Question 7.14: Use an energy balance to analyze the steady, fully developed...

As discussed in Example 7.7, the laminar flow velocity profile in the pipe is parabolic (Figure 7.26A), while the turbulent flow velocity profile is nearly uniform (Figure 7.26B). We will analyze each case by assuming that the fully developed, axisymmetric velocity field is given in cylindrical coordinates by v_r = 0, v_θ = 0, and v_z(r), and inserting the appropriate function for laminar and turbulent flow. Consider a CV containing all the fluid in a section of pipe of length L. Applying a mass balance, we find that \dot{M} = ρ_1A_1 \bar V_1 = ρ₂A₂ \bar V_2. Since the density is constant and the areas are the same, this result shows that the average velocity at each port is the same. The steady process energy balance is given in general by Eq. 7.34 as

\int_{CS}\rho \left(u+\frac{p}{\rho } +\frac{1}{2}u  •  u+gz\right)(u  •  n)dS=\dot W_{power} +\dot W_{shaft} +\dot Q_C +\dot S

In this problem there is no fluid or shaft power input or energy addition; and because the pipe is horizontal, the effects of gravity cancel. Thus, the energy balance can be written as

\int_{CS}\rho \left(u+\frac{p}{\rho } \right)(u  •  n)dS+\int_{CS}\rho \left(\frac{1}{2}u • u \right)(u  •  n)dS=\dot Q_C                     (A)

Note how we have written the flux integral in two parts. Since we are told that the internal energy and pressure are uniform on the port surfaces in both the laminar and turbulent flow, we can evaluate the first integral to obtain

\int_{CS}\rho \left(u+\frac{p}{\rho } \right)(u  •  n)dS=\dot M\left[\left(u_2+\frac{p_2}{\rho } \right)-\left(u_1+\frac{p_1}{\rho } \right)\right]

=\dot M\left[\left(u_2-u_1 \right)+\frac{p_2-p_1}{\rho } \right]                      (B)

The value of the second integral in (A) depends in general on the density and the form of the velocity profile at a port. For turbulent flow, the velocity profile may be assumed to be uniform, and the integral defining the kinetic energy flux at the inlet and exit port gives

\int_{CS} ρ\left(\frac{1}{2}u  •  u\right) (u  •  n)dS= \frac{1}{2}ρ_2A_2 \bar V^3 _2 − \frac{1}{2}ρ_1A_1 \bar V^3 _1 =0

Here we have made use of the fact that when the density is constant and the areas are the same, the mass balance shows that the average velocities are also the same. For laminar flow the velocity profile is not uniform but is parabolic. However, since the flow in the pipe is fully developed, the velocity profile is the same at the inlet and exit ports. Furthermore, the density is constant, so we can conclude that for laminar flow the integral is also zero without bothering with the calculation. Thus we have in both cases

\int_{CS} ρ\left(\frac{1}{2}u  •  u\right) (u  •  n)dS=0                                        (C)

Inserting (B) and (C) into the energy balance (A), we find \dot M[(u₂ − u₁) + ( p₂ − p₁)/ρ]= \dot Q_C, which after rearrangement and dividing by the mass flowrate becomes

\frac{p_1-p_2}{\rho }=(u_2-u_1)-\frac{\dot Q_C}{\dot M}                                         (D)

We can interpret this result by noting that the pressure drop down the pipe due to friction results in a loss of pressure potential energy per unit mass in the amount (p₁ − p₂)/ρ. The energy balance shows that this energy loss appears as a combination of an increase in the internal energy of the fluid per unit mass (u₂ − u₁), and a heat transfer per unit mass out of the pipe -\dot Q_C/ \dot M.

Recall that in the case study of Section 3.3.1 (flow in a round pipe), the pressure drop, \Delta p = p₁ − p₂, is given in terms of the friction factor by Eq. 3.15 as \Delta p = ρf (L/D)( \bar V^2 /2). Upon using this to substitute for the pressure drop in the energy balance (D), we find

f\frac{L}{D}\frac{\bar V^2}{2}=(u_2-u_1)-\frac{\dot Q_C}{\dot M}                            (E)

This result can be used to predict the increase in temperature of a liquid flowing in an insulated pipe, i.e., one for which -\dot Q_C = 0, by making use of the fact that the temperature change is related to the change in internal energy by the specific heat: u₂ − u₁ = c(T₂ − T₁). Then according to (E) we have f (L/D)( \bar V^2 /2) = c(T₂ − T₁). Similarly, the temperature change in a low speed gas flow in an insulated pipe could be estimated using the perfect gas relation Eq. 2.21a to obtain u₂ − u₁ = c_V(T₂ − T₁), with the result f (L/D)( \bar V^2 /2) = c_V(T₂ − T₁).