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CU-Boulder ASEN 5519 - Homework #2

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Homework 2 - Thermal DesignWe will design a sealed plant growth chamber for use aboard the International Space Station. The sealed and insulated green house has the following performance requirements:Qp = Qfoam + Qlid ____________ WattThe plant chamber wall (not including the clear lid) is a multi-layered wall, consisting of the foam insulation in contact with the ambient temperature of 35°C and a thickness of t1=0.50” (Pyrell, k1=0.035 W m-1s-1). The foam is in contact with a t2=0.065” thick fiberglass face sheet (k2=01.7 W m-1s-1), followed by a honeycomb core (t3=0.25”, k3=3 W m-1s-1), followed by the internal aluminum face sheet (t4=0.065”,k4=170 W m-1s-1).The thermal conductivity of this multi-layered wall can be calculated using:Qfoam ml = ____________ WattIn all cases, we purposely neglected the contribution of heat transfer to the wall by means of convection. Let’s consider convective heat transfer for the lid only, since the foam-insulated walls have no convective transfer – they are inside a box with Tbox = Toutside):Convective heat transfer is described by Q = h * A * T. If we use thermal resistance to describe the heat transfer to a wall (convection) and through a wall (multi-layer, conduction) and back to air (convection), we get:Qint = ____________ WattQout = ____________ WattQdt = ____________ WattQHP = ____________ WattT = ____________ CQHX = ____________ wattRequired Air flow = ____________ CFMASEN5519 Space Hardware Design – Homework #2 Page - 1Name: ___________________ SSN: __________________ Due Date: Tu 10/12/99Homework 2 - Thermal DesignWe will design a sealed plant growth chamber for use aboard the International Space Station. The sealed and insulated green house has the following performance requirements:Internal Geometry of greenhouse (usable space): 0.275m x 0.325m x 0.30mInsulation: Baseline: Pyrell foam on outside of 5 walls k=0.035 W m-2 K-1, 0.0125 m thick (0.5”)Clear Lexan (Polycarbonate) lid on top (0.275x0.325m)Heat Sources inside plant chamber:2 internal circulation fans, 2 watt each, 4 Wattsensor assembly 3 Wattradiated energy from lights: 50 Watt/m2, all absorbed inside chamberRequired rate of cooling 0.05°C/minInternal Temperature 18°CExternal Temperature 35°CAir Inlet Temperature 35°CMax. Air Outlet Temperature 49°CCabin Pressure 14.7 psia nominal / 10.2 psia during spacewalk1. Calculate the Parasitic Heat (heat transferred from outside to inside of chamber due to thermal conductivity)basic equation for thermal conductivity through flat plate with thermal conductivity k and area A, thickness t: Qp = k/t * A * (Toutside – Tinside):(http://www.colorado.edu/ASEN/asen5519/08Incubator-heat.htm)-calculate Qp using the foam-covered surface area Afoam and foam insulation on five walls, and the clear Lexan wall Alid on the 6th wall (Qp = Qfoam + Qlid):Qfoam = kfoam/tfoam * Afoam * (Toutside – Tinside) Qfoam = ___________ WattQlid = klid/tlid * Alid * (Toutside – Tinside) Qlid = ___________ WattQp = Qfoam + Qlid ____________ WattT.int = 18°CQ.i = 3 WattFoam InsulationT.external = 35°CP.external = 14.7 psiaPolycarbonate LidRadiated EnergyHeat PumpQ.cFan: Q.c + PelT.in = T.ambientT.out < 49°CASEN5519 Space Hardware Design – Homework #2 Page - 2However, for multi-layer walls, it may be more convenient to use a notation using ‘thermalresistance’ or R- in units of [°C/watt] (http://www.colorado.edu/ASEN/asen5519/09fans-heat.htm)- The plant chamber wall (not including the clear lid) is a multi-layered wall, consisting of the foam insulation in contact with the ambient temperature of 35°C and a thickness of t1=0.50” (Pyrell, k1=0.035 W m-1s-1). The foam is in contact with a t2=0.065” thick fiberglass face sheet (k2=01.7 W m-1s-1), followed by a honeycomb core (t3=0.25”, k3=3 W m-1s-1), followed by the internal aluminum face sheet (t4=0.065”,k4=170 W m-1s-1).The thermal conductivity of this multi-layered wall can be calculated using:Qfoam.ml = 1 / (k1/t1*A1 + k2/t2*A2 + k3/t3*A3 + k4/t4*A4) * (Toutside – Tinside),Simplify with A1=A2=A3=A4 (all 5 walls), and calculate the heat conducted through this multi-layered wall.Using thermal resistances Ri, this equation can be re-written as:Qfoam.ml = (-Ri)-1 * (Toutside – Tinside), with Ri = ti / (ki * Ai)4321inside outside444333222111inside outsidefoam.mlRRRR)T -(T*t*t*t*t)T- (T QAkAkAkAkQfoam ml = ____________ WattCompare that to the calculated single foam insulation Qfoam = ___________ WattWhy is there hardly any difference ?_________________________________________________________________________- Calculate the new parasitic heat into the chamber using the multi-layer insulation:-Qlid = ___________ WattQp.ml = Qfoam.ml + QlidQp.ml = Qfoam.ml + Qlid = ___________ Watt- In all cases, we purposely neglected the contribution of heat transfer to the wall by means of convection. Let’s consider convective heat transfer for the lid only, since the foam-insulated walls have no convective transfer – they are inside a box with Tbox = Toutside):Convective heat transfer is described by Q = h * A * -T. If we use thermal resistance to describe the heat transfer to a wall (convection) and through a wall (multi-layer, conduction) and back to air (convection), we get:Rconvection = 1 / (h * A)54321inside lamp243221inside lampchamber-lampRRRRR)T -(T*1*t*t*t*1)T -(T QlidlidlidkidlidlidlidAhAkAkAkAhASEN5519 Space Hardware Design – Homework #2 Page - 3The lamp box has internal circulation fans and we assume from experience that the convective heat transfer coefficient between the lamp box air and the Lexan lamp lid is h1 = 25 watt m-2 K-1.The plant chamber has internal circulation fans at higher speed and we assume from experience that the convective heat transfer coefficient between the chamber lid and the chamber air is h2 = 50 watt m-2 K-1.Since the lid transferred so much heat in our previous calculations, we build a double wall lid, with an insulating air gap in between. The thickness of each of the Lexan lids is 0.0675”, i.e. two thinner lids of 0.0675” make the same thickness as the previous single lid with 0.125” thickness. The two Lexan lids are separated by a 0.125” thin layer of ‘stagnant’ air. Obviously, in microgravity, this layer of air has very little convection.The lamp lid is t2= t4=0.065” thick, Lexan (k2= k4=0.19W m-1 K-1). plant chamber has internal circulation fans at


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