Design a heat pipe cooling system for a spherical satellite that dissipates 5000 W/m3 , has a surface area of 5 m2 , and cannot exceed a temperature of 120°C. All the heat must be dissipated by radiation into space. State all your assumptions.

GIVEN



FIND

(a) A design for a heat pipe cooling system

ASSUMPTIONS

? Temperature drop between the satellite interior and the heat pipe evaporator is < 20°C

? Neglect vapor pressure drop in the heat pipe


Since the satellite has a 5 m2 surface area, the radius of the spherical satellite is



from which the satellite volume is



and the total power dissipated by the satellite, and therefore by the heat pipe cooling system, is



Since we have assumed that the temperature drop between the satellite interior and the heat pipe evaporator is less than 20°C, we can safely operate the heat pipe at 100°C.

From Figure 9.24, water has the highest figure of merit, M, at the desired temperature of 373 K and it should operate satisfactorily at the desired temperature. From that figure we find



Since we have neglected vapor neglected vapor pressure drop, and since there is no gravitational head, Equation (9.40) simplifies to



Rearranging the previous equation to solve for the geometry of the heat pipe







For a reasonably sized heat pipe, let Leff = 30 cm. Then the total wick cross-sectional area required is Aw = 30 cm/ 2.4 cm–1 = 12.5 cm2. Assuming that we need N pipe of diameter D and thickness t, we have



If the exterior surface of the heat pipe condenser section is black, the following equation gives the required surface area of the condenser section



This equation assumes that the heat pipes are separated sufficiently that they all have a near-unity shape factor with respect to space.

Using Tcond = 100°C = 373 K and assuming Tspace = 0 K and we can solve for the condenser surface area as follows



If the condenser length is Lcond, then



Let Lcond = 10 cm giving N?D = 47.9 m = 4790 cm. Since we found previously that ?DtN = 12.5 cm2, we can solve for the wick thickness t



Choose a pipe diameter of 3 cm, then



Since Leff = L + (Lcond + Levap)/2, letting the evaporator and condenser lengths be the same, 10 cm, we

find L = 20 cm and the overall length is 20 + 10 + 10 = 40 cm. We summarize the cooling system below

Pipe diameter = 3 cm

Pipe length = 40 cm

Condenser length = 10 cm

Evaporator length = 10 cm

Adiabatic section length = 20 cm

Wick: 0.0026 cm thickness of 200 mesh nickel powder

Working fluid: water

Pressure: atmospheric

Number of pipes required: 508

Physics & Space Science

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