Modern Engineering Thermodynamics

176 CHAPTER 6: First Law Open System Applications
For an incompressible substance such as a liquid flowing through these systems, Eq. (3.34) gives the specific
enthalpy change as
h in − h out = cðT in − T out Þ + vðp in − p out Þ (6.19)
where c is the specific heat of the material and v is its specific volume. Combining Eq. (6.19) with Eqs. (6.15)
and (6.16), we obtain
V
out
=
n2g nozzle with
c Q _ / _m + cðT in − T out Þ + vðp in − p out Þ o 1/2
(6.20)
incompressible
fluid
and
V out
adiabatic
nozzle with
incompressible
fluid
p
= ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2g c ½cðT in − T out Þ + vðp in − p E ÞŠ
Combining Eq. (6.19) with Eqs. (6.17) and (6.18), we can solve for the diffuser outlet pressure:
(6.21)
p out
diffuser
incompressible
fluid
= p in + ð1/vÞ½cðT in − T out Þ + V 2 in /2g c + _Q / _m Š
and
p out
adiabatic
diffuser
incompressible
fluid
= p in + ð1/vÞ½cðT in − T out Þ + V 2 in /2g cŠ
Foranidealgaswithconstantspecificheats(suchasair at atmospheric pressure and temperature), Eq. (3.42)
gives 4 h in − h out = c p ðT in − T out Þ (6.22)
where c p is the constant pressure specific heat. Then, Eqs. (6.15) and (6.16) become
V out
nozzle with
ideal gas
=
n2g c Q _ / _m + c p ðT in − T out Þ o 1/2
and
and Eqs. (6.17) and (6.18) become:
V out
adiabatic
nozzle with
ideal gas
=
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2g c c p ðT in − T out Þ
T out
diffuser with
ideal gas
= T in + ð1/c p ÞðV 2 in /2g c + _Q / _m Þ
and
T out
adiabatic
diffuser with
ideal gas
= T in + V 2 in /ð2g cc p Þ
4 Note that this formula can be used here even though this is not a constant pressure process, because the enthalpy of an ideal gas
depends only on temperature and is therefore independent of pressure (see Chapter 3).