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Zorluk: OrtaComparing Multiple Data Sources

Geothermal geologists analyzed fluid samples collected from 4 distinct wells (Well W, Well X, Well Y, and Well Z). Table 1 lists the measured reservoir temperature in degrees Celsius (C^\circ\text{C}) and the dissolved silica (SiO2SiO_2) concentration in milligrams per liter (mg/L\text{mg/L}) for each well.

Table 1:
WellReservoir Temperature (C^\circ\text{C})Dissolved SiO2SiO_2 (mg/L\text{mg/L})
Well W140180
Well X200400
Well Y170250
Well Z230520

Figure 1 displays the fluid viscosity (η\eta, in centipoise, cP\text{cP}) as a function of dissolved SiO2SiO_2 concentration at four different temperature curves:
- At 140C140^\circ\text{C}: η=0.005×(dissolved SiO2)+0.10\eta = 0.005 \times (\text{dissolved } SiO_2) + 0.10
- At 170C170^\circ\text{C}: η=0.003×(dissolved SiO2)+0.20\eta = 0.003 \times (\text{dissolved } SiO_2) + 0.20
- At 200C200^\circ\text{C}: η=0.002×(dissolved SiO2)+0.40\eta = 0.002 \times (\text{dissolved } SiO_2) + 0.40
- At 230C230^\circ\text{C}: η=0.001×(dissolved SiO2)+0.50\eta = 0.001 \times (\text{dissolved } SiO_2) + 0.50

Based on Table 1 and Figure 1, rank the geothermal wells in order from lowest fluid viscosity to highest fluid viscosity.

  1. 1Well Y
  2. 2Well W
  3. 3Well Z
  4. 4Well X

Cevap

The correct order from lowest to highest fluid viscosity is Well Y, Well W, Well Z, and Well X.
By looking up each well's specific reservoir temperature and dissolved SiO2SiO_2 concentration in Table 1 and substituting those values into the corresponding temperature equation from Figure 1, the calculated viscosities are 0.95 cP0.95\text{ cP} for Well Y, 1.00 cP1.00\text{ cP} for Well W, 1.02 cP1.02\text{ cP} for Well Z, and 1.20 cP1.20\text{ cP} for Well X. Sorting these from lowest to highest yields the sequence: Well Y, Well W, Well Z, Well X.

Adım Adım Çözüm

1
Extract the temperature and dissolved SiO2SiO_2 concentration for each well from Table 1.
Well W: 140C140^\circ\text{C}, 180 mg/L180\text{ mg/L}; Well X: 200C200^\circ\text{C}, 400 mg/L400\text{ mg/L}; Well Y: 170C170^\circ\text{C}, 250 mg/L250\text{ mg/L}; Well Z: 230C230^\circ\text{C}, 520 mg/L520\text{ mg/L}.
Both temperature and concentration are necessary to select the corresponding line equation from Figure 1.
2
Apply the viscosity formulas from Figure 1 corresponding to each well's temperature.
Well W (140C140^\circ\text{C}): η=0.005(180)+0.10=0.90+0.10=1.00 cP\eta = 0.005(180) + 0.10 = 0.90 + 0.10 = 1.00\text{ cP}; Well X (200C200^\circ\text{C}): η=0.002(400)+0.40=0.80+0.40=1.20 cP\eta = 0.002(400) + 0.40 = 0.80 + 0.40 = 1.20\text{ cP}; Well Y (170C170^\circ\text{C}): η=0.003(250)+0.20=0.75+0.20=0.95 cP\eta = 0.003(250) + 0.20 = 0.75 + 0.20 = 0.95\text{ cP}; Well Z (230C230^\circ\text{C}): η=0.001(520)+0.50=0.52+0.50=1.02 cP\eta = 0.001(520) + 0.50 = 0.52 + 0.50 = 1.02\text{ cP}.
Synthesizing data from both sources determines the numerical fluid viscosity for each well.
3
Arrange the resulting viscosity values from smallest to largest.
0.95 cP (Well Y)<1.00 cP (Well W)<1.02 cP (Well Z)<1.20 cP (Well X)0.95\text{ cP (Well Y)} < 1.00\text{ cP (Well W)} < 1.02\text{ cP (Well Z)} < 1.20\text{ cP (Well X)}.
The question specifies ranking from lowest viscosity to highest viscosity.

Anahtar Kavram

Synthesizing values from a data table with temperature-dependent functional relationships in a graph.
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