All categories

3 years ago

Why is the combined cycle power generation system so much more efficient that the straight steam cycle?

Chemistry

1 answer:

rewona [7]3 years ago

8 0

Answer:

Because it uses the residual energy of the fluid used by the first engine.

Explanation:

A combined cycle power generation counts with two heat engines that work in tandem from the same source of heat. The engines turn the energy into mechanical energy.

The cycle is much more efficient than the other, almost 60% more.

I hope this answer helps you.

You might be interested in

Choose the term that best describes the following statement: a system in equilibrium will oppose a change in a way that helps re

Scorpion4ik [409]

Le Chatelier's principle (D)

3 0

3 years ago

Consider the following multistep reaction:

chubhunter [2.5K]

The question is incomplete, here is the complete question:

Consider the following multistep reaction:

C+D⇌CD (fast)

CD+D→CD₂ (slow)

CD₂+D→CD₃ (fast)

C+3D→CD₃

Based on this mechanism, determine the rate law for the overall reaction.

Answer: The rate law for the reaction is $\text{Rate}=k'[C][D]^2$

Explanation:

Rate law is the expression which is used to express the rate of the reaction in terms of the molar concentration of reactants where each term is raised to the power their stoichiometric coefficient respectively from a balanced chemical equation.

In a mechanism of the reaction, the slow step in the mechanism determines the rate of the reaction.

For the given chemical reaction:

$C+3D\rightarrow CD_3$

The intermediate reaction of the mechanism follows:

Step 1: $C+D\rightleftharpoons CD;\text{ (fast)}$

Step 2: $CD+D\rightarrow CD_2;\text{(slow)}$

Step 3: $CD_2+D\rightarrow CD_3;\text{(fast)}$

As, step 2 is the slow step. It is the rate determining step

Rate law for the reaction follows:

$\text{Rate}=k[CD][D]$ ......(1)

As, [CD] is not appearing as a reactant in the overall reaction. So, we apply steady state approximation in it.

Applying steady state approximation for CD from step 1, we get:

$K=\frac{[CD]}{[C][D]}$

$[CD]=K[C][D]$

Putting the value of [CD] in equation 1, we get:

$\text{Rate}=k.K[C][D]^2\\\\\text{Rate}=k'[C][D]^2$

Hence, the rate law for the reaction is $\text{Rate}=k'[C][D]^2$

5 0

3 years ago

How can you tell which compound in a solution is the solvent? (own words please)

Free_Kalibri [48]

Answer:

The way you can know which compound in a solution is a solvent is by usually just determining which substance is a solute and which is a solvent. The solute is when it is dissolved and it takes on the characteristics of the solvent.

Explanation:

Hope this helps!

3 0

3 years ago

Scientists learn about the properties of some particles from the collisions that occur inside a blank

Thepotemich [5.8K]

Answer:

Gases consist of very large numbers of tiny spherical particles that are far apart from one another compared to their size. Gas particles are in constant rapid motion in random directions. Collisions between gas particles and between particles and the container walls are elastic collisions.

6 0

3 years ago

Read 2 more answers

Tin(IV) sulfide, SnS2, a yellow pigment, can be produced using the following reaction.

Maksim231197 [3]

The theoretical yield of $SnS_2$ will be 4.20 grams while the percent yield will be 7.93%

<h3>How is yield calculated?</h3>

From the equation of the reaction, the mole ratio of $SnBr_4$ to $Na_2S$ is 1:2.

Mole of 48.1 mL, 0.478 M $SnBr_4$ = 0.478 x 48.1/100 = 0.023 mols

Mole of 48.8 mL, 0.160 M $Na_2S$ = 0.160 x 48.8/1000 = 0.0078 moles

$SnBr_4$ $Na_2S$ is the limiting reactant.

Mole ratio of $SnBr_4$ and $SnS_2$ = 1:1

Equivalent mole of $SnS_2$ = 0.023 moles

Mass of 0.023 noles $SnS_2$ = 0.023 x 182.81 = 4.20 grams

With 0.0333 g of $SnS_2$ recovered, percent yield = 0.333/4.2 x 100 = 7.93%

More on yields of reactions can be found here: brainly.com/question/17042787

#SPJ1

Tin(IV) sulfide, SnS2, a yellow pigment, can be produced using the following reaction.

SnBr4(aq)+2Na2S(aq)⟶4NaBr(aq)+SnS2(s)

Suppose a student adds 48.1 mL of a 0.478 M solution of SnBr4 to 48.8 mL of a 0.160 M solution of Na2S.

1) Calculate the theoretical yield of SnS2. ;

2) The student recovers 0.333 g of SnS2. Calculate the percent yield of SnS2 that the student obtained.

7 0

2 years ago