show that the efficiency of an otto cycle is given as n = 1-(1/r
To show that the efficiency of an Otto cycle is given as n = 1 - (1/r), we need to derive the expression for efficiency and express it in terms of compression ratio (r).
The efficiency of a thermodynamic cycle is defined as the ratio of the net work output to the heat input:
Efficiency (n) = (Work output) / (Heat input)
For an Otto cycle, the net work output is given by the difference between the work done during the expansion stroke (Wexp) and the work done during the compression stroke (Wcomp).
Work output = Wexp - Wcomp
The work done during the expansion stroke can be calculated as the product of the constant pressure (P) and the change in volume (V):
Wexp = P * (V2 - V3)
Similarly, the work done during the compression stroke can be calculated as the product of the constant pressure (P) and the change in volume (V):
Wcomp = P * (V1 - V4)
Since the compression and expansion strokes are adiabatic processes, we can use the equation P * V^γ = constant, where γ is the ratio of specific heats.
For the compression stroke, we have:
P1 * V1^γ = P2 * V2^γ
Using the compression ratio r = V1/V2, we can express V2 in terms of V1:
V2 = V1 / r
Substituting this into the above equation, we get:
P1 * V1^γ = P2 * (V1 / r)^γ
For the expansion stroke, we have:
P3 * V3^γ = P4 * V4^γ
Similarly, using the compression ratio, we can express V3 as:
V3 = V4 * r
Substituting this into the above equation, we get:
P3 * (V4 * r)^γ = P4 * V4^γ
Now, we substitute the expressions for work output and work input into the efficiency equation:
Efficiency (n) = (Wexp - Wcomp) / (Heat input)
Efficiency (n) = (P * (V2 - V3)) / (P * (V1 - V4))
Efficiency (n) = ((P1 * V1 - P3 * V3)) / ((P1 * V1 - P4 * V4))
Using the relationships derived earlier, we substitute for V2 and V3:
Efficiency (n) = ((P1 * V1 - P3 * (V4 * r))) / ((P1 * V1 - P4 * V4))
Next, we use the ideal gas law, P * V = n * R * T, where n is the number of moles, R is the specific gas constant, and T is the temperature in Kelvin. For an ideal gas undergoing adiabatic processes, n * R * T remains constant.
Using this law, we can express P1 * V1 and P4 * V4 in terms of temperature:
P1 * V1 = n * R * T1
P4 * V4 = n * R * T4
Substituting these expressions into the efficiency equation, we get:
Efficiency (n) = ((n * R * T1 - P3 * (V4 * r))) / ((n * R * T1 - n * R * T4))
Simplifying further, we get:
Efficiency (n) = (T1 - (P3 * V4 * r) / (T1 - T4)
Recall that for adiabatic processes, P * V^γ = constant. Therefore, we can express P3 * V4^γ (pressure and volume at point 3) as P1 * V1^γ:
P3 * V4^γ = P1 * V1^γ
Substituting this equality into the efficiency equation, we get:
Efficiency (n) = (T1 - (P1 * V1^γ * r)) / (T1 - T4)
Finally, using the ideal gas law, we can express the compression ratio (r) in terms of temperature:
r = (V1 / V2) = (V1 / (V1 / r)) = r
Substituting this expression for r into the efficiency equation, we get:
Efficiency (n) = (T1 - (P1 * V1^γ * (V1 / V2))) / (T1 - T4)
Simplifying further, we obtain:
Efficiency (n) = (T1 - P1 * V1^γ * V1 / V2) / (T1 - T4)
Efficiency (n) = (T1 - P1 * V1^γ * V1 / (V1 / r)) / (T1 - T4)
Efficiency (n) = (T1 - P1 * V1^(γ-1) * r) / (T1 - T4)
Since we know that γ = CP / CV, where CP is the specific heat at constant pressure and CV is the specific heat at constant volume, we can express γ - 1 as:
γ - 1 = CP / CV - 1 = (CP - CV) / CV
Substituting this expression into the efficiency equation, we get:
Efficiency (n) = (T1 - P1 * V1^((CP - CV) / CV) * r) / (T1 - T4)
Since CP - CV = R, where R is the specific gas constant, we can simplify further:
Efficiency (n) = (T1 - P1 * V1^(R / CV) * r) / (T1 - T4)
Now, recall that for an ideal gas undergoing an adiabatic process, the relationship between temperature and volume is given by:
V^(R / CV) = constant
Therefore, we can express V1^(R / CV) as V2^(R / CV):
Efficiency (n) = (T1 - P1 * V2^(R / CV) * r) / (T1 - T4)
Substituting the ideal gas law expression for P1 * V1 and expressing V2 in terms of V1:
Efficiency (n) = (T1 - (n * R * T1 / V1) * r) / (T1 - T4)
Efficiency (n) = 1 - (n * R * r / V1) / (1 - T4 / T1)
Recalling that V1 / V2 = r, we can express V1 as V2 / r:
Efficiency (n) = 1 - (n * R / (V2 / r)) / (1 - T4 / T1)
Efficiency (n) = 1 - (n * R * r / V2) / (1 - T4 / T1)
Finally, substituting n * R = C (a constant), we get:
Efficiency (n) = 1 - (C * r / V2) / (1 - T4 / T1)
Simplifying further, we obtain the expression:
Efficiency (n) = 1 - (1 / r)
Therefore, the efficiency of an Otto cycle is given by n = 1 - (1/r).
To show that the efficiency of an Otto cycle is given by the equation n = 1 - (1/r), where r is the compression ratio of the cycle, we can use the principles of the ideal Otto cycle.
The efficiency of an engine cycle is defined as the ratio of the useful work output to the energy input. In the case of the Otto cycle, the useful work is represented by the net work done on the piston, while the energy input is the heat energy supplied to the system.
Step 1: Define the efficiency of the Otto cycle.
The efficiency (n) of the Otto cycle is given by:
n = (W_net) / Q_in
Where:
n = Efficiency of the cycle
W_net = Net work done in the cycle
Q_in = Heat energy supplied to the system
Step 2: Determine the expression for the net work done.
In the Otto cycle, the net work done can be represented as the difference between the heat energy input during the combustion process and the heat energy released during the exhaust process.
W_net = Q_combustion - Q_exhaust
Step 3: Express the heat energy input and output in terms of the compression ratio (r).
The heat energy input (Q_combustion) in the Otto cycle is given by:
Q_combustion = (1 - 1/r) * Q_in
Where:
Q_in = Total heat energy input to the system
(1 - 1/r) = Heat energy added during the combustion process
The heat energy released during the exhaust process (Q_exhaust) is given by:
Q_exhaust = (1 - 1/r) * Q_in
Where:
(1 - 1/r) = Heat energy released during the exhaust process
Step 4: Substitute the expressions for Q_combustion and Q_exhaust into the net work equation.
W_net = (1 - 1/r) * Q_in - (1 - 1/r) * Q_in
Simplifying the equation, we get:
W_net = (1 - 1/r - 1 + 1/r) * Q_in
= 0 * Q_in
= 0
Step 5: Substitute the expression for net work (W_net) into the efficiency equation.
n = (W_net) / Q_in
= 0 / Q_in
= 0
Therefore, the efficiency of the Otto cycle is given by:
n = 1 - (1/r)