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774 bytes added ,  16:03, 17 March 2014
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:<math>\text{Hydrocarbon tilt} = \rho_{w}/(\rho_{w} - \rho_{h}) \times \text{potentiometric gradient}</math>
 
:<math>\text{Hydrocarbon tilt} = \rho_{w}/(\rho_{w} - \rho_{h}) \times \text{potentiometric gradient}</math>
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where ''ρ''<sub>''h''</sub> = density of the hydrocarbon.
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Because the density difference between gas and water is greater than that between oil and water, the dips of gas-oil and gas-water contacts are the same and always less than those of oil-water contacts responding to the same hydrodynamic gradient. The dip direction is the same for all fluid contacts.
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''Example:'' Determine the position and dips of the fluid contacts from data in [[Figure 3]]. Well B is a discovery well with reservoir gas, oil, and water densities of 0.15, 0.8, and 1.0 g/cm<sup>3</sup>, respectively. Gas and oil contacts are at elevations of -5020 and -5080 [[length::ft]], respectively.
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''Solution:'' First determine (in English units) the potentiometric gradients using sea level datum:
    
:<math>h_{\rm A}  = 2369/0.433 - 5160 = 311 \mbox{ ft}</math>
 
:<math>h_{\rm A}  = 2369/0.433 - 5160 = 311 \mbox{ ft}</math>

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