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vlabodo [156]
3 years ago
10

Consider an infinitely thin flat plate of chord c at an angle of attack α in a supersonic flow. The pressure on the upper and lo

wer surfaces are different but constant over each surface; that is, pu(s) = C1 and pl(s) = C2, where C1 and C2 are constants and C2 > C1. Ignoring the shear stress, calculate the location of the center of pressure.

Engineering
1 answer:
amm18123 years ago
4 0

Answer:

X_cp = c/2

Explanation:

We are given;

Chord = c

Angle of attack = α

p u (s) = c 1

​p1(s)=c2,

and c2 > c1

First of all, we need to find the resultant normal force on the plate and the total moment about leading edge.

I've attached the solution

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typedef struct bitNode { int data; struct bitNode *left; struct bstNode *right; } bstNode; int solve(bstNode* root) { if (root =
Sergio039 [100]

Answer:

The ten numbers to be filled in the blanks are: 18, 7, 7, 11, 18, 36, 3, 8, 13, 50.

Explanation:

keeps on going to the left node until node->left == NULL;

now at Node 18;

left = right = 0; hence condition is not satisfied

18 is printed first.

the value 18 is returned .

Then we reach at 4;

from there we move to 7;

just like 18, similar things happen with 7 and 7 is printed, the value 7 is returned.

Now coming to Node 4,

left = 0, right = 7 ; hence the condition is satisfied & res = 7; 7 is printed.

For Node 16, left = 7 ; right = 11(but for this we visit 11 first and 11 is printed)

for 16; condition is satisfied; res = 7 + 11 = 18 ; 18 is printed

Now for 5; left = right = 18; the condition is satisfied; so res = 18 + 18 = 36; 36 is printed

Next we visit Node 3; 3 is printed & 3 is returned

Then Node 8 ; 8 is printed & 8 is returned

for Node 13; left = 3, right = 8 ; condition is not satisfied, 13 is printed.

For Node 50; left = 36 right = 13 ; condition is not satisfied hence 50 is printed.

So the order of printing is  18 7 7 11 18 36 3 8 13 50.

4 0
3 years ago
Ignition for heavy fuel oil?
ipn [44]

Answer:

What heavy fuel oil?

Heavy Fuel Oil (HFO) is a category of fuel oils of a tar-like consistency. Also known as bunker fuel, or residual fuel oil, HFO is the result or remnant from the distillation and cracking process of petroleum.

Explanation:

7 0
2 years ago
Read 2 more answers
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
Close to 16 billion pounds of ethylene glycol (EG) were produced in 2013. It previously ranked as the twenty-sixth most produced
bekas [8.4K]

Answer:

a) 0.684

b) 0.90

Explanation:

Catalyst

EO + W → EG

<u>a) calculate the conversion exiting the first reactor </u>

CAo = 16.1 / 2   mol/dm^3

Given that there are two stream one  contains 16.1 mol/dm^3 while the other contains   0.9 wt% catalyst

Vo = 7.24 dm^3/s

Vm = 800 gal = 3028 dm^3

hence Im = Vin/ Vo = (3028 dm^3) / (7.24dm^3/s) = 418.232 secs = 6.97 mins

next determine the value of conversion exiting the reactor ( Xai ) using the relation below

KIm = \frac{Xai}{1-Xai}  ------ ( 1 )

make Xai subject of the relation

Xai = KIm / 1 + KIm  ---  ( 2 )

<em>where : K = 0.311 ,  Im = 6.97   ( input values into equation 2 )</em>

Xai = 0.684

<u>B) calculate the conversion exiting the second reactor</u>

CA1 = CA0 ( 1 - Xai )

therefore CA1 = 2.5438 mol/dm^3

Vo = 7.24 dm^3/s

To determine the value of the conversion exiting the second reactor  ( Xa2 ) we will use the relation below

XA2 = ( Xai + Im K ) / ( Im K + 1 ) ----- ( 3 )

<em> where : Xai = 0.684 , Im = 6.97,  and K = 0.311  ( input values into equation 3 )</em>

XA2 = 0.90

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4 0
3 years ago
Technician a s ays both an ohmmeter and a self-powered test light may be used to test for continuity. technician b says both may
amm1812

Both A and B technicians are correct because both might be used to test fuses, according to technician B.

<h3>What is continuity?</h3>

The behavior of a function at a certain point or section is described by continuity. The limit can be used to determine continuity.

From the question:

We can conclude:

The technician claims that you may check for continuity using both an ohmmeter and a self-powered test light. Both might be used to test fuses, according to technician B.

Thus, both A and B technicians are correct because both might be used to test fuses, according to technician B.

Technician A says both an ohmmeter and a self-powered test light may be used to test for continuity. Technician B says both may be used to test fuses. Who is correct?

Learn more about the continuity here:

brainly.com/question/15025692

#SPJ1

5 0
1 year ago
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