Recent content by Tethys

Bump! It would be great if I could get some verification on how I worked this problem before I use the same methods to solve the original version (which has different temperature, pressure, and mass flow numbers). I don't mean to complain, but I've had this problem up for four days now, and I...

Part 3 This part is calculated in a similar fashion as the previous two, with the given relevant equation simplifying to: -w = h4 - h3 h4 is calculated from the given quality (90%) and given pressure (15 kPa) rather than T and P from the steam tables. -w = 2362 - 3434.7 = -1072.7...

Part 2 Further simplification of the given equation gives the following relation for Part 2: q + h2 = w + h3 h2 = 230.99 kJ/kg h3 = 3434.7 kJ/kg There is no work in this step, so: q = h3 - h2 = 3434.7 - 230.99 = 3203.7 kJ/kg The heat flow rate can be found using the mass flow rate: q(.m.)...

Part 1: The given requation, when all equivalent terms are cancelled, and irrelivent terms removed simplifies to: -w = h2 - h1 where all terms are per unit mass. From steam tables, h2 = 230.1 kJ/kg h1 = 225.9 kJ/kg So: -w = 230.1 - 225.9 = 4.2 kJ/kg And the work flow rate can be found...

Part 1 From the given information, it appears that this is an adiabatic process as well as an isentropic process. H = U + pV ---> h = u + pv (expressed per unit mass) U = KE + PE = .5mV^2 + mgh Qcv = 0 mi = me KE is constant so (V^2)/2 can be canceled from both sides (?) The potential...

Bump! I could really use some help on this. Any input is appreciated.

Hello! I've been a long-time follower of the Physics Forums and have found it to be a great resource for a variety of physics and engineering topics, although I have not actually felt compelled to create an account on the boards until today. I have encountered a difficult thermodynamics problem...