Study Union- Thermodynamics Problems

Ch1-5 Gage press, work sign convention, specifc heat relations cp=cv+R, k=cp/cv (when to use)

1) Air at 150 kPa and 300 K is compressed steadily to 500 kPa and 600 K. The mass flow rate of the air is 0.017 kg/s. A heat loss of 25 kJ/kg occurs during the process. Assume the changes in kinetic and potential energies are negligible. What is the necessary power input to the compressor? (cp=1.028 kJ/kg-K)

Water enters an electrical water heater at a temperature and pressure of 20oC and 1 bar, respectively. The flow rate of water is 25 kg/min and the electrical power to heater is 6.3 MJ/min. What is the exit temperature of the water? Kinetic energy and potential energy changes of the water are negligible.

PolytropicA piston cylinder device contains 0.15 kg of air initially at 2MPa and 350*C. Air is first expanded isothermally to 500kPa, then compressed polytropically with a polytropic exponent of 1.2 to the initial pressure, and finally compressed at the constant pressure to the initial state. Determine boundary work for each process and the net work of the cycle.

Ch6

Ch7 Tds, measurable phenonmenon, Pr2/Pr1=P1/P2 for isentropic, compression ratio Vmax/Vmin

0.5 kg of Nitrogen is compressed in a piston-cylinder from its initial state of 101 kPa and 15oC to its final state of 2.0 MPa and 215oC. Calculate the entropy change of Nitrogen in this process. (cp,N2,avg=1.048 kJ/kg-K , cv,N2,avg=0.748 kJ/kg-K)

Even if temperature change is zero, entropy change need not be zero, as it can be observed

in gases that entropy change can also be due to pressure change.s2 – s1 = cp ln(T2/T1) – R ln(P2/P1).

Even if temperature change is zero and pressure change is zero, phase change occurs due

addition or removal of heat. Entropy is measure of degree of disorderliness which changes in

phase change process. Entropy change of a system can be positive or negative unless the

system in consideration is universe. Entropy generated cannot be negative.

A 2-kg block of steel at 600 K is thrown into a very large water tank with water at a temperature of 300 K. After some time, the steel block reaches thermal equilibrium. Due to the large amount of water in the tank, you may assume the water temperature remains essentially the same even as heat flows from the steel block into the water. The average specific heat of steel is 0.5 kJ/kg.K. a) Calculate the entropy change of the steel block. b) Calculate the entropy change of the water tank. c) And entropy generated during the process.

A) Given, mass of steel block = 2 kg.Initial temperature = 600KFinal temperature = 300KHeat lost by steel block = 2*0.5*(600-300) = 300 kJEntropy change in block = 2*0.5*ln(300/600) = -0.693 kJ/K

B) Heat gained by water = 2*0.5*(600-300) = 300 kJEntropy change in lake water = 300/300 = 1 kJ/K

C) Entropy generated is Sgen = ΔSsource + ΔSsink = -0.693 +1 = 0.31 kJ/K

Ch10

A) Determine pump work b) quality of sat liq-vap mixture c)thermal efficiency

Similar to Ex. 9-6