The motors that drive airplane propellers are, in some cases, tuned by using sound beats. The whirring motor produces a sound wa
1 answer:
Complete Question
The motors that drive airplane propellers are, in some cases, tuned by using sound beats. The whirring motor produces a sound wave of the same frequency as the propeller. Consider a plane with 2 engines driving 2 propellers. You want to tune them to turn at identical frequencies.
Required:
a. If one single-bladed propellor is turning at 575 rpm and your hear 2.0 Hz beats when you run the second propeller, what are the two possible frequencies of the second propeller in Hz and rpm?
b
Suppose you increase the speed of the second propeller slightly and find that the beat frequency changes to 2.10 Hz . In part (A), which of the two answers was the correct one for the frequency of the second single-bladed propeller ?
c. How do you know the answer in part (B) to be correct?
Answer:
a
The two possible frequencies of the second propeller in Hz
The two possible frequencies of the second propeller in rpm
b
The correct answer for the frequency of the second single-bladed propeller is
c
The above answer is correct because when the beat frequency of the second propeller increases(i.e from 2.0 Hz to 2.10 Hz) the frequency of the second propeller becomes much greater than that of the first propeller so looking at the two possible value of frequency of the second propeller (i.e 695 rpm and 455 rpm ) we see that it is 695 rpm that is showing that increase of the second propeller compared to the first propeller
Explanation:
From the question we are told that
The number of engines is n = 2
The number of propellers is m = 2
The angular frequency of the single-bladed propellor
The frequency of the beat heard at this velocity is
Converting the beat frequency to rpm
Generally the the two possible frequencies of the second propeller in rpm is
=>
=>
And
=>
=>
Converting the angular frequency of the single-bladed propellor to rpm
Generally the the two possible frequencies of the second propeller in
Hz is
=>
=>
And
=>
=>
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<u>Answer :</u>
(a) The Initial temperature of the steam is 247.557°C
(b) The enthalpy change per unit mass of the steam is -1771 kJ/kg
(c) The final pressure is 1555 kPa, and the liquid vapor mixture contains small mass of vapor.
<u>Explanation :</u>
From the given statements, we know
Initial steam pressure () = 3.5 MPa
Degree of Super heat = 5°C
Saturation Temperature () = 242.557°C [From Steam Table]
Now, the Initial temperature () of steam is given by-
=
+ Degree of Superheat
= 242.557 + 5 = 247.557°C
(b)To find: The enthalpy change per unit mass of the steam is
We know enthalpy (h) change is given by
Δh = –
From steam Table, properties of the gas at state 1 and state 2
= 2821.1 kJ/kg
= 1049.7 kJ/kg
On substituting the values, we get
or, Δh = 1049.7 - 2821.1 = -1771 kJ/kg
(c) To Find: The final pressure of the liquid vapor mixture contains small mass of vapor.
We find that the gas at the final position, i.e., at position 3
Final Temperature () = 200°C [Given] ……………………………..(1)
Specific Volume () = Specific Volume (
) = 0.00123
/kg ….....(2)
Using steam table for corresponding values of (1) and (2), we get
Final Pressure = 1555 kPa
Dryness fraction () = 0.0006.
Answer:0.478 c
Explanation:
Given
mass of lighter Particle
mass of heavier Particle
speed of lighter particle
Let speed of heavier particle
and Momentum of the particle is given by
as momentum is conserved therefore