Energy Analysis of Closed Systems Chapter 4

Recall that a closed system does not include mass transfer  Heat can get in or out  Work can get in or out  Matter does not cross the system boundaries

Mechanical Work  There are many kinds of mechanical work  The most important for us will be moving boundary work W b Sometimes called PdV work  The primary form of work involved in automobile engines is moving boundary work produced by piston cylinder devices

The mass of the substance contained within the system boundary causes a force, the pressure times the surface area, to act on the boundary surface and make it move. Boundary Work

The area under the curve is the work

The work depends on the process path

Consider some special cases  Constant volume  Constant pressure  Isothermal for an ideal gas  Polytropic

Constant Volume P V 1 2

Constant Pressure P V 12 W

Isothermal for an ideal gas P V 1 2

Polytropic P V 1 2

n=1 PV 1 = constant equivalent to the isothermal case for an ideal gas PV= mRT P V

Lets go through the polytropic case integration step-wise 2 1

But… And…. WbWb So…

So far, we have not assumed an ideal gas in this derivation, if we do, then….

Energy Balance for a Closed System (Chapter 2) How can energy get in and out of a closed system? Heat and Work Total Energy entering the system Total Energy leaving the system The change in total energy of the system - =

Rate form

E = U + KE + PE 0 0 If the system isn’t moving

In a cyclic process, one where you end up back where you started: You convert heat to work, or vise versa

The net work is the area inside the figure 1 2 Pressure Volume W net If we can calculate the work done for each step in this process, we can find the net work produced or consumed by the system

Calculating Properties Chapter 4b

We know it takes more energy to warm up some materials than others For example, it takes about ten times as much energy to warm up a pound of water, as it does to warm up the same mass of iron.

Specific Heats – Cp and Cv  Also called the heat capacity  Energy required to raise the temperature of a unit mass one degree  Units kJ/(kg 0 C) or kJ/(kg K) cal/(g 0 C) or cal/(g K) Btu/(lb m 0 F) or Btu/(lb m 0 R)

Consider a stationary constant volume system E=U+KE +PE d  dU  U  mC v dT du  C v dT Q-W=ΔU First Law

Consider a stationary constant pressure system  It takes more energy to warm up a constant pressure system, because the system boundaries expand  You need to provide the energy to increase the internal energy do the work required to move the system boundary

Consider a stationary constant pressure system E=U+KE +PE Q=ΔU+PΔV  ΔH  U  mC p dT dh  C p dT Q-W=ΔU Q-PΔV=ΔU dH

Cp is always bigger than Cv h includes the internal energy and the work required to expand the system boundaries

C p and C v are properties  Both are expressed in terms of u or h, and T, which are properties  Because they are properties, they are independent of the process!!  The constant volume or constant pressure process defines how they are measured, but they can be used in lots of applications

Ideal Gases Joule determined that internal energy for an ideal gas is only a function of temperature

Which means that h is also only a function of temperature for ideal gases!! For non ideal gases both h and u vary with the state

Specific Heats vary with temperature – but only with temperature –for an ideal gas Note that the Noble gases have constant specific heats Why is water on this chart?

If C v and C p are functions of temperature – how can we integrate to find  U and  H? Use an average value, and let the heat capacity be a constant

That only works, if the value of heat capacity changes linearly in the range you are interested in. OK Approximation Crummy Approximation Sometimes the best you can do is the room temperature value

What if you need a better approximation? All of these functions have been modeled using the form C p = a + bT + cT 2 + dT 3 The values of the constants are in the appendix of our book – Table A-2c

This is a pain in the neck!! Only do it if you really need to be very accurate!! Isn’t there a better way?

Use the Ideal Gas Tables  Table A-17 pg 910 (air)  Both u and h are only functions of T – not pressure  Relative values have been tabulated for many ideal gases  If the gas isn’t ideal – then it’s a function of both T and P and these tables don’t work!!

C p is modeled in the Appendix as a function of temperature – so you could calculate  h, but what if you want to calculate  u? You’d need C v There is no corresponding C v table !! C p = C v + R

Three Ways to Calculate u

Specific Heat Ratio k does not vary as strongly with temperature as the heat capacity k = 1.4 for diatomic gases (like air) k = for noble gases Use in Chapter 7

Solids and Liquids  Treat as incompressible fluids C p = C v = C

But dv is 0 if the system is incompressible 0 small

Summary  Boundary work  Looked at four different special cases Constant volume Constant pressure Constant temperature Polytropic

Summary  First Law for a closed system

Summary  Defined Constant pressure heat capacity Constant temperature heat capacity By performing a first law analysis of a closed system

Summary  The three ways to calculate changes in Internal energy (u) Enthalpy (h)  Look up properties at state one and two in the tables  Assume constant values of specific heat, then integrate  Use the curve fit equation for specific heat, then integrate