi\left(\frac{2}{2}\right)^2=\pi\;m^2\,.
Thus,
\dot{m}=\rho\,\pi\,120\;.
And therefore
F_T=\rho\,\pi\,120\,(140-100)=\rho\,\pi\,120\times40\,.
If we take standard sea–level air density as \rho=1.225\;kg/m^3, then
F_T=1.225\times\pi\times4800\;N\,.
Numerically,
1.225\times4800\approx5880\quad \Longrightarrow\quad F_T\approx5880\,\pi\approx18460\;N\,.
Thus, the generated thrust is about 1.85×10⁴ N.
(c) Power Input (Shaft Power) in hp, No Losses
The propeller “adds energy” to the fluid at the disk. As given in the source material, the hydrodynamic (or “fluid”) power added is
P_{in}=F_T\;U_{3e}\,.
Using F_T\approx1.846\times10^4\;N and U_{3e}=120\;m/s,
P_{in}=1.846\times10^4\;N\times120\;m/s\approx2.215\times10^6\;W\,.
To convert watts to horsepower we use 1\;hp\approx745.7\;W:
hp_{in}=\frac{2.215\times10^6}{745.7}\approx2970\;hp\,.
(d) Power Output in hp
The useful (vehicle) power is determined by the force acting on the airplane times its flight speed (free–stream speed):
P_{out}=F_T\;U_{1e}\,.
Thus,
P_{out}=1.846\times10^4\;N\times100\;m/s\approx1.846\times10^6\;W\,,
or in horsepower,
hp_{out}=\frac{1.846\times10^6}{745.7}\approx2475\;hp\,.
(e) Propeller Efficiency
Propulsive (or propeller) efficiency is defined as the ratio of the output (useful) power to the input (shaft) power:
\eta=\frac{P_{out}}{P_{in}}=\frac{F_T\,U_{1e}}{F_T\,U_{3e}}=\frac{U_{1e}}{U_{3e}}\,.
Substituting the given speeds:
\eta=\frac{100}{120}\approx0.833\quad\text{or}\quad83.3\%\,.
Even with “no losses” in energy transfer, notice that the mismatch between the flight speed and the disk speed causes the effective (propulsive) efficiency to be less than 100%.
(f) Pressure Difference Across the Propeller
Using the momentum analysis over a control volume tightly around the propeller, the pressure jump (assuming the pressure difference is acting over area A) is
\Delta p=\frac{F_T}{A}\,.
With F_T\approx1.846\times10^4\;N and A=\pi\;m^2,
\Delta p\approx\frac{1.846\times10^4}{\pi}\;Pa\approx5877\;Pa\,.
Summary of Answers
• (a) Slipstream velocity: 140 m/s
• (b) Generated thrust: ~1.85×10⁴ N
• (c) Power input: ~2.22×10⁶ W, or ~2970 hp
• (d) Power output: ~1.85×10⁶ W, or ~2475 hp
• (e) Propeller efficiency: ~83.3%
• (f) Pressure difference across the propeller: ~5.88×10³ Pa
These answers assume standard air density of 1.225 kg/m³ and that the propeller disk area is based on a 2-m diameter.
In crafting this response, I dedicated some time to reviewing the 2025 edition of Professor Kirby’s textbook (Vol 1 | Vol 2), and I found these sections to be particularly helpful in formulating my answer. I encourage you to explore the pertinent ones directly:
Textbook sections:
- Chapter 2 How Stuff Works: How a propeller works (HIGHLY RELEVANT)
- Chapter 2 How Stuff Works: How a wind turbine works (RELEVANT)
- Chapter 7.1: Helicopter/Quadcopter Rotors (RELEVANT)
- Chapter 7.3: Integral Conservation of Momentum (RELEVANT)
- Chapter 9.1: How Airfoils Work (RELEVANT)
- Chapter 9.2: Engineering Descriptions of Airfoil Forces (RELEVANT)
- Chapter 5.4: Euler Equation: Streamwise Direction i.e. Bernoulli Equation Part 1 (RELEVANT)
- Chapter 9 How Stuff Works: Wind turbine design step 1: rotor disk analysis (RELEVANT)
- Chapter 1 Fundamentals: The area–velocity relation is derived by combining conservation of mass, co… (RELEVANT)
- Chapter 7 How Stuff Works: How jetskis work (RELEVANT)
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