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

Chapter 5 | 241

EXAMPLE 5–9 Mixing of Hot and Cold Waters in a Shower

Consider an ordinary shower where hot water at 140°F is mixed with cold
water at 50°F. If it is desired that a steady stream of warm water at 110°F
be supplied, determine the ratio of the mass flow rates of the hot to cold
water. Assume the heat losses from the mixing chamber to be negligible and
the mixing to take place at a pressure of 20 psia.

Solution In a shower, cold water is mixed with hot water at a specified
temperature. For a specified mixture temperature, the ratio of the mass flow
rates of the hot to cold water is to be determined.
Assumptions 1 This is a steady-flow process since there is no change with
time at any point and thus mCV0 and ECV0. 2 The kinetic and
potential energies are negligible, ke pe 0. 3 Heat losses from the system
are negligible and thus Q

.
0. 4 There is no work interaction involved.
Analysis We take the mixing chamberas the system (Fig. 5–33). This is a
control volumesince mass crosses the system boundary during the process.
We observe that there are two inlets and one exit.
Under the stated assumptions and observations, the mass and energy bal-
ances for this steady-flow system can be expressed in the rate form as follows:

Mass balance:

Energy balance:

Rate of net energy transfer Rate of change in internal, kinetic,

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

Combining the mass and energy balances,

Dividing this equation by m. 2 yields

where ym. 1 /m. 2 is the desired mass flow rate ratio.
The saturation temperature of water at 20 psia is 227.92°F. Since the tem-
peratures of all three streams are below this value (TTsat), the water in all
three streams exists as a compressed liquid (Fig. 5–34). A compressed liquid
can be approximated as a saturated liquid at the given temperature. Thus,

Solving for yand substituting yields

Discussion Note that the mass flow rate of the hot water must be twice the
mass flow rate of the cold water for the mixture to leave at 110°F.

y

h 3 
h 2

h 1 
h 3



78.02
18.07

107.99
78.02

2.0

h 3 hf @ 110°F78.02 Btu/lbm

h 2 hf @ 50°F18.07 Btu/lbm

h 1 hf @ 140°F107.99 Btu/lbm

yh 1 h 2 (y1)h 3

m

#

1 h 1 m

#

2 h 2 (m

#

1 m

#

2 )h 3

m# 1 h 1 m# 2 h 2 m# 3 h 3 1 since Q

#

0, W

#

0, kepe 02

E

#

inE

#

out

E

#

in
E

#

out^ ^ dEsystem>dt^ ^0

m#in m#outSm# 1 m# 2 m# 3

m

#

in
m

#

out^ dmsystem>dt^ ^0

0 (steady)

¡

0 (steady)

⎭⎪⎪⎬⎪⎪⎫ ⎭⎪⎪⎪⎬¡⎪⎪⎪⎫

FIGURE 5–33

Schematic for Example 5–9.

Compressed

liquid states

P = const.

T

Tsat

v

FIGURE 5–34

A substance exists as a compressed

liquid at temperatures below the

saturation temperatures at the given

pressure.

T 1 = 140°F

T 2 = 50°F T 3 = 110°F

m· 2 m· 3

m· 1

P = 20 psia

Mixing

chamber