CHAPTER 8 ENTROPY. Blank

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1 CHAPER 8 ENROPY Blank

2 SONNAG/BORGNAKKE SUDY PROBLEM 8-8. A heat engine efficiency from the inequality of Clausius Consider an actual heat engine with efficiency of η working between reservoirs at and L. Use the second law in the form of Eq.8. for the whole heat engine to prove that a. If the heat engine is ideal then η = L b. If the heat engine is non-ideal then η η Carnot HE C.V. Heat engine out to the reservoirs at and L. he heat engine whether ideal or not does not store any energy or entropy so the energy equation is Energy Eq.: 0 = Q H Q L W Clausius : dq = Q H Q L 0 Q L Q L H L Substitute this into the energy equation and solve for the work W = Q H Q L Q H [ L ] Case a. Ideal, the equal sign applies and the work is W = Q H [ L ] = Q H η and the efficiency is equal to the Carnot cycle efficiency. Case b. Non-ideal, the inequality applies and the efficiency η = W/Q H becomes η [ L ] = η Carnot HE And we see that the efficiency is smaller than the Carnot efficiency. Remark: After writing the entropy equation as Eq.8. we can express how much smaller the efficiency is due to S gen as Entropy Eq.8.: 0 = dq + δs gen = Q H Q L L + S gen ; S gen 0 0 = Q H Q L + S gen Q L = Q L H + L L S gen H Notice that Q L is larger than the ideal case by an amount proportional to S gen W = Q H Q L L S gen = Q H [ L ] L S gen = Q H η η = [ L ] L S gen Q H

3 SONNAG/BORGNAKKE SUDY PROBLEM 8-8. Change in s from the steam tables Water at 00 o C is heated to 50 o C at constant P. What are changes in u and s when the starting state is at a kpa b. 500 kpa c. 00 kpa d. 0 kpa a. From able B..4 compressed liquid states at kpa, we get u u = =.3 kj/kg s s = = 0.53 kj/kg K b. From able B..4 compressed liquid states at 500 kpa, we get u u = =.9 kj/kg s s = = kj/kg K c. From able B..3 superheated vapor states at 00 kpa, we get 50 u u = ( ) = kj/kg 50 s s = ( ) = kj/kg K d. From able B..3 superheated vapor states at 0 kpa, we get u u = = 7. kj/kg s s = = 0.40 kj/kg K P kpa C 00 C v s he changes in u and s are larger for the liquid than for the gas. Within the liquid phase the influence of P is very modest. Water vapor at 00 kpa is close to an ideal gas (Z = Pv g /R sat = 0.985). For the vapor phase changing P from 00 kpa to 0 kpa changes s, but u is nearly constant. he influence of P becomes much stronger as the state approaches the dense fluid region near the critical point.

4 SONNAG/BORGNAKKE SUDY PROBLEM A Carnot cycle heat pump Consider a Carnot-cycle heat pump with R-34a as the working fluid. Heat is absorbed into the R- 34a at 0 o C, during which process it changes from a two phase state to saturated vapor. he heat is rejected from the R-34a at 60 o C so it ends up as saturated liquid. Find the pressure after compression, before the heat rejection process and determine the coefficient of performance for the cycle. From the definition of the Carnot cycle we have two constant temperature (iso-thermal) processes that involve heat transfer and two adiabatic processes where the temperature changes. he variation in s follows from Eq. 8. ds = dq / and the Carnot cycle is shown in Figure 8.8. We therefore have State 4 able B.5.: s 4 = s 3 = s f@60deg =.857 kj/kg K State able B.5.: s = s = s g@0deg =.76 kj/kg K State able B.5.: 60 o C, s = s = s g@0deg Interpolate between 400 kpa & 600 kpa in able B P = ( ) = 487. kpa P C 0 C v s From the fact that it is a Carnot cycle the COP becomes from Eq.7.3 β = q H = = w IN L 60 = 5.55 Remark: Notice how the pressure varies during the heat rejection process. his process is very difficult to accomplish in a real device so no heat pump or refrigerator is designed to approach a Carnot cycle.

5 SONNAG/BORGNAKKE SUDY PROBLEM Direction of work and heat transfer Ammonia at 00 kpa, x =.0 is compressed in a piston/cylinder to MPa, 00 o C in a reversible process. Find the sign for the work and the sign for the heat transfer. he directions of work and heat transfer are given by variations in v and s as: Work Eq.4.3: w = P dv so sign w follows sign of dv Heat transfer Eq.8.: q = ds so sign q follows sign of ds Properties from the ammonia tables: State : able B..: v = m 3 /kg ; s = kj/kg K State : able B..: v = m 3 /kg ; s = kj/kg K Changes in v and s gives: dv < 0 w is negative ds > 0 q is positive P v s

6 SONNAG/BORGNAKKE SUDY PROBLEM An isothermal reversible process A cylinder containing L of saturated liquid refrigerant R- at 0 o C. he piston now slowly expands maintaining constant temperature to a final pressure of 400 kpa in a reversible process. Calculate the required work and heat transfer to accomplish this process. C.V. he refrigerant R- which is a control mass. Continuity Eq.: m = m = m ; Energy Eq.5.: m(u u ) = Q W Entropy Eq.8.3: m(s s ) dq/ Process: = constant, reversible so equal sign applies in entropy Eq. State (, P) able B.3.: u = kj/kg, s = kj/kg K m = V/v = 0.00/ =.33 kg State (, P) able B.3.: u = kj/kg, s = kj/kg K P v s As is constant we have dq/ = Q /, so from the entropy equation Q = m(s s ) = ( ) = 00 kj he work is then from the energy equation W = m(u u ) + Q =.33 ( ) + 00 = 3.3 kj Notice that it would be a little difficult to get work as the area in the P-v diagram due to the shape of the process curve.

7 SONNAG/BORGNAKKE SUDY PROBLEM An internally reversible, externally irreversible process Ammonia in a piston-cylinder that maintains constant P is at 0 o C, 600 kpa and is now heated to 60 o C in an internally reversible process. he external heat source is at a constant 70 o C. We want to find the specific heat transfer in the process and the external (total) specific entropy generation. ake as a control volume the ammonia only, this is a control mass. Neglect storage of energy and entropy in the cylinder walls and piston mass. Energy Eq.: u u = q w Entropy Eq.: s s = dq + s gen ammonia Process: P = C w = P dv = P (v v ) Reversible s gen ammonia = 0 State : able B.. Approximate compressed liquid state with saturated same v = m 3 /kg, u = 7.89 kj/kg, s =.0408 kj/kg K State : able B.. Superheated vapor v = m 3 /kg, u = kj/kg, s = kj/kg K Work from the process equation is w = P (v v ) = 600 kpa ( ) m3 /kg = 40.6 kj/kg Heat transfer is from the energy equation (we could also have done it as h h ) q = u u + w = = 57 kj/kg he external entropy generation is a q over a so we can do it as in Eq.8.8: s gen tot = s CV + s SUR = s s + [ q source ] = = kj/kg K P NH 3 F = C 70 o C v s Choosing a CV that is the ammonia plus walls out to the source has the heat transfer crossing the CS at source not the ammonia so the entropy equation becomes s s = dq + s gen tot = q + S s gen tot source and as the space where changes (s is generated) is included in the CV, the entropy generation term includes that too becoming the same as before.

SONNAG/BORGNAKKE SUDY PROBLEM 8-7 8.7 An irreversible mixing process A piston cylinder with a constant force on it contains 0.5 kg R-34a saturated vapor in volume A and 0.

8 SONNAG/BORGNAKKE SUDY PROBLEM An irreversible mixing process A piston cylinder with a constant force on it contains 0.5 kg R-34a saturated vapor in volume A and 0. kg of R-34a at 60 o C in volume B both at 600 kpa. he two masses mix in an adiabatic process. We want to know the final temperature and the entropy generation for the process. Continuity Eq.: m m A m B = 0 Energy Eq.5.: m u m A u A m B u B = W Entropy Eq.8.4: m s m A s A m B s B = dq/ + S gen Process: P = Constant => W = PdV = P(V - V ) Q = 0 Substitute the work term into the energy equation and rearrange to get m u + P V = m h = m A u A + m B u B + PV = m A h A + m B h B where the last rewrite used PV = PV A + PV B. State A: able B.5. h A = kj/kg ; s A =.779 kj/kg K State B: able B.5. h B = kj/kg ; s B =.8379 kj/kg K Energy equation gives: State : (P, h ) h = m A m h A + m B With the zero heat transfer we have h m B = s =.7536 kj/kg K; = 3.4 C S gen = m s m A s A m B s B = 4.4 kj/kg = = kj/k B A A B

9 SONNAG/BORGNAKKE SUDY PROBLEM Cooling of a solid with water Due to a sensor malfunction a 00 kg steel tank has been heated to 50 o C. o cool it down we put x kg of liquid water at 0 o C into it and wait for the masses to come to final uniform temperature of maximum 80 o C. Assume the tank is open so P = 00 kpa and neglect any external heat transfer. Find the required water mass x and the total entropy generation in the process assuming no water evaporates. C.V. Steel tank and contents, constant pressure process Energy Eq.: m steel (u - u ) steel + m H O(u - u ) H O = Q - W Entropy Eq.: m steel (s - s ) steel + m H O(s - s ) H O = dq/ + S gen Process Eq.: P = constant and Q = 0 W = -P(V - V ) Substitute the work term into the energy equation and get: m steel (h - h ) steel + m H O(h - h ) H O = 0 For this problem we may also say that the work is nearly zero as the steel and the liquid water will not change volume to any measurable extent. Now we get changes in u's instead of h's. For these phases we have C V = C P = C from ables A.3 and A.4. able A.3: C steel = 0.46 kj/kg K he energy equation with m H O = x becomes able A.4: C H O = 4.8 kj/kg K (80-50) + x 4.8 (80-0) = 0 x = 3.8 kg he entropy generation from the entropy equation is, using Eq.8.0, S gen = m steel(s - s ) steel + m H O(s - s ) H O = ln ln = 6.9 kj/k Conceptual schematics:

10 SONNAG/BORGNAKKE SUDY PROBLEM An isothermal expansion of air A mass of kg of air contained in a cylinder at.5 MPa, 000 K, expands in a reversible isothermal process to a pressure 5 times smaller. Calculate the heat transfer during the process and the change of entropy of the air. C.V. Air, which is a control mass. Energy Eq.5.: m(u - u ) = Q - W Entropy Eq.8.3: m(s s ) = dq/ + S gen Process: = constant so dq/ = Q / reversible implies S gen = 0 For an ideal gas constant => u = u then Q = W P P P v s From the process equation and ideal gas law PV = mr = constant we can calculate the work term as in Eq.4.5 Q = W = PdV = P V ln (V /V ) = mr ln (V /V ) = ln (5) = 46.9 kj he change of entropy from the entropy equation Eq.8.3 is S air = m(s - s ) = Q / = 46.9 = kj/k If instead we use Eq.8.6 we would get S air = m(s - s ) = m[c vo ln + R ln v v ] consistent with the above result. = [ ln(5)] = 0.46 kj/k

11 SONNAG/BORGNAKKE SUDY PROBLEM A light bulb filled with argon gas A light bulb has been on a while so 0.5 g of Argon and 50 g of glass (we disregard the rest of bulb) is at 80 o C. he bulb is turned off and now cools to ambient temperature of 5 o C. he Argon was at 0 kpa while warm. Find the total entropy generation in the process. C.V. Argon gas and the glass out to the ambient. Constant volume so W = 0. Energy Eq.5.: U - U = m argon (u - u ) + m glass (u - u ) = Q Entropy Eq.8.4: S - S = dq/ + S gen = S gen + Q / Process: v = constant => For argon gas: P / P = / he heat transfer becomes the change in energy of the argon and the glass Q = m argon (u - u ) + m glass (u - u ) = m argon C argon ( ) + m glass C glass ( ) = (-65) (-65) = =.6 kj Evaluate changes in s from Eq.8.6 or 8.8 for the argon gas m argon (s - s ) = mc p ln ( / ) mr ln ( / ) = mc v ln( / ) = ln [ 88.5 ] = kj/k Evaluate the change in s for the solid glass, Eq.8.0 m glass (s - s ) = mc ln = ln [ 88.5 ] = kj/k Now the entropy generation becomes S gen = S - S - Q / = m argon (s - s ) + m glass (s - s ) - Q / = (-.6 / 88.5) = kj/k he heat transfer from the argon goes out through the glass and the total heat transfer from both masses goes out through a convection layer in the air. he entropy generation is distributed but our lumped (overall) analysis cannot give this information.

12 SONNAG/BORGNAKKE SUDY PROBLEM 8-8. Entropy generation and its location due to heat transfer Hot combustion gases at 500 K loses 3000 W to the steel walls at 700 K which in turn delivers the 3000 W to a flow of glycol at 00 o C and the glycol has heat transfer to atmospheric air at 0 o C. Assume steady state and find the rates of entropy generation and where it is generated. For a C.V at steady state we have the entropy equation as a rate form as Eq.8.43 ds c.v. = 0 = dq. / + Ṡ dt gen C.V. From combustion gases at 500 K to the steel at 700 K. Q. goes through but enters and leaves at two different s Ṡ gen = dq. / = [ 3000 ] = W/K C.V. From the steel wall at 700 K to the glycol at 00 o C. Q. goes through but enters and leaves at two different s Ṡ gen = dq. / = [ ] [ 3000 ] = W/K C.V. From the glycol at 00 o C to the atmospheric air at 0 o C. Q. goes through but enters and leaves at two different s Ṡ gen = dq. / = [ ] [ 3000 ] = W/K Notice the biggest Ṡ gen is for the largest change [/] Gases Steel Glycol Air flow Radiator Remark: he flux of S is Q. / flowing across a surface. Notice how this flux increases as Q. flows towards lower and lower. [K] Q. / [kw/k]

+ m B1 = 1. u A1. u B1. - m B1 = V A. /v A = , u B1 + V B. = 5.5 kg => = V tot. Table B.1.

+ m B1 = 1. u A1. u B1. - m B1 = V A. /v A = , u B1 + V B. = 5.5 kg => = V tot. Table B.1. 5.6 A rigid tank is divided into two rooms by a membrane, both containing water, shown in Fig. P5.6. Room A is at 200 kpa, v = 0.5 m3/kg, VA = m3, and room B contains 3.5 kg at 0.5 MPa, 400 C. The membrane