1、 (1) Example Boundary Work during a Constant-Volume A rigid tank contains air at 500kPa and 150. As a result of heat transfer to the surroundings, the temperature and pressure inside the tank drop to 65 and 400kPa, respectively. Show this process on the p-V diagram. Determine the boundary work done
2、during this process. Solution The boundary work can be determined to be 0d 2 1 VpW This is expected since a rigid tank has a constant volume and dV= 0 in the above equation. Therefore, there is no boundary work done during this process. That is, the boundary work done during a constant-volume proces
3、s is always zero. This is also evident from the p-V diagram of the process (the area under the process curve is zero). (2) Example Cooling of a Hot Fluid in a Tank A rigid tank contains a hot fluid that is cooled while being stirred by a paddle wheel. Initially, the internal energy of the fluid is 8
4、00kJ. During the cooling process, the fluid loses 500kJ of heat, and the paddle wheel does 100kJ of work on the fluid. 1 2 p, kPa 500 400 Determine the final internal energy of the fluid. Neglect the energy stored in the paddle wheel. Solution Take the contents of the tank as the system. This is a c
5、losed system since no mass crosses the boundary during the process. We observe that the volume of a rigid tank is constant and thus there is no boundary work and V2=V1 . Also, heat is lost from the system and shaft work is done in the system. Assumption The tank is stationary and thus the kinetic and potential energy changes are zero. 0 EE PK Therefore, UE and internal energy is the only form of the systems energy that may change during this process. systemoutin EEE 12outinpw, UUUQW 100kJ-500kJ = U2 - 800kJ , U2 = 400kJ Therefore, the final internal energy of the system is 400kJ. (3) (4)
