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1173018735新蟲 (初入文壇)
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[求助]
fluent模擬流化床甲烷干重整反應(yīng)udf使用浮點(diǎn)溢出
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我在復(fù)現(xiàn)一篇用歐拉雙流體方法復(fù)現(xiàn)流化床甲烷干重整反應(yīng)的論文,我找不出來有什么問題,udf中公式的表達(dá)是沒問題的,這個(gè)模型不使用UDF可以正常計(jì)算,但是一使用了UDF就會(huì)報(bào)錯(cuò)。而這個(gè)UDF在另外一個(gè)簡(jiǎn)單的模型里使用沒有問題。有大佬幫我看看什么問題嗎? # include "udf.h" # include "mem.h" DEFINE_VR_RATE(rate, c, t, r, wk, yk, rate, rr_t) { #define UGC 8.314 real T = C_T(c, t); real k1 = 1.29e6 * exp(-102065.0 / UGC / C_T(c, t)); real k2 = 0.35e6 * exp(-81030.0 / UGC / C_T(c, t)); real k3 = 6.95e3 * exp(-58893.0 / UGC / C_T(c, t)); real k4 = 5.55e9 * exp(-166397.0 / UGC / C_T(c, t)); real k5 = 1.34e15 * exp(-243835.0 / UGC / C_T(c, t)); real Kco2_1 = 2.61e-2 * exp(37641.0 / UGC / C_T(c, t)); real Kch4_1 = 2.60e-2 * exp(40684.0 / UGC / C_T(c, t)); real Kco2_2 = 0.5771 * exp(9262.0 / UGC / C_T(c, t)); real Kh2_2 = 1.494 * exp(6025.0 / UGC / C_T(c, t)); real Kch4_3 = 0.21 * exp(-567.0 / UGC / C_T(c, t)); real Kh2_3 = 5.18e7 * exp(-133210.0 / UGC / C_T(c, t)); real Kh2_4 = 1.83e13 * exp(-216145.0 / UGC / C_T(c, t)); real Kh2o_4 = 4.73e-6 * exp(97770.0 / UGC / C_T(c, t)); real Kch4_4 = 3.49 * exp(-0 / UGC / C_T(c, t)); real Kco_5 = 7.34e-6 * exp(100395.0 / UGC / C_T(c, t)); real Kco2_5 = 2.81e7 * exp(-104085.0 / UGC / C_T(c, t)); real Kp1 = 6.78e14 * exp(-259660.0 / UGC / C_T(c, t)); real Kp2 = 56.4971 * exp(-36580.0 / UGC / C_T(c, t)); real Kp3 = 2.95e5 * exp(-84400.0 / UGC / C_T(c, t)); real Kp4 = 1.3827e7 * exp(-125916.0 / UGC / C_T(c, t)); real Kp5 = 1.9393e9 * exp(-168527.0 / UGC / C_T(c, t)); real mCO = C_YI(c, t, 0); real mH2 = C_YI(c, t, 1); real mH2O = C_YI(c, t, 2); real mCH4 = C_YI(c, t, 3); real mCO2 = C_YI(c, t, 4); real pOp = 101325; real sum = (mCH4 / 16 + mCO2 / 44 + mCO / 28 + mH2 / 2 + mH2O / 18); real PCH4 = mCH4 / 16 / sum * pOp; real PCO2 = mCO2 / 44 / sum * pOp; real PCO = mCO / 28 / sum * pOp; real PH2 = mH2 / 2 / sum * pOp; real PH2O = mH2O / 18 / sum * pOp; if (!strcmp(r->name, "reaction-1" ){ *rate = k1 * Kco2_1 * Kch4_1 * (PCH4 / 101325) * (PCO2 / 101325) / pow((1 + Kco2_1 * (PCO2 / 101325) + Kch4_1 * (PCH4 / 101325)), 2) * (1 - pow(((PCO / 101325) * (PH2 / 101325)), 2) / (Kp1 * (PCH4 / 101325) * (PCO2 / 101325))); C_UDMI(c, t, 0) = *rate; } else if (!strcmp(r->name, "reaction-2" ){ *rate = k2 * Kco2_2 * Kh2_2 * (PCO2 / 101325) * (PH2 / 101325) / pow((1 + Kco2_2 * (PCO2 / 101325) + Kh2_2 * (PH2 / 101325)), 2) * (1 - (PCO / 101325) * (PH2O / 101325) / (Kp2 * (PCO2 / 101325) * (PH2 / 101325))); C_UDMI(c, t, 1) = *rate; } else if (!strcmp(r->name, "reaction-3" ){ *rate = k3 * Kch4_3 * ((PCH4 / 101325) - (pow((PH2 / 101325), 2) / Kp3)) / pow((1 + Kch4_3 * (PCH4 / 101325) + 1 / Kh2_3 * pow((PH2 / 101325), 1.5)), 2); C_UDMI(c, t, 2) = *rate; } else if (!strcmp(r->name, "reaction-4" ){ *rate = k4 / Kh2o_4 * ((PH2O / 101325) / (PH2 / 101325) - (PCO / 101325) / Kp4) / pow((1 + Kch4_4 * (PCH4 / 101325) + ((PH2O / 101325) / (PH2 / 101325)) / Kh2o_4 + pow((PH2 / 101325), 1.5) / Kh2_4), 2); C_UDMI(c, t, 3) = *rate; } else if (!strcmp(r->name, "reaction-5" ){ *rate = k5 / Kco_5 / Kco2_5 * ((PCO2 / 101325) / (PCO / 101325) - (PCO / 101325) / Kp5) / pow((1 + Kco_5 * (PCO / 101325) + ((PCO2 / 101325) / (PCO / 101325)) / (Kco_5 * Kco2_5)), 2); C_UDMI(c, t, 4) = *rate; } *rr_t = *rate; }@wuming524 |
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