Energy Conservation

2. Energy Conservation

a. Integral Form In a steam turbine the term E c.v. must be evaluated for the entire turbine in which the properties are spatially nonuniform. Therefore, the energy varies within the c.v. The methalpy crossing the cs is

r r – ∫ in ρe V dA T ⋅ ,

(69) r r

where ⋅ V dA < 0 for the incoming flow and ⋅ V dA > 0 for the exiting flow. A negative sign is added to the value in order to be consistent with our sign convention for the mass inflow and

outflow. Assuming the c.v. boundary to be rigid, W ˙ d = 0. Therefore,

W ˙ cv = ∫ cv w dV cv ′′′ , (70) where w ′′′ cv denotes the work done per unit volume. The heat crossing the system boundary

Q ˙ cv is given by

Q ˙ cv =− ∫ cv q dA ′′ ⋅ , (71) r

The negative sign associated with Eq. (71) is explained as follows. The vector r r dA is outward normal to the surface, i.e., dA = ndA . The incoming heat flux vector q r ″ enters the surface in a direction that is opposite to the outward normal vector n . Therefore, the dot prod- The negative sign associated with Eq. (71) is explained as follows. The vector r r dA is outward normal to the surface, i.e., dA = ndA . The incoming heat flux vector q r ″ enters the surface in a direction that is opposite to the outward normal vector n . Therefore, the dot prod-

(72 ) The term d/dt( ∫ρedV) in Eq. (72) denotes the rate of change of energy in the entire control vol-

/ d dt ( ∫ cv ρ edV ) = ∫ q ′′ ⋅− dA ∫ w dV ′′′ cv − ∫ ρ T e V dA ⋅ .

ume.

b. Differential Form Applying the Gauss divergence theorem to Eq. (72), and converting the surface inte- gral into a volume integral, we obtain

/ d dt ( ∫ cv ρ edV ) + ∇ ⋅= − ∇ ⋅ ∫ ( ρ T e V dV ) ∫ q dV ′′ − ∫ w dV cv ′′′ .

rr

(73 ) Simplifying the result

r r ∂ρ ()/ e ∂ t + ∇ ⋅= − ∇ ⋅ ( ρ eV T ) Q ˙" − ′′′ w cv .

(74 ) If the heat transfer in the c.v. occurs purely through conduction and the Fourier law applies,

rr Q ˙" =−∇ λ T .

(75 ) Furthermore, if no work is delivered, and the kinetic and potential energies are negligible

(namely, e T = h, and e = u), Eq. (74) may be expressed in the form

r r ∂ρ ()/ u ∂ t +∇⋅=∇⋅ ( ρ Vh ) ( ∇ T ) .

(76 ) Using the relation u = h – P/ ρ, Eq. (76) may be written as

r r ∂ρ ()/ h ∂ t +∇⋅= ( ρ Vh ) ∂ P / ∂ t +∇⋅ ( ∇ T ) .

(77 ) These differential forms of the energy conservation equation are commonly employed

in analyses involving heat transfer, combustion, and fluid mechanics.

c. Deformable Boundary Examples of a deforming boundary include the surface of a balloon while it is being filled, and the leakage of gases past a piston. The above formulations have accounted for de- formation work, but assumed that there is no flow at the deforming boundary. If mass flow occurs at the deforming boundary, the input and exit mass and energy flows will be influenced. Consider the leakage of air past a piston in addition to the mass otherwise entering and leaving r

a cylinder. If the absolute velocity of the leaking fluid adjacent to the piston is r r r r r V and defor- mation velocity is V d , then leakage flow rate will be zero if V = V d . If V > V d , then the leak- age flow rate of fluid past the deforming boundary is

m ˙ = ∫ ρ r ⋅ V dA ,

where the relative velocity V r = V – V d . The RHS of Eq. (34), written in terms of the mass flow rates, can now be expressed in terms of the relative velocity. Furthermore, the integrals

(79) Therefore, the mass and energy conservation equations can be written in the forms

d/dt( ∫ρdV) → ( ∫ ∂ρ ∂t dV /) , and d/dt( ∫ρedV) → ( ( ) / ) ∫ ∂ρ e ∂ t dV .

∫ r ( ∂ρ ∂ /) t dV = ∫ ρ r ⋅ V dA , and ∫ (()/) ∂ρ e ∂ t dV Q = ˙ cv − W ˙ cv + ∫ ρ T e V dA r ⋅ . (80)

Rigorous proof of the formulation of Eqs. (80) is contained in the Appendix to this chapter. If a balloon releases gas (as shown in Figure 18 ) at an absolute velocity of 8 m s –1 and it shrinks at the rate of –2 m s –1 , the magnitude of the relative velocity V

r = 8–(–2) = 10 m s . If the cross-sectional area through which leakage occurs is 1 mm 2 , assuming the density of air

×1×10 to be 1.1 kg m –9 , the mass flow exiting the deforming balloon equals 10 ×1.1 = 1.1x10 –8 kg s –1 .