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Triss [41]
2 years ago
10

Why the efficiency of heat engine is always less than 100%​

Physics
2 answers:
valina [46]2 years ago
7 0

Answer:

because some waste heat is always produced produced in a heat engine.

Explanation:

the fraction of heat input that is converted to a net work output is a measure of the performance of the heat engine and is called the thermal efficiency.Since the machine does not contain a source of energy ,nor can it store energy ,from this conversation of energy the power output of machine can never be greater than its input,so efficiency can be less than 100%.

lukranit [14]2 years ago
6 0

Explanation:

The efficiency of heat engine is always less than 100% most of the times because of waste energy in form of heat.

  • Different forms of heating occurs during the working of an engine
  • As the mechanical workings of the engine sets in motion, heat is generated.
  • One of the most common type of the heat is the frictional heating.
  • Therefore, as energy is transformed from one form to another, some are lost as heat.
  • In this regard, the efficiency of the engine suffers.
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Eq-45 what type of boat requires navigation lights?
irina [24]

All the boats operating at night requires Navigation light.

Navigation light helps prevent collisions between boats and see if visibility is poor. The types of boat are : Rowboats, Tug boats, Vessels, Sailboat etc.

Various boats have different lightning color to show its use and side of the boat.

8 0
3 years ago
A heat engine accepts 200,000 Btu of heat from a source at 1500 R and rejects 100,000 Btu of heat to a sink at 600 R. Calculate
diamong [38]

To solve the problem it is necessary to apply the concepts related to the conservation of energy through the heat transferred and the work done, as well as through the calculation of entropy due to heat and temperatra.

By definition we know that the change in entropy is given by

\Delta S = \frac{Q}{T}

Where,

Q = Heat transfer

T = Temperature

On the other hand we know that by conserving energy the work done in a system is equal to the change in heat transferred, that is

W = Q_{source}-Q_{sink}

According to the data given we have to,

Q_{source} = 200000Btu

T_{source} = 1500R

Q_{sink} = 100000Btu

T_{sink} = 600R

PART A) The total change in entropy, would be given by the changes that exist in the source and sink, that is

\Delta S_{sink} = \frac{Q_{sink}}{T_{sink}}

\Delta S_{sink} = \frac{100000}{600}

\Delta S_{sink} = 166.67Btu/R

On the other hand,

\Delta S_{source} = \frac{Q_{source}}{T_{source}}

\Delta S_{source} = \frac{-200000}{1500}

\Delta S_{source} = -133.33Btu/R

The total change of entropy would be,

S = \Delta S_{source}+\Delta S_{sink}

S = -133.33+166.67

S = 33.34Btu/R

Since S\neq   0 the heat engine is not reversible.

PART B)

Work done by heat engine is given by

W=Q_{source}-Q_{sink}

W = 200000-100000

W = 100000 Btu

Therefore the work in the system is 100000Btu

4 0
3 years ago
What is the frequency of a wave that has a speed of 5 m/s and a wavelength of 0.5meter?
Debora [2.8K]

Explanation:

frequency =speed/wavelength

=5/0.5=10Hz

5 0
2 years ago
If 2 objects had the same momentum, what must be true about the mass of the object that traveled the fastest?
julsineya [31]

Yes, the above-given statement is true

<u>Explanation:</u>

  • The product of the mass x the velocity will be the same for both. Momentum is the action of a body with a particular mass through space and there is the conservation of momentum.
  • Momentum is described as the mass of the object multiplied by its velocity.
  • <u>Momentum (p) = Mass (M) * Velocity (v)</u>
  • Therefore for two objects with many masses to have a similar momentum, then the lighter one has to be moving quicker than the heavier object.

4 0
2 years ago
Blood in a carotid artery carrying blood to the head is moving at 0.15 m/s when it reaches a section where plaque has narrowed t
sp2606 [1]

Answer:

26.9 Pa

Explanation:

We can answer this question by using the continuity equation, which states that the volume flow rate of a fluid in a pipe must be constant; mathematically:

A_1 v_1 = A_2 v_2 (1)

where

A_1 is the cross-sectional area of the 1st section of the pipe

A_2 is the cross-sectional area of the 2nd section of the pipe

v_1 is the velocity of the 1st section of the pipe

v_2 is the velocity of the 2nd section of the pipe

In this problem we have:

v_1=0.15 m/s is the velocity of blood in the 1st section

The diameter of the 2nd section is 74% of that of the 1st section, so

d_2=0.74d_1

The cross-sectional area is proportional to the square of the diameter, so:

A_2=(0.74)^2 A_1=0.548 A_1

And solving eq.(1) for v2, we find the final velocity:

v_2=\frac{A_1 v_1}{A_2}=\frac{A_1 (0.15)}{0.548 A_1}=0.274 m/s

Now we can use Bernoulli's equation to find the pressure drop:

p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho v_2^2

where

\rho=1025 kg/m^3 is the blood density

p_1,p_2 are the initial and final pressure

So the pressure drop is:

p_1 - p_2 = \frac{1}{2}\rho (v_2^2-v_1^2)=\frac{1}{2}(1025)(0.274^2-0.15^2)=26.9 Pa

8 0
3 years ago
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