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svp [43]
3 years ago
14

Acetylene gas (C2H2) at 25°C is burned during a steady-flow combustion process with 30 percent excess air at 27°C. It is observe

d that 75,000 kJ of heat is being lost from the combustion chamber to the surroundings per kmol of acetylene. Assuming combustion is complete, determine the exit temperature of the product gases.

Engineering
1 answer:
Hunter-Best [27]3 years ago
7 0

Answer:

Final answer should be TP = 2062.1 K

Explanation:

Assumptions:

Combustion is complete.

Air and combustion gases are ideal gases.

Kinetic and  potential energies are negligible.

There are no work interactions.

Analysis:

The fuel is burned completely with the excess air, and thus the products will contain only CO2,  H2O, N2, and some free O2. Considering 1 kmol of C2H2, the combustion equation can be written as:

(<em><u>the rest of the solution is attached</u></em><em><u>)</u></em>

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Answer:

Fa = 57.32 N

Explanation:

given data

mass = 5 kg

acceleration = 4 m/s²

angular velocity ω = 2 rad/s

solution

first we take here moment about point A that is

∑Ma = Iα +  ∑Mad    ...............1

put here value and we get

so here I = ( \frac{1}{12} ) × m × L²    ................2

I = ( \frac{1}{12} ) × 5 × 0.8²

I = 0.267 kg-m²  

and

a  is =  r × α    

a  = 0.4 α

so now put here value in equation is 1

0 = 0.267 α + m r α (0.4) - m A (0.4)

0 = 0.267 α + 5 (0.4α) (0.4 ) - 5 (4) 0.4

so angular acceleration α = 7.5 rad/s²

so here force acting on x axis will be

∑ F(x) = m a(x)    ..............3

a(x) = m a - m rα

put here value

a(x) = 5 × 4 -  5 × 0.4 × 7.5

a(x) = 5 N

and

force acting on y axis will be

∑ F(y) = m a(y)    .............. 4

a(y) - mg = mrω²

a(y) - 5 × 9.81  = 5 × 0.4 × 2²

a(y) = 57.1 N

so

total force at A will be

Fa = \sqrt{a(x)^2+a(y)^2}    ...............5

Fa = \sqrt{(57.1)^2+(5)^2}  

Fa = 57.32 N

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An inverted tee lintel is made of two 8" x 1/2" steel plates. Calculate the maximum bending stress in tension and compression wh
Kamila [148]

Answer:

hello your question lacks some information attached is the complete question

A) (i)maximum bending stress in tension = 0.287 * 10^6 Ib-in

    (ii) maximum bending stress in compression =  0.7413*10^6 Ib-in

B) (i)  The average shear stress at the neutral axis = 0.7904 *10 ^5 psi

    (ii)  Average shear stress at the web = 18.289 * 10^5 psi

    (iii) Average shear stress at the Flange = 1.143 *10^5 psi

Explanation:

First we calculate the centroid of the section,then we calculate the moment of inertia and maximum moment of the beam( find attached the calculation)

A) Calculate the maximum bending stress in tension and compression

lintel load = 10000 Ib

simple span = 6 ft

( (moment of inertia*Y)/ I ) = MAXIMUM BENDING STRESS

I = 53.54

i) The maximum bending stress (fb) in tension=

= \frac{M_{mm}Y }{I}  = \frac{6.48 * 10^6 * 2.375}{53.54} =  0.287 * 10^6 Ib-in

ii) The maximum bending stress (fb) in compression

= \frac{M_{mm}Y }{I} = \frac{6.48 *10^6*(8.5-2.375)}{53.54} = 0.7413*10^6 Ib-in

B) calculate the average shear stress at the neutral axis and the average shear stresses at the web and the flange

i) The average shear stress at the neutral axis

V = \frac{wL}{2} = \frac{1000*6*12}{2} = 3.6*10^5 Ib

Ay = 8 * 0.5 * (2.375 - 0.5 ) + 0.5 * (2.375 - \frac{0.5}{2} ) * \frac{(2.375 - (\frac{0.5}{2} ))}{2}

= 5.878 in^3

t = VQ / Ib  = ( 3.6*10^5 * 5.878 ) / (53.54 8 0.5) = 0.7904 *10 ^5 psi

ii) Average shear stress at the web ( value gotten from the shear stress at the flange )

t = 1.143 * 10^5 * (8 / 0.5 )  psi

  = 18.289 * 10^5 psi

iii) Average shear stress at the Flange

t = VQ / Ib = \frac{3.6*10^5 * 8*0.5*(2.375*(0.5/2))}{53.54 *0.5}

= 1.143 *10^5

4 0
4 years ago
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