Hay, this thing works like a Stirling cycle engine. Looks like the same calculations. So instead of turbines pistons can be used. Little tricks here and there and high efficiency can be achieved. Advantage of Stirling being use on lower power applications. Like when there is a source of heat think Brayton or Stirling to generate work.
Based on my own analysis regarding why Cp(dT) is used on turbine and compressor work, Enthalpy in each state is considered. For a cycle which uses heat as a source of energy, enthalpy is used thus, the turbine and compressor work is produced because of heat. We all know in an adiabatic process, enthalpy H=CpT so considering the whole cycle itself (not individually e.g. Compressor only) it is safe to use Wt=Wc=Cp(dT). I still studying so correct me if am mistaken.
You are always helpful in every sub. That I need to clear my concepts. Thank you very much sir I just love the way you teach. All the concepts are cleared in just one watch.
I know Im randomly asking but does any of you know of a way to get back into an Instagram account?? I was dumb forgot my account password. I would appreciate any help you can give me.
For an open system, using second TdS relation(TdS= dH-VdP) , Reversible steady flow work= - integral(VdP) - (change in KE) - (change in PE). This may help to understand why CpdT is used for workdone
In first law of thermodynamics Q=W+∆E, ∆E which is a state function zero for the whole cycle making Q=W. Now for the whole cycle net work done by turbine and compressor (isentropic parts) are equal to net heat absorbed during the cycle (constant pressure parts). That's why he took Cp∆T for Qin and Qout. Hope I could answer you.
In brayton cycle we use open system analysis which means, we use enthalpy difference to calculate the work or heat interactions,moreover the working substance is air which is an ideal gas......so for an ideal gas enthalpy is dependent on Cp... since enthalpy could be written as (cp*dt) he has written so.
@@sagargoswami5385 let me elaborate,by first law of thermodynamics Q=W+∆E,since this is open system there is no change in internal energy,since we are interested in finding Workdone(W) which is equal to Q. As heat transfer is accompanied by change in enthalpy, i said so in the previous comment.
Based on my own analysis regarding why Cp(dT) is used on turbine and compressor work, Enthalpy in each state is considered. For a cycle which uses heat as a source of energy, enthalpy is used thus, the turbine and compressor work is produced because of heat. We all know in an adiabatic process, enthalpy H=CpT so considering the whole cycle itself (not individually e.g. Compressor only) it is safe to use Wt=Wc=Cp(dT). I still studying so correct me if am mistaken.
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Thank you very much. You literally explained a concept which would have taken days for me to understand in 10 minutes.
Hay, this thing works like a Stirling cycle engine. Looks like the same calculations. So instead of turbines pistons can be used. Little tricks here and there and high efficiency can be achieved. Advantage of Stirling being use on lower power applications. Like when there is a source of heat think Brayton or Stirling to generate work.
Based on my own analysis regarding why Cp(dT) is used on turbine and compressor work, Enthalpy in each state is considered. For a cycle which uses heat as a source of energy, enthalpy is used thus, the turbine and compressor work is produced because of heat. We all know in an adiabatic process, enthalpy H=CpT so considering the whole cycle itself (not individually e.g. Compressor only) it is safe to use Wt=Wc=Cp(dT).
I still studying so correct me if am mistaken.
You are always helpful in every sub. That I need to clear my concepts.
Thank you very much sir
I just love the way you teach.
All the concepts are cleared in just one watch.
I know Im randomly asking but does any of you know of a way to get back into an Instagram account??
I was dumb forgot my account password. I would appreciate any help you can give me.
For an open system, using second TdS relation(TdS= dH-VdP) , Reversible steady flow work= - integral(VdP) - (change in KE) - (change in PE). This may help to understand why CpdT is used for workdone
Sir, Work done by the turbine and the compressor are isentropic work and those aren't constant pressure and constant volume processes.
Please explain?
In first law of thermodynamics Q=W+∆E, ∆E which is a state function zero for the whole cycle making Q=W. Now for the whole cycle net work done by turbine and compressor (isentropic parts) are equal to net heat absorbed during the cycle (constant pressure parts). That's why he took Cp∆T for Qin and Qout. Hope I could answer you.
Sir thank you for your simple explanation
Compressors type is Axial , not centrifugal as shown.
why is it not in FHD? I love it but it would be great if the resolutions was higher
Hello Sir, why gas turbine exhaust temperature, was still high, more than 450 Celcius ? limited by what condition
sir you wrote that turbine work out put is cp *temperature difference but it is not constant pressure heat rejection.plz explain
same doubt
In brayton cycle we use open system analysis which means, we use enthalpy difference to calculate the work or heat interactions,moreover the working substance is air which is an ideal gas......so for an ideal gas enthalpy is dependent on Cp... since enthalpy could be written as (cp*dt) he has written so.
@@vigneshwaranrm6277 irrelevant
@@sagargoswami5385 let me elaborate,by first law of thermodynamics Q=W+∆E,since this is open system there is no change in internal energy,since we are interested in finding Workdone(W) which is equal to Q. As heat transfer is accompanied by change in enthalpy, i said so in the previous comment.
@@vigneshwaranrm6277 so we can use efficiency=Qin-Qout/Qin, ... instead of W1-W2/W1,....
Simple n clear....great
How can I identify the efficiency visually using the T-S Graph??
how is Wc=Wt for maximum pressure ratio.? and how to define maximum pressure ratio
discuss the constant pressure gas turbine power plants with neat sketch
good explanation sir, but apne khud me he last last me complex kr diya.
Sir bahot badya teacher hai aap bahot achha padha te ho
But thoda bold and dark likha karen sir.
super explanation.....sir. thank u sir....
Sir why we could not use hot aor direct in compressor from turbine??
great explanation
Sir a great lecture
you are teaching great but it will be more better if you explain details with proper enthusiasm
Sir process ka T2 by T1 kese aaya pressure ke form vo to samja na chahiye khali likha huaa to hamko bhi miljai ga
Have u heard of Phillip Ligrani?
Thank you so much Sir
Sir thoda bada likho pls
tq sir
The professor at UAH
this is wrong but correct...!
Based on my own analysis regarding why Cp(dT) is used on turbine and compressor work, Enthalpy in each state is considered. For a cycle which uses heat as a source of energy, enthalpy is used thus, the turbine and compressor work is produced because of heat. We all know in an adiabatic process, enthalpy H=CpT so considering the whole cycle itself (not individually e.g. Compressor only) it is safe to use Wt=Wc=Cp(dT).
I still studying so correct me if am mistaken.