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p(t) = p_i + \sum_{\alpha = 1}^{N} \big(left[ q^{(\alpha)}_k - q^{(\alpha-1)}_k \big)right] \cdot p^up_u(t - t_t^{\alpha k}) |
where
| 1 | | _n | pressure at -th well at arbitrary moment of time |
| { | ,n} | initial pressure at -the well | 3 |
| q^{( | )}_nrate value of | index number of a transient (period of time where rate was constant) |
| \alpha | -th transient at | total number of transients |
| n | | th well
| 4 | | pressure transient response in -th wel to unit-rate production from -th well |
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body | t_{\alpha k} | -uriencoded--t%5e%7B\alpha%7D |
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| starting point of the -th transient in |
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| k | -th well6| --uriencoded--q%5e%7B(\alpha)%7D |
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| 7 | | N | number of wells in the test | -th transient which starts at the time moment | | LaTeX Math Inline |
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| body | --uriencoded--t%5e%7B(\alpha)%7D |
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| N_k | number of transients in | k | | pressure transient response to the unit-rate production | -th well |
with assumption:
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| body | q^{--uriencoded--q%5e%7B(-1)}_k %7D = 0 |
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– for any well , which means that well was shut-in before it started the first transient
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| body | p^up_{nk}(\tauu(t) = 0 |
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at for any pair of wells | LaTeX Math Inline |
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body | n, k = 1.. \ N which means pressure drop is zero before the well starts unit-rate production
Hence, convolution is using initial formation pressure
, unit-rate transient responses of wells and cross-well intervals
and rate histories
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| body | \{ q_k (t) \}_{k = 1 .. N} |
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to calculate pressure bottom-hole pressure response as function time
:
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