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HW_6_Solution - Thus 0< P< 4 kPa(VAPOR 0.004 MPa<...

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Homework Assignment 06 Solution 1. Conversion 20 hp = 20 × 0.7457 kW = 14.91 kW Efficiency: η = Power output / Power used = 14.91/35 = 0.43 2. At 13,000 ft, atmospheric pressure is ~9 psia. The corresponding saturation temperature for this pressure is ~187 ºF. No matter how long the potatoes cooked, they were not going to soften because the water temperature was too cold. 3. Solution: The phases can be seen in the phase diagram for water, T = 27 °C = 300 Κ From the phase diagram: P VL 4 × 10 3 MPa = 4 kPa, (vapor liquid) P LS = 103 MPa (Liquid-solid)
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Unformatted text preview: Thus: 0 < P < 4 kPa (VAPOR) 0.004 MPa < P < 1000 MPa (LIQUID) P > 1000 MPa SOLID (ICE) 4. ρ ICE = 920 kg/m 3 ΔP = ρgH = 920 kg/m 3 × 9.80665 m/s2 × 1000 = 9022 118 Pa P = Po + ΔP = 101.325 + 9022 = 9123 kPa See the phase diagram for water: liquid solid interphase => T LS = − 1 ° C 5. Absorption of energy 6. Release of energy Please Note: The students are using a diagram to solve questions 03 & 04. Their values may not be exactly the same as the solution. Please take this into consideration when evaluating their answers....
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