A student engineer is given a summer job to find the drag force on a new unmaned aerial vehicle that travels at a cruising speed
of 320 km/h in air at 15 C. The student decided to build a 1/50 scale model and test it in a water tunnel at the same temperature.
(a) What conditions are required to ensure the similarity of scale model and prototype.
(b) Find the water speed required to model the prototype.
(c) If the drag force on the model is 5 kN, determine the drag force on the prototype
(a) What conditions are required to ensure the similarity of scale model and prototype.
(b) Find the water speed required to model the prototype.
(c) If the drag force on the model is 5 kN, determine the drag force on the prototype
1 answer:
Answer:
b. 1232.08 km/hr
c. 1.02 kn
Explanation:
a) For dynamic similar conditions, the non-dimensional terms R/ρ V2 L2 and ρVL/ μ should be same for both prototype and its model. For these non-dimensional terms , R is drag force, V is velocity in m/s, μ is dynamic viscosity, ρ is density and L is length parameter.
See attachment for the remaining.
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Answer:
The answer is "15 N".
Explanation:
Please find the complete question in the attached file.
In frame B:
For just slipping:
Answer:
(a). The value of temperature at the end of heat addition process = 2042.56 K
(b). The value of pressure at the end of heat addition process = 1555.46 k pa
(c). The thermal efficiency of an Otto cycle = 0.4478
(d). The value of mean effective pressure of the cycle = 1506.41
Explanation:
Compression ratio = 8
Initial pressure = 95 k pa
Initial temperature = 278 °c = 551 K
Final pressure = 8 ×
= 8 × 95 = 760 k pa
Final temperature =
×
Final temperature = 551 ×
Final temperature = 998 K
Heat transferred at constant volume Q = 750
(a). We know that Heat transferred at constant volume =
⇒ 1 × 0.718 × ( - 998 ) = 750
⇒ = 2042.56 K
This is the value of temperature at the end of heat addition process.
Since heat addition is constant volume process. so for that process pressure is directly proportional to the temperature.
⇒ P ∝ T
⇒ =
⇒ =
× 760
⇒ = 1555.46 k pa
This is the value of pressure at the end of heat addition process.
(b). Heat rejected from the cycle
For the compression and expansion process,
⇒
⇒
⇒ = 1127.7 K
Heat rejected = 1 × 0.718 × ( 1127.7 - 551)
⇒ = 414.07
Net heat interaction from the cycle =
-
Put the values of &
we get,
⇒ = 750 - 414.07
⇒ = 335.93
We know that for a cyclic process net heat interaction is equal to net work transfer.
⇒ =
⇒ = 335.93
This is the net work output from the cycle.
(c). Thermal efficiency of an Otto cycle is given by
Put the values of &
in the above formula we get,
⇒ = 0.4478
This is the thermal efficiency of an Otto cycle.
(d). Mean effective pressure :-
We know that mean effective pressure of the Otto cycle is given by
=
---------- (1)
where is the swept volume.
=
---------- ( 2 )
From ideal gas equation
= m × R ×
Put all the values in above formula we get,
⇒ 95 × = 1 × 0.287 × 551
⇒ = 0.6
From the same ideal gas equation
= m × R ×
⇒ 760 × = 1 × 0.287 × 998
⇒ = 0.377
Thus swept volume = 0.6 - 0.377
⇒ = 0.223
Thus from equation 1 the mean effective pressure
⇒ =
⇒ = 1506.41
This is the value of mean effective pressure of the cycle.