ME 300 Thermodynamics II Exam 2 November 13, :00 p.m. 9:00 p.m.

Transcription

1 ME 300 Therodynaics II Exa 2 Noveber 3, 202 8:00 p.. 9:00 p.. Nae: Solution Section (Circle One): Sojka Naik :30 a.. :30 p.. Instructions: This is a closed book/notes exa. You ay use a calculator. You ust start fro the ost basic for of the governing equations and siplify as necessary to solve each proble. You ust show all work for full credit. Please keep your eyes on your own exa. If you are caught cheating you will get a zero for the exa and your nae will be turned over to the Dean of Students. Please feel free to use the tables/equation sheet provided with the exa. If you need to interpolate any property data, feel free to use the closest value to save tie. State clearly the source of property data. Question Total Score Total 00

2 . An air-conditioning syste operates at a total pressure of at and consists of a heating section and an evaporative cooler. Air enters the heating section at 0C and 70% relative huidity (state ) with a voluetric flow rate of 30 3 /in. The heated air fro the heating section (state 2) enters the evaporative cooler and leaves at 20C and 60% relative huidity (state 3). (a) Show the air-conditioning process on the attached psychroetric chart. Clearly label states, 2, and 3 and indicate the appropriate processes. (8 points) (b) Deterine the rate of heat transfer (kj/in) in the heating section using the psychroetric chart. (5 points) (c) Calculate the aount of water added to the air (kg/in) in the evaporative cooler using the psychroetric chart. (7 points) Assuptions. Steady state 2. Neglect KE and PE effects 3. No work interaction 4. Air and water vapor behave as ideal gases 3 V 30 in Solution (a) Heating process (-2) occurs at constant huidity ratio i.e. = 2 and the evaporative cooling process occurs at constant wet bulb teperature i.e. h = h 2. The heating (-2) and evaporative cooling (2-3) processes are shown on the psychroetric chart. (b) Considering energy balance for the control volue around the heating section: deheating Q heating W heating air hh2kepeq heating air h2 h dt State : v = /kg dry air, h = 24 kj/kg dry air State 2: h 2 = 42 kj/kg dry air 3 30 V in kg dry air air v in 0.8 kg dry air kg dry air kj Q heating air h2 h in kg dry air kj Rate of heat addition: Q heating in

3 3 2

4 Use this as extra space for Proble. (c) Considering ass balance for the control volue around the evaporative cooling section: Dry Air: air,2 air,3 air Water Vapor:,2,3,3,2 3,3 vapor water vapor water vapor vapor air 2 air,2 water air 3 2 State 2: 2 = = kg water/kg dry air State 3: 3 = kg water/kg dry air Aount of water added to the air in the evaporative cooling section: kg dry air kg water water in kg dry air kg water in

5 2. (a) Liquid octane (C 8 H 8 ) is burned copletely with air in a steady-flow reactor operating at a pressure of bar and at an equivalence ratio of.. Calculate the dew point teperature (C) of the products. (5 points) Solution For coplete cobustion of octane with 00% theoretical air: C 8 H 8(g) + a(o N 2 ) xh 2 O (g) + yco 2 + zn 2 H Balance: 8 = 2x x = 9 C Balance: 8 = y O Balance: 2a = x + 2y = 25 a = 25/2 = 2.5 N Balance: 3.76a = z z = 47 Balanced cheical reaction is: C 8 H 8(g) + 2.5(O N 2 ) 9H 2 O (g) + 8CO N 2 % Actual Air = / = /(.) = 90% For coplete cobustion of octane with 90% theoretical air: C 8 H 8(g) + ( )(O N 2 ) wh 2 O (g) + xco 2 + yco + zn 2 H Balance: 8 = 2w w = 9 C Balance: 8 = x + y O Balance: 22.5 = w + 2x + y 2x + y = 3.5 x = 5.5 and y = 2.5 N Balance: 42.3 = z Balanced cheical reaction is: C 8 H 8(g) +.25(O N 2 ) 9H 2 O (g) + 5.5CO CO N 2 Partial pressure of water vapor in the products: nho 2 ( ) 9 p g vapor ptotal bar 0.58 bar n total Dew point teperature of products: Tdewpoint Tsat p vapor Using Table A-2: Tdewpoint 55 C

6 2. (b) kol of carbon onoxide (CO) is burned copletely with 00% excess air in a closed rigid tank. Initially, the reactants (CO and air) are at 25C and bar; finally, the products are at 927C. Calculate the pressure (bar) of the products. (0 points) Assuption Reactants and products behave as ideal gases Solution For coplete cobustion of carbon onoxide with 00% theoretical air: CO + a(o N 2 ) xco 2 + yn 2 C Balance: = x O Balance: + 2a = 2x 2a = a = 0.5 N Balance: 3.76a = y y =.88 Balanced cheical reaction is: CO + 0.5(O N 2 ) CO N 2 For coplete cobustion of carbon onoxide with 00% excess air: CO + (2 0.5)(O N 2 ) xco 2 + yo 2 + zn 2 C Balance: = x O Balance: + 2 = 2x + 2y y = 0.5 N Balance: 3.76 = z Balanced cheical reaction is: CO + (O N 2 ) CO O N 2 For reactants: PV For products: PV n RT n RT R R R R P P P P PV K P P nprtp np TP P PR bar PV n RT n T K R R R R R R Pressure of the products: PP 3.68 bar

7 3. kol of water vapor enters a steady-flow reactor at at and an equilibriu ixture of H 2 O (g), H 2, O 2, H, and OH exits the reactor at 2500 K and at. Measureents indicate that the equilibriu ixture contains kol of O 2, kol of H, and kol of OH. (a) Calculate the nuber of kol of H 2 O (g) and H 2 in the equilibriu ixture. (0 points) (b) Verify the equilibriu coposition at 2500 K and at using the definition of equilibriu constant for all three relevant reactions. (20 points) (c) Explain whether concentration of O atos can be considered negligibly sall so that it need not be included in the equilibriu ixture. No calculation is required. (5 points) Solution (a) The cheical reaction is: H 2 O (g) ah 2 O (g) + bh 2 + co 2 + dh + eoh O Balance: = a + 2c + e a = a = H Balance: 2 = 2a + 2b + d + e 2b = 2 2a d e b = n kol and n kol HO 2 H2 (b) There are five products in the equilibriu ixture and two ato balances three reactions need to be considered in equilibriu at 2500 K and at. i ni Products P Kp T i n i ntotalpref Reactants Considering equilibriu of the reaction: H 2 O H O n H n O P 2 Kp T nho ntotalpref log K T p 3 Considering equilibriu of the reaction: H 2 O OH + 0.5H n n P Kp T log K T OH H nh.032 2O ntotalpref p Considering equilibriu of the reaction: H 2 2H n H P Kp T nh ntotalpref log K T p Using Table A-27, the values of equilibriu constant are consistent for 2500 K and at. 3

8 Use this as extra space for Proble 3. (c) Considering equilibriu of the reaction: O 2 2O Using Table A-27: log0 K p 2500 K3.684 Since the equilibriu constant is nearly sae for H 2 2H and O 2 2O at 2500 K concentration of O atos cannot be considered negligibly sall.

9 4. Answer the following questions. Justify your answers with equations. (a) Consider the dissociation of hydroperoxyl radical in reacting systes described by HO 2 H + O 2. This reaction is endotheric (H > 0) and its equilibriu constant is K at T and K 2 at T 2 (T 2 > T ). (5 points) K > K 2 K < K 2 K = K 2 Insufficient Inforation K2 H ln K R T T2 At constant pressure, the value of equilibriu constant increases when teperature increases for endotheric reaction K 2 > K (b) Consider the Haber process for industrial production of aonia described by N 2 + 3H 2 2NH 3. At constant teperature, what is the effect of increasing pressure on the nuber of oles of NH 3? (5 points) Increase No Effect Decrease Insufficient Inforation K p T n i 2 23 i Products P nnh P 3 3 i n i ntotalpref nn n 2 H2 ntotalpref Reactants At constant teperature, the value of equilibriu constant reains the sae The nuber of oles of NH 3 ust increase if pressure increases at constant teperature