Question 3.1: A fixed control volume has three one-dimensional boundary se...

• System sketch: Figure E3.1 shows two inlet flows, 1 and 2, and a single outlet flow, 3.

• Assumptions: Steady flow, fixed control volume, one-dimensional inlet and exit flows.

• Approach: Apply Eq. (3.17) with energy as the property, where B = E and β = dE/dm = e. Use the one-dimensional flux approximation and then insert the data from the table.

\frac{d}{dt}(B_{syst}) = \int_{CV}{\frac{\partial}{\partial t}(\beta \rho) d^{\circ} \mathcal{V}} +\int_{CS}{\beta \rho(V \cdot n) dA}             (3.17)

• Solution steps: Outlet 3 contributes a positive term, and inlets 1 and 2 are negative. The appropriate form of Eq. (3.12) is shown in Figure 3.1a.

\frac{d}{dt}(B_{syst}) = \frac{d}{dt}\left(\int_{CV}{\beta \rho  d^{\circ} \mathcal{V}}\right) +\int_{CS}{\beta \rho(V \cdot n) dA}             (3.12)

Since the flow is steady, the time-derivative volume integral term is zero. Introducing (ρAV)_i as the mass flow grouping, we obtain

Introducing the numerical values from the table, we have

Thus the system is losing energy at the rate of 0.24 MJ/s = 0.24 MW. Since we have accounted for all fluid energy crossing the boundary, we conclude from the first law that there must be heat loss through the control surface, or the system must be doing work on the environment through some device not shown. Notice that the use of SI units leads to a consistent result in joules per second without any conversion factors. We promised in Chap. 1 that this would be the case.

• Comments: This problem involves energy, but suppose we check the balance of mass also. Then B = mass m, and β = dm/dm = unity. Again the volume integral vanishes for steady flow, and Eq. (3.17) reduces to

Thus the system mass does not change, which correctly expresses the law of conservation of system mass, Eq. (3.1).

m_{syst} = const

or

\frac{dm}{dt} = 0         (3.1)