The indicated power developed by an engine with compression ratio of 8, is calculated using an air-standard Otto cycle {constant properties). The rate of heat addition is 10 kW. The ratio of specific heats at constant pressure and constant volume is 1.4. The mechanical efficiency of the engine is 80 percent. The brake power output of the engine is ________ kW (round off to one decimal place)

The correct answer is 4.5.

Step-by-Step Explanation:

First, let's identify the given values from the problem statement:
Compression ratio, r = 8
Rate of heat addition (supplied heat rate), Q_supplied = 10 kW
Ratio of specific heats, γ = 1.4
Mechanical efficiency of the engine, η_mechanical = 80% = 0.8

Step 1: Calculate the thermal efficiency of the air-standard Otto cycle
The thermal efficiency (η₀) of an air-standard Otto cycle is given by the formula:
$${\eta }_0=1-\frac{1}{r^{\gamma -1}}$$
Substituting the given values:
$${\eta }_0=1-\frac{1}{8^{1.4-1}}=1-\frac{1}{8^{0.4}}$$
Calculating the value of 8^0.4:
8^0.4 ≈ 2.2974
Therefore, the thermal efficiency is:
$${\eta }_0=1-\frac{1}{2.2974}\approx 1-0.4353=0.5647$$
This gives a thermal efficiency of 56.47%.

Step 2: Calculate the net work output (Indicated Power)
The indicated power (I.P.) or net work output rate (W_net) of the cycle is calculated using the definition of thermal efficiency:
$${\eta }_0=\frac{I.P.}{Q_{\text{supplied}}}$$
Rearranging the formula to solve for Indicated Power:
$$I.P.={\eta }_0\times Q_{\text{supplied}}$$
$$I.P.=0.5647\times 10\text{kW}=5.647\text{kW}$$

Step 3: Calculate the Brake Power output
The mechanical efficiency (η_mechanical) relates the brake power (B.P.) to the indicated power (I.P.) as follows:
$${\eta }_{\text{mechanical}}=\frac{B.P.}{I.P.}$$
Rearranging the equation to solve for Brake Power:
$$B.P.={\eta }_{\text{mechanical}}\times I.P.$$
$$B.P.=0.8\times 5.647\text{kW}=4.5176\text{kW}$$
Rounding off to one decimal place as requested in the question, we get:
Brake Power ≈ 4.5 kW