mid_term11-20-06-solution

mid_term11-20-06-solution - Mid Term 105A Question 1 A...

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Unformatted text preview: Mid Term 105A 11/20/06 Question 1 A solar panel receives energy from the Sun and converts it into heat. The panel is 5 meter high and 2 meter wide. It is inclined 40 with the horizontal. This panel receives 40 100 W/m2 of energy, which warms liquid water from 20 C to 50 C. We will suppose that water is incompressible and its density equal to 996 kg/m3. The area of the pipes is considered constant throughout the system. First we ignore both the kinetic and potential energy of water. a- What is the mass flow rate of water at steady state in kg/s? b- How many liters of water at 50 C one collector can provide in one hour? c- How many collectors will be needed to produce the same power as a regular power plant delivering 500MW? What will be the total surface covered by the collectors, supposing there is no free space around them? We now take into account the potential energy of water. d- Supposing the flow of water goes from the top of the panel to the bottom, what is the new mass flow rate of water at steady state in kg/s? We also take into account the fact that the specific volume of water changes as it heats up inside the collector. The water velocity at the outlet is 0.8 m/s. e- Still supposing the flow of water goes from the top of the panel to the bottom, what is the new mass flow rate of water at steady state in kg/s? f- What is the area of the pipe? g- What is the water velocity at the inlet? Using the mass conservation equation to our i Qcv control volume at steady state we get 40 e dmcv dt 0 = mi - me . Hence mi = me = m (1). Now using the energy conservation equation at steady state we have Ve 2 Vi 2 = Qcv -Wcv + mi hi + + gzi - me he + + gze . Using the equation 2 2 2 2 V V above we find -Qcv = m (hi - he ) + i - e + g ( zi - ze ) (2) since the 2 2 collector does not produce or receive work from the surroundings. We will use equations (1) and (2) in the whole problem. a- (10 pts) We neglect in this part the kinetic and potential energies. Using (2) we find m = we get m = dEcv dt 0 (209.32 - 83.95 ) Qcv . Using the tables given and approximating h with hf (he - hi ) 0.1x 5x 2 = 7.98 kg/s or 28.7 kg/h . 1/4 Mid Term 105A 11/20/06 b- (5 pts) The number of liters per hour is V = m / = 28.8 l/h . c- (5 pts) Since one collector receives 100 W/m2, the total number of collectors is N=500,000,000/(100x10) or 500,000 collectors. The surface on the ground will be 500,000x5x2xcos40 = 3.8 Mm2 or 3.8 km2. Note : while this area seems reasonable, remember that the best efficiency of this power plant would be =1-Tc/Th or 9% if used in a power cycle (and nothing during the night). d- (10 pts) m = Qcv or (he - hi ) + g (ze - zi ) 0.1x 5x 2 = 7.97 kg/s . m= (209.32 - 83.95 ) - ( 9.8x 5x sin 40 ) x 10 -3 The potential energy change is indeed negligible. e- (10 pts) We now take into account the kinetic energy and the fact that there is a change in specific volume. Using (1) we get iVi A = eVe A or Vi = e Ve . So using (2) we find m = i 1 = 7.97 kg/s . 125.37 - 6.6x 10 -6 - 3.2x 10 -2 negligible and can be ignored. Qcv . Ve 2 i2 (he - hi ) + 1 - 2 + g ( z e - zi ) 2 e The kinetic energy is m= f- (5 pts) The area of the pipe is m = A or 10-2 m2. eVe g- (5 pts) The velocity at the inlet is Vi = e Ve or 0.79 m/s. i Question 2 A waste incinerator exhausts a large 1 amount of heat as a byproduct of the incineration process. A waste Water Air management company wishes to 2 recover this heat and produce Turbine B electricity to cut down their operating 3 costs. Hot air enters a heat exchanger at TA = 200 C, pA = 1 bar, with a volumic velocity (AV)A = 2 m3/s. This air comes out at a temperature TB = 150 C. The heat from the air is used to boil water. The liquid enters with a pressure p1 = 3 bar and a temperature T1 = 100 C. The mass flow rate of water is 1.7 kg/min. It exits the exchanger with no pressure drop and enters a turbine. At the exit of the turbine the pressure is p3 = 6x10-2 bar and a quality x3 = 80 %. A 2/4 Mid Term 105A 11/20/06 We will neglect the effect of gravity and kinetic energy for both air and water. We suppose the whole system at steady state. Air is considered an ideal gas and its molar mass is 29 g/mol. We also suppose there is no heat loss to the surroundings in the whole system. a- Determine the temperature and quality of the water at the exit of the heat exchanger. b- Determine the temperature of the water at the exit of the turbine. c- Find the power developed by the turbines, in kW. The incinerator is running 12 hours a day all year long. We suppose all the power generated by the turbine is transformed in electricity with an efficiency of 95%. d- If the power company buys the electricity at 5 cents/kWh, find the profit made by the company in one year. Using the mass conservation equation A 1 for the air part of the heat exchanger Water Air control volume at steady state we get 2 B Turbine 3 Wcv dmcv dt 0 = mA - mB . Hence mB = mA = mAir (1). Using the mass conservation equation for the water part of the heat exchanger control volume at steady state we get dmcv dt 0 = m1 - m2 . Hence m1 = m2 = mW (2). Now using the energy conservation equation at steady state for the heat exchanger we have 0 V 2 V 2 dEcv = Qcv - cv + mAir hA - hB + A - B + g [z A - z B ] W dt 2 2 V 2 V 2 h1 - h2 + 1 - 2 + g [z1 - z 2 ] +mW 2 2 Since we can ignore the potential and kinetic energies and there is no work in the heat exchanger we have mAir (hB - hA ) + mW (h2 - h1 ) = 0 (3). Using the the energy 0 conservation for the turbine V 2 V 2 dEcv = Qcv -Wcv + mW h2 - h3 + 2 - 3 + g [z 2 - z 3 ] . 2 dt 2 turbine is well insulated we have Wcv = mW (h2 - h3 ) (4). we Since get the a- (20 pts) Since there is no heat loss in the heat exchanger equation (3) gives h2 = mAir (hA - hB ) + h1 . Since the air is an ideal gas we have mW 3/4 Mid Term 105A 11/20/06 mAir = (AV )A vA = (AV )A MAir pA = 1.47 kg/s . Furthermore p1 > psat(100 C) so we RTA have compressed liquid water and we approximate h1 by hf(100 C). So 1.47 h2 = ( 475.5 - 424.5 ) + 419.04 = 3065 kJ/kg . At p2 = 3 bar with 0.028 h2 = 3065 kJ/kg we have superheated steam so the quality is not defined and T2 = 297.9 C. b (10 pts) Since we have a saturated liquid-vapor substance T3 = Tsat(p3) = 36.16 C. c (10 pts) Steam exits with a quality x3 = 0.8 at p3 = 0.06 bar. Since we have a saturated liquid-vapor substance h3 = hf(p3)x3(hg(p3)-hf(p3)) = 2084.2 kJ/kg and using equation (4) we find Wcv = mW (h2 - h3 ) = 0.028 (3065 -2084.2 ) = 27.8 kW d (10 pts) Since the turbine runs 365 days and 12 hours a day the turbine produces power for 4380 hours per year. So the work produced is 122x103 kWh and the electricity produced is 116 kWh. The profit is 5795 $/year. 4/4 ...
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This note was uploaded on 03/08/2008 for the course MAE MAE 105A taught by Professor Pierregourdain during the Fall '06 term at UCLA.

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