Control Volumes

Thermodynamics

Professor Lee Carkner

Lecture 9

PAL # 8 Specific Heat Piston of N2 compressed polytropically

Can find work from polytropic eqn.

W = mR(T2-T1)/(n-1) Final pressure:

P1Vn1=P2Vn

2

P2 = (V1/V2)1.3 (P1) = (21.3)(100) = Final temperature

P1V1/T1 = P2V2/T2

T2 = (P2/P1)(V2/V1)T1 = (246.2/100)(0.5)(300) =

PAL # 8 Specific Heat Work is:

W = (0.8)(0.2968)(369.3-300) / (1-1.3) =

Can get E from cv

Average T = (369+300)/2 = 335 K

U = mu = mcvT U = (0.8)(0.744)(369.3-300) =

Q = 54.8 – 41.2 =

Control Volume

For a control volume, mass can flow in and out Mass flow rate depends on:

Velocity normal to Ac: Vn

Note that:

V = velocity V =volume

Flow

Mass flowing through a pipe

Not easy to solve

Velocity is not

Vavg = V = (1/Ac) ∫ Vn dAc

We can now write:

Volume

The volume flow rate is just:

Related to the mass flow rate just by the

density:

In m3/s

Conservation of Mass

We examine: Mass streams flowing out = mout

So then: m’in – m’out = dmcv/dt

Common Cases For the steady flow case:

m’in = m’out

For systems with just a single stream:

1V1A1 = 2V2A2

For incompressible fluids, 1 = 2

V’1 = V’2

Flow Work

The amount of energy needed to push fluid into a control volume is the flow work:

Can think of it as a property of the fluid itself

Energy

Our previous result for fluid energy was:

We can add the flow work, but note:

We designate the total fluid energy per unit

mass by :

= h + V2/2 + gz Now we don’t have to worry about the flow work

Energy and Mass Flow

The total is:

E’mass = m’ = m’(h + V2/2 +gz)

If V and z are small:

True for most systems

Steady-Flow Systems

Once they are up and running the

properties of the fluid at a given point don’t change with time

q – w = = h + V2/2 + gz Energy balance for a steady flow system per

unit mass

Steady Flow Heat and Work

Heat This is the net heat Q’in-Q’out

Work Boundary work = 0, flow work part of enthalpy Remaining work:

Steady Flow Energies

Enthalpy h is difference between inlet and exit

Kinetic energy

Often very small

Potential energy

10’s or 100’s of meters

Automotive Cooling System

Example: Radiator

Assume cooling fluid is water

Flow velocity about 1 m/s

Maximum height differential about 1 meter gz = (9.8)(1) /(1000) = 0.0098 kJ/kg

Next Time

Read: 5.4-5.5 Homework: Ch 5, P: 17, 21, 36, 75