You ask, lexaal answers.

[LANS]

me and 3 of my friends independently got the same answer to this question, but all of us have the same wrong answer according to the textbook's solution. We can't figure out what we did wrong. (There were a few questions dealing with entropy, the rest were on Rankine cycles and those I've solved correctly)

A 12-kg steel container contains 0.2-kg superheated water vapor at 1000 kPa and 200-C. Total mass is cooled to an ambient temperature of 30-C. How much heat transfer occurs? Find the steel and water entropy change.

Our answer
Heat transfer: 1434.77 kJ out of the system
Entropy of steel: -2.458 kJ/K
Entropy of water: -1.14 kJ/K

Textbook answer
1353 kJ out
Total entropy change of +0.979 kJ/K

I'll post my full solution later, I need to rewrite it legibly first.

[LANS]

Its not really a question I need answered lexaal, it was more of a response to your "shouldn't you be studying instead?" comment.

Thanks though.

[lexaal]

Typical mistakes for entropy calculations (From 7th to 9th semester i was assistant teacher for 1-3rd semester engineers in thermodynamics)
1) Is there a phase change involved?
2) Did you take c_v and not c_p (volume is constant!)
3) Did you use kelvin temperatures for calculating the entropy integral (1/T dQ) ?
4) Of there is a phase change did you correctly calculate the heat and entropy change?

Currently I am thinking that my job in electrical car engineering will bore me soon and I should switch to chemical eng, where i can do stuff like above all day long.

[LANS]

1) partially, first state is superheated, second state is saturated mixture
2) yes
3) yes
4) dunno, I think so but that's probably where I screwed up.

Well, I found the problem with heat transfer. I was using specific internal energy for water, not specific enthalpy (oops).

I still can't figure out where I screwed up the entropy calculation. I did use C_v and kelvin (for steel at least)
Steel: S = m* C_v * ln (T2/T1) = 0.46 * ln (303/473) = -2.46 kJ/K

For water I used the tables at the back of the textbook. I know initial state is 1000kPa, 200C, and final state is defined by constant specific volume (v_2 = v_1, find v_1 in the tables at 1000kPa, 200C) and T2 = 30C. From that I can find quality and specific entropy for both states, find the difference and get -1.14 kJ/K.

These numbers don't add up to +0.979 kJ/K. Ideas?

[lexaal]

Can the total entropy change be negative?

[LANS]

In this case, no. But it doesn't help me figure out where I screwed up. Oh well, I'll ask the prof in class tomorrow.

[Naboki]

i think you used the the c_p from iron (thats 0.45) (my physics book says so)
another thing: how is the composition from steel? the c_v varries for different compositions

[LANS]

i think you used the the c_p from iron (thats 0.45) (my physics book says so)
another thing: how is the composition from steel? the c_v varries for different compositions

That's exactly the problem, the question doesn't say - it just says steel. The textbook also has one table "Properties of Selected Solids at 25C" which lists the C_p (not C_v) of standard 304-grade steel at 0.46 kJ/kg*K. That's the only mention of C_p pertaining to steel I can find in the whole book, and C_v isn't for steel isn't mentioned anywhere. If I knew what kind of steel the question had in mind, I could look up the properties. If the question gave me the C_v, I could solve the question.

That's not the only issue - my heat values are way off still. The container is rigid, so no work is done, so its heatremoved = \delta(enthalphy)

Here's my current attempt:

Given:
P_1 = 1000kPa
T_1 = 200C
T_2 = 30C
0.2 kg water
12 kg steel
Rigid container - constant volume, no work

Looking in the superheated tables, I find specific volume for initial conditions is v_1 = 0.20596 m^3/kg, specific enthalpy h_1 = 2827.86 kJ/kg
The volume stays constant so I look in the saturated tables for 30C, find specific volumes for that temp
v_liq = 0.001004 m^3/kg v_gas = 32.8922 m^3/kg

Using the lever rule I find quality X_2 = 6.23 * 10^-3

In the same table, for 30C, h_liq = 125.77 kJ/kg, h_gas = 2340.48 kJ/kg

Using above quality, h_2 = 140 kJ/kg

From the first law, q = h_2 - h_1 in this case q = 2687.5 kJ/kg

I know the mass of the water is 0.2kg, so Q = 537.5 kJ

There's nothing I see wrong with this section, that Q value seems reasonable.

Problem is with steel. If I use Q = m*c*ΔT, with m as 12 and ΔT as -170, that c value of 0.46 gives me the wrong answer.

If I solve backwards using the known correct solution Q = 1353 kJ, I find the required C_v to be 0.3997 ~ 0.4.

So I'll just use that 0.4 value, claim I found it in the textbook and get the question right.


This doesn't help with entropy however.

For water, looking in the tables I got the data in the first section from, I find s_1 = 6.69 kJ/kg-K, s_2 = 0.6855 kJ/kg-K.
Thus S = m*Δs = -1.2 kJ/K

Using S = m*c*ln(T_2/T_1) for steel, I get S = -2.13 kJ/K

If the entropy of steel is positive then I get the right answer, but its negative. I'm mixed up somewhere here, just not sure where.

[Devilian]

Could it be that the txt book  is worng?

[lexaal]

Can the total entropy change be negative?

In this case, no.

It is never negative (cause 2nd fundamental law says so)
What you forgot is the entropy exchange with environement.

[Naboki]

i am such a noob
http://www.grc.nasa.gov/WWW/K-12/airplane/specheat.html

[Dangeruss]

this thread makes me feel autistic 

[LANS]

i am such a noob
http://www.grc.nasa.gov/WWW/K-12/airplane/specheat.html

R = 0.287 for the case in this question, so that still means the textbook's C_p value (4.6) is wrong because that would result in C_v or 4.4, when it needs to be 4.0 to get the right answer.

Or I'm just looking in the wrong table (most likely)

[Naboki]

R=8.314472 * J K−1 mol−1

in all cases

[LANS]

R=8.314472 * J K−1 mol−1

in all cases

Right, need to convert R to kJ/K-kg for steel, but its too late now. Assignment was handed in before. Besides, that's not in my work, just a screwup in the last post.