Isentropic process

From Wikipedia, the free encyclopedia

(Redirected from Isentropic flow)
Jump to: navigation, search

In thermodynamics, an isentropic process (iso = "equal" (Greek); entropy = "disorder") is one during which the entropy of the system remains constant.

Contents

Second law of thermodynamics states that,

\delta Q \le TdS

where δQ is the amount of energy the system gains by heating, T is the temperature of the system, and dS is the change in entropy. The equal sign will hold for a reversible process. For a reversible isentropic process, there is no transfer of heat energy and therefore the process is also adiabatic. For an irreversible adiabatic process, the entropy will increase. Hence removal of heat from the system (cooling) is necessary to maintain a constant internal entropy for an irreversible process. Thus an irreversible isentropic process is not adiabatic.

For reversible processes, an isentropic transformation is carried out by thermally "insulating" the system from its surroundings. Temperature is the thermodynamic conjugate variable to entropy, and so the conjugate process would be an isothermal process in which the system is thermally "connected" to a constant-temperature heat bath.

An isentropic flow is a flow that is both adiabatic and reversible. That is, no energy is added to the flow, and no energy losses occur due to friction or dissipative effects. For an isentropic flow of a perfect gas, several relations can be derived to define the pressure, density and temperature along a streamline.

For a closed system, the total change in energy of a system is the sum of the work done and the heat added,

dU = dQ + dW

The work done on a system by changing the pressure is,

dW = − pdV

where p being the pressure and V the specific volume. The change in enthalpy (H = U + pV) is given by,

dH = dU + pdV + Vdp

Since a reversible process is adiabatic (i.e. no heat transfer occurs), so dQ = 0. This leads to two important observations,

dU = − pdV, and
dH = Vdp or dQ = dHVdp = 0

The heat capacity ratio can be written as,

\gamma = \frac{C_p}{C_V} = -\frac{dp/p}{dV/V}

For an ideal gas γ is constant. Hence on integrating the above equation, we get

 pV^{\gamma} = \mbox{constant} \, i.e.
\frac{p_2}{p_1} = \left( \frac{V_1}{V_2} \right)^{\gamma}

Using the equation of state for an ideal gas, pV = RT,

 TV^{\gamma-1} = \mbox{constant} \,
 \frac{p^{\gamma -1}}{T^{\gamma}} = \mbox{constant}



Advanced Search
Included Web Search Engines


Safe Search

close

Top Matching Results

Occasionally Search.com will highlight specialized results that are based on the context of your query. Examples of specialized results include specific links to news, images, or video.

Top Matching Results may highlight information from other Search.com pages, content from the CNET Network of sites, or third party content. The listings are based purely on relevance. Search.com does not receive payment for listings in this section but our partners that provide this data may get paid for listing these products.

Sponsored Links

This section contains paid listings which have been purchased by companies that want to have their sites appear for specific search terms and related content. These listings are administered, sorted and maintained by a third party and are not endorsed by Search.com.

Search Results

Search.com sends your search query to several search engines at one time and integrates the results into one list which has been sorted by relevance using Search.com's proprietary algorithm. You can customize the list of search engines included in your metasearch from the preferences.

The search engines that are used in your metasearch may allow companies to pay to have their Web sites included within the results. To view the Paid Inclusion policy for a specific search engine, please visit their Web site. Search.com does not accept payment or share revenue with any search engine partner for listings in this section.