CHAPTER 4
The First Law of Thermodynamics
65
We can now apply the first law to systems involving working fluids with tabulated property values. Before we apply the first law to systems involving substances such as ideal gases or solids, we will introduce several additional properties that will simplify that task in the next three articles, but first some examples. EXAMPLE 4.2 A 5-hp fan is used in a large room to provide for air circulation. Assuming a well insulated, sealed room, determine the internal energy increase after 1 hour of operation. Solution By assumption, Q = 0. With ΔPE = ΔKE = 0 the first law becomes −W = ΔU.
The work input is W = (−5 hp )(1 hr)(746 W/hp )(3600 s/hr) = −1.343 × 107 J where W = J/s. The negative sign results because the work is input to the system. Finally, the internal energy increase is ΔU = − (−1.343 × 107) = 1.343 × 107 J EXAMPLE 4.3 A frictionless piston is used to provide a constant pressure of 400 kPa in a cylinder containing steam originally at 200°C with a volume of 2m3. Calculate the final temperature if 3500 kJ of heat is added. Solution The first law of thermodynamics, using ΔPE = ΔKE = 0, is Q − W = ΔU. The work done during the motion of the piston is W = ∫ PdV = P (V2 − V1 ) = 400(V2 − V1 ) The mass before and after remains unchanged. Using steam table C.3, this is expressed as m=
2 V1 = = 3.744 kg v1 0.5342