This is my first ever post, so bear with me. I am a Mechanical engineering student with a pretty good grasp on thermo, but my new teacher is throwing me for a loop, so I come to you good people for some assistance. My question is:
a 0.4m diameter hollow steel sphere with a 3mm wall thickness is filled with water at 1.6 MPa and at 250 C. The system(Steel sphere and water) are cooled to 25 C. What is the entropy change of the system and the amount of entropy generation.
I have used constant specific heats, and what I got isn’t the correct answer. I also have tried using the conservation of energy route to find the heat loss, then plug that in for the general entropy formula. That answer was close to the correct answer but it still was not correct. I plead for help!!
It’s been a while, but I think you need the volume of the water which would be 4/3PI(0.4-.003)^3. From that, you can get the net weight of the water using its density. Then, you can use the specific heat of water to find its internal energy. Then you find the internal energy of the water when its cooled down to 25 C. Then, you find the change in internal energy and the entropy is simply the change in internal energy divided by the initial temperature.
What are the answers you get, and what is the “correct” answer? We generally don’t like to provide full solutions to homework problems on this board, but if you post your work thus far, we might be able to point you in the right direction.
State 1:
Superheated Water
P_1=1.6MPa From Tables: v_1=0.14190 m^3/kg
T_1=250C u_1=2692.9 KJ/kg
s_1=6.6753 KJ/Kg*k
State 2:
2-phase vapor x =(v-v_f)/(v_fg) = 0.00325
v2=v1 (Closed system logic) u_2 = u_f + xu_fg = 112.31 KJ/kg
T_2=25C s_2 = s_f + xs_fg = .39381 KJ/Kg*k
Using the relationship m= V/v, m(h2o)=1.846 kg
m(steel)=row*V=63.657 kg
Conservation of energy simplifies to:
Q_total=m_steel(Delta U_steel)+m_h2o(Delta U_h2o)
Q_total=m_steel(cp(T2-T1)+m_h2o(Delta U_h2o) //Note:This may be a source of error because I am assuming the steel and water are in thermo eq. if not I would have to find T2
Cp_steel=0.235 KJ/KgK
Q_Total = 63.657Kg((0.235KJ/KgK)((25+273)k - (250+273)k)) + (1.846Kg)(112.29 - 2692.9)KJ/Kg)
Q_Total = -3365.863KJ + -4763.80 KJ = -8129.663 KJ //Note: This is a departure from what I had the first time
Well, from there I said the Delta S = Q/T, but my new numbers give me almost double the answer, so I don’t know where to go from here.
By the way, the answers my professor gave us are -4.7 KJ/k for change and S_gen = 1.7 KJ/k
Do you need to take into account the coefficient of thermal expansion for the water and the steel, and figure out what the the size of the steel sphere and the pressure and volume of the water are at 25 C? Or are you just neglecting the steel CTE? Gotta believe that would make a difference, but if you’re not neglecting it, the type of steel will make a difference. So the problem isn’t completely specified.
Oops, that’s an embarrassing error. Yes the inner and outer radii should be cubed not squared. That brings the V(steel) to 0.00149 m^3, pretty much what you calculated the 2nd time.
Where are you getting your Rho(steel) from? It’s higher than I’m finding online, but I don’t have access to my Perry’s atm.
Oh man, is my face red. lol, I keep telling you all that the numbers are fine with the density and the Csp, but I had been using the incorrect numbers. The new numbers are rho_steel = 7,830 kg/m^3 and Csp_Steel = 0.500 KJ/Kg*K, I will let you all know if they make a major difference.
