150 ENGINEERING THERMODYNAMICS

dharm

/M-therm/Th4-3.pm5

=

F −

HG

I

KJ

= F +

HG

I

KJ

F −

HG

I

KJ

A

S

ppA

S

(^222122) pp pp21 21
222

F +
HG
I
KJ −
A
S
ppS
A
VV
(^221)
2 2 ()^21
or W
=Fpp+ VV
HG
I
KJ −
(^2121)
2 () ...(ix)
= (Mean pressure) × (Change in volume)
W=FH40 45 30. 2 + KI×−(. )2209 45
= 6196 N-cm or 61.96 N-m
It may be noted that work done does not cross the system boundary when spring and
cylinder are considered system.
Now, to find T 2 , using the relation,
pV
T
pV
T
11
1
22
2

∴ T 2 pVTpV^221
11
40 45 2209 293
==30 45
××
×
.. = 1939.3 K
Also, m==RTpv^11 ×××
11 3
30 45
(.0 287 10) 293 = 0.0001605 kg
Now, change in internal energy,
∆ U = m × cv × (T 2 – T 1 )
= 0.0001605 × 0.71 × (1939.3 – 293) = 0.1876 kJ
According to first law,
Q1–2 = ∆ U + W
= 0.1876 + 61.96 × 10–3 = 0.2495 kJ
∴ Amount of heat added to the system = 0.2495 kJ. (Ans.)

4.10. Application of First Law to Steady Flow Process

Steady Flow Energy Equation (S.F.E.E.)

In many practical problems, the rate at which the fluid flows through a machine or piece of

apparatus is constant. This type of flow is called steady flow.

Assumptions :

The following assumptions are made in the system analysis :

(i) The mass flow through the system remains constant.

(ii) Fluid is uniform in composition.

(iii) The only interaction between the system and surroundings are work and heat.

(iv) The state of fluid at any point remains constant with time.

(v) In the analysis only potential, kinetic and flow energies are considered.