CHEMICAL ENGINEERING

FLOW OF FLUIDS — ENERGY AND MOMENTUM RELATIONSHIPS 17

Solution

See Volume 1, Example 2.2.

PROBLEM 2.3

Calculate the energy stored in 1000 cm^3 of gas at 80 MN/m^2 at 290 K using STP as the

datum.

Solution

The key to this solution lies in the fact that the operation involved is an irreversible

expansion. TakingCvas constant betweenT 1 andT 2 ,UDWDnCv

T 2 T 1 where

nis the kmol of gas andT 2 andT 1 are the final and initial temperatures, then for a constant

pressure process, the work done, assuming the ideal gas laws apply, is given by:

WDP 2 

V 2 V 1 DP 2 [

nRT 2 /P 2 

nRT 1 /P 1 ]

Equating these expressions forWgives:Cv

T 2 T 1 DP 2

(

RT 2

P 2



RT 1

P 1

)

In this example:

P 1 D80000 kN/m^2 ,P 2 D 101 .3kN/m^2 ,V 1 D
1 ð 10 ^3 m^3 ,RD 8 .314 kJ/kmol K, and

T 1 D290 K

Hence:Cv

T 2  290 D 101. 3 R[

T 2 / 101. 3  

290 / 80 , 000 ]

By definition,DCp/CvandCpDvCR(from equation 2.27) or:CvDR/

 1

Substituting: T 2 D 174 .15 K.

PVDnRTandnD 

80000 ð 10 ^3 /

 8. 314 ð 290 D 0 .033 kmol

∴ UDWDCvn

T 2 T 1 D 

1. 5 ð 8. 314 ð 0. 033

174. 15  290 D 47 .7kJ

PROBLEM 2.4

Compressed gas is distributed from a works in cylinders which are filled to a pressure

Pby connecting them to a large reservoir of gas which remains at a steady pressureP

and temperatureT. If the small cylinders are initially at a temperatureTand pressureP 0 ,

what is the final temperature of the gas in the cylinders if heat losses can be neglected

and if the compression can be regarded as reversible? Assume that the ideal gas laws are

applicable.

Solution

From equation 2.1, dUDυqυW. For an adiabatic operation,qD0andυqD0and

υWDPdvor dUDPdv. The change in internal energy for any process involving an