The Study of Thermodynamics - Essay Example

ASSIGNMENT FACULTY OF: APPLIED DESIGN AND ENGINEERING OF PROGRAMME: BEng YEAR OR LEVEL: Year OF MODULE/UNIT/EXAM PAPER: Engineering Science II Thermodynamics Date set: 12th March 2014 Date to submit: 1sr May 2014 QUESTION ONE (25 marks): Closed system A well-insulated piston-cylinder device contains 5 L of saturated liquid water at a constant pressure P of 175 kPa. Water is stirred by a paddle wheel while a current of 8 A flows for 45 min through a resistor placed in the water. One-half of the liquid is evaporated during this constant-pressure process and the paddle-wheel work amounts to 400 kJ. Notes: Kinetic and potential energy changes are negligible. For a constant-pressure process, the boundary work is given as: The electrical work can be determined from: In a saturated liquid-vapour mixture, the mass fraction of vapour is called the quality and is expressed as: Quality may have values between 0 (saturated liquid) and 1 (saturated vapour). It has no meaning in the compressed liquid or superheated vapour regions. In the saturated mixture region, the average value of any intensive property y is determined from: where f stands for saturated liquid, g for saturated vapour and y stands for v, u or h. a. Show that for a closed system the boundary work Wb and the change in internal energy ∆U in the first-law relation can be combined into one term, ∆H, for a constant-pressure process. Proving: Energy Balance (in general): Q + W = ΔU + ΔKE + ΔPE Wint negligible changes in potential and kinetic energies, then ΔKE = ΔPE = 0 so that the equation simplifies as -- ΔU = Q + W (1st Law of Thermodynamics) where work (W) is made of electrical work (We,in), paddle-wheel work (Wpw,in), and boundary work (Wb) and for a piston – cylinder setup with insulation, ‘Q’ zeroes out (since there would be ‘no heat transfer’ between the system and its surroundings). Hence, ΔU = We,in + Wpw,in Wb Then ΔH = V*I*Δt + Wpw,in so that on substituting this for ΔH, the previous equation becomes ΔU = ΔH Wb or ΔH = ΔU + Wb b. Using the given tables A-5, determine the voltage of the source. Calculations: Because the pressure is constant, then P1 = P2 = 175 kPa For a saturated liquid @ P1 = 175 kPa, --- h1 = hf = 487.01 kJ/kg and Vf = 0.001057 m3/kg (from Table A – 5) then applying y = yf + x * yfg where x = 0.5 (since half of the saturated liquid evaporated), the associated enthalpy is – h2 = hf + x2 * hfg --- hfg = 2213.1 kJ/kg (via Table A – 5) and substituting values to solve for h2– h2 = 487.01 kJ/kg + (0.5)*(2213.1 kJ/kg) = 1593.56 kJ/kg also mass, m = = = 4.7304 kg Considering now ΔH = V*I*Δt + Wpw,in = m (h2 - h1) --- m (h2 - h1) = V*I*Δt + Wpw,in (4.7304 kg) (1593.56 kJ/kg 487.01 kJ/kg) = (V) (8A) (45 min) ( ) + 400 kJ Solving for voltage (V): V = --- V = 223.82 volts QUESTION TWO (20 marks): Power generation by a steam Turbine (Steady-flow system) The power output of an adiabatic steam turbine is 5 MW, and the inlet and the exit conditions of the steam are as indicated in the following figure. a. Calculate, on a unit mass basis: the change in enthalpy h (use table A-6), P1 = 2 MPa and T1 = 400°C --- h1 = 3248.4 kJ/kg (at the inlet) (using Table A – 6) given h2 = 2361.01 kJ/kg (at the outlet), then ΔH = h2 - h1 = 2361.01 kJ/kg - 3248.4 kJ/kg so that ΔH = - 887.39 kJ/kg the change in kinetic energy KE ΔKE = ( ) = [ ] * ( ) so that ΔKE = 14.95 kJ/kg the change in potential energy PE. ΔPE = g ( ) = (9.81 m/s2) (6 m - 10 m) ( ) so that ΔPE = - 0.03924 kJ/kg What observations can be made from these results? Based on the results, the energy balance around the steam turbine (under steady – state condition) may be shown as follows: E1 E2 = 0 (energy in energy out = 0) and h1 + + g*z1 [ h2 + + g*z2 ] + Q W = 0 then combining like terms: - (h2 - h1) - ( - ) - ( g*z2 - g*z1 ) - W = 0 (no heat ‘Q’) b. Determine the work done per unit mass of the steam flowing through the turbine. Using the steady-state energy equation obtained earlier: W = ΔH ΔKE ΔPE = - (-887.39 kJ/kg) 14.95 kJ/kg (-0.03924 kJ/kg) W = 872.48 kJ/kg c. Calculate the mass flow rate of the steam. Ẇout = 5 MW * ( ) = 5000 kJ/s Therefore, mass flow rate of the steam ṁ = = so that ṁ = 5.731 kg/s QUESTION THREE (30 marks): Cooling of refrigerant-134a by water (Steady-flow system) Hot exhaust gases of an internal combustion engine are to be used to produce saturated water vapour at 2 MPa pressure. The exhaust gases enter the heat exchanger at 400C at a rate of 32 kg/min while water enters at 15C. The heat exchanger is not well insulated, and it is estimated that 10% of heat given up by the exhaust gases is lost to the surroundings. If the mass flow rate of the exhaust is 15 times that of water, determine: Use the constant specific heat properties of air for the exhaust gases. Notes: Kinetic and potential energy changes are negligible. There are no work interactions. Exhaust gases are assumed to have air properties with constant specific heats, that is: You will need to use table A-2, A-4 and A-5. Calculations: Granted that the exhaust gases lost 10% of heat to the surroundings, this means that 90% of the total heat given off is absorbed by the water. Mathematically, Qexhaust gases + Qwater Qsurr = 0 (so that energy is conserved) For the cold side: Qwater = ṁH2O (h4 - h3) where h3 (@ 15°C) = 62.982 kJ/kg from Table A – 4 and h4 (@ 2 MPa) = 2798.3 kJ/kg for a saturated vapour from Table A – 5 15 * ṁH2O = 32 kg/min --- ṁH2O = 2.1333 kg/min Upon substitution for Qwater = - 0.90*Qexhaust Qexhaust = so that Qexhaust = 6483.62 kJ/min For the hot side: Qexhaust = ṁexhaust (h2 - h1) = ṁexhaust * Cp * ( - ) Interpolating via Table A – 2 @ 400°C --- Cp = 1.0686 kJ/kg*K Calculating for T2 (plugging known values into the Q equation): (32 kg/min) (1.0686 kJ/kg*K) (T2 - 400°C) = - 6483.62 kJ/min T2 = 210.39°C QUESTION FOUR (25 marks): Air conditioning system (Steady-flow system) An air-conditioning system involves the mixing of cold air and warm outdoor air before the mixture is routed to the conditioned room in steady operation. Cold air enters the mixing chamber at 5C and 105 kPa at a rate of 1.25 m3/s while warm air enters at 34C and 105 kPa. The air leaves the room at 24C. The ratio of the mass flow rates of the hot to cold air streams is 1.6. Using variable specific heats, determine: Notes: Kinetic and potential energy changes are negligible. There are no work interactions. The device is adiabatic and thus heat transfer is negligible. Air is considered as an ideal gaz. You will need to use table A-1 and A-17. a. The mixture temperature at the inlet of the room. According to Table A – 17 (ideal gas properties for air), hcold (at 5°C) = 278.13 kJ/kg and hwarm (at 34°C) = 307.23 kJ/kg also, hexit (at 24°C) = 297.18 kJ/kg Doing mass – balance @ steady – flow (due to mass transfer across the boundary): entering mass exiting mass = 0 Using the mass flowrate ratio of 1.6 = with mcold + mwarm = mTot becoming mcold + 1.6*mcold = mTot = 2.6*mcold Doing energy – balance (where and ΔPE are assumed negligible): mcold*hcold + mwarm*hwarm = mTot*hTot mcold*hcold + 1.6mcold*hwarm = 2.6mcold*hTot hTot = Then by substitution of enthalpy values found in previous steps, --- hTot = = 296.04 kJ/kg This corresponds to Tmixture = 296 K or 23°C from the enthalpy table for air b. The rate of heat gain of the room. V (specific) = = = 0.7599 m3/kg mcold = = = 1.645 kg/s mmixture = 2.6*mcold = 2.6*(1.645 kg/s) = 4.277 kg/s Consequently, the heat which the room absorbed is Q = mmixture (hexit hTot) = (4.277 kg/s)*(297.18 kJ/kg 296.04 kJ/kg) so that Q = 4.88 kJ/s (gained heat) Read More