EGR 334 Thermodynamics Chapter 6: Sections 11-13 Paper Topics Due Lecture 27: Isentropic Processes and Isentropic Efficiencies Quiz Today?

Today’s main concepts: Understand definition and how to use isentropic turbine efficiency Understand definition and how to use isentropic nozzle efficiency Understand definition and how to use Compressor and Pump Isentropic efficiencies Introduce equation to calculate work of an isentropic process for a control volume. Reading Assignment: Read Chapter 8, Sections 1-2 Homework Assignment: Problems from Chap 6: 114, 124, 151,164

Review: For a Control Volume: Mass Rate Balance: Energy Rate Balance: New: For a Control Volume: Entropy Rate Balance:

1st Law Efficiency: Qin : Heat in from hot body Back in Chapter 2, you were introduced to efficiency of the power cycle. Qin : Heat in from hot body Qout : Heat out to cold body Wcycle = Qcycle = Qin - Qout Power Cycle Efficiency This efficiency is a 1st Law of Thermodynamics concept. It represents a description of how much useable work was extracted from the system knowing how much heat/energy was put into the system and how much was dumped to the lower temperature reservoir. The 1st Law efficiency, implies you cannot get more work out than the amount of heat put in.

Isentropic Efficiencies 6.12 : Isentropic Device Efficiency 2nd Law Efficiency: The requirements of the 2nd Law mean that the ability to get useable work out a system is generally much less then the amount limited by the 1st Law. It implies that the maximum ability to get usable work out from a cycle, depends upon the difference in the Hot and Cold reservoirs associated with heat transfer. Not all the heat flow can be successfully converted to useful work, no matter how well designed the system may be. Isentropic Efficiencies In Section 6.12, Isentropic Efficiencies are introduced which compare the actual behavior of some thermodynamics devices to the very best performance possible as defined by the 2nd Law. Instead of Cycle Efficiencies, these will be Component Efficiencies.

Isentropic Device Efficiencies: 6.12 : Isentropic Device Efficiency Isentropic Device Efficiencies: The following devices have defined isentropic efficiencies: Turbine: Nozzle: Compressor: Pump:

For a turbine, the energy balance is: 6.12 : Isentropic Device Efficiency Consider the Turbine. To make any process more efficient, we want to eliminate entropy creation. We would like to operate as close to isentropic as possible. WCV . min mexit For a turbine, the energy balance is: For a turbine, the entropy balance is: The maximum theoretical 1st Law efficiency, η, is 1.0 Which equals 1 only when Wcycle = Qin

6.12 : Isentropic Device Efficiency A 2nd Law efficiency for the turbine might compare the compare the actual Work out to the most possible Work Out. Write two energy balances: - actual - isentropic

Isentropic Turbine Efficiency The isentropic turbine efficiency is the ratio of the actual turbine work to the maximum theoretical work, each per unit of mass flowing:

Example 1: Water vapor enters a turbine at p1 = 5 bar, T1 = 320oC and exits at p2 = 1 bar. The work developed is 271 kJ/kg. Determine the isentropic turbine efficiency and final enthalpy. 1) Determine state properties: 2 From Table A-4: State 1: Given p1 = 5 bar, T1 = 320oC then h1 = 3105.6 kJ/kg and s1 = 7.5308 kJ/kg. 1 State 2: Given p2 = 1 bar and s2s = s1= 7.5308 kJ/kg then h2s = 2743.0 kJ/kg 2) Calculate Isentropic Efficiency: 3) Calculate h2:

Generally, for a nozzle, the energy balance reduces to 6.12 : Isentropic Device Efficiency Consider a Nozzle: Vi Ve Generally, for a nozzle, the energy balance reduces to then for the actual nozzle: and for an ideal nozzle: The 2nd Law nozzle efficiency is:

Isentropic Compressor and Pump Efficiencies Consider the Energy Balance for a pump or compressor: 1 2 If the change in kinetic energy of flowing matter is negligible, ½(V12 – V22) drops out. If the change in potential energy of flowing matter is negligible, g(z1 – z2) drops out. If the heat transfer with surroundings is negligible, drops out. then

Isentropic Compressor and Pump Efficiencies The isentropic compressor efficiency is the ratio of the minimum theoretical work input to the actual work input, each per unit of mass flowing: Isentropic pump efficiency is defined similarly.

Example 2: Refrigerant 134A at 20 psi, 0°F enters a compressor operating at steady state with a mass flow rate of 250 lbm/hr and exits at 140 psi, 200°F. Heat transfer occurs from the compressor to its surroundings, which are at 70°F. Changes in KE and PE can be ignored. The power input is claimed to be 2 HP. Determine whether this claim can be correct. 2 1 Step 1: Determine State Properties (Tables A10E-A12E) State (134a) 1 2 T (°F) 200 p (psi) 20 140 h (BTU/lb) s (BTU/lb·°R) State (134a) 1 2 T (°F) 220 p (psi) 20 140 h (BTU/lb) 101.88 146.18 s (BTU/lb·°R) 0.2238 0.2667 State 1: Given p1 = 20 psi and T1 = 0 deg F. then h1 = 101.88 Btu/lbm s1= 0.2238 Btu/lbm-R State 2: Given p2 = 140 psi and T1 = 200 deg F. then h2 = 146.18 Btu/lbm s1= 0.2667 Btu/lbm-R

Reduces to: Example 2 continued: … Heat transfer occurs from the compressor to its surroundings, which are at 70°F. Changes in KE and PE can be ignored. The power input is claimed to be 2 horsepower, Determine whether this claim can be correct. State (134a) 1 2 T (°F) 220 P (psi) 20 140 h (BTU/lb) 101.88 146.18 s °(BTU/lb·°R) 0.2238 0.2667 Check out the claim based on 1st Law requirements: Reduces to: Heat flows from the environment to the compressor…this seems unlikely, but doesn’t violate the 1st Law.

Check out the claim based on the 2nd Law requirements: Example 2: … Heat transfer occurs from the compressor to its surroundings, which are at 70°F. Determine whether this claim can be correct. State (134a) 1 2 T (°F) 220 P (psi) 20 140 h (BTU/lb) 101.88 146.18 s °(BTU/lb·°R) 0.2238 0.2667 Check out the claim based on the 2nd Law requirements: then Since σ is negative, it violates the 2nd Law requirement… this process is not possible.

Evaluating heat transfer for a Control Volume: 6.13 : Heat Transfer and Work in Internally Reversible Processes Evaluating heat transfer for a Control Volume: For a control volume/open system that is isothermal and internally reversible (ie. a boiler) (Area under the T-s curve)

and using the 2nd Tds equation in the form of : 6.13 : Heat Transfer and Work in Internally Reversible Processes Evaluating work: then and using the 2nd Tds equation in the form of : Thus, And if PE and KE are negligible, then Area to the left of the p-v diagram

It can be used to evaluate the work done over some p 6.13 : Heat Transfer and Work in Internally Reversible Processes What does this mean? Work is related to the magnitude of the specific volume of the gas as it flows through a control volume from inlet to outlet. It can be used to evaluate the work done over some p Consider the case of applying this to incompressible liquid in a pump. For a constant specific volume: Consider when it is applied to a nozzle or diffuser for which W = 0, recognizable as the Bernoulli Equation

Wait a Minute… At this point in the discussion, you should be scratching your head a little bit in confusion. In the past we defined work due to compression of a gas as Now work per unit mass of gas flowing through a CV is defined as How can these both be right? They represent different areas on the p-v diagram!!! or

Class Discussion:

(a) the velocity at the nozzle exit, in ft/s. Example 3: Helium gas at 810°R, 45 psi and a velocity of 10 ft/s enters an isolated nozzle operating at steady state and exits at 670°R, 25 psi. Modeling helium as an ideal gas with k=1.67, determine (a) the velocity at the nozzle exit, in ft/s. (b) the isentropic nozzle efficiency (c) the rate of entropy production within the nozzle in BTU/°R per lb of He State (He) 1 2 T (°R) 810 670 p (psi) 45 25 V (ft/s) 10 Helium behaving as Ideal Gas Step 1: Identify State Properties Step 2: Identify Ideal Gas Values k = 1.67 R = 0.49613 Btu/lbm-R

Example (6. 140): a) the velocity at the nozzle exit, in ft/s Example (6.140): a) the velocity at the nozzle exit, in ft/s. b) the isentropic nozzle efficiency c) the rate of entropy production within the nozzle in BTU/°R per lb of He State (He) 1 2 T (°R) 810 670 p (psi) 45 25 V (ft/s) 10 2945 Apply the Energy Balance:

b) the isentropic nozzle efficiency, Example 3: b) the isentropic nozzle efficiency, c) the rate of entropy production within the nozzle in BTU/°R per lb of He For an ideal gas polytropic process State (He) 1 2 2s T (°R) 810 670 639.8 P (psi) 45 25 v (ft/s) 10 2945 For isentropic process: n = k

b) the isentropic nozzle efficiency, Example 3: b) the isentropic nozzle efficiency, c) the rate of entropy production within the nozzle in BTU/°R per lb of He State (He) 1 2 2s T (°R) 810 670 639.8 p (psi) 45 25 V (ft/s) 10 2945 3252 Using the Energy Balance again, this time for the isentropic process. Therefore Isentropic Nozzle Efficiency:

Using the Entropy Balance Example 3: c) the rate of entropy production within the nozzle in BTU/°R per lb of He State (He) 1 2 2s T (°R) 810 670 639.8 p (psi) 45 25 V (ft/s) 10 2945 3252 Using the Entropy Balance Using the ideal gas model

Assume the ideal gas model for air and neglect changes in KE and PE. Example 4: An air compressor operates at steady state with air entering at p1=15 psi, T1=60°F. The air undergoes a polytropic process, and exits at p2=75psi, T2=294°F. (a) Evaluate the work and heat transfer, each in BTU per lb of air flow. (b) Sketch the process on p-v an T-s diagrams and associate areas on the diagrams with work and heat transfer, respectively. Assume the ideal gas model for air and neglect changes in KE and PE. State (Air) 1 2 T (°F) 60 294 p (psi) 15 75 Draw Process on p-v and T-s diagrams. p v T s T S 2 1 p v 2 1 PVn=const.

Example (6.154): (a) Evaluate the work and heat transfer, each in BTU per lb of air flow State (Air) 1 2 T (°F) 520 754 p (psi) 15 75 h(Btu/lbm) 124.27 180.63 For an ideal gas and polytropic process ( for h use Table A22E ) Work is defined as where

Example (6.154): (a) Evaluate the work and heat transfer, each in BTU per lb of air flow State (Air) 1 2 T (°F) 520 754 p (psi) 15 75 h(BTU/lb) 124.27 180.63 applying Ideal Gas: then

Example (6.154): (a) Evaluate the work and heat transfer, each in BTU per lb of air flow State (Air) 1 2 T (°F) 520 754 p (psi) 15 75 h(BTU/lb) 124.27 180.63 To find heat transfer apply energy balance

End of Lecture 28 Slides