Thermodynamics

222 | Thermodynamicsm in = 50 kgWater∆m bathtub = m in – m out = 20 kgFIGURE 5–5Conservation of mass principle for anordinary bathtub.Conservation of Mass PrincipleThe conservation of mass principle for a control volume can be expressedas: The net mass transfer to or from a control volume during a time intervalt is equal to the net change (increase or decrease) in the total mass withinthe control volume during t. That is,orTotal mass entering Total mass leavinga b athe CV during ¢t the CV during ¢t b a Net change in masswithin the CV during ¢t bm in m out ¢m CV 1kg2(5–8)where m CV m final m initial is the change in the mass of the control volumeduring the process (Fig. 5–5). It can also be expressed in rate form asm # in m # out dm CV >dt1kg>s2(5–9)where ṁ in and ṁ out are the total rates of mass flow into and out of the controlvolume, and dm CV /dt is the time rate of change of mass within the control volumeboundaries. Equations 5–8 and 5–9 are often referred to as the mass balanceand are applicable to any control volume undergoing any kind of process.Consider a control volume of arbitrary shape, as shown in Fig. 5–6. Themass of a differential volume dV within the control volume is dm r dV.The total mass within the control volume at any instant in time t is determinedby integration to beTotal mass within the CV: m CV r dV(5–10)CVThen the time rate of change of the amount of mass within the control volumecan be expressed asdVdmControlvolume (CV)dAControl surface (CS)FIGURE 5–6The differential control volume dV andthe differential control surface dA usedin the derivation of the conservation ofmass relation.→nu→Vdm CVRate of change of mass within the CV: d r dV(5–11)dt dtCVFor the special case of no mass crossing the control surface (i.e., the controlvolume resembles a closed system), the conservation of mass principlereduces to that of a system that can be expressed as dm CV /dt 0. This relationis valid whether the control volume is fixed, moving, or deforming.Now consider mass flow into or out of the control volume through a differentialarea dA on the control surface of a fixed control volume. Let n →bethe outward unit vector of dA normal to dA and V → be the flow velocity at dArelative to a fixed coordinate system, as shown in Fig. 5–6. In general, thevelocity may cross dA at an angle u off the normal of dA, and the mass flowrate is proportional to the normal component of velocity V → n V → cos u rangingfrom a maximum outflow of V → for u 0 (flow is normal to dA) to a minimumof zero for u 90° (flow is tangent to dA) to a maximum inflow of V →for u 180° (flow is normal to dA but in the opposite direction). Making useof the concept of dot product of two vectors, the magnitude of the normalcomponent of velocity can be expressed asNormal component of velocity: V n V cos u S V # Sn (5–12)The mass flow rate through dA is proportional to the fluid density r, normalvelocity V n , and the flow area dA, and can be expressed asDifferential mass flow rate: dm # rV n dA r 1V cos u2 dA r 1V S # S n 2 dA (5–13)