Hi, for the second question (#9), how do we find the specific enthalpy and entropy values? I went to A-5 and found that at state one the temperature is much higher than the saturated temperature so it must be a superheated liquid. But when I go to table A-6, I don't know how to interpolate between 0.1MPa and 0.2MPa because they don't share the same temperature ranges (0.2MPa actually doesn't list 120 deg. Celcius at all).
just replying for anyone whos interested in the future ur probably done with it; find the enthalpy in the table for 0.1MPa using interpolation between temps 100 and 150, then use that value to interpolate again with the value of enthalpy from the 0.2MPa table where its temperature at saturation is almost 120 so its safe to use, (Make the temperature constant (120C) and interpolate using pressures and enthalpy) same thing for entropy
The pressure of 0.15MPa @120C is not in my table A-6 for superheated water.. When I look at the T-s diagram (figure A-9) you can kinda approximate the values (120C @ 1.5bar).. They look to be what he has, but since on a chart its hard to get an exact value.
Can someone explain why on number 8, when we try and find h2, why do we interpolate between 250 c and 300 c ? Because we don't have any given temperatures except for 450 c but that's for state 1 and at 5 MPa or 5000 kPa. And for 1.4 MPa the phase is superheated vapor still, but we don't have any given temperatures except for the t-sat found at 1400 kPa or 1.4 MPa for state 2. So how do we know to use 250 c and 300 c ? EDIT: still need help, but the only connection I see that could make us interpolate between 250 c and 300 c is that in state 1 the t sat is 263.94. but even then that's for state one not state 2. So where is the connection to use those numbers ?
Hey i was having the same problem. But you need to use the fact that the process is isentropic. Such that s1=s2. You then find where the entrolpy lies on the tables for me it was superheated water 1.4MP between T250-300. Given that the s = 6.8210 it lays between the 2 values. You can use an equation to find the enthalpy given you know the upper limits, lower limits and the target value. in basic form it is Y = YL + [X-XL/XU-XL]*[YU-YL] where Y is the what you want, YL is the lower limit of what you want, YU is upper limit. XL is lower limit of what you know X is what you know and XU is the upper limit. so for this problem the equations comes out to be H2 = 2927.9 (Lower limit) + [6.8210 (what we know) - 6.7488 (lower limit) / 6.9553 (upper limit) - 6.7488] * [3040.9 ( Upper limit) - 2927.9 (lower limit) ] = 2967.41 kJ/kg. Hope this helps :)
Quick question, why is the first law written with ΔH sometimes and other times with ΔU? I saw this occurring in previous chapters but my book did not explain it at all; it only states that it is used for a constant pressure process. That being said, This process is NOT constant pressure... How I derived the formula for work: E_in - E_out = dE_sys / dt E_in - E_out = 0 E_in = E_out (m\dot)h_1 = (W\dot)_out + (m\dot)h_2 Work out rate = mass flow rate(h_1 - h_2) ∴ Work_out = h_1 - h_2
I have a video about when to use U and when to use H. In general, for flowing fluids (inlets and outlets) use H because H includes both U and flow work. For closed, non-moving processes, use H when it is a constant pressure process because H includes both U and constant pressure boundary work. For other processes use U.
Thank you for all your video that I find really useful. One thing I don't understand at 12:30: assuming the process were irreversible but still adiabatic, then it wouldn't be isoentropic and s2 would be greater than s1. Looking on the table of superheated steam for p=300 kPa any point with s2 > s1 = s2_rev has also h2 > h2_rev. This lead me to the conclusion that if the process were irreversible h2 would be greater than h2_rev (2845.7 kJ/kg). In turn v2 would be lower than v2_rev from the first equation on conservation of energy that means that v2_rev is NOT the minimum velocity value at the outlet of the diffuser. What am I missing? Can someone help? Thanks
If you write your equation as Q-W, then yes, work out is positive. If you write your equation as Q+W, then work out is negative. It is just a matter of preference. Most thermodynamicists like Q-W and doing it like you said. Most physicists (and me!) like Q+W. Do you see that we would end up the the same equation?
hello doc. can you please give quick explanation about how you found h1 at 11:23 ??? the 150 kPa isn't there in the tables
You probably have to interpolate the values.
@@Johnnyy832 is it double interpolate
That is so well said in the beginning, light bulb!!
at 8:05, how to get h2? .I can't find the values of h2 on my tablele
I also stuck on this part!
Hi, for the second question (#9), how do we find the specific enthalpy and entropy values? I went to A-5 and found that at state one the temperature is much higher than the saturated temperature so it must be a superheated liquid. But when I go to table A-6, I don't know how to interpolate between 0.1MPa and 0.2MPa because they don't share the same temperature ranges (0.2MPa actually doesn't list 120 deg. Celcius at all).
just replying for anyone whos interested in the future ur probably done with it;
find the enthalpy in the table for 0.1MPa using interpolation between temps 100 and 150, then use that value to interpolate again with the value of enthalpy from the 0.2MPa table where its temperature at saturation is almost 120 so its safe to use,
(Make the temperature constant (120C) and interpolate using pressures and enthalpy)
same thing for entropy
@@kk2937 use the 120 almost at saturation did not get the answer
The pressure of 0.15MPa @120C is not in my table A-6 for superheated water.. When I look at the T-s diagram (figure A-9) you can kinda approximate the values (120C @ 1.5bar).. They look to be what he has, but since on a chart its hard to get an exact value.
this is so helpful. Thanks a lot!
Can someone explain why on number 8, when we try and find h2, why do we interpolate between 250 c and 300 c ? Because we don't have any given temperatures except for 450 c but that's for state 1 and at 5 MPa or 5000 kPa. And for 1.4 MPa the phase is superheated vapor still, but we don't have any given temperatures except for the t-sat found at 1400 kPa or 1.4 MPa for state 2. So how do we know to use 250 c and 300 c ?
EDIT: still need help, but the only connection I see that could make us interpolate between 250 c and 300 c is that in state 1 the t sat is 263.94. but even then that's for state one not state 2. So where is the connection to use those numbers ?
Hey i was having the same problem. But you need to use the fact that the process is isentropic. Such that s1=s2. You then find where the entrolpy lies on the tables for me it was superheated water 1.4MP between T250-300. Given that the s = 6.8210 it lays between the 2 values. You can use an equation to find the enthalpy given you know the upper limits, lower limits and the target value. in basic form it is Y = YL + [X-XL/XU-XL]*[YU-YL] where Y is the what you want, YL is the lower limit of what you want, YU is upper limit. XL is lower limit of what you know X is what you know and XU is the upper limit. so for this problem the equations comes out to be H2 = 2927.9 (Lower limit) + [6.8210 (what we know) - 6.7488 (lower limit) / 6.9553 (upper limit) - 6.7488] * [3040.9 ( Upper limit) - 2927.9 (lower limit) ] = 2967.41 kJ/kg. Hope this helps :)
@@Eggobaco thank you!
Thanks. That was really helpful
Quick question, why is the first law written with ΔH sometimes and other times with ΔU?
I saw this occurring in previous chapters but my book did not explain it at all; it only states that it is used for a constant pressure process. That being said, This process is NOT constant pressure...
How I derived the formula for work:
E_in - E_out = dE_sys / dt
E_in - E_out = 0
E_in = E_out
(m\dot)h_1 = (W\dot)_out + (m\dot)h_2
Work out rate = mass flow rate(h_1 - h_2)
∴ Work_out = h_1 - h_2
I have a video about when to use U and when to use H. In general, for flowing fluids (inlets and outlets) use H because H includes both U and flow work. For closed, non-moving processes, use H when it is a constant pressure process because H includes both U and constant pressure boundary work. For other processes use U.
Thank you for all your video that I find really useful. One thing I don't understand at 12:30: assuming the process were irreversible but still adiabatic, then it wouldn't be isoentropic and s2 would be greater than s1. Looking on the table of superheated steam for p=300 kPa any point with s2 > s1 = s2_rev has also h2 > h2_rev. This lead me to the conclusion that if the process were irreversible h2 would be greater than h2_rev (2845.7 kJ/kg). In turn v2 would be lower than v2_rev from the first equation on conservation of energy that means that v2_rev is NOT the minimum velocity value at the outlet of the diffuser. What am I missing? Can someone help? Thanks
5:40 why does it equal to delta h ?
Thank you so much!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Amazing video thank you
Hello, please is there a comprehensive steam table that has temerature in the superheated range.. I am in dire need of one
I dİd not find last question h1 please sir answr my question I have exam tomorrow how could find h1 I look all tables dont find it is superheated ?
Do you ever solve for Entropy change with Cpln(T2/T1) - Rln(P2/P1)??? Or is it further down the list?
do we have to interpolate?
yes i did interpolation to have h2
but one sec...I can't find the values of h1 and s1 on my table
Sameee did u figure it out?
@@ezez9213 State 1 is super heated table A-6. Then it will guide you to P=1.4 MPA and T=450 Celsuis . Then you will find h1 and s1
@@sokundavannchriv3967 how find h2?
ola why do use q+w = deltah is in't q+w = mdot delta h for turbine
make sure as it is for specific h (which means all terms were divided by m and gave us for specific info for all)
my superheated tables skip from 4mpa to 6mpa
This is a turbine, Work OUT of the system is positive, Q is negative
If you write your equation as Q-W, then yes, work out is positive. If you write your equation as Q+W, then work out is negative. It is just a matter of preference. Most thermodynamicists like Q-W and doing it like you said. Most physicists (and me!) like Q+W. Do you see that we would end up the the same equation?