Physics 27 First Law of Thermodynamics (9 of 22) Constant Volume (Isovolumetric) (Isochoric)

[@tylerlivingston7445]

pablopenguin, C sub v is the heat capacity of the object. If it's monatomic, it's 3R/2, diatomic is 5R/2. R is the gas constant.

[@Peter_1986]

The term "isochoric" comes from the Greek words ἴσος, which means "equal", and χώρα, which means "space". That name does make sense, however most people are much more familiar with the word "volume", so I guess that this was one reason why they changed that name to "isovolumetric".

[@MichelvanBiezen]

Yes, I prefer the old term as well. Thanks for explaining it. It is very interesting to see the origin of words.

[@historyisthebest5831]

Another way to understand Isochoric process is that there's no change in volume so delta V = 0. And W = P * delta V is therefore 0.

[@ItsThatMilkshake]

Bloody great video mate.

[@PunmasterSTP]

Isochoric? More like "Interesting videos that are lit!" 🔥

[@MichelvanBiezen]

Thank you.

[@johannwegmann4365]

Dear Professor van Biezen, Excellent class ,as ever, but I am strugling with Lenoir cycle,for valved pulsejet, we have stoichiometric adiabatic flame with 2150 K, and LHV=40 MJ/kg,my concern is combustion chamber displacement adiabatic walls,closed valves also exhaust,initial conditions p1=1 bar, T1=300 K,, Thrust Force N= function of Chamber displacement V, empirical, ????,Thankyou Professor,I enjoy all lectures of you. Johann Wegmann, yr in line student of 70 years old.If MJ/kg is changed in mass flow rate of fuel and air mxture,15,1/1, nobody has a theoretical equations to say this eq.satisfies Thrust en Chamber dislacement!!!Regards .Johann Wegmann

[@justinnacu4502]

Thankyou very much! easy to understand.

[@MichelvanBiezen]

Glad it helped!

[@justinnacu4502]

@@MichelvanBiezen do you have lessons about pumps, sir?

[@jacekmichalski9202]

What about liquids pumping ? It looks that we can increase pressure in liquids "for free" i.e. without work spending. Please explain how internal energy will be changed in closed system (a cylinder) completely filled with liquid during pressure increase (piston pressing). Please find that there is no thermal efect during liquids compression.

[@carultch]

Good question. An incompressible liquid is a simplification, and not something that exists in reality. All liquids have a finite bulk modulus, and will compress with enough pressure. There will be work required. As an example, consider 1 liter of water being pressurized from 100 kPa to 1 MPa. Assuming water's bulk modulus of 2.1 GPa is constant, I get -0.9002 kJ of work. In the limit of infinite bulk modulus, I get -0.9 kJ. This is negative, because work is done on the liquid. Accounting for assumed-constant bulk modulus (B), pressure vs volume is given by: P = P0 - B*(V-V0)/V0 Integrating to find work, we get the following formula for compressing from (V0, P0) to Vf: W = (P0 + B)*(Vf - V0) - B/V0*(Vf^2 - V0^2) This might seem like it should be zero, if there is no volume change from V0 to Vf. But, since incompressibility requires B=infinity, we have an indeterminate form, and need to reconcile it, to find out what really happens. Finding an expression for Vf, and substituting, we get: Vf = V0*(1-(P - P0)/B) W = -V0/(2*B) * (Pf^2 - P0^2) - V0*(Pf - P0) In the limit of infinite B for an incompressible liquid, we get the following: W = -V0*(Pf - P0) And this directly relates to the equation of power, for a pump operation: W_dot = -m_dot/rho * (Pf - P0)

[@jacekmichalski9202]

Jacek Michalski

[@jacekmichalski9202]

Thank you very much for the perfect derivation. It is only worth noting that the work obtained (differential) dW = V0 dp is equal to the so-called technical work. Another thing is that this expression is used in the definition of enthalpy, not internal energy. I suppose that this term is a consequence of the equality of molar heats for the constant volume and constant pressure characterizing an incompressible liquid. Please take a look at my papers on Researchgate.

[@carultch]

@@jacekmichalski9202 I took another look at this calculation. It must've just been bad luck how my earlier math came out to match the pump work by enthalpy, that I didn't second guess it. Indeed, just as you expected, infinite bulk modulus should mean zero work, and it does. That is, assuming we pressurize the sample in a closed system. For my example, accounting for water's finite bulk modulus, I get 0.225 Joules (not kJ, but J) to pressurize 1 liter of water from 100 kPa to 1 MPa, in a closed system. This is consistent with the increase in internal energy that I looked up in EES (0.2247 J/kg), which is negligible compared to the increase in enthalpy (0.9014 kJ/kg) that governs pumping power. Since H = U + P*V, we see that pumping power is dominated by the increase in P*V, and the change in U is negligible, since the closed-system work is barely anything at all. The pump continues to pressurize a new kilogram of water each time, and has to make room for the outlet kilogram, thus the enthalpy increase applies instead of the internal energy increase. This explains why open system pumping power is directly proportional to the increase in pressure. Here's the setup: We start with bulk modulus's definition, as a diffEQ: B = -dP/(dV/V) Solve the diffEQ for P as a function of V, assuming B is a constant: P = P0 - B*ln(V/V0) Then, we integrate P dV from V0 to Vf, to find work (which will evaluate as negative): W = Vf*(B*log(V0/Vf) + B + P0) - V0*(B + P0) Revisit equation for P from earlier, and solve for Vf in terms of Pf: Vf = V0*e^(-(Pf-P0)/B) Replace Vf: W = V0*(B + Pf)*e^(-(Pf - P0)/B) - V0*(B + P0) In the limit as B approaches infinity: W approaches 0. The exponential term approaches 1, and each of the remaining terms approach V0*B, which cancel out.

[@traianbasescu3598]

isn't there also a politropic change of state?

[@carultch]

It's polytropic, and yes there is such a process. It isn't so much as one specific kind of process, by rather a general form that encompasses all the processes of the ideal gas, and everything in between. It comes in the form of P*V^n = constant. The n-value is a constant called the polytropic index, which could be any positive real number. For isobaric, n=0 For isothermal, n=1 For adiabatic/isentropic, n=gamma For isochoric, n=infinity

[@xolanimavuso6989]

than you

[@kingplunger1]

in german it is still called isochor :)

[@byTrecz]

Michael van Biezen is so much better than Khan Academy!

[@pablopenguin24]

how do you find C sub v?

[@spicykiller7779]

Sir can u be my teacher plz I will give u a gift

[@BruceWayne-et2xp]

Jk