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# Chapter_2 - Chapter 2 Solutions Engineering and Chemical...

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Chapter 2 Solutions Engineering and Chemical Thermodynamics Wyatt Tenhaeff Milo Koretsky Department of Chemical Engineering Oregon State University [email protected]

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2.1 There are many possible solutions to this problem. Assumptions must be made to solve the problem. One solution is as follows. First, assume that half of a kilogram is absorbed by the towel when you dry yourself. In other words, let [ ] kg 5 . 0 2 = O H m Assume that the pressure is constant at 1.01 bar during the drying process. Performing an energy balance and neglecting potential and kinetic energy effects reveals h q ˆ ˆ = Refer to the development of Equation 2.57 in the text to see how this result is achieved. To find the minimum energy required for drying the towel, assume that the temperature of the towel remains constant at K 298.15 C º 25 = = T . In the drying process, the absorbed water is vaporized into steam. Therefore, the expression for heat is l O H v O H h h q 2 2 ˆ ˆ ˆ - = where is v O H h 2 ˆ is the specific enthalpy of water vapor at bar 01 . 1 = P and K 15 . 298 = T and l O H h 2 ˆ is the specific enthalpy of liquid water at bar 01 . 1 = P and K 15 . 298 = T . A hypothetical path must be used to calculate the change in enthalpy. Refer to the diagram below P = 1 [atm] liquid vapor liquid vapor ˆ h 1 P ˆ h 2 ˆ h 3 ˆ h P sat 1 atm 3.17 kPa By adding up each step of the hypothetical path, the expression for heat is ( 29 ( 29 [ ] ( 29 ( 29 [ ] ( 29 ( 29 [ ] C º 25 ˆ bar 01 . 1 C, º 25 ˆ C º 25 ˆ C º 25 ˆ bar 1.01 C, º 25 ˆ C º 25 ˆ ˆ , , , , 3 2 1 2 2 2 2 2 2 sat v O H v O H sat l O H sat v O H l O H sat l O H h h h h h h h h h q - + - + - = + + = However, the calculation of heat can be simplified by treating the water vapor as an ideal gas, which is a reasonable assumption at low pressure. The enthalpies of ideal gases depend on 2
temperature only. Therefore, the enthalpy of the vapor change due to the pressure change is zero. Furthermore, enthalpy is weakly dependent on pressure in liquids. The leg of the hypothetical path containing the pressure change of the liquid can be neglected. This leaves ( 29 ( 29 kPa 3.17 C, º 25 ˆ kPa .17 3 C, º 25 ˆ ˆ 2 2 l O H v O H h h q - = From the steam tables: = kg kJ 2 . 2547 ˆ , 2 sat v O H h (sat. H 2 O vapor at 25 ºC) = kg kJ 87 . 104 ˆ , 2 sat l O H h (sat. H 2 O liquid at 25 ºC) which upon substitution gives = kg kJ 3 . 2442 ˆ q Therefore, [ ] ( 29 [ ] kJ 2 . 1221 kg kJ 442.3 2 kg 5 . 0 = = Q To find the efficiency of the drying process, assume the dryer draws 30 A at 208 V and takes 20 minutes (1200 s) to dry the towel. From the definition of electrical work, [ ] ( 29 [ ] ( 29 [ ] ( 29 [ ] kJ 7488 s 1200 V 208 A 30 = = = IVt W Therefore, the efficiency is [ ] [ ] % 3 . 16 % 100 kJ 7488 kJ 221.2 1 % 100 = × = × = W Q η There are a number of ways to improve the drying process. A few are listed below.

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