Recent content by michealyap

oh alright.. thanks.. i able to solve all the questions and understand the theory behind. thanks

i thought Cv can be found in steam table, as it is different temperature.

ok delt U = n CvDeltaT .. so what temperature should we use to for Cv? point 2 or point 3

Question d.. for heat transfer of adibiatic compression, I can use what formula ya.. the cp cv , and k that ?

Question B) am i right got question B? can i take it as dry air? for question C) i use W= nRTln(v2/v1). so i got v1=0.238m^3

pv=mrt. dry air = 28.97 g/m so i take 1000g/28.97g = 34.518 mol .. 560000 [pa] * 0.3 [m^3] = 34.518mol * 8.314 * T. solve the equation i got .. T1 = 312C then using carnot cycle efficiecy 0.6=1−T1/T4. i got 124.8C. For question C) how to do? i know p1v1=p2v2 . we can use w=pdeltaV to get the...

state 1 has the highest temperature. yeah i know t1=t2 and t3=t4. i understand that that i am stuck. we know at point 2 pressure is 5.6bar and volume is 0.3m^3. isothermal means same temperature same pressure as well? W=-pdeltv.

Homework Statement how to know the maximum and minimum temperature for cycle in. Carnot thermal efficiency 60% heat transfer during isothermal expansion 40kj pressure is 5.6bar and volume is 0.3m^3 Homework Equations One kilogram of air as an ideal gas executes a Carnot power cycle having a...

erm.. that is awesome. thank you.. really have a clearer picture.

it did not work so well for the method 1. change of U = Q - Pdelta V . so, 2135.7 kj/kg - 631.68/kj/kg = Q - 4.758 *10^5 * (0.3066-0.001091(m^3/kg)) , therefore .143.857kj/kg = Q , this the Q i obtained.. For second method how to get the heat of vaporiztion?

ok.. since at point 3 all the water vapor has condensed to become water. the specific internal energy is 631.68 kj/kg for saturated liquid. pressure is 4.758bar. what is the delta V at here? as the formula U2-U3 = Q -PdeltaV . Specific internal energy = 2135.7 kj/kg at point 2 Specific...

Q=heat transfer out. how do we know that it is condensed at constant pressure. In the same time, how do we unable to find the internal energy at point 3, which only contain water.

for process between point 2 and 3, 1st law of thermodynamics , the change of internal energy is equal to zero. so Q=W.

Hello, yea i thought the question is asking the whole system first is constant volume , then is condensed isothermally. so the specific workdone is on the process of condensation.

hi, after that i do not know how to proceed. 1st law of thermodynamics change of internal energy = Q- W. for the process of 1 to 2, constant volume so work done on this is equal to zero. therefore internal energy of procee 1 - process 2 = 2597.3 kj/kg - 2135.310 kj/kg = 462kj/kg. heat transfer...