Thermodynamics Problem
Solutions
Elaboration of the Given Text
1. Kinetic Energy and Speed of a 3220-lb Body
Assuming there are no heat effects and no frictional effects, the given
problem aims to find the kinetic energy and speed of a 3220-lb body after it
falls 778 ft from rest. The solution starts with the steady flow equation,
deleting energy terms that are irrelevant.
Given: - No heat transfer: Q = 0 - Height of fall: h = 778 ft - Mass of the
body: m = 3220 lb - Kinetic energy: KE = ? - Speed: V = ?
Since there are no heat effects and no frictional effects, the potential energy
(PE) is converted to kinetic energy (KE). Therefore, the solution is: PE = KE
V = 224 ft/s
2. Work on Air in a Reciprocating Compressor
A reciprocating compressor draws in 500 cubic feet per minute of air with a
density of 0.079 lb/cu.ft and discharges it with a density of 0.304 lb/cu.ft.
The pressure at the suction is 15 psia, and at the discharge, it is 80 psia. The
increase in the specific internal energy is 33.8 Btu/lb, and the heat
transferred from the air by cooling is 13 Btu/lb. The change in kinetic
energy is neglected.
Solution: - Work on the air: W = -2384.378 Btu/min - Work on the air: W =
56.235 hp
3. Work in a Steam Turbine
Steam enters a turbine with an enthalpy of 1292 Btu/lb and leaves with an
enthalpy of 1098 Btu/lb. The transferred heat is 13 Btu/lb. The problem asks
to find the work in Btu/min and in hp for a flow of 2 lb/sec.
Solution: - Work: W = 21719.994 Btu/min - Work: W = 512.264 hp
4. Work in a Thermodynamic Steady Flow System
A thermodynamic steady flow system receives 4.56 kg per minute of a fluid
where the inlet conditions are: p1 = 137.90 kPa, v1 = 0.0388 m³/kg, V1 =
122 m/s, and u1 = 17.16 kJ/kg. The fluid leaves the system at a boundary
where the conditions are: p2 = 551.6 kPa, v2 = 0.193 m³/kg, V2 = 183 m/s,
and u2 = 52.80 kJ/kg. During the passage through the system, the fluid
receives 3000 J/s of heat.
