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I was solving a problem on turbine.the steam works on turbine adiabatic reversibly .Is change in enthalpy or is it change in internal energy which equal this work?

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Let a given mass of fluid (e.g. steam) have internal energy $U_1$, volume $V_1$ and pressure $p_1$ as it passes into the turbine and internal energy $U_2$, volume $V_2$ at pressure $p_2$ as it flows out. The external forces pushing the fluid in do work $p_1 V_1$ and the fluid coming out does work $p_2 V_2$ on its environment as this mass of fluid moves through the turbine. (The formulae are simple here because each pressure is constant as the fluid flows). The conservation of energy for this process is $$ U_1 + p_1 V_1 + Q = U_2 + p_2 V_2 + W_s $$ where $Q$ is heat flowing into the fluid and $W_s$ (called shaft work) is work done by the fluid on other things, such as turning the shaft of a turbine. On the left of this equation you see all the energy going into the turbine as a given mass of material flows in. On the right you see all the energy coming out.

The case you want to treat has $Q=0$. You then find that $$ (U_1 + p_1 V_1) - (U_2 + p_2 V_2) = W_s $$ which tells you that the work done to turn the turbine is equal to the drop in enthalpy of the fluid.

(Appendix. Sometimes you see more complicated-looking expressions with things like kinetic energy of the flow, and gravitational potential energy. All of these are included in the internal energy so the above is quite general. Also, if you are not familiar with $p V$ for the work done pushing a flowing fluid along, then focus your thought on unit mass of fluid and imagine a piston pushing it along. At the input end the piston does work $$ \int_{V_1}^0 -p dV = -p_1\int^0_{V_1} dV = p_1 V_1 $$ which we have used above.)

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Internal Energy = Heat - Work, or U = Q - W.

By definition an adiabatic process has constant heat, so when finding the rate of change of each in a process by taking the time derivative:

dU = dQ(0) - dW

Q goes to zero, and U = -W.

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  • $\begingroup$ But the book gives it as enthalpy not internal energy.moreover you are saying q of system is constant but as I know there is nothing like q of system it is given or take and remain in system as it's internal energy $\endgroup$ – user62184 Oct 17 '14 at 11:43
  • $\begingroup$ I was thinking the same but correct answer is enthalpy change $\endgroup$ – user62184 Oct 17 '14 at 11:44
  • $\begingroup$ @Alecg_O your sign convention for W is not consistent. $\endgroup$ – velut luna Jan 19 '16 at 17:10
  • $\begingroup$ @Kyson good catch. I took thermo a couple years ago and haven't done anything more with it since, but I'm curious about the reasoning behind user62184's comments - have any light to shed on the problem? $\endgroup$ – Alecg_O Jan 19 '16 at 21:02
  • $\begingroup$ You can define a boundary for the system so that dU = -dW + energy due to fluid entering - energy due to fluid leaving $\endgroup$ – ignacio Jan 19 '16 at 21:12
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Turbine is a device with steady flow process and it has one inlet and one exit. First law of thermodynamics for a steady flow process with one inlet and one exit is as below $$q-w=(h_e-h_i)+\frac 12\left(V_e^2-V_i^2\right)+g(z_e-z_i)\;\tag 1$$ Potential energy change is negligible for this device. Usually, kinetic energy change isn't negligible but it seems in your case, it has been assumed negligible. As you have mentioned in question title, heat transfer from turbine is usually negligible. So, above equation turns to $$-w=h_e-h_i\;\tag 2$$

If you want a explanation about equation $(1)$, you can see this.

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The actual frst law of thermodyanamics is dQ=dE+dW E here refers to internal energy U + potential energy + kinetic energy W refers to all kinds of work involved in the process..not just PdV W in flow process includes PdV and d(PV) d(PV) is the work required for the flow of fluid PV=mRT U=mCvT Substitue these values in the first law equation KE and PE are neglegible in your case...so dQ=dU+dW(turbine work)+d(PV) dQ=mCvdT + mRdT +dW 0=mdT(Cv+R)+dW 0=mCpdT+dW 0=dH+dW dw=-dH

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