- 2071 Magh · 8 marks
Newton interpolation: steam table
Short answer
Equal spacing h=10. At 142°C, forward formula (p=0.2): P≈3.899 kgf/cm². At 175°C, backward formula (p=−0.5): P≈9.100 kgf/cm².
The following data are taken from the steam table.
| Temp. °C | 140 | 150 | 160 | 170 | 180 |
|---|---|---|---|---|---|
| Pressure kgf/cm² | 3.685 | 4.854 | 6.302 | 8.076 | 10.225 |
| Find the pressure at the temperature T = 142°C and T = 175°C using Newton's interpolation. |
The temperatures are equally spaced (h = 10). T = 142 is near the start of the table (forward formula) and T = 175 is near the end (backward formula).
Difference table
| x | y | Δy | Δ²y | Δ³y | Δ⁴y |
|---|---|---|---|---|---|
| 140 | 3.685 | 1.169 | 0.279 | 0.047 | 0.002 |
| 150 | 4.854 | 1.448 | 0.326 | 0.049 | |
| 160 | 6.302 | 1.774 | 0.375 | ||
| 170 | 8.076 | 2.149 | |||
| 180 | 10.225 |
Pressure at T = 142°C (Newton's forward formula)
T₀ = 140, p = (142 − 140)/10 = 0.2. Top diagonal: Δy₀ = 1.169, Δ²y₀ = 0.279, Δ³y₀ = 0.047, Δ⁴y₀ = 0.002.
| Term | Coefficient | Difference | Value |
|---|---|---|---|
| y₀ | 1 | 3.685 | 3.6850 |
| Δy | p = 0.2000 | 1.169 | 0.2338 |
| Δ²y | p(p−1)/2! = -0.0800 | 0.279 | -0.0223 |
| Δ³y | p(p−1)(p−2)/3! = 0.0480 | 0.047 | 0.0023 |
| Δ⁴y | p(p−1)(p−2)(p−3)/4! = -0.0336 | 0.002 | -0.0001 |
P(142) ≈ 3.8987 kgf/cm²
Pressure at T = 175°C (Newton's backward formula)
Tₙ = 180, p = (175 − 180)/10 = -0.5. Last diagonal: ∇yₙ = 2.149, ∇²yₙ = 0.375, ∇³yₙ = 0.049, ∇⁴yₙ = 0.002.
| Term | Coefficient | Difference | Value |
|---|---|---|---|
| yₙ | 1 | 10.225 | 10.2250 |
| ∇y | p = -0.5000 | 2.149 | -1.0745 |
| ∇²y | p(p+1)/2! = -0.1250 | 0.375 | -0.0469 |
| ∇³y | p(p+1)(p+2)/3! = -0.0625 | 0.049 | -0.0031 |
| ∇⁴y | p(p+1)(p+2)(p+3)/4! = -0.0391 | 0.002 | -0.0001 |
P(175) ≈ 9.1005 kgf/cm²
Answer: P(142°C) ≈ 3.899 kgf/cm², P(175°C) ≈ 9.100 kgf/cm²
✓ Answer checked by Claude Opus 5.5 (high effort)
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