Mathcad - CHME 333 Spring 2010 Final Solutions

Mathcad - CHME 333 Spring 2010 Final Solutions - 1. Open...

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ρ As 1 11.06 kg m 3 := ρ As 0.09 kg m 3 = Air temp should be used to calculate conc of water in the air ρ Asatinf 1 39.13 kg m 3 := ρ Asatinf 0.026 kg m 3 = ρ Ainf 0.52 ρ Asatinf := ρ Ainf 0.013 kg m 3 = D AB 0.26 10 4 m 2 s := assume this doesn't change with temperature Sc ν air D AB := Sc 0.711 = Radiation Q rad εσ A T water 4 T surr 4 := Q rad 757.232 W = Convection Q conv hA T water T air () := Q conv 550 W = 1. Open hot water baths are used as a testing facility for pressurized products (cans of spray paint, lubricants, cleaners, hairspray, etc.). The products are submerged in the bath for a period of time to ensure that they can tolerate conditions during shipping and storage. Consider a hot water bath that is 1 m wide by 4 m long by 40 cm deep, with water at 50 o C (well mixed). Its top surface is open to the ambient air so that workers can easily observe the process, and the bottom and sides are well insulated. The conditions in the testing facility are as follows: air temperature of 25 o C, 52% relative humidity. The surroundings are at 20 o C. Calculate the heat loss components of the bath by radiation, convection, and evaporation. Also calculate how much water must be replaced each day to maintain the bath’s water level. Assume the convection coefficient for heat transfer is 5.5 W/m 2 •K, and the emissivity of water is 0.95. The following properties may be used: Air: k = 0.02644 W/m•K, Pr = 0.7262, α = 2.546 x 10 -5 m 2 /s; ν = 1.849 x 10 -5 m 2 /s ρ = 1.16 kg/m 3 ; c p = 1.007 kJ/kg•K kJ 1000J := k air 0.02644 W mK := Pr air 0.7262 := α air 2.546 10 5 m 2 s := ν air 1.849 10 5 m 2 s := ε 0.95 := h 5.5 W m 2 K := T air 298K := T surr 293K := RH 0.52 := T water 323K := A4 m 2 := σ 5.67 10 8 W m 2 K 4 := ρ air 1.16 kg m 3 := c pair 1.007 kJ kg K := Evaluate water properties at water temp ~325K h fg 2378 kJ kg :=
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Evaporation - need to calculate the evaporation rate First get hm using the heat/mass transfer analogy n 1 3 := Le α air D AB := Le 0.979 = h m h ρ air c pair Le 1n := h m 4.775 10 3
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Mathcad - CHME 333 Spring 2010 Final Solutions - 1. Open...

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