2011+powerplant+cooling

2011+powerplant+cooling - Cooling Devices 1 Why do we have...

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1 Cooling Devices
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Why do we have to cool Efficiency says we need to get rid of twice the amount of electrical energy produced 2
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3 Introduction Pump and Boiler add energy to the working fluid Turbine takes out only a fraction Cooling system typically needs to remove majority of added energy Normally two steps – Condenser transfers energy from working fluid to cooling water (called circulating water) – Cooling water transfer energy to the environment
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4 Introduction Sometimes condenser rejects heat directly to the atmosphere – Where water is expensive or not available – Linden Cogen Plant Cooling tower would make dangerous fog Preferred method is using a water source to dump heat into – Heat exchanger design – “down by the river” If air is used, evaporative cooling is a better play
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5 Psychrometrics
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What should be learned Application of Dalton’s laws Dew point Wet bulb/adiabatic saturation Psychrometrics Humidity control in buildings Evaporative cooling and cooling towers 6
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7 Air Cooling If we want to cool anything with atmospheric air, the limit of cooling is the dry bulb temperature Dry-Bulb Temperature – The ordinary temperature of atmospheric air (to differentiate it from wet bulb) If we evaporate water, energy needed to change liquid to vapor must come from somewhere – Evaporative cooling – (Condensation heating also occurs)
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Constant-Pressure Cooling and Dew-Point Temperature Note – if pressure does not change, no mass transfer occurs Dew point is found by cooling air without any mass transfer of water 8
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9 Inclusion of mass transfer Alternately, we can add water without external cooling – Mass transfer changes partial pressure – Latent heat changes temperature
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10 Adiabatic Saturator Eliminates heat transfer - only mass transfer in play Temp. reduction entirely latent heat e c d
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11 Adiabatic Saturator
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12 Adiabatic Saturator - analysis Energy Equation: Note, air represents dry air. Mass Conservation Equations: O H O H air air O H O H O H O H air air h m h m h m h m h m 2 2 2 2 2 2 3 3 3 3 2 2 1 1 1 1 & & & & & + = + + O H O H O H air air m m m m m 2 2 2 3 2 1 3 1 & & & & & = + =
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13 Adiabatic Saturator - analysis Utilize definitions of humidity ratio (absolute humidity) Connect definitions with mass conservation Scale energy equation by the air mass flow rate and solve for w 1 : air O H air O H m m w m m w 3 3 3 1 1 1 2 2 ; & & & & = = air O H m m w w 1 2 1 3 2 & & = O H O H O H O H air air O H air O H O H air h h h h w h h w h w h h w w h w h 2 2 2 2 2 2 2 2 1 2 3 3 1 3 1 3 3 3 2 1 3 1 1 1 ) ( ) ( ) ( + = + = + +
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14 Adiabatic Saturator - analysis Most of these terms are easily evaluated: (h 3 -h 1 )air = c p (T 3 -T 1 ) w 3 is assumed saturated (=p g /(p total -p g ) at T 3 ) (h 3 -h 1 ) water h fg at T 3 (assumes that T 2 =T 3 ) – Normally T 2 will be a little bit cooler than T 3 (h 1 -h 2 ) water (h 1g at T 1 ) - (h 2f at T 2 or T 3 ) – We evaluate h 1 at a saturated state (even though it is
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This note was uploaded on 01/17/2012 for the course ME 462 taught by Professor Muller during the Fall '11 term at Rutgers.

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2011+powerplant+cooling - Cooling Devices 1 Why do we have...

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