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Zolol [24]
4 years ago
6

The entrance of the space module into atmosphere is one of the very critical phases of any space mission. Due to the formation o

f bow shock (Detached) in front of the module, the air temperature drastically increases to thousands of Kelvin. One suggested in order to find the amount of heat transfer to the space module at the reentry phase, we can set up an experiment on the ground with a scaled model in the wind tunnel. The proposed scaled model is 1/x times and is located in a wind tunnel with a maximum air velocity 100 times less than the actual velocity of the module. The density of the air at the reentry zone is almost 1000 times less than air density at the sea level, where the wind turbine is located. The see level air Prandtl number is approximately the same as the reentry zone. But the dynamic viscosity of air at the reentry condition is almost 3.59 times of the sea level air viscosity. The Thermal conductivity of air in the reentry condition is 4.356 times the sea level condition.
In the wind tunnel for the air temperature difference of 100°C between air and the scaled model, the energy transfer due to the heat was measured to be about 400W. At the reentry condition with 1500C Temperature how much is the energy transfer due to the heat to the module? For consistency modeling find the 1/x scaling ration.

Engineering
1 answer:
Hatshy [7]4 years ago
5 0

Answer:

Check the explanation

Explanation:

The transferred energy calculations can be done using the equation:

This is when:

• power is calculated in watts (W)

• energy is calculated in joules (J)

• time is calculated in seconds (s)

it can also be calculated by utilizing the same equation but with different units:

• when energy is measured in kilowatt-hours (kWh)

• when power is measured in kilowatts (kW)

• and when time is measured in hours (h)

kindly check the comprehensive step by step calculation to the question in the attached images below

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The uniform beam is supported by two rods AB and CD that have crosssectional areas of 10 mm2 and 15 mm2 , respectively. Determin
ddd [48]

Answer:

w=2.25

Explanation:

It is necessary to determine the maximum w so that the normal stress in the AB and CD rods does not exceed the permitted normal stress.  

The surface of the cross-section of the stapes was determined:  

A_ab= 10 mm^2

A-cd=  15 mm^2

The maximum load is determined from the condition that the normal stresses is not higher than the permitted normal stress σ_allow.

σ_ab = F_ab/A_ab\leqσ_allow

σ_cd =  F_cd/A_cd\leqσ_allow

In the next step we will determine the static size: Picture b).  

We apply the conditions of equilibrium:  

∑F_x=0

∑F_y=0

  ∑M=0

∑M_a=0 ==> -w*6*0.5*6*0.75*F_cd*6 =0

              ==> F_cd = 2*w*k*N

∑F_y=0 ==> F_cd+F_ab - 6*w*0.5 ==>2*w+F_ab -6*w*0.5 =0

              ==> F_ab = w*k*N

Now we determine the load w  

<u>Sector AB:  </u>

σ_ab = F_ab/A_ab\leq σ_allow=300 KPa

         = w/10*10^-6\leq σ_allow=300 KPa

w_ab = 3*10^-3 kN/m

<u>Sector CD:  </u>

σ_cd = F_cd/A_cd\leq σ_allow=300 KPa

         = 2*w/15*10^-6\leq σ_allow=300 KPa

w_cd = 2.25*10^-3 kN/m

w=min{w_ab;w_cd} ==> w=min{3*10^-3;2.25*10^-3}

                                ==> w=2.25 * 10^-3 kN/m

<u>The solution is:  </u>

                                w=2.25 N/m

note:

find the attached graph

6 0
3 years ago
A shaft made of stainless steel has an outside diameter of 42 mm and a wall thickness of 4 mm. Determine the maximum torque T th
fiasKO [112]

Answer:

Explanation:

Using equation of pure torsion

\frac{T}{I_{polar} }=\frac{t}{r}

where

T is the applied Torque

I_{polar} is polar moment of inertia of the shaft

t is the shear stress at a distance r from the center

r is distance from center

For a shaft with

D_{0} = Outer Diameter

D_{i} = Inner Diameter

I_{polar}=\frac{\pi (D_{o} ^{4}-D_{in} ^{4}) }{32}

Applying values in the above equation we get

I_{polar} =\frac{\pi(0.042^{4}-(0.042-.008)^{4})}{32}\\I_{polar}= 1.74 x 10^{-7} m^{4}

Thus from the equation of torsion we get

T=\frac{I_{polar} t}{r}

Applying values we get

T=\frac{1.74X10^{-7}X100X10^{6}  }{.021}

T =829.97Nm

7 0
3 years ago
Is a gas turbine a heat engine?
Nookie1986 [14]
Gas turbines extract the energy from combustion gas

and heat engines convert thermal and chemical energy to mechanical energy

gas is considered a chemical so i'm pretty sure it is considered one
4 0
3 years ago
The inside temperature of a wall in a dwelling is 15°C. If the air in the room is at 21°C, what is the maximum relative humidity
Sveta_85 [38]

Given:

Wall's inside temperature, T_{wa} = 15^{\circ}C

Room air temperature,  T_{air} = 21^{\circ}C

Solution:

To calculate percentage max relative humidity, we make use of steam table for saturated pressure of wall and air:

From steam table:

At T_{wa} = 15^{\circ}C:

P_{wa(sat)} = 1.7057 kPa

At  T_{air} = 21^{\circ}C:

P_{air(sat)} = 2.487 kPa

Now,

% Relative Humidity (RH)_{max} = \frac{P_{wa(sat)}}{P_{air(sat)}} \times 100

(RH)_{max} = \frac{1.7057}{2.487}\times 100

(RH)_{max} = 68.58%

Therefore, max Relative Humidity of the air before the occurrence of condensation in wall is 68.85%

8 0
3 years ago
Determine the maximum volume in gallons​ [gal] of olive oil that can be stored in a closed cylindrical silo with a diameter of 3
Olin [163]

Answer:

V=1601gal

Explanation:

Hello! This problem is solved as follows,

First we must raise the equation that defines the pressure at the bottom of the tank with the purpose of finding the height that olive oil reaches.

This is given as the sum due to the atmospheric pressure (1atm = 101.325kPa), and the pressure due to the weight of the olive oil, taking into account the above, the following equation is inferred.

P=Poil+Patm

P=total pressure or absolute pressure=26psi=179213.28Pa

Patm= the atmospheric pressure =101325Pa

Poil=pressure due to the weight of olive oil=0.86αgh

α=density of water=1000kg/m^3

g=gravity=9.81m/s^2

h= height that olive oil reaches

solving

P=Poil+Patm

P=Patm+0.86αgh

h=\frac{P-Patm}{0.86\alpha g } =\frac{179214.28-101325}{(0.86)(1000)(9.81)} \\h=9.23m[/tex]

Now we can use the equation that defines the volume of a cylinder.

V=V=\frac{\pi }{4} D^{2} h

D=3ft=0.9144m

h=9.23m

solving

V=\frac{\pi }{4} (0.9144)^{2} 9.23=6.06m^3

finally we use conversion factors to find the volume in gallons

V=6.06m^3\frac{1000L}{1m^3} \frac{1gal}{3.785L} =1601gal

3 0
4 years ago
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