Thermodynamics

y the normal and shear components of viscous forces on the control surface,and W other is the work done by other forces such as electric, magnetic, andsurface tension, which are insignificant for simple compressible systems andare not considered in this text. We do not consider W viscous either since it isusually small relative to other terms in control volume analysis. But it shouldbe kept in mind that the work done by shear forces as the blades shearthrough the fluid may need to be considered in a refined analysis of turbomachinery.Work Done by Pressure ForcesConsider a gas being compressed in the piston–cylinder device shown in Fig.5–52a. When the piston moves down a differential distance ds under theinfluence of the pressure force PA, where A is the cross-sectional area of thepiston, the boundary work done on the system is dW boundary PA ds. Dividingboth sides of this relation by the differential time interval dt gives thetime rate of boundary work (i.e., power),Chapter 5 | 253Pds AdW # pressure dW # V pistonboundary PAV pistonwhere V piston ds/dt is the piston velocity, which is the velocity of the movingboundary at the piston face.Now consider a material chunk of fluid (a system) of arbitrary shape, whichmoves with the flow and is free to deform under the influence of pressure, asshown in Fig. 5–52b. Pressure always acts inward and normal to the surface,and the pressure force acting on a differential area dA is PdA. Again notingthat work is force times distance and distance traveled per unit time is velocity,the time rate at which work is done by pressure forces on this differential partof the system isdW # pressure P dA V n P dA 1 V S # n S 2(5–50)since the normal component of velocity through the differential area dA isV n V cos u V → · n → . Note that n → is the outer normal of dA, and thus thequantity V → · n →is positive for expansion and negative for compression. Thetotal rate of work done by pressure forces is obtained by integrating dW . pressureover the entire surface A,W # pressure,net out AP 1V S # n S 2 dA APr r 1VS # n S 2 dAIn light of these discussions, the net power transfer can be expressed asW # net,out W # shaft,net out W # pressure,net out W # shaft,net out AP 1V S # n S 2 dA(5–51)(5–52)Then the rate form of the conservation of energy relation for a closed systembecomesQ # net,in W # shaft,net out W # pressure,net out dE sysdt(5–53)To obtain a relation for the conservation of energy for a control volume,we apply the Reynolds transport theorem by replacing the extensive propertyB with total energy E, and its associated intensive property b with totalSystem(gas in cylinder)dV(a)dmSystem(b)dAPu→V→nSystem boundary, AFIGURE 5–52The pressure force acting on (a) themoving boundary of a system in apiston–cylinder device, and (b) thedifferential surface area of a system ofarbitrary shape.