ME 343 Exam 2 November 24, 2014

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1 Name Time of lecture (circle) 11:00 am or 1:00 pm ME 343 Exam 2 November 24, ) /50 pts 2) /50 pts Total /100

2 Please! Be neat, write out equations before inserting numbers, and circle your answers. If you cannot figure out one part, assume an answer and carry it through the other parts. Place your answers next to your work. 1. (50%) As we discussed in our class Thermonews, the standardized test procedure for measuring fuel volatility is the Reid vapor pressure measurement. In this test 25 o C fuel is placed in the bottom volume of the device, while 25 o C moist air at 101 kpa and relative humidity φ = 50% is above it, occupying a volume that is 4 times as large. This is state 1 at which point no fuel has yet vaporized. In this test pure toluene (C 7 H 8 ), which is the most abundant hydrocarbon found in gasoline, is the fuel being tested. By coincidence, toluene has nearly the same saturation vapor pressure as water, being 1.00 kpa, higher than water at any given temperature; so the water tables can be used to find its vapor pressure (saturation pressure) versus temperature. The loaded test rig is now placed in a 35 o C water bath and comes to equilibrium at 35 o C (state 2). There is no significant change in the volume of the upper chamber due to fuel vaporization. a) At 35 o C, after the toluene has had sufficient time to vaporize and the system comes to equilibrium, what is the reading on the pressure gauge (kpa gauge)? b) Find the mole fractions of toluene, water and air in the upper chamber at equilibrium at 35 o C. Now assume in a similar, but separate, test the air is dry air and the toluene and air are held at equilibrium chilled to 10 o C where the total pressure is 99 kpa absolute. c) What are the mole fractions of toluene and air in the upper chamber (bomb)? d) Write a chemical balance equation for the upper chamber based on 1 kmole of fuel e) What is the equivalence ratio (φ) in the upper chamber? f) If this mixture were ignited and burned to complete combustion what would be the molar specific internal energy (kj/kmole) of the CO 2 only if the products were at 1000 o C?

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5 2. (50%) On a winter day, the fuel cell in Toyota s new fuel cell car takes in ambient air at T 1 = 0 o C, and relative humidity 20% along with a supply of hydrogen to generate electric power. Exiting the fuel cell is a mixture of water vapor and nitrogen at a pressure of 1 atm. The properties of the nitrogen are close enough to air such that the psychrometric chart can be used to find flow properties. Exiting the fuel cell is this moist air at T 2 = 30 o C and φ 2 = 95%. This warm air is used to heat a separate flow of dry cabin air that warms the car s interior. The air enters at 0 o C and exits at 25 o C (properties from air tables (variable specific heat). The two streams do not mix. The flow from the fuel cell exits the cabin heater saturated at 25 o C; liquid water condensate leaves, also at 25 o C. The flow is then mixed adiabatically with air (T= 10 o C and φ = 20%) to prevent condensation in the exhaust. In a final step the flow is heated with liquid coolant exiting from the fuel cell used to cool it. The total mass flow rate of the flow exiting the fuel cell at state 2 is 0.05 kg/s. The total volumetric flow rate of ambient air into the mixer is m 3 /s, and the heat delivery to the flow by the heater from the coolant is 1.5 kw. Solve using the psychometric chart (attached). Label your state ponts 2-5 on the psychometric chart. a) What is the mass flow rate of the dry air exiting the fuel cell at state 2 (kg/s)? b) What is the mass flow rate of the cabin air (kg/s)? c) What is the mass flow rate of liquid water exiting the cabin heater (kg/s)? d) What is T 4 and humidity ratio at state 4? e) What is T 5 and relative humidity at state 5? f) What is the dew point temperature of state 5? T=0 o C 1 φ 1 =20% T 1 =0 o C hydrogen Fuel cell φ 2 =95% T 2 =30 o C Cabin heater 2 3 Heated air to cabin T=25 o C T 3 =25 o C Liquid water out At 25 o C Ambient air mixer φ=20% T=10 o C Heater 4 5 Q = 1.5 kw

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