Compressor & Condenser [Thermodynamics]?

May 2021 3 19 Report

Compressor & Condenser [Thermodynamics]?

A common problem in the design of chemical processes is the steady-state compression of gasses from a low pressure P1 to a much higher pressure P2. We can gain some insight about optimal design of this process by considering adiabatic reversible compression of ideal gasses with stagewise intercooling, as shown in the figure below. If the compression is to be done in two stages, first compressing the gas from P1 to P*, then cooling the gas at constant pressure down to the compressor temperature T1, and then compressing the gas to P2, what would the value of the intermediate pressure be to accomplish the compression with minimum work?

Thanks!

Update:

Link to figure:

http://i.imgur.com/bmPtu.jpg

Science & Mathematics / Engineering

gintable

Verified answer

FYI: the middle device ISN'T a "condenser" if it is handling gasses that don't condense. It is better called a heat exchanger for a more general term.

Work input of an adiabatic compression process, first compressor:

Win1 = m_dot*cp*(Tstar - T1)

Work input of an adiabatic compression process, first compressor:

Win2 = m_dot*cp*(T2 - T1).

Net work input:

Winnet = m_dot*cp*((T2 - T1) + (Tstar - T1))

Adiabatic Relation between T1 and Tstar:

Tstar = T1*(Pstar/P1)^((k-1)/k)

Adiabatic Relation between T2 and T1:

T2 = T1*(P2/Pstar)^((k-1)/k)

k is the adiabatic index of the gas flavor

KEEP ALL TEMPERATURES IN KELVIN

To minimize typing, let's define E, the exponent:

E = ((k-1)/k)

Update work equation:

Winnet = m_dot*cp*T1*(((P2/Pstar)^E - 1) + ((Pstar/P1)^E - 1))

Simplify:

Winnet = m_dot*cp*T1*((P2/Pstar)^E + (Pstar/P1)^E - 2)

Express as:

Winnet = m_dot*cp*T1*(P2^E*Pstar^(-E) + Pstar^E/P1^E - 2)

Take derivative of dWinnet/dPstar, as per optimization strategy:

dWinnet/dPstar = m_dot*cp*T1*(-E*P2^E*Pstar^(-E-1) + E*Pstar^(E-1)/P1^E)

Equate to zero, as per optimization strategy:

0 = m_dot*cp*T1*(-E*P2^E*Pstar^(-E-1) + E*Pstar^(E-1)/P1^E)

To hell with the leading factors:

0 = -E*P2^E*Pstar^(-E-1) + E*Pstar^(E-1)/P1^E

Solve for Pstar:

E*P2^E*Pstar^(-E-1) = E*Pstar^(E-1)/P1^E

P2^E*Pstar^(-E-1) = Pstar^(E-1)/P1^E

P2^E/Pstar^(E+1) = Pstar^(E-1)/P1^E

P2^E * P1^E = Pstar^(E-1) * Pstar^(E+1)

Add exponents:

P2^E * P1^E = Pstar^(2*E)

Take the 2Eth root of both sides:

Pstar = (P2^E * P1^E)^(1/(2*E))

Power raised to power means multiply exponents:

Pstar = P2^(E/(2*E)) * P1^(E/(2*E))

Cancel E:

Pstar = P2^(1/2) * P1^(1/2)

Or in otherwords:

Pstar = sqrt(P1*P2)

This is what mathematicians call the "geometric mean".

It turns out, that the optimal midway pressure is the geometric mean of the initial and final pressure.

Interestingly enough, this conclusion is independent of flavor of gas, as long as ideal gas approximation is reasonable.

gatorbait

Theoretically, the minimum power with perfect intercooling and no pressure loss between stages is obtained by making the ratio of compression the same in all stages.

rs = the best compression ratio per stage

rt = overall compression ratio = (Pfinal/Pinitial)

s = the number of stages.

Then rs = s root(rt)

For 2 stages rs = 2nd root(rt)

so if p1 = 14.7 psi and p2 = 114.7 psi, rt = 114.7/14.7 = 7.8027

and rs = (7.8027)^0.5 = 2.8

Stage1 = 14.7 x 2.8 = 41.6 psi and stage 2 = 41.6 x 2.8 = 116.5 psi

For 3 stages rs = 3rd root(rt)

For 4 stages rs = 4th root(rt)