Microsoft Word - Cengel and Boles TOC _2-03-05_.doc

Chapter 4 | 175

Use actual data from the experiment

shown here to verify the first law of

thermodynamics.See end-of-chapter

problem 4 –177.

© Ronald Mullisen

Use actual data from the experiment

shown here to verify the first law of

thermodynamics.See end-of-chapter

problem 4 –176.

© Ronald Mullisen

Analysis We take the contents of the cylinder, including the resistance wires,
as the system(Fig. 4–13). This is a closed systemsince no mass crosses the
system boundary during the process. We observe that a piston–cylinder device
typically involves a moving boundary and thus boundary work Wb. The pres-
sure remains constant during the process and thus P 2
P 1. Also, heat is lost
from the system and electrical work Weis done on the system.

(a) This part of the solution involves a general analysis for a closed system
undergoing a quasi-equilibrium constant-pressure process, and thus we con-
sider a general closed system. We take the direction of heat transfer Qto be
to the system and the work Wto be done by the system. We also express the
work as the sum of boundary and other forms of work (such as electrical and
shaft). Then the energy balance can be expressed as

Net energy transfer Change in internal, kinetic,

by heat, work, and mass potential, etc., energies

For a constant-pressure process, the boundary work is given as Wb

P 0 (V 2 V 1 ). Substituting this into the preceding relation gives

However,

Also H
UPV, and thus

(4–18)

which is the desired relation (Fig. 4–14). This equation is very convenient to
use in the analysis of closed systems undergoing a constant-pressure quasi-
equilibrium process since the boundary work is automatically taken care of
by the enthalpy terms, and one no longer needs to determine it separately.

QWother
H 2 H 1 ¬¬ 1 kJ 2

P 0
P 2
P 1 ¬S¬QWother
1 U 2 P 2 V 22  1 U 1 P 1 V 12

QWotherP 01 V 2 V 12 
U 2 U 1

QWotherWb
U 2 U 1

QW
¢U¢KE¢PE

EinEout¬


¬

¢Esystem

Qout = 3.7 kJ

H 2 O

5 min

120 V

0.2 A

2

P, kPa

300

1

P 1 = P 2 = 300 kPa

m = 25 g

Sat. vapor

v

FIGURE 4 –13

Schematic and P-vdiagram for Example 4 –5.

¡

0

¡

0

⎭⎪⎬⎪⎫ ⎭⎪⎬⎪⎫

EXPERIMENT

EXPERIMENT