E234-Assignment2-S-F2008 (1)

# E234-Assignment2-S-F2008 (1) - Stevens Institute of...

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Unformatted text preview: Stevens Institute of Technology E234 - Thermodynamics Assignment – Week 2 1. My weight is kg =(lbs)x0.453 My height in m = 2.54x10-2x(inches) 2. Convert the formula for water density to be for T in degrees Kelvin. ρ = 1008 – TC/2 [kg/m3] ρ = 1008 – TC/2 [kg/m3] We need to express degrees Celsius in degrees Kelvin TC = TK – 273.15 and substitute into formula ρ = 1008 – TC/2 = 1008 – (TK – 273.15)/2 = 1144.6 – TK/2 3. How much mass is there approximately in 1 L of mercury (Hg)? Atmospheric air? What is the specific volume of mercury? of air Check tables at end of book Mercury: density = 13 560 kg/m3 1 L has a mass of 13.56 kg, v =1/density = 7.4x10-5 m3/kg Water: density = 1000 kg/m3 1 L has a mass of 1.0 kg, v = 1/density = 0.001 m3/kg 4. You exercise on a stairmaster. Each step is 25 cm high. You do 120 steps/minute. You weigh xx kg (see 1. above). How much energy will you expand in one hour? What is your average power? Results are to be given in cal/minute and SI units. You work against gravity; Each step requires an energy E = mgh. Assume a mass of 80 kg. 120 steps per minute is 2 steps per second. You power is thus P = mghx2 = 80x10x.25x2 = 400 W Energy in one hour = Pxtime = 400x3600 = 1.44x106 J Energy calorie per hour = 1.44x106/(60x4.180) = 5740 cal/min To lose one pound of fat, you would not to exercise for t = 3500/5.74 = 609 minutes = 10 hours! 5. A piston/cylinder with cross sectional area of 0.01 m2 has a piston mass of 100 kg resting on the stops, as shown below. With an outside atmospheric pressure of 100 kPa, what should the water pressure be to lift the piston? Fall 2008 The force acting down on the piston comes from gravitation and the outside atmospheric pressure acting over the top surface. Force balance: F↑ = F↓ = PA = mpg + P0A Now solve for P (divide by 1000 to convert to kPa for 2nd term) P = Po + mpgA = 100 kPa + 100 × 9.806650.01 × 1000 kPa = 100 kPa + 98.07 kPa = 198 kPa 6. The difference in height between the columns of a manometer is 200 mm with a fluid of density 900 kg/m3. What is the pressure difference? What is the height difference if the same pressure difference is measured using mercury, density 13600 kg/ m3, as manometer fluid? ΔP = ρ1gh1 = 900 kg/m3 × 9.807 m/s2 × 0.2 m = 1765.26 Pa = 1.77 kPa hHg = ΔP/ (ρhg g) = (ρ1 gh1) / (ρhg g) = 90013600 × 0.2 = 0.0132 m = 13.2 mm 7. Cengel and Boles 6th edition problem 2-72 Assumptions 1 The wind is blowing steadily at a constant uniform velocity. 2 The efficiency Wind of the wind turbine is independent of the wind speed. 12 m/s Properties The density of air is 3 given to be = 1.25 kg/m . Analysis Kinetic energy is the only form of mechanical energy the wind possesses, and it can be converted to work entirely. Therefore, the power potential of the wind is its kinetic energy, which is V2/2 per unit mass, and mV 2 / 2 for a given mass flow rate: emech ke Wind turbine V 2 (12 m/s)2 1 kJ/kg 0.072 kJ/kg 2 2 2 2 1000 m /s m VA V D 2 4 (1.25 kg/m3 )(12 m/s) (50 m)2 29,450 kg/s 4 Wmax Emech memech (29,450 kg/s)(0.072 kJ/kg) 2121 kW The actual electric power generation is determined by multiplying the power generation potential by the efficiency, Welect wind turbineWmax (0.30)(2121 kW) 636 kW Fall 2008 50 m Therefore, 636 kW of actual power can be generated by this wind turbine at the stated conditions. Discussion The power generation of a wind turbine is proportional to the cube of the wind velocity, and thus the power generation will change strongly with the wind conditions. 2-78 Assumptions 1 The pump operates steadily. 2 The elevations of the reservoirs remain constant. 3 The 2 changes in kinetic energy are negligible. Reservoi Properties We take the density of 3 r water to be = 1000 kg/m . 45 m Pump Analysis The elevation of water and thus its potential energy changes 1 during pumping, but it experiences no changes in its velocity and Reservoir pressure. Therefore, the change in the total mechanical energy of water is equal to the change in its potential energy, which is gz per unit mass, and mgz for a given mass flow rate. That is, E mech me mech mpe mgz Vgz 1N (1000 kg/m 3 )(0.03 m 3 /s)(9.81 m/s 2 )(45 m) 1 kg m/s 2 1 kW 1000 N m/s 13.2 kW Then the mechanical power lost because of frictional effects becomes Wfrict W pump, in E mech 20 13.2 kW 6.8 kW Discussion The 6.8 kW of power is used to overcome the friction in the piping system. The effect of frictional losses in a pump is always to convert mechanical energy to an equivalent amount of thermal energy, which results in a slight rise in fluid temperature. Note that this pumping process could be accomplished by a 13.2 kW pump (rather than 20 kW) if there were no frictional losses in the system. In this ideal case, the pump would function as a turbine when the water is allowed to flow from the upper reservoir to the lower reservoir and extract 13.2 kW of power from the water. Fall 2008 ...
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