Problem 3 - University Physics with Modern Physics

Question

A Gasoline Engine. Agasoline engine takes in \(1.61 \times 10^{4} \mathrm{J}\) of heat and delivers 3700 \(\mathrm{J}\) of work per cycle. The heat is obtained by burning gasoline with a heat of combustion of \(4.60 \times 10^{4} \mathrm{J} / \mathrm{g}\) . (a) What is the thermal efficiency? (b) How much heat is discarded in each cycle? (c) What mass of fuel is burned in each cycle? (d) If the engine goes through 60.0 cycles per second, what is its power output in kilowatts? In horsepower?

Step-by-Step Solution

Answer

(a) 22.98% efficiency. (b) 1.24 × 10⁴ J heat discarded. (c) 0.35 g fuel burned. (d) 222 kW, 297.32 hp power.

Step 1: Calculating Thermal Efficiency

The thermal efficiency (\(\eta\)) is calculated using the formula: \[\eta = \frac{\text{work done}}{\text{heat input}}\times 100 = \left(\frac{3700 \, \text{J}}{1.61 \times 10^4 \, \text{J}}\right) \times 100\]Let's perform the calculation:\[\eta = \left(\frac{3700}{16100}\right) \times 100 \approx 22.98\%\]

Step 2: Calculating Heat Discarded

The amount of heat discarded (\(Q_c\)) in each cycle can be found using the energy conservation equation, \[Q_h = W + Q_c\]where \(Q_h = 1.61 \times 10^4 \, \text{J}\) is the input heat and \(W = 3700 \, \text{J}\) is the work done. Solving for \(Q_c\):\[Q_c = Q_h - W = 1.61 \times 10^4 - 3700 = 1.24 \times 10^4 \, \text{J}\]

Step 3: Calculating Mass of Fuel Burned

The heat of combustion is \(4.60 \times 10^4 \, \text{J/g}\). To find the mass of fuel burned \(m\) in each cycle, use the formula:\[Q_h = m \times \text{heat of combustion}\]Solve for \(m\):\[m = \frac{1.61 \times 10^4 \, \text{J}}{4.60 \times 10^4 \, \text{J/g}} \approx 0.35 \, \text{g}\]

Step 4: Calculating Power Output

Power (\(P\)) is the work done per unit time. If the engine goes through 60 cycles per second, calculate the work done per second:\[P = 3700 \, \text{J/cycle} \times 60 \, \text{cycles/s} = 222000 \, \text{J/s}\]Convert the result to kilowatts:\[P = \frac{222000}{1000} = 222 \, \text{kW}\]Convert to horsepower using 1 horsepower = 746 watts:\[P = \frac{222000}{746} \approx 297.32 \, \text{hp}\]

Key Concepts

Thermal Efficiency

Thermal efficiency is a measure of how effectively an engine converts heat from fuel into work. This concept is crucial in thermodynamics as it determines how much of the input energy is transformed into useful mechanical energy that an engine can use. To calculate thermal efficiency, we use the formula:\[\eta = \frac{\text{work done}}{\text{heat input}} \times 100\]In our example, the engine takes in approximately 16,100 J of heat and performs 3,700 J of work. By substituting these values into the formula, we find the thermal efficiency to be about 22.98%. This percentage tells us that around 23% of the heat energy is converted into work, while the rest is lost, mainly as waste heat.

• A higher thermal efficiency means a more effective engine.
• It's typical for internal combustion engines to have efficiencies ranging from 20% to 30%.
• Improving thermal efficiency is key for reducing fuel consumption and emissions.

Heat of Combustion

The heat of combustion is the amount of energy released when a specific amount of fuel is burned completely. This value indicates how much heat energy a fuel can potentially provide. In the problem, gasoline has a heat of combustion of \(4.60 \times 10^4 \, \text{J/g}\). This is a high value, which is why gasoline is a preferred fuel in many applications.Knowing the heat of combustion allows us to determine the mass of fuel burned. For a cycle input heat of \(1.61 \times 10^4 \, \text{J}\), the mass can be calculated:\[m = \frac{1.61 \times 10^4 \, \text{J}}{4.60 \times 10^4 \, \text{J/g}} \approx 0.35 \, \text{g}\]This shows that 0.35 grams of gasoline are required per cycle. Understanding this helps in analyzing the fuel efficiency and consumption rate of the engine.

Power Output

Power output is a crucial concept related to how much work an engine can perform over time. It reflects the engine's ability to produce work continuously, influencing factors like vehicle speed and performance. In our exercise, the engine completes 60 cycles every second, and each cycle provides 3,700 J of work.Computing the power output involves calculating the total work done per second:\[P = \text{Work per cycle} \times \text{Number of cycles per second}\]\[P = 3700 \, \text{J/cycle} \times 60 \, \text{cycles/s} = 222000 \, \text{J/s} \, \text{or} \, 222 \, \text{kW}\]When converted to horsepower (1 hp = 746 watts), this yields about 297.32 hp, a unit often used in automotive specifications. High power output translates to greater capacity to perform work rapidly, a key advantage in many applications.

Energy Conservation

Energy conservation is a fundamental principle in thermodynamics. It states that energy cannot be created or destroyed, only transformed from one form to another. In the context of our exercise, this principle is illustrated by the distribution of input heat energy (\(Q_h\)) into work done (\(W\)) and heat discarded (\(Q_c\)).The relationship is expressed as:\[Q_h = W + Q_c\]In simpler terms:

• All input energy must be accounted for within the system.
• Whatever isn't converted into work (3700 J in this case) is released as waste heat (\(1.24 \times 10^4 \, \text{J}\) here).

This exercise underlines the importance of maximizing the work portion to enhance efficiency. By understanding and applying energy conservation, engineers can design more efficient engines, minimizing wastage and improving performance. This foundational concept is reflected in all engineering and scientific disciplines.