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3 years ago

What are the parameters that affect life and drag forces on an aerofoil?

Engineering

1 answer:

Vinil7 [7]3 years ago

6 0

Answer:

1.The velocity of fluid

2.Fluid properties.

3.Projected area of object(geometry of the object).

Explanation:

Drag force:

Drag force is a frictional force which offered by fluid when a object is moving in it.Drag force try to oppose the motion of object when object is moving in a medium.

Drag force given as

$F_D=\dfrac{1}{2}\rho\ A\ V^2$

So we can say that drag force depends on following properties

1.The velocity of fluid

2.Fluid properties.

3.Projected area of object(geometry of the object).

You might be interested in

A fair die is thrown, What is the probability gained if you are told that 4 will

Semmy [17]

Answer:

1/6

Explanation:

A dice has 6 sides, the probability of 4 appearing is 1/6.

7 0

2 years ago

Consider a system with two tasks, Task1 and Task2. Task1 has a period of 200 ms, and Task2 has a period of 300 ms. All tasks ini

Murrr4er [49]

<u>Explanation:</u>

Task 1 time period = 200ms, Task 2 time period = 300ms

Task ticked = $\frac{1000ms}{200ms}= 5$ → 5 times

Task 2 ticked = $\frac{1000ms}{300ms} = 3.33$ → 3 times

At 600 ms → 200ms 200ms 200ms

300ms → $\frac{30ms}{60ms}$

Largest time period = H.C.M of (200ms, 300ms)

= 600ms

4 0

2 years ago

Water vapor at 6 MPa, 600 degrees C enters a turbine operating at steady state and expands to 10kPa. The mass flow rate is 2 kg/

kirill115 [55]

Answer:

Explanation:

Obtain the following properties at 6MPa and 600°C from the table "Superheated water".

$h_1=3658.8KL/Kg\\s_1=7.1693kJ/kg.k$

Obtain the following properties at 10kPa from the table "saturated water"

$h_{f2}=191.81KJ/Kg.K\\h_{fg2}=2392.1KJ/Kg\\s_{f2}=0.6492KJ/Kg.K\\s_{fg2}=7.4996KJ/Kg.K$

Calculate the enthalpy at exit of the turbine using the energy balance equation.

$\frac{dE}{dt}=Q-W+m(h_1-h_2)$

Since, the process is isentropic process $Q=0$

$0=0-W+m(h_1-h_2)\\h_2=h_1-\frac{W}{m}\\\\h_2=3658.8-\frac{2626}{2}\\\\=2345.8kJ/kg$

Use the isentropic relations:

$s_1=s_{2s}\\s_1=s_{f2}+x_{2s}s_{fg2}\\7.1693=6492+x_{2s}(7.4996)\\x_{2s}=87$

Calculate the enthalpy at isentropic state 2s.

$h_{2s}=h_{f2}+x_{2s}.h_{fg2}\\=191.81+0.87(2392.1)\\=2272.937kJ/kg$

a.)

Calculate the isentropic turbine efficiency.

$\eta_{turbine}=\frac{h_1-h_2}{h_1-h_{2s}}\\\\=\frac{3658.8-2345.8}{3658.8-2272.937}=0.947=94.7%$

b.)

Find the quality of the water at state 2

since $h_f$ at 10KPa < $h_2$ < $h_g$ at 10KPa

Therefore, state 2 is in two-phase region.

$h_2=h_{f2}+x_2(h_{fg2})\\2345.8=191.81+x_2(2392.1)\\x_2=0.9$

Calculate the entropy at state 2.

$s_2=s_{f2}+x_2.s_{fg2}\\=0.6492+0.9(7.4996)\\=7.398kJ/Kg.K$

Calculate the rate of entropy production.

$S=\frac{Q}{T}+m(s_2-s_1)$

since, Q = 0

$S=m(s_2-s_1)\\=2\frac{kg}{s}(7.398-7.1693)kJ/kg\\=0.4574kW/k$

6 0

3 years ago

Introduce JTA and JT

Vinil7 [7]

JT2 oaoaosnforeneomdocdmsnsksmsmsodnfnfj

7 0

2 years ago

100 kg of R-134a at 200 kPa are contained in a piston–cylinder device whose volume is 12.322 m3. The piston is now moved until t

LekaFEV [45]

Answer:

T=151 K, U=-1.848*10^6J

Explanation:

The given process occurs when the pressure is constant. Given gas follows the Ideal Gas Law:

pV=nRT

For the given scenario, we operate with the amount of the gas- n- calculated in moles. To find n, we use molar mass: M=102 g/mol.

Using the given mass m, molar mass M, we can get the following equation:

pV=mRT/M

To calculate change in the internal energy, we need to know initial and final temperatures. We can calculate both temperatures as:

T=pVM/(Rm); so initial T=302.61K and final T=151.289K

Now we can calculate change of U:

U=3/2 mRT/M using T- difference in temperatures

U=-1.848*10^6 J

Note, that the energy was taken away from the system.

5 0

3 years ago