IN general view the efficiency formula can be written as follows: n = A*100%/Q. In this formula, the symbol n is used to denote efficiency, the symbol A represents the amount of work done, and Q is the amount of energy expended. It is worth emphasizing that the unit efficiency measurements are interest. Theoretically, the maximum value of this coefficient is 100%, but in practice it is almost impossible to achieve such an indicator, since in the operation of each mechanism there are certain energy losses.
Engine efficiency
Engine internal combustion(ICE), which is one of the key components of the mechanism modern car, is also a variant of a system based on the use of a resource - gasoline or diesel fuel. Therefore, the efficiency value can be calculated for it.
Despite everything technical advances automotive industry, the standard efficiency of internal combustion engines remains quite low: depending on the technologies used in the design of the engine, it can range from 25% to 60%. This is due to the fact that the operation of such an engine is associated with significant energy losses.
So, greatest losses The efficiency of the internal combustion engine is due to the operation of the cooling system, which takes up to 40% of the energy generated by the engine. A significant part of the energy - up to 25% - is lost in the process of exhaust gas removal, that is, it is simply carried away into the atmosphere. Finally, approximately 10% of the energy generated by the engine is spent on overcoming friction between the various parts of the internal combustion engine.
Therefore, technologists and engineers involved in the automotive industry make significant efforts to increasing efficiency engines by reducing losses in all listed items. Thus, the main direction of design developments aimed at reducing losses related to the operation of the cooling system is associated with attempts to reduce the size of the surfaces through which heat transfer occurs. Reduction of losses in the gas exchange process is carried out mainly using a turbocharging system, and reduction of losses associated with friction is carried out through the use of more technologically advanced and modern materials when designing an engine. According to experts, the use of these and other technologies can raise the efficiency of internal combustion engines to 80% and higher.
In reality, the work done with the help of any device is always more useful work, since part of the work is performed against the frictional forces that act inside the mechanism and when moving it individual parts. So, using a movable block, they perform extra work, lifting the block itself and the rope and overcoming the frictional forces in the block.
Let us introduce the following notation: we denote useful work by $A_p$, full time job- $A_(full)$. In this case we have:
Definition
Efficiency factor (efficiency) called the ratio of useful work to complete work. Let us denote the efficiency by the letter $\eta $, then:
\[\eta =\frac(A_p)(A_(poln))\ \left(2\right).\]
Most often, the efficiency is expressed as a percentage, then its definition is the formula:
\[\eta =\frac(A_p)(A_(poln))\cdot 100\%\ \left(2\right).\]
When creating mechanisms, they try to increase their efficiency, but mechanisms with efficiency equal to one(and especially more than one) does not exist.
So, the efficiency factor is physical quantity, which shows the share that useful work makes up of all work produced. Using efficiency, the efficiency of a device (mechanism, system) that converts or transmits energy and performs work is assessed.
To increase the efficiency of mechanisms, you can try to reduce friction in their axes and their mass. If friction can be neglected, the mass of the mechanism is significantly less than the mass, for example, of the load that lifts the mechanism, then the efficiency is slightly less than unity. Then the work done is approximately equal to the useful work:
The golden rule of mechanics
It must be remembered that winning at work cannot be achieved using a simple mechanism.
Let us express each of the works in formula (3) as the product of the corresponding force and the path traveled under the influence of this force, then we transform formula (3) to the form:
Expression (4) shows that using a simple mechanism, we gain in strength as much as we lose in travel. This law called the “golden rule” of mechanics. This rule was formulated in ancient Greece Heron of Alexandria.
This rule does not take into account the work of overcoming friction forces, therefore it is approximate.
Energy transfer efficiency
Efficiency can be defined as the ratio of useful work to the energy expended on its implementation ($Q$):
\[\eta =\frac(A_p)(Q)\cdot 100\%\ \left(5\right).\]
To calculate the efficiency of a heat engine, use the following formula:
\[\eta =\frac(Q_n-Q_(ch))(Q_n)\left(6\right),\]
where $Q_n$ is the amount of heat received from the heater; $Q_(ch)$ - the amount of heat transferred to the refrigerator.
The efficiency of an ideal heat engine that operates according to the Carnot cycle is equal to:
\[\eta =\frac(T_n-T_(ch))(T_n)\left(7\right),\]
where $T_n$ is the heater temperature; $T_(ch)$ - refrigerator temperature.
Examples of efficiency problems
Example 1
Exercise. The crane engine has a power of $N$. In a time interval equal to $\Delta t$, he lifted a load of mass $m$ to a height $h$. What is the efficiency of a crane?\textit()
Solution. The useful work in the problem under consideration is equal to the work of lifting a body to a height $h$ of a load of mass $m$; this is the work of overcoming the force of gravity. It is equal to:
We find the total work done when lifting a load using the definition of power:
Let's use the definition of efficiency to find it:
\[\eta =\frac(A_p)(A_(poln))\cdot 100\%\left(1.3\right).\]
We transform formula (1.3) using expressions (1.1) and (1.2):
\[\eta =\frac(mgh)(N\Delta t)\cdot 100\%.\]
Answer.$\eta =\frac(mgh)(N\Delta t)\cdot 100\%$
Example 2
Exercise. Ideal gas performs a Carnot cycle, while Cycle efficiency equals $\eta $. What is the work in the gas compression cycle at constant temperature? The work done by the gas during expansion is $A_0$
Solution. We define the efficiency of the cycle as:
\[\eta =\frac(A_p)(Q)\left(2.1\right).\]
Let's consider the Carnot cycle and determine in which processes heat is supplied (this will be $Q$).
Since the Carnot cycle consists of two isotherms and two adiabats, we can immediately say that in adiabatic processes(processes 2-3 and 4-1) there is no heat exchange. IN isothermal process 1-2 heat is supplied (Fig. 1 $Q_1$), in an isothermal process 3-4 heat is removed ($Q_2$). It turns out that in expression (2.1) $Q=Q_1$. We know that the amount of heat (the first law of thermodynamics) supplied to the system during an isothermal process goes entirely to doing work by the gas, which means:
The gas performs useful work, which is equal to:
The amount of heat that is removed in the isothermal process 3-4 is equal to the work of compression (the work is negative) (since T=const, then $Q_2=-A_(34)$). As a result we have:
Let us transform formula (2.1) taking into account the results (2.2) - (2.4):
\[\eta =\frac(A_(12)+A_(34))(A_(12))\to A_(12)\eta =A_(12)+A_(34)\to A_(34)=( \eta -1)A_(12)\left(2.4\right).\]
Since by condition $A_(12)=A_0,\ $we finally get:
Answer.$A_(34)=\left(\eta -1\right)A_0$
