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3 years ago

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A gas turbine receives a mixture having the following molar analysis: 10% CO2, 19% H2O, 71% N2 at 720 K, 0.35 MPa and a volumetr

ic flow rate of 3.2 m3 /s. The mixture exits the turbine at 380 K, 0.11 MPa. For adiabatic operation with negligible kinetic and potential energy effects, determine the power developed at steady state, in kW.

1 answer:

Sliva [168]3 years ago

4 0

Answer:

2074.2 KW

Explanation:

<u>Determine power developed at steady state </u>

First step : Determine mass flow rate ( m )

m / Mmax = ( AV )₁ P₁ / RT₁ -------------------- ( 1 )

<em> where : ( AV )₁ = 8.2 kg/s, P₁ = 0.35 * 10^6 N/m^2, R = 8.314 N.M / kmol , </em>

<em> T₁ = 720 K . </em>

insert values into equation 1

m = 0.1871 kmol/s ( mix )

Next : calculate power developed at steady state ( using ideal gas tables to get the h values of the gases )

W( power developed at steady state )

W = m [ Yco2 ( h1 - h2 )co2

Attached below is the remaining part of the detailed solution

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Consider a single crystal of nickel oriented such that a tensile stress is applied along a [001] direction. If slip occurs on a

Elena L [17]

Answer:

$\mathbf{\tau_c =5.675 \ MPa}$

Explanation:

Given that:

The direction of the applied tensile stress =[001]

direction of the slip plane = [ $\bar 1$ 01]

normal to the slip plane = [111]

Now, the first thing to do is to calculate the angle between the tensile stress and the slip by using the formula:

$cos \lambda = \Big [\dfrac{d_1d_2+e_1e_2+f_1f_2}{\sqrt{(d_1^2+e_1^2+f_1^2)+(d_2^2+e_2^2+f_2^2) }} \Big]$

where;

$[d_1\ e_1 \ f_1]$ = directional indices for tensile stress

$[d_2 \ e_2 \ f_2]$ = slip direction

replacing their values;

i.e $d_1$ = 0 , $e_1$ = 0 $f_1$ = 1 & $d_2$ = -1 , $e_2$ = 0 , $f_2$ = 1

$cos \lambda = \Big [\dfrac{(0\times -1)+(0\times 0) + (1\times 1) }{\sqrt{(0^2+0^2+1^2)+((-1)^2+0^2+1^2) }} \Big]$

$cos \ \lambda = \dfrac{1}{\sqrt{2}}$

Also, to find the angle $\phi$ between the stress [001] & normal slip plane [111]

Then;

$cos \ \phi = \Big [\dfrac{d_1d_3+e_1e_3+f_1f_3}{\sqrt{(d_1^2+e_1^2+f_1^2)+(d_3^2+e_3^2+f_3^2) }} \Big]$

replacing their values;

i.e $d_1$ = 0 , $e_1$ = 0 $f_1$ = 1 & $d_3$ = 1 , $e_3$ = 1 , $f_3$ = 1

$cos \ \phi= \Big [ \dfrac{ (0 \times 1)+(0 \times 1)+(1 \times 1)} {\sqrt {(0^2+0^2+1^2)+(1^2+1^2 +1^2)} } \Big]$

$cos \phi= \dfrac{1} {\sqrt{3} }$

However, the critical resolved SS(shear stress) $\mathbf{\tau_c}$ can be computed using the formula:

$\tau_c = (\sigma )(cos \phi )(cos \lambda)$

where;

applied tensile stress $\sigma =$ 13.9 MPa

∴

$\tau_c =13.9\times ( \dfrac{1}{\sqrt{2}} )( \dfrac{1}{\sqrt{3}})$

$\mathbf{\tau_c =5.675 \ MPa}$

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3 years ago

Compare a series circuit powered by six 1.5-volt batteries to a series circuit powered by a single 9-volt battery. Make sure the

lana [24]

Answer:

Both series circuits provide a total voltage of 9 volts to the two bulbs connected in series and the voltage will be equally divided among two bulbs and they will have same brightness. Therefore, both circuits will have same characteristics.

Explanation:

We are asked to compare two series circuits having equal number of light bulbs.

1st circuit is powered by 6 batteries each having a voltage of 1.5V

2nd circuit is powered by a single battery having a voltage of 9V.

The six batteries in the 1st circuit can be connected together in series or in parallel.

When the batteries are connected in series (positive terminal of one battery connected to negative terminal of another battery) their voltage gets added which means

Voltage of pack = number of batteries*voltage of each battery

Voltage of pack = 6*1.5

Voltage of pack = 9 volts

But the current remains same in the series connection since there is only path for the current to flow.

On the other hand, when the batteries are connected in parallel, the voltage remains same but the current increases.

Circuit 1:

In this circuit, we have 6 batteries each of 1.5 volts connected in series to provide a voltage of 9 volts.

We have connected 2 bulbs in this series circuit.

The voltage will be equally divided between two bulbs if both bulbs are identical in construction.

So there will be 4.5 volts across each bulb and both bulbs will have same brightness.

Circuit 2:

In this circuit, we have 1 battery which provide a voltage of 9 volts.

We have connected 2 bulbs in this series circuit just like in circuit 1.

The voltage will be equally divided between two bulbs if both bulbs are identical in construction.

So there will be 4.5 volts across each bulb and both bulbs will have same brightness.

Conclusion:

Both series circuits provide a total voltage of 9 volts to the two bulbs connected in series and the voltage will be equally divided among two bulbs and they will have same brightness. Therefore, both circuits will have same characteristics.

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The technician is true

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almond37 [142]

Answer:

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MicroNutrients are simply nutrients the body needs but in little amount e.g Minerals.

Explanation:

So for further breakdown:

What are nutrients? Nutrients are essential elements that nourish the body in different capacities. We as humans get most of out nutrients from the food and water we ingest.

Now about Macro Nutrients: From the prefix "Macro" which means large, we can infer that macro nutrients are elements need by the body for the fundamental processes of the body, deficiency in this nutrients are very easy to spot. Examples are: Carbohydrates, Protein, Fats amd Water.

Micro Nutrients: In relation to macro nutrients this are elements that the body needs but are not needed in Large quantities. They mostly work like supporting nutrients. Most chemical activities like reaction that occur in the body are a function of micro nutrients. Defiencies in micrp nutrients may take some time to spot e.g Minerals and Vitamins

In regards to exercise: Macro nutrients are the essential ones here since they are the ones that generate energy. PS: micro nutrients dont generate energy.

In regards to rest: Both the Macro and Micro Nutrients are essentail for the overall well being of the body.

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