ASEN5519-002 Fall 2003 Space Hardware Design – Homework #2 Page - 1 Name: ___________________ Unique Identifier: __________________ Due Date: Tu 10/21/03 Homework 2 - Thermal Design We 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): L=0.275m x W=0.325m x H0.30m Insulation: Baseline: Pyrell foam on outside of 5 walls k=0.035 W m-1 K-1, 0.0125 m thick (0.5”) chamber is 0.30m tall, 0.275 x 0.325 foot print) Clear Lexan (Polycarbonate) lid on top (0.275x0.325m), 0.125” thick, k = 0.19 W m-1 K-1 Heat Sources inside plant chamber: 2 internal circulation fans, 2 watt each, 4 Watt sensor assembly (CO2, temp., humidity) 3 Watt radiated energy from lights: 50 Watt/m2, all absorbed inside chamber Water to condense m.dot.H2O = 300 ml/day; Heat of condensation L=2,256,000 Joule/kg Required rate of cooling 0.05°C/min Internal Temperature 18°C External Temperature 30°C Air Inlet Temperature 30°C Max. allowed Air Outlet Temperature 49°C Cabin Pressure 14.7 psia nominal / 10.2 psia during spacewalk 1. 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/engineering/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 = ___________ Watt Qlid = klid/tlid * Alid * (Toutside – Tinside) Qlid = ___________ Watt T.int = 18°C Q.i = 3 Watt Foam Insulation, k=0.035 W m-1 K-1, 0.0125 m (0.5”) T.external = 30°C P.external = 14.7 psia Polycarbonate Lid, 0.125”, k=0.19W m-1 K-1Lid = 11” x 13” (0.275 x 0.325m) Radiated Energy from lamp Heat Pump Q.c Fan: Q.c + Pel T.in = T.ambient T.out < 49°CASEN5519-002 Fall 2003 Space Hardware Design – Homework #2 Page - 2 Qp = Qfoam + Qlid ____________ Watt However, for multi-layer walls, it may be more convenient to use a notation using ‘thermal resistance’ or RΘ 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 30°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=0.17 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 = (ΣRi)-1 * (Toutside – Tinside), with Ri = ti / (ki * Ai) 4321inside outside444333222111inside outsidefoam.mlRRRR)T -(T*t*t*t*t)T- (T Q+++=+++=AkAkAkAk Qfoam ml = ____________ Watt Compare that to the calculated single foam insulation Qfoam = ___________ Watt Why is there hardly any difference ? _________________________________________________________________________ • Calculate the new parasitic heat into the chamber using the multi-layer insulation: • Qlid = ___________ Watt Qp.ml = Qfoam.ml + Qlid Qp.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: Using Rconvection = 1 / (h * A)ASEN5519-002 Fall 2003 Space Hardware Design – Homework #2 Page - 3 54321inside lamp54433221inside lampchamber-lampRRRRR)T -(T*1*t*t*t*1)T -(T Q++++=++++=lidlidkidlidlidAhAkAkAkAh h1: The 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. h5: 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. k2=k4=lid: The lamp lid is t2= t4=0.0675” thick, Lexan (k2= k4=0.19W m-1 K-1). k3 air gap: The air layer is t3=0.125” thick, air (k3=0.0257 W m-1 K-1). Tlamp is unknown. But since it is cooled by convection to ambient air, lets assume that the air inside the lamp is about 5°C warmer than ambient air. We can verify that later. Tlamp = Tambient + 5°C = 30°C + 5°C = 35°C The heat from the lamp air to chamber air is: Qlamp-chamber = _________ watt Now, that the chamber is so much better insulated, we use the previously calculated parasitic heat through the multi-layered foam insulated chamber (Qfoam.ml) and the heat conducted through the double-pane lid (Qlamp-chamber) to calculate the amount of parasitic heat to pump out of the chamber: Qp.ml = Qfoam.ml + Qlamp-chamber = ____________________ watt Calculate the additional heat that has to be removed from the greenhouse to maintain steady state temperature, i.e., heat in = heat out. Internally dissipated heat: Qi = Σ Pel , where Pel = fans + sensor ______ watt Heat by light source: Qrad = hν *