Physics College

Transverse thrusters are used to make large ships fully maneuverable at low speeds without tugboat assistance. A transverse thruster consists of a propeller mounted in a duct; the unit is then mounted below the waterline in the bow or stern of the ship. The duct runs completely across the ship. Calculate the thrust developed by a 1900 kW unit supplied to the propeller if the duct is 2.6 m in diameter and the ship is stationary.

Answers

Answer 1

Answer:

Thrust developed = 212.3373 kN

Explanation:

Assuming the ship is stationary

Determine the Thrust developed

power supplied to the propeller ( Punit ) = 1900 KW

Duct distance ( diameter ; D  ) = 2.6 m

first step : calculate the area of the duct

A = π/4 * D^2

   =  π/4 * ( 2.6)^2  = 5.3092 m^2

next : calculate the velocity of propeller

Punit = (A*v*β ) / 2  * V^2     ( assuming β = 999 kg/m^3 ) also given V1 = 0

∴V^3 = Punit * 2 / A*β

         = ( 1900 * 10^3 * 2 ) / ( 5.3092 * 999 )

hence V2 = 8.9480 m/s

Finally determine the thrust developed

F = Punit / V2

  = (1900 * 10^3) / ( 8.9480)

  = 212.3373 kN

Related Questions

15 points!

a. Calculate the electric potential energy stored in a 1.4 x 10-7 F capacitor
that stores 3.40 x 10-6 C of charge at 24.0 V.

Answers

Answer:

[tex]4.12\times 10^{-5}\ J[/tex].

Explanation:

Given that,

Capacitance, [tex]C=1.4\times 10^{-7}\ F[/tex]

Charge stored in the capacitor, [tex]Q=3.4\times 10^{-6}\ C[/tex]

We need to find the electric potential energy stored in the capacitor. The formula for the electric potential energy stored in the capacitor is given by :

[tex]E=\dfrac{Q^2}{2C}[/tex]

Put all the values,

[tex]E=\dfrac{(3.4\times 10^{-6})^2}{2\times 1.4\times 10^{-7}}\\\\=4.12\times 10^{-5}\ J[/tex]

So, the required electric potential eenergy is equal to [tex]4.12\times 10^{-5}\ J[/tex].

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