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Unformatted text preview: dm dt = mm sys in inflows out outflows & & ∑ ∑ We recognize that we have only one inflow (air intake) and just one outflow (exhaust). We can explode the derivative in to a difference and rewrite the equation as m ( t)  m (0) t = m m sys sys in out Δ Δ & & Solving for Δ t Δ Δ t = m ( t)  m (0) m m sys sys in out & & ME 201 Thermodynamics Spring 2006 2 Now evaluating our masses and mass flows & m = 1.26 x 10 kg / s in6 ( ) & & m = (0.85)m = (0.85) 1.26 x 10 = 1.07 x 10 kg / s out in66 & m (0) = 0 kg sys (assume no air initially stored in the lungs) & m ( t) = 75% lung capacity = 75% density of air volume capacity of lungs = (0.75)(1.3)(0.006) = 5.85 x 10 kg sys3 Δ × × Now substituting we find hr 8.6 = s 30,952 = 10 x 1.0710 x 1.2610 x .85 5 = t 663 Δ...
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This note was uploaded on 07/25/2008 for the course ME 201 taught by Professor Somerton during the Spring '06 term at Michigan State University.
 Spring '06
 SOMERTON

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