(Too Simple?) Cost Scaling

1. (Too Simple?) Cost Scaling LArTPC Engineering Meeting David Finley June 22, 2006

2. NuSAG Submission Costs 15 kton

3. NuSAG LArTPC Cost Pie 15 kton

4. To Do • Perhaps the most important thing to do next is to get actual engineering estimates or vendor quotes for these items for various designs.

5. Purpose • The purpose of this scaling is to get an estimate of the costs of the argon and the tank* for a LArTPC. • These two parts of the LArTPC are chosen because they seem to be the largest costs and, because the design options for them are few, they appear to be the most straightforward costs. *Here, the “tank” means the LNG tank “as delivered” from a vendor.

6. Context • Assumption: The Ultimate LArTPC Detector requires a demonstration with a Penultimate LArTPC detector. • The three purposes of the Penultimate Detector are: • Establish the scalability of the technology and costs to multi-tens of kilotons. • Demonstrate the physics potential of the technology, and better yet do forefront physics. • Establish functional international coordination of funding, and a functioning international collaboration that can eventually do the Ultimate Detector.

7. The Simple Scaling Method and Definition of Terms • H = height of argon containment tank • D = diameter of argon containment tank • H only extends to the surface of the pool of liquid argon; the costs required for people to work safely above the argon and the costs for including the TPC are included elsewhere. • The volume of argon to be purchased to fill the containment tank is • V_argon = (pi/4) D^2 H • The area of steel to be purchased for the containment tank is • A_steel = pi D H + 2 (pi/4) D^2

8. Reduced Volume • A reduced volume V_reduced is defined as being smaller than V_argon by • removing the volume within a distance d of the wall, and • removing the volume above a height h below the roof of the containment tank, and • removing the volume below a height h above the floor of the containment tank. • V_reduced = (pi/4) [ D – 2 d ]^2 [H – 2 h]

9. D = H and d = h simplifying assumptions • For D = H and d = h • V_argon = (pi/4) D^3 • A_steel = pi D^2 + (pi/2) D^2 = (3 pi/2) D^2 • V_reduced = (pi/4) [ D – 2 d]^3 • The ratio of argon mass in the reduced volume to argon mass in the containment tank is • M_reduced / M_argon = • V_reduced / V_argon = • { (D – 2 d) / D}^3 = [ 1 – ( 2 d / D ) ]^3

10. Costing Argon and the Tank • The “commodity costs” are the cost of delivered liquid argon with commercial purity, and the cost of an assembled steel LNG tank with no extras associated with the experiment. • Parameterize the costs as a function of argon mass. • Assume the cost of argon is proportional to the volume of argon. • Assume the cost of tank is proportional to the area of the steel. • The NuSAG report shows • an argon cost of $13.3M for 15 kton of argon, and • the tank cost as $13.0M for a tank to contain 15 kton of argon.

11. Costs of Argon, Tank and Both Argon and Tank • The cost of argon as a function of argon mass is • C_argon = $13.0M (M_argon / 15 kton) • Derive the cost of the tank as a function of argon mass. • M_argon = rho_A V_argon • V_argon = M_argon / rho_A = (pi/4) D^3 • D = [ 4 M_argon / (pi*rho_A)]^(1/3) • A_steel = [3 pi / 2] { 4 M_argon / (pi*rho_A)}^(2/3) • The cost of the tank as a function of argon mass is • C_tank = $13.3M (M_argon / 15 kton)^(2/3) • C_both = C_argon + C_tank

12. Efficiency of using argon • Parameterize the efficiency of using the total argon as a function of argon mass. • M_reduced / M_argon = [ 1 – ( 2 d / D ) ]^3 • D = [ 4 M_argon / (pi*rho_A)]^(1/3) • Use rho_A = 1.4 g / cm^3 and • choose d = 0.5 m. 0.75m and 1.0m.

13. Costs for purchased argon and tank

14. Costs for purchased argon and tank

15. Costs for Reduced Masses

16. Costs for Reduced Masses

17. Conclusion • To put this simple scaling in perspective • Need to dig up 50 kton costs (which might be lower than these scaled costs) • Need to create 3 kton costs