Delft University of Technology DEPARTMENT OF AEROSPACE ENGINEERING

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Cathleen Rodgers
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1 Delft University of Technology DEPRTMENT OF EROSPCE ENGINEERING Course: Physics I (E-04) Course year: Date: Time: 4:00-7:00 Student name and itials (capital letters): Student number:. You have attended the lectures given (cross as appropriate) Dutch English Instructions Write your answers usg the blank space on the page where the problem is given. Or use the back of the page, if necessary. Deliver ONLY the present booklet at the end of the exam. (do not clude scrap paper or any other sheet) Problem : (0 pots) True or false exercise (mark the appropriate box after the statement) Statement True False Carnot heat enge is an isolated system X When compressg a gas adiabatically its energy creases X modern diesel enge can be more efficient than a Carnot heat enge X 4 No device can steadily produce work out of a sgle temperature reservoir X 5 diabatic compression of a gas will crease its entropy X 6 n object that absorbs no energy from the cident radiation is termed a blackbody X 7 Heatg superheated vapor will crease its quality X 8 ir at standard conditions can be considered a pure substance X 9 Convective heat transfer a boilg fluid is more efficient than a gas flow X 0 Enthalpy and ternal energy of an ideal gas have the same units X When you mix together hot and cold water entropy is produced X The ketic energy is the most important term the analysis of gas turbes X

Problem (5 pots) Solution (-7) hydraulic turbe-generator is generatg electricity from the water of a large reservoir. The combed turbe-generator efficiency and the turbe efficiency are to be determed.

2 Problem (5 pots) Solution (-7) hydraulic turbe-generator is generatg electricity from the water of a large reservoir. The combed turbe-generator efficiency and the turbe efficiency are to be determed. ssumptions The elevation of the reservoir remas constant. The mechanical energy of water at the turbe exit is negligible. nalysis We take the free surface of the reservoir to be pot and the turbe exit to be pot. We also take the turbe exit as the reference level (z = 0), and thus the potential energy at pots and are pe = gz and pe = 0. The flow energy P/ at both pots is zero sce both and are open to the atmosphere (P = P = P atm). Further, the ketic energy at both pots is zero (ke = ke = 0) sce the water at pot is essentially motionless, and the ketic energy of water at turbe exit is assumed to be negligible. The potential energy of water at pot is pe kj/kg gz (9.8m/s )(70 m) 000 m /s kj/kg Then the rate at which the mechanical energy of the fluid is supplied to the turbe become E mech,fluid m ( e mpe mech, (500 kg/s)(0.687 kj/kg) 0 kw e mech,out ) m ( pe 0) 70 m 750 kw The combed turbe-generator and the turbe efficiency are determed from their defitions, turbe-gen W elect,out Emech, fluid 750 kw 0kW 0.77 or 7.7% Turbe Generator turbe W shaft,out Emech, fluid 800 kw 0 kw or 77.6% Therefore, the reservoir supplies 0 kw of mechanical energy to the turbe, which converts 800 kw of it to shaft work that drives the generator, which generates 750 kw of electric power. Discussion This problem can also be solved by takg pot to be at the turbe let, and usg flow energy stead of potential energy. It would give the same result sce the flow energy at the turbe let is equal to the potential energy at the free surface of the reservoir.

3 Problem (5 pots) Solution (-80) Two rigid tanks connected by a valve to each other conta air at specified conditions. The volume of the second tank and the fal equilibrium pressure when the valve is opened are to be determed. ssumptions t specified conditions, air behaves as an ideal gas. Properties The gas constant of air is R = 0.87 kpa.m /kg.k (Table -). nalysis Let's call the first and the second tanks and. Treatg air as an ideal gas, the volume of the second tank and the mass of air the first tank are determed to be Thus, mrt P m P RT (500 kpa)(.0 m ) kg (0.87 kpa m /kg K)(98 K) (5 kg)(0.87 kpa m /kg K)(08 K). m 00 kpa m m m.0.. m kg Then the fal equilibrium pressure becomes ir = m T = 5C P = 500 kpa ir m = 5 kg T = 5C P = 00 kpa mrt P (0.846 kg)(0.87 kpa m /kg K)(9 K). m 84. kpa

4 Problem 4 (5 pots) Solution (4-5) piston-cylder device contas air gas at a specified state. The air undergoes a cycle with three processes. The boundary work for each process and the net work of the cycle are to be determed. Properties The properties of air are R = 0.87 kj/kg.k, k =.4 (Table -a). nalysis For the isothermal expansion process: mrt (0.5 kg)(0.87 kj/kg.k)(50 7 K) 0.04 m P (000 kpa) mrt (0.5 kg)(0.87 kj/kg.k)(50 7 K) m P (500 kpa) m P ln (000 kpa)(0.04m )ln 0.04m W b, For the polytropic compression process: P 7.8kJ n n.. P (500 kpa)( m ) (000 kpa) m W b, P P n For the constant pressure compression process: W b (000 kpa)( m ) (500 kpa)( m., P ( ) (000 kpa)( )m The net work for the cycle is the sum of the works for each process W net W W W 7.8 ( 4.86) ( 6.97 b, b, b, ) -6.97kJ -4.65kJ ir MPa 50C ) -4.86kJ

5 Problem 5 (5 pots) Solution (5-5) ir is compressed at a rate of 0 L/s by a compressor. The work required per unit mass and the power required are to be determed. ssumptions This is a steady-flow process sce there is no change with time. Ketic and potential energy changes are negligible. ir is an ideal gas with constant specific heats. Properties The constant pressure specific heat of air at the average temperature of (0+00)/=60 C=4 K is c p =.08 kj/kg K (Table -b). The gas constant of air is R = 0.87 kpam /kgk (Table -). nalysis (a) There is only one let and one exit, and thus m m m. We take the compressor as the system, which is a control volume sce mass crosses the boundary. The energy balance for this steady-flow system can be expressed the rate form as E 0 (steady) Eout Esystem 0 Rate of net energy transfer by heat, work,and mass Thus, W E mh W w p Rate of change ternal,ketic, potential,etc. energies E out mh m ( h (sce ke pe 0) h ) mc p ( T T ) c ( T T ) (.08 kj/kg K)(00 0)K 85.0kJ/kg (b) The specific volume of air at the let and the mass flow rate are v RT ( 0.87 kpa m /kg K)(0 7 K) m P 0 kpa m v 0.00 m /s kg/s m /kg Then the power put is determed from the energy balance equation to be W mc ( T T) (0.047 kg/s)(.08 kj/kg K)(00 0)K 4.068kW p /kg MPa 00 C Compressor 0 kpa 0 C 0 L/s

6 Problem 6 (5 pots) 9-94 simple ideal rayton cycle with air as the workg fluid operates between the specified temperature and pressure limits. The net work and the thermal efficiency are to be determed. ssumptions Steady operatg conditions exist. The air-standard assumptions are applicable. Ketic and potential energy changes are negligible. 4 ir is an ideal gas with constant specific heats. Properties The properties of air at room temperature are c p =.005 kj/kg K and k =.4 (Table -a). nalysis Usg the isentropic relations for an ideal gas, T ( k) / k 0.4/.4 P 000 kpa T (00 K) P 00 kpa Similarly, T ( k) / k 0.4/.4 P4 00 kpa 4 T (000 K) P 000 kpa 706.K 44.9 K pplyg the first law to the constant-pressure heat addition process - produces Similarly, q h h c p qout h4 h c p The net work production is then w net ( T T ) (.005 kj/kg K)( )K 95.4 kj/kg ( T T ) (.005 kj/kg K)( )K 5.5 kj/kg 4 q q kJ/kg out and the thermal efficiency of this cycle is th w q net 69.9 kj/kg 95.4 kj/kg T 000 K 00 K q q out 4 s PPENDIX Universal gas constant: R = 8.4 kj/kmol K, Gas constant for air: R air = 87 kj/kg K Constant pressure specific heat for air: c p = 008 kj/kg K Ratio of specific heats for air: k =.4

ENERGY TRANSFER BY WORK: Electrical Work: When N Coulombs of electrical charge move through a potential difference V

ENERGY TRANSFER BY WORK: Electrical Work: When N Coulombs of electrical charge move through a potential difference V Weight, W = mg Where m=mass, g=gravitational acceleration ENERGY TRANSFER BY WOR: Sign convention: Work done on a system = (+) Work done by a system = (-) Density, ρ = m V kg m 3 Where m=mass, V =Volume