Entropy of a liquid or a solid

Entropy of a liquid or a solid 8.75 A piston cylinder has constant pressure of 000 kpa with water at 0 o C. It is now heated up to 00 o C. Find the heat transfer and the entropy change using the steam tables. Repeat the calculation using constant heat capacity and incompressibility. C.V. Water. Constant pressure heating. Energy Eq.5.: u - u = q w Entropy Eq.8.3: s - s = q / T SOURCE + s gen Process: P = P => w = P(v - v ) The energy equation then gives the heat transfer as q = u - u + w = h - h Steam Tables B..4: h = 85.8 kj/kg; s = 0.96 kj/kg K h = 40.45 kj/kg; s =.3053 kj/kg K q = h - h = -85.8 + 40.45 = 334.63 kj/kg s - s =.3053 0.96 =.009 kj/kg K Now using values from Table A.4: Liquid water C p = 4.8 kj/kg K h - h C p (T T ) = 4.8 80 = 334.4 kj/kg s - s C p ln(t /T ) = 4.8 ln 373.5 93.5 =.0086 kj/kg K Approximations are very good

8.90 Consider a small air pistol with a cylinder volume of cm 3 at 50 kpa, 7 C. The bullet acts as a piston initially held by a trigger. The bullet is released so the air expands in an adiabatic process. If the pressure should be 00 kpa as the bullet leaves the cylinder find the final volume and the work done by the air. C.V. Air. Assume a reversible, adiabatic process. Energy Eq.5.: u - u = 0 w ; Entropy Eq.8.4: s - s = dq/t + s gen = 0/ State : (T,P ) State : (P,?) So we realize that one piece of information is needed to get state. Process: Adiabatic q = 0 Reversible s gen = 0 With these two terms zero we have a zero for the entropy change. So this is a constant s (isentropic) expansion process giving s = s. From Eq.8.3 k- k T = T ( P / P ) = 300 0.4 00 50.4 0.8575 = 300 = 30.9 K The ideal gas law PV = mrt at both states leads to V = V P T /P T = 50 30.9/00 300 =.9 cm 3 The work term is from Eq.8.38 or Eq.4.4 with polytropic exponent n = k W = - k (P V - P V ) = -.4 (00.9-50 ) 0-6 = 0.45 J

8.04 A rigid container with volume 00 L is divided into two equal volumes by a partition, shown in Fig. P8.04. Both sides contain nitrogen, one side is at MPa, 00 C, and the other at 00 kpa, 00 C. The partition ruptures, and the nitrogen comes to a uniform state at 70 C. Assume the temperature of the surroundings is 0 C, determine the work done and the net entropy change for the process. C.V. : A + B no change in volume. W = 0 m A = P A V A /RT A = (000 0.)/(0.968 473.) =.44 kg m B = P B V B /RT B = (00 0.)/(0.968 373.) = 0.806 kg P = m TOT RT /V TOT = (.6046 0.968 343.)/0. = 87 kpa From Eq.8.5 S SYST =.44[.04 ln 343. 87 473. - 0.968 ln 000 ] + 0.806[.04 ln 343. 87 373. - 0.968 ln 00 ] = -0.894 kj/k Q = U - U =.44 0.745(70-00) + 0.806 0.745(70-00) = -4.95 kj From Eq.8.8 S SURR = - Q /T 0 = 4.95/93. = +0.484 kj/k S NET = -0.894 + 0.484 = +0.947 kj/k

8. The power stroke in an internal combustion engine can be approximated with a polytropic expansion. Consider air in a cylinder volume of 0. L at 7 MPa, 800 K, shown in Fig. P8.. It now expands in a reversible polytropic process with exponent, n.5, through a volume ratio of 8:. Show this process on P v and T s diagrams, and calculate the work and heat transfer for the process. C.V. Air of constant mass m = m = m. Energy Eq.5.: m(u u ) = Q W Entropy Eq.8.4: m(s s ) = dq/t + S gen = dq/t Process: PV.50 = constant, V /V = 8 State : P = 7 MPa, T = 800 K, V = 0. L m = P V 7000 0. 0-3 RT = =.7 0 0.87 800-3 kg State : (v = V /m,?) Must be on process curve so Eq.8.37 gives T = T (V /V ) n- = 800 (/8) 0.5 = 636.4 K Table A.7: u = 486.33 kj/kg and interpolate u = 463.05 kj/kg P V T S Notice: n =.5, k =.4 n > k Work from the process expressed in Eq.8.38 W = PdV = mr(t - T )/( - n) =.7 0-3 0.87(636.4-800) -.5 Heat transfer from the energy equation =.8 kj Q = m(u - u ) + W =.7 0-3 (463.05-486.33) +.8 = -0.963 kj

9.7 Atmospheric air at -45 C, 60 kpa enters the front diffuser of a jet engine with a velocity of 900 km/h and frontal area of m. After the adiabatic diffuser the velocity is 0 m/s. Find the diffuser exit temperature and the maximum pressure possible. C.V. Diffuser, Steady single inlet and exit flow, no work or heat transfer. Energy Eq.6.3: h i + V i / = h e + V e /, and h e h i = C p (T e T i ) Entropy Eq.9.8: s i + dq/t + s gen = s i + 0 + 0 = s e (Reversible, adiabatic) Heat capacity and ratio of specific heats from Table A.5: C Po =.004 k =.4, the energy equation then gives:.004[ T e - (-45)] = 0.5[(900 000/3600) - 0 ]/000 = 3.05 kj/kg => T e = 4.05 C = 59. K Constant s for an ideal gas is expressed in Eq.8.3: k P e = P i (T e /T i ) k- = 60 (59./8.) 3.5 = 93.6 kpa kj kg K, Fan P v T s

9.43 A turbo charger boosts the inlet air pressure to an automobile engine. It consists of an exhaust gas driven turbine directly connected to an air compressor, as shown in Fig. P9.43. For a certain engine load the conditions are given in the figure. Assume that both the turbine and the compressor are reversible and adiabatic having also the same mass flow rate. Calculate the turbine exit temperature and power output. Find also the compressor exit pressure and temperature. CV: Turbine, Steady single inlet and exit flows, Process: adiabatic: q = 0, reversible: s gen = 0 EnergyEq.6.3: w T = h 3 h 4, Engine 3 W 4 Entropy Eq.9.8: s 4 = s 3 Compressor Turbine The property relation for ideal gas gives Eq.8.3, k from Table A.5 k- s 4 = s 3 T 4 = T 3 (P 4 /P 3 ) k = 93. 00 70 0.86 = 793. K The energy equation is evaluated with specific heat from Table A.5 w T = h 3 h 4 = C P0 (T 3 - T 4 ) =.004(93. - 793.) = 30.5 kj/kg. W T = ṁw T = 3.05 kw C.V. Compressor, steady inlet and exit, same flow rate as turbine. Energy Eq.6.3: -w C = h h, Entropy Eq.9.8: s = s Express the energy equation for the shaft and compressor having the turbine power as input with the same mass flow rate so we get -w C = w T = 30.5 = C P0 (T - T ) =.004(T - 303.) T = 433. K The property relation for s = s is Eq.8.3 and inverted as k P = P (T /T ) k- = 00 433. 303. 3.5 = 348.7 kpa

9.53 A condenser in a power plant receives 5 kg/s steam at 5 kpa, quality 90% and rejects the heat to cooling water with an average temperature of 7 C. Find the power given to the cooling water in this constant pressure process and the total rate of enropy generation when condenser exit is saturated liquid. C.V. Condenser. Steady state with no shaft work term. Energy Eq.6.: ṁ h i + Q. = ṁh e Entropy Eq.9.8: ṁ s i + Q. /T + Ṡ gen = ṁ s e Properties are from Table B.. h i = 5.9 + 0.9 373.4 = 36.74 kj/kg, h e = 5.9 kj/kg s i = 0.7548 + 0.9 7.536 = 7.83 kj/kg K, s e = 0.7548 kj/kg K Q. out = Q. = ṁ (h i h e ) = 5(36.74 5.9) = 0679 kw Ṡ gen = ṁ (s e s i ) + Q. out /T = 5(0.7548 7.83) + 0679/(73 + 7) = 3.64 + 36.84 = 4.83 kw/k

9.57 In a heat-driven refrigerator with ammonia as the working fluid, a turbine with inlet conditions of.0 MPa, 70 C is used to drive a compressor with inlet saturated vapor at 0 C. The exhausts, both at. MPa, are then mixed together. The ratio of the mass flow rate to the turbine to the total exit flow was measured to be 0.6. Can this be true? Assume the compressor and the turbine are both adiabatic. C.V. Total: Compressor Turbine Continuity Eq.6.: ṁ 5 = ṁ + ṁ 3 Energy Eq.6.0: ṁ 5 h 5 = ṁ h + ṁ 3 h 3 3 4 5 Entropy: ṁ 5 s 5 = ṁ s + ṁ 3 s 3 + Ṡ C.V.,gen s 5 = ys + (-y)s 3 + Ṡ C.V.,gen /ṁ 5 Assume y = ṁ /ṁ 5 = 0.6 State : Table B.. h = 54.7 kj/kg, s = 4.98 kj/kg K, State 3: Table B.. h 3 = 48. kj/kg, s 3 = 5.66 kj/kg K Solve for exit state 5 in the energy equation h 5 = yh + (-y)h 3 = 0.6 54.7 + ( - 0.6)48. = 495.4 kj/kg State 5: h 5 = 495.4 kj/kg, P 5 = 00 kpa s 5 = 5.056 kj/kg K Now check the nd law, entropy generation Ṡ C.V.,gen /ṁ 5 = s 5 - ys - (-y)s 3 = -0.669 Impossible The problem could also have been solved assuming a reversible process and then find the needed flow rate ratio y. Then y would have been found larger than 0.6 so the stated process can not be true.