<jupyter_start><jupyter_text>Thermal anchoring of wiresWire is thermally anchored at four stages- At 300 K - At 150K (approximately) - At 80K (approximately) - At 4K  MW wire- Outer diameter = 1.40 mm- Inner conductor diamter = 0.26 mm- PTFE diameter = 1 mm- Copper braid diameter = 1.1 mm- SS braid = 1.25 mm<jupyter_code>display(Image(filename="stormflex.jpg", width=300))<jupyter_output><empty_output><jupyter_text>Copper wiringThe thermal conductivity depends heavily on the Residual  Resistive  Ratio  (RRR)- for  ordinary  coppers,  5 <  RRR <  150;- for  Oxygen-Free High  thermal  Conductivity(OFHC) coppers, 100 < RRR < 200;- for very pure coppers,  200 < RRR < 5000.We assume an RRR = 40 (similar to value used by Montana Instruments)Conductivity curve from https://arxiv.org/ftp/arxiv/papers/1501/1501.07100.pdf<jupyter_code>display(Image(filename="k_copper.jpg", width=600))<jupyter_output><empty_output><jupyter_text>Manganin and PhBr wiresThermal conductivity values from Lakeshore website:https://www.lakeshore.com/products/categories/specification/temperature-products/cryogenic-accessories/cryogenic-wire  Thermal resistance$$ R = \frac{L}{kA} $$In series, $$R = R_1 + R_2 ...$$In parallel, $$\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} ...$$To find heat transferred,  $$ Q = \frac{\Delta T}{R} $$  Considering all wires in system<jupyter_code># Thermal conductivity Copper (W/(m*K))
k_Cu_80K = 557
k_Cu_4K = 300

# Thermal conductivity Manganin (W/(m*K))
k_Mn_80K = 13
k_Mn_4K = 0.5

# Thermal conductivity PhBr (W/(m*K))
k_Ph_80K = 25
k_Ph_4K = 1.6

# Distance from 80 K to 4 K (m)
L_80_4 = 0.8 

# Areas (m^2)
A_Cu1 = np.pi * (0.18e-3/2)**2 # Cu wiring 0.18 mm diameter
A_Cu2 = np.pi * (0.224e-3/2)**2 # Cu wiring 0.224 mm diameter
A_Mn = np.pi * (0.15e-3/2)**2 # Mn wiring 0.15 mm diameter
A_Ph = np.pi * (0.127e-3/2)**2 # PhBr wiring 0.127 mm diameter
# Microwave cable area (m^2) (inner conductor and outer braid, both Copper)
A_inner = np.pi * (0.26e-3/2)**2
A_braid = np.pi * (1.1e-3/2)**2 - np.pi * (1e-3/2)**2
A_MW = A_inner + A_braid
# Total series resistance of each wire
R_Cu1 = 1/A_Cu1 * L_80_4/(k_Cu_80K - k_Cu_4K)
R_Cu2 = 1/A_Cu2 * L_80_4/(k_Cu_80K - k_Cu_4K)
R_Mn = 1/A_Mn * L_80_4/(k_Mn_80K - k_Mn_4K)
R_Ph = 1/A_Ph * L_80_4/(k_Ph_80K - k_Ph_4K)
R_MW = 1/A_MW * L_80_4/(k_Cu_80K - k_Cu_4K)

# Taking all similar wires in parallel
R_Cu1_parallel = R_Cu1/20
R_Cu2_parallel = R_Cu2/22
R_Mn_parallel = R_Mn/3
R_Ph_parallel = R_Ph/18
R_MW_parallel = R_MW/4

# Total resistance of parallel combination of similar wires
R_total = 1/(1/R_Cu1_parallel + 1/R_Cu2_parallel + 1/R_Mn_parallel + 1/R_Ph_parallel + 1/R_MW_parallel)

# Heat load from total resistance
Q_total = (80 - 4)/R_total

print(f"Total heat load = {int(Q_total*1e3)} mW")<jupyter_output>Total heat load = 55 mW
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