F2008-E234-S-Exam1

F2008-E234-S-Exam1 - E234 THERMODYNAMICS Exam 1 –...

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Unformatted text preview: E234 THERMODYNAMICS Exam 1 – September 22, 2008 Data: R = 8.314 J/mol.K For air R = 0.287 kJ/kg.K M(air) = 29 kg/kmol 1. A tank has two rooms at 80.3 oF separated by a membrane. Room A has 1 kg air and pressure of 1 bar, room B has 750 L air with a specific volume of 0.83 m3/kg. The membrane is broken and the air comes to a uniform state. Determine a. The total mass of air b. The final density of the air c. The final pressure in bar and in mmHg T = 80.3 oF = 26.83 oC = 300 K 1 bar = 105 Pa Initial state: Tank A: PA = 1atm mA = 1kg TA = 300 K 3 Ideal gas law: VA = mARairTA/PA = 0.861 m (watch your units) Tank B: VB = 0.75 m3 vB = 0.83 m3/kg m = VB/vB = 0.904 kg TB = 300 K a. m = 0.904 + 1 =1.904 kg b. density = m/V = 1.904/(0.861 + 0.75) =1.18 kg/m3 c. Ideal gas law: P = mRairT/V = 101.75 kPa 2. A water heater tank is cylindrical, 1 m tall, and diameter 0.5 m. It is made of a thin steel sheet covered by a 1.5 cm thick foam insulation with k = 0.025 W/m.K. The combined convection and radiation effects to the surrounding air are expressed in a convection coefficient of h = 12 W/m2.K. The hot water inside is at 72 oC and the outside room air is at 20 oC. Neglect the steel sheet, the top and bottom surfaces and assume plane geometry. Determine a. The power needed to maintain the temperature at 72 oC b. The outside surface temperature c. Why is it reasonable to neglect the steel sheet in the calculation? Fall 2008 3. One mole of air at T = 1000 K and P = 2 bar is cooled at constant volume to T = 500 K. The air is then heated at constant pressure until the temperature reaches 1000 K. If this two-stage process is replaced with a single isothermal expansion of air from 1000 K and 2 bar to some final pressure P, what is the value of P that makes the work of the two processes the same? Assume mechanical reversibility. P 1 First we need to determine the various states 1 2 State 1: T1 = 1000 K, P1 = 2 bar Ideal gas law PV = nRT V1 = 1x8.314x1000/200,000 = 0.04157 m3 State 2 : At constant V, P2/T2 = P1/T1 P2 = (500/1000)x2 = 1 bar 3 1 V Fall 2008 State 3 : At constant P, V3/T3 = V2/T2 V3 = (1000/500)x0.04157 = 0.0.8314 m3 Process 1-2 dW = PdV but the volume is constant so W12 = 0 Process 2-3 dW = PdV but the pressure is constant W23 = P2(V3-V2) = 100x0.04157 = 4.157 kJ The total work is the area under the process curve. For an isothermal process (see class notes and book) W = - nRT ln(Pfinal/Pinitial) or 8.314x1000 ln(P/2) = - 4,154 in Joules ln(P/2) = -0.5 and thus P = 1.21bar Fall 2008 ...
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This note was uploaded on 05/25/2010 for the course E-234 E-234 taught by Professor Gallois during the Spring '09 term at Stevens.

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