Stage (B) DCB pre-charged on surface. Fluid is empty.

Stage (C) Fluid being charged into DCB

Stage (D1) DCB fully charged on surface.

Stage (D2) DCB fully charged on seabed.

Stage (F1) DCB fluid used through Casing Shear supply.

Stage (F2) DCB fluid used through Blind Shear supply.

Stage (G) DCB hydraulic fluid emptied, on Seabed.

Stage (B) Gas : VolN2 = V1+V2 V2 = Constant V3 = minimal Liquid : VBF = maximal VHF = minimal Vbarrier = Minimal

Stage (C) Gas : VolN2 = V1+V2 V1 = decreased, V3 = increased Liquid : VBF = decreased VHF = increased Vbarrier=increased

Stage (D1) Gas : VolN2 = V1+V2 V1 = minimal, V3 = maximal Liquid : VBF = decreased VHF = maximal Vbarrier=maximal

Stage (D2) Gas : VolN2 = V1+V2 V2 = Constant V3 = minimal Liquid : VBF = maximal VHF = minimal Vbarrier = Minimal

Stage (F1) Gas : VolN2 = V1+V2 V2 = Constant V3 = minimal Liquid : VBF = maximal VH F = minimal Vbarrier = Minimal

Stage (F2) Gas : VolN2 = V1+V2 V2 = Constant V3 = minimal Liquid : VBF = maximal VHF = minimal Vbarrier = Minimal

Stage (G) Gas : VolN2 = V1+V2 V2 = Constant V3 = minimal Liquid : VBF = maximal VHF = minimal Vbarrier = Minimal

NITROGEN PRESSURE SIDE: Calculate precharge parameters for subsea “deployment”

At subsea seabed temperature: 34.18 °F, let’s work through the NIST table to determine the density required to have the desired pressure: 3300 psig

Isothermal Data for T = 34.180 F

Temperature (F) Pressure (psia) Density (lbm/ft3) Volume (ft3/lbm) Internal Energy

(Btu/lbm) Enthalpy (Btu/lbm) Entropy (Btu/lbm*R)

34.18000 3312.500 16.53991 0.06045983 67.03213 104.1175 1.193123

34.18000 3325.000 16.59032 0.06027610 66.97188 104.0841 1.192772

𝜌𝑠𝑒𝑎𝑏𝑒𝑑 = 16.53991 + (3300 + 14.7 − 3312.5)(16.59032−16.53991)

(3325−3312.5)= 16.54879 lbm/ft3

So for nitrogen at seabed, targeted pressure is:

PN2Seabed = 3300 psig = 3314.7 psia

Back to the NIST table, for the same density (Isochoric) and 90°F temperature at surface:

Isochoric Data for D = 16.549 lbm/ft3

Temperature (F) Pressure (psia) Density (lbm/ft3) Volume (ft3/lbm) Internal Energy (Btu/lbm) Enthalpy (Btu/lbm) Entropy (Btu/lbm*R)

89.000 3876.3 16.549 0.060427 77.380 120.75 1.2130

90.000 3886.5 16.549 0.060427 77.568 121.06 1.2133

91.000 3896.7 16.549 0.060427 77.757 121.36 1.2136

Provide PN2Surface = 3886.5 psia

So charging the DCB’s with PN2Surface at 90 °F (psig) will ensure we have the required N2 density when deployed at seabed depth.

State of DCB’s on seabed:

N2 vol = V1 + V2 at PN2Seabed = 208.34 US Gal - 𝜌𝑠𝑒𝑎𝑏𝑒𝑑 = 16.54878 lbm/ft3 - 1.193061 Btu/lbm entropy

A bit of Thermodynamic:

1→2: Isentropic Expansion: Constant entropy (s), Decrease in pressure (P), Increase in volume (v), Decrease in temperature (T)

2→3: Isochoric Cooling: Constant volume(v), Decrease in pressure (P), Decrease in entropy (S), Decrease in temperature (T)

3→4: Isentropic Compression: Constant entropy (s), Increase in pressure (P), Decrease in volume (v), Increase in temperature (T)

4→1: Isochoric Heating: Constant volume (v), Increase in pressure (P), Increase in entropy (S), Increase in temperature (T)

To ease the process, we will square this to consider about ideal cycle:

An ideal cycle is constructed out of:TOP and BOTTOM of the loop: a pair of parallel isobaric processes

1. LEFT and RIGHT of the loop: a pair of parallel isochoric processes

Internal energy of a perfect gas undergoing different portions of a cycle:

Isothermal:

Isochoric:

Isobaric:

We now need to walk along the cycle to determine N2 behavior and parameters during operations of the DCB’s.

It considers once on seabed, only fluid hydrostatic is there to start with, no seawater. Phyd = (WD + AirGap) x ρHydFluid x 0.052 = 5230.407 psig = 5245.107 psia So condition 0, pre-charged and empty DCB then have: N2 vol = V1 + V2 at PN2Seabed = 208.34 US Gal - 𝜌𝑠𝑒𝑎𝑏𝑒𝑑 = 16.54878 lbm/ft3 - 1.193061 Btu/lbm entropy - PN2Seabed = 3300 psig = 3314.7 psia

Once fluid is allowed in the bottle:

Phyd = Psup + (WD + AirGap) x ρHydFluid x 0.052 = 10230.407 psig = 10245.107 psia Balanced on opposite rod side by PSeaWater P SeaWater = WD x ρSeaWater x 0.052 = 5328.96 psig = 5343.66 psia

Giving:

P N2Seabed𝐿𝑜𝑎𝑑𝑒𝑑=

Δ(P𝐻𝑦𝑑 − 𝑃𝑆𝑒𝑎𝑊𝑎𝑡𝑒𝑟)

Δ𝐴𝑟𝑒𝑎𝐶𝑦𝑙/𝑅𝑜𝑑=

4901.447

1.17= 4189.27 𝑝𝑠𝑖𝑎

Which would only apply providing piston does not fully stroke up. Continuing for PN2Seabed-Loaded = 4189.27 psia, and checking NIST tables for Isobaric properties at 4189.27 psia:

Isobaric Data for P = 4189.3 psia

Temperature (F) Pressure (psia) Density (lbm/ft3) Volume (ft3/lbm) Internal Energy (Btu/lbm) Enthalpy (Btu/lbm) Entropy (Btu/lbm*R) Phase

34.18000 4189.270 19.80005 0.05050492 63.14779 102.3267 1.171400 supercritical

This would provide a liquid volume of:

Vol liquid= (1 x 208.47) x (1- (16.54878/19.80005)) / AR = 34.2314/1.17 = 29.25 US Gal > 28 US Gal ⇔ Piston stroked up fully. So PN2Seabed-Loaded is limited by the minimum N2 volume available in the DCB = 175.47 US Gal.

Working through the density ratio now that V1V2 is known (principle of mass conservation of N2). So 𝜌𝑠𝑒𝑎𝑏𝑒𝑑−𝑙𝑜𝑎𝑑𝑒𝑑 = unknown lbm/ft3

𝜌𝑠𝑒𝑎𝑏𝑒𝑑 = 16.54878 lbm/ft3 N2 Vol seabed = N2Vol Max = 208.34 US Gal N2 Vol seabed-loaded = N2Vol Min = 175.47 US Gal

𝝆𝒔𝒆𝒂𝒃𝒆𝒅−𝒍𝒐𝒂𝒅𝒆𝒅 = 𝜌𝑠𝑒𝑎𝑏𝑒𝑑 x N2 Vol seabed

N2 Vol seabed−loaded = 16.54878 x

208.34

175.47 = 19.64878 lbm/ft3

Isothermal Data for T = 34.180 F

Temperature (F) Pressure (psia) Density (lbm/ft3) Volume (ft3/lbm) Internal Energy (Btu/lbm) Enthalpy (Btu/lbm) Entropy (Btu/lbm*R) Phase

34.18000 4145.000 19.64832 0.05089493 63.32807 102.3923 1.172375 supercritical

Closing CSR: Starting from N2 Vol seabed-loaded = N2Vol Min = 175.47 US Gal, and with fluid requirement for CSG shear (highest of either shear pressure).

P CSG Shear = 4229 psig at surface (including mud, shear ratio and Hyd head pressure), which with hydrostatic ⇔ 9468 psia

P N2Seabed𝐶𝑆𝐺 𝑆ℎ𝑒𝑎𝑟=

Δ(P𝐶𝑆𝐺 𝑆ℎ𝑒𝑎𝑟 − 𝑃𝑆𝑒𝑎𝑊𝑎𝑡𝑒𝑟)

Δ𝐴𝑟𝑒𝑎𝐶𝑦𝑙−𝑅𝑜𝑑=

4901.447

1.17= 3550 𝑝𝑠𝑖𝑎

From there we get through an Isobaric table for 3550 psia, with constant Entropy

Isobaric Data for P = 3550 psia

Temperature (F) Pressure (psia) Density (lbm/ft3) Volume (ft3/lbm) Internal Energy (Btu/lbm) Enthalpy (Btu/lbm) Entropy (Btu/lbm*R) Phase

12.50000 3550.000 18.47278 0.05413370 60.58756 96.17330 1.171463 supercritical

We selected only the row showing a constant entropy S = 1.171463 Btu/lbm This results in final temperature of T CSG Shear = 12. 5 °F and 𝝆 𝑪𝑺𝑮 𝑺𝒉𝒆𝒂𝒓 = 18.47278 lbm/ft3

Empty bottle condition for which it assumes N2 returning to original precharge density for an empty DCB, still with constant Entropy of S = 1.171463 Btu/lbm

Isochoric Data for D = 16.549 lbm/ft3

Temperature (F) Pressure (psia) Density (lbm/ft3) Volume (ft3/lbm) Internal Energy (Btu/lbm) Enthalpy (Btu/lbm) Entropy (Btu/lbm*R) Phase

-16.75000 2788.233 16.54878 0.06042742 57.33272 88.53189 1.172354 supercritical

T seabed empty = -16.75 °F and P seabed empty = 2788.23 psia To summarize at this stage we obtained: 𝜌𝑠𝑒𝑎𝑏𝑒𝑑 = 16.54878 lbm/ft3 = ρ0 𝜌𝑠𝑒𝑎𝑏𝑒𝑑−𝑙𝑜𝑎𝑑𝑒𝑑 = 19.64878 lbm/ft3 = ρ1 𝜌 𝐶𝑆𝐺 𝑆ℎ𝑒𝑎𝑟 = 18.47278 lbm/ft3= ρ2 Going back to DCB status after CSG Shear, we have: P CSG Shear = 9468 psia ; P N2Seabed𝐶𝑆𝐺 𝑆ℎ𝑒𝑎𝑟

= 3550 𝑝𝑠𝑖𝑎 ; with Vol N2 = 186.5 US Gal and Vol Hyd = 18.77 US Gal

Now Closing BSR:

P𝐵𝑆𝑅 𝑆ℎ𝑒𝑎𝑟 = 3371 psig ⇔ 8610 psia

P N2Seabed𝐶𝑆𝐺 𝑆ℎ𝑒𝑎𝑟=

Δ(P𝐵𝑆𝑅 𝑆ℎ𝑒𝑎𝑟 − 𝑃𝑆𝑒𝑎𝑊𝑎𝑡𝑒𝑟)

Δ𝐴𝑟𝑒𝑎𝐶𝑦𝑙−𝑅𝑜𝑑= 2817 𝑝𝑠𝑖𝑎

From there we get through NIST Table looking for constant Entropy (1.171470…) within an Isobaric table for 2817 psia

Isobaric Data for P = 2817.0 psia

Temperature (F) Pressure (psia) Density (lbm/ft3) Volume (ft3/lbm) Internal Energy (Btu/lbm) Enthalpy (Btu/lbm) Entropy (Btu/lbm*R) Phase

-16.60000 2817.000 16.68412 0.05993723 57.19568 88.46104 1.171470 supercritical

So for BSR closure, we have 𝝆 𝑩𝑺𝑹 𝑺𝒉𝒆𝒂𝒓 = 𝟏𝟔. 𝟔𝟖𝟒𝟏𝟐 𝐥𝐛𝐦/𝐟𝐭𝟑= ρ2, P N2Seabed𝐵𝑆𝑅 𝑆ℎ𝑒𝑎𝑟

= 3550 𝑝𝑠𝑖𝑎

This give a Vol N2 = 206.5 US Gal and Vol Hyd = 1.688 US Gal And then fully depleted with return to seabed density of 𝜌𝑠𝑒𝑎𝑏𝑒𝑑 = 16.54878 lbm/ft3 = ρ0

Then constant Entropy (1.171470…) within an Isochoric table for 𝜌𝑠𝑒𝑎𝑏𝑒𝑑 = 16.54878 lbm/ft3 = ρ0

Isochoric Data for D = 16.549 lbm/ft3

Temperature (F) Pressure (psia) Density (lbm/ft3) Volume (ft3/lbm) Internal Energy (Btu/lbm) Enthalpy (Btu/lbm) Entropy (Btu/lbm*R) Phase

-18.77500 2767.201 16.54878 0.06042742 56.94566 87.90948 1.171478 supercritical

34.18000 3314.698 16.54878 0.06042742 67.02153 104.1116 1.193061 supercritical

From there we can verify that for same density, the return to about seabed temperature closes the cycle at PN2Seabed = 3314.7 psia and 1.193061 Btu/lbm entropy

Determining 𝑉𝐸𝑝 and 𝑉𝐸𝑣

Then 𝑉𝐸𝑝 =(

ρ0

ρ2−

ρ0

ρ1)

1.1 ×𝐴𝑅 and 𝑉𝐸𝑣 =

(1−ρ0

ρ1)

1.1 ×𝐴𝑅 VEpCSG = 0.04166

For the above 𝜌𝑠𝑒𝑎𝑏𝑒𝑑 = 16.54878 lbm/ft3 = ρ0 𝜌𝑠𝑒𝑎𝑏𝑒𝑑−𝑙𝑜𝑎𝑑𝑒𝑑 = 19.64878 lbm/ft3 = ρ1 𝜌 𝐵𝑆𝑅 𝑆ℎ𝑒𝑎𝑟 = 16.68412 lbm/ft3= ρ2

Summary for 1 x 28 US Gal DCB Required providing χ vol of fluid at 4229 psig for Casing shear pressure, followed by ϒ vol of fluid at 3371 psig for BSR shearing… Started with 28 US Gal, used 9.23 US Gal available at minimum 4229 psig, then 17.09 US Gal at minimum 3371 psig So if we needed for CSG shear 37.3 USGal with 10% reserve (already accounted for in𝑉𝐸𝑝 𝑎𝑛𝑑 𝑉𝐸𝑣), we would require (37.3)/9.23 = 4.041 5 DCB

Similarly for BSR shear, (37.3)/17.09 = 2.182 3 DCB The volume is driven by the N2 gas volume required to supply high pressure volume above CSG Shear MOP. So 5 DCB’s would provide enough fluid volume (5x9.23=46.15 US Gal) above 4229 psi to complete CSG Shear, and still have enough (5x17.09=85.45 US Gal) to then complete BSR shear at 3371 psig. Using VEpCSG = 0.04166, we would have VolN2 = FVR / VEpCSG = (37.3)/0.04166 = 895.34 US Gal VolN2 / N2 Vol seabed = 984.86 / 208.47 = 4.295 5 DCB’s