CONSERVATION: MASS, MOMENTUM, AND ENERGY BALANCES - Biomass as a Sustainable Energy Source for the Future - page 84

Based on the aforementioned assumptions, the first two terms on the
right-hand side of the equation are zero, and all v i =

and gz i terms are

2

canceled, which gives

φ m , 3
φ m , 1

f 2 380 ð − 318 Þ + 400 g

= =8 : 36

f 4 : 18 308 ð − 293 Þ g

b. When considering the control volume of the condensing stream only, the
energy rate balance at steady state is

φ m , 1 h 1 + v 1 − φ m , 2 h 2 + v 2

0= Q cv − W cv +

2 +g z 1 2 +g z 2

With no work done by this subsystem and changes in KE and PE
neglected, it follows that

Q cv
φ m , 1

Q cv =

kg −1 φ m , 1 h 2 − ð h 1 Þ) =h 2 − ð h 1 Þ = − f 400 + 2 ð 380 − 318 Þ g kJ

kg −1

= 524 kJ

Example 3.4 Energy balance (microscopic)

A cylindrical tube made of copper is used for heating up water to a final temper-
ature of 30 C. The tube has an internal radius, r, of 1.5 cm. Water enters the tube at
an ambient temperature of 20 C. The tube wall is kept at a constant temperature of
40 C. Consider the water density and heat capacity (c p ) to be constant with values of
1000 kg K −1 , respectively. Assume that the heat transfer from

the tube wall is linearly dependent on the temperature difference and the surface
area over which the heat is transferred, with the overall heat transfer coefficient ( h )
having a value of 3450 W

m −3 and 4.18 J g −1

K −1 .
With a required flow of 2232 l

m −2

h −1 , what is the tube length (L)?

Solution
For this problem, one needs to set up a microscopic energy balance over an infin-
itesimally small tube segment. The process is considered to be in steady state. The
energy balance then becomes

0= φ V ρ c p T x − φ V ρ c p T x+ Δ x + h T 0 − T x 2 π r Δ x

Now, T x+ Δ x can be written as a Taylor series, T x+ Δ x = T x + dT

dx ×

Δ x+ , with

Δ x ! 0 subsequent terms tending to zero. Then it follows that

Δ

dT x
dx

0= φ V ρ c p − x+ h T 0 − T x 2 π r Δ x