Problem 30 - Single Variable Calculus

Question

Refer to Example \(6 .\) The pressure \(P\) and volume \(V\) of the steam in a steam engine are related by the equation \(P V^{1.2}=k\), where \(k\) is a constant. If the initial pressure of the steam is \(P_{0} \mathrm{lb} / \mathrm{in} .{ }^{2}\) and its initial volume is \(V_{0}\) in. \(^{3}\), find an expression for the work done by the steam as it expands to a volume of four times its initial volume.

Step-by-Step Solution

Answer

The work done by the steam as it expands to a volume of four times its initial volume is given by the expression: \(W = \frac{-P_0 V_0^{1.2}}{0.2} \left( (4V_0)^{-0.2} - V_0^{-0.2} \right)\)

Step 1: Write the pressure-volume relationship

Given the pressure-volume relationship for the steam engine, we have: \(P V^{1.2} = k\)

Step 2: Find an expression for the work done by the steam

The work done by the steam can be found using the equation: \(W = \int_{V_0}^{4V_0} P \, dV\) We need to express the pressure \(P\) in terms of the volume \(V\), using the pressure-volume relationship given in the problem statement. From the relationship we have \(P = \frac{k}{V^{1.2}}\) Now let's substitute this expression for \(P\) into the work equation: \(W = \int_{V_0}^{4V_0} \frac{k}{V^{1.2}} \, dV\)

Step 3: Evaluate the integral

The integral can be evaluated as follows: \(W = k \int_{V_0}^{4V_0} V^{-1.2}\, dV\) To evaluate the integral, use the power rule (add 1 to the exponent and divide by the new exponent): \(W = k \left[\frac{V^{-1.2 + 1}}{-1.2 + 1}\right]_{V_0}^{4V_0}\) Which simplifies to: \(W = k \left[\frac{V^{-0.2}}{-0.2}\right]_{V_0}^{4V_0}\)

Step 4: Substitute the limits of integration and simplify

Plug in the volume limits and evaluate: \(W = k\left( \frac{(4V_0)^{-0.2}}{-0.2} - \frac{V_0^{-0.2}}{-0.2} \right)\) Factor out the common factor \(\frac{-k}{0.2}\): \(W = \frac{-k}{0.2} \left( (4V_0)^{-0.2} - V_0^{-0.2} \right)\)

Step 5: Substitute the initial pressure and volume in terms of the constant

Recall that \(P_0 V_0^{1.2} = k\), so we can write \(k\) in terms of the initial pressure and volume: \(k = P_0 V_0^{1.2}\) Substitute this expression for \(k\) into the equation for the work: \(W = \frac{-P_0 V_0^{1.2}}{0.2} \left( (4V_0)^{-0.2} - V_0^{-0.2} \right)\) This expression represents the work done by the steam as it expands to a volume of four times its initial volume in terms of the initial pressure \(P_0\) and initial volume \(V_0\).

Key Concepts

Integration

Integration is a fundamental concept in calculus, used to compute areas, volumes, and other quantities that accumulate. In essence, integration is the process that allows us to add up infinite tiny pieces or slices of a quantity to get a whole.

For example, when calculating work done by steam in an engine, we use integration to sum up the small pieces of work done as steam expands through its volume change. Mathematically, finding the integral of a function is akin to finding the antiderivative of that function.

• The integral of a constant is just the constant times the variable of integration.
• The power rule of integration tells us how to integrate functions of the form \(x^n\), where \(x\) is the variable and \(n\) is a constant exponent.

By setting up an integral with given limits, such as from initial to final volume, we efficiently calculate variables like work done by steam.

Pressure-Volume Relationship

The pressure-volume relationship in steam engines is a critical formula that ties the pressure and volume of gases or steam together. It helps in understanding how an engine performs work as the steam expands.

The relationship given by \(PV^{1.2} = k\), where \(k\) is a constant, helps derive other useful expressions. This equation suggests that for a given constant \(k\), as volume increases or decreases, the pressure will adjust inversely so that together they maintain this specific relationship.

When steam pressure drops while the volume increases, the product \(PV^{1.2}\) stays the same, allowing us at each stage of expansion to know the precise relationship between pressure and volume. Understanding this dynamic is essential when determining how much work is performed during such processes.

Work Done by Steam

Work done by steam relates to the amount of energy transferred by the steam engine when steam expands. It is a crucial measure in thermodynamics to understand how efficiently a system like a steam engine operates.

We calculate the work performed by integrating pressure over a change in volume: \(W = \int_{V_0}^{4V_0} P \, dV\). By substituting for pressure using our pressure-volume relationship \(P = \frac{k}{V^{1.2}}\), we can perform this integral to find the work done during the steam expansion.

• Relates closely to energy conversion processes in steam engines.
• Measures how much mechanical work can be derived as steam moves and undergoes volume change.
• Helps in understanding engine efficiency and operational parameters.

Power Rule

The power rule is an essential method used in calculus to find the antiderivative of a function in the form of \(x^n\).

When integrating a function, the power rule guides us by telling us to add 1 to the power and then divide the expression by this new exponent. It's an effective rule for simplifying the integration process, especially with polynomial expressions or any expressions where the variable is raised to a power.

In the context of the integration problem in steam work, the power rule helps simplify the integral \(\int V^{-1.2} \, dV\) by transforming it into \([V^{-0.2}/-0.2]_{V_0}^{4V_0}\). This sort of manipulation is foundational when applying calculus to real-world engineering problems.

• Reduces the complexity of solving integrals in polynomial-like forms.
• Provides a quick approach for handling powers of variables in integration tasks.
• Simplifies the understanding and calculation of complex expressions.