What components are included in a basic engine block?

Q1. (20 marks) Entropy Analysis of the heat engine: consider a 35% efficient heat engine operating between a large, high- temper

Anvisha [2.4K]

The rate of gain for the high reservoir would be 780 kj/s.

A. η = 35%

$\frac{w}{Q1} = \frac{35}{100}$

W = $1.2*\frac{35}{100}*1000kj/s$

W = 420 kj/s

Q2 = Q1-W

= 1200-420

= 780 kJ/S

<h3>What is the workdone by this engine?</h3>

B. W = 420 kj/s

= 420x1000 w

= 4.2x10⁵W

The work done is 4.2x10⁵W

c. 780/308 - 1200/1000

= 2.532 - 1.2

= 1.332kj

The total enthropy gain is 1.332kj

D. Q1 = 1200

T1 = 1000

$\frac{1200}{1000} =\frac{Q2}{308} \\\\Q2 = 369.6 KJ$

<h3>Cournot efficiency = W/Q1</h3>

= 1200 - 369.6/1200

= 69.2 percent

change in s is zero for the reversible heat engine.

Read more on enthropy here: brainly.com/question/6364271

6 0

2 years ago

In tropical climates, the water near the surface of the ocean remains warm throughout the year as a result of solar energy absor

mote1985 [20]

Answer:

7.07%

Explanation:

Thermal efficiency can be by definition seen as the ratio of the heat utilized by a heat engine to the total heat units in the fuel consumed.

We will determine the thermal efficiency of the given problem at the attached file.

7 0

3 years ago

Applying the Entropy Balance: Closed Systems Five kg of carbon dioxide (CO2) gas undergoes a process in a well-insulated piston–

Mrrafil [7]

Answer:

a) the amount of energy produced in kJ/K is 0.73145 kJ/K

b) the amount of energy produced in kJ/K is 0.68975 kJ/K

The value for entropy production obtained using constant specific heats is approximately 6% higher than the value obtained when accounting explicitly for the variation in specific heats.

Explanation:

Draw the T-s diagram.

a)

$C_p$ = 0.939 kJ/kg.K , m = 5 kg , T₂ = 520 K , T₁ = 280

R = [8.314 kJ / 44.01 kg.K] , P₂ = 20 bar , P₁ = 2 bar

Δs = $m[c_p ln(\frac{T_2}{T_1}) - Rln(\frac{P_2}{P_1})]$

Substitute all parameters in the equation

Δs = $5[(0.939) ln(\frac{520}{280}) - (\frac{8.314}{44.01})ln(\frac{20}{2})]$

Δs = 5 kg × 0.14629 kJ/kg.K

= 0.73145 kJ/K

b)

Δs = $m[\frac{s^0(T_2) - s^0(T_1)}{M} - Rln(\frac{P_2}{P_1})]$

Where T₁ = 280 K , s°(T₁) = 211.376 kJ/kmol.K

T₂ = 520 K , s°(T₂) = 236.575 kJ/kmol.K

R = [8.314 kJ / 44.01 kg.K] , M = 44.01 kg.K , P₂ = 20 bar , P₁ = 2 bar

Δs = $5[\frac{236.575 - 211.376}{44.01} - (\frac{8.314}{44.01})ln(\frac{20}{2})]$

= 5 kg (0.13795 kJ/kg.K)

= 0.68975 kJ/K

The value for entropy production obtained using constant specific heats is approximately 6% higher than the value obtained when accounting explicitly for the variation in specific heats.

7 0

3 years ago

Darin cheers at a football game when his hometown team scores, and his Dad gives him a high five. Darin later begins cheering wi

Kruka [31]

Answer:

generalizing

Explanation:

We all have a generalization system that operates as an autopilot, allowing us to be fast and consistent with our own identity. Thanks to these we package and label all the information with which we are bombed every second, to immediately think and act.

Otherwise, if we pay attention to each data individually, however tiny, every minute of our lives would become an exhausting and extremely slow process of analyzing and digesting, leaving us so overloaded to the point of collapse and not being able to function more mentally.

4 0

2 years ago

A heat engine absorbs 2500 J of heat from a hot reservoir and expels 1000 J to a cold reservoir. When it is run in reverse, with

Alexxandr [17]

To solve this problem, we must simply use the concept of Total Energy transferred both in terms of work and heat. This is basically conjugated in the first law of thermodynamics.

If we take the heat absorbed as positive and the expelled as negative we have that the total work done in the heat engine is

$W_1 = 2500-1000$

$W_1 = 1500J$

For the case of the engine pumps the Energy absorbed is

$W_2 = 1500J$

In this way the ratio between the two would be

$Ratio = \frac{W_1}{W_2} = \frac{1500}{1500} = 1$

So it is reversible, because the state of efficiency of the body is totally efficient.

3 0

3 years ago