Source: Midterm Make Up Assignment S26-4-1-1
Tags: forward finite difference, Fourier's law, thermal conduction, numerical differentiation, mass balance, mixing problem, steady-state concentration, sugar concentration
Forward finite difference approximates a derivative at a point using the value at that point and the next point in the dataset. Mass balance problems for a continuously stirred tank use the ratio of total solute inflow to total mass inflow to find steady-state concentration. Both are standard engineering tools for turning discrete data or flow rates into usable answers.
Forward finite difference
A numerical method that approximates the first derivative at point i using the next data point:
dT/dx |ᵢ ≈ (Tᵢ₊₁ − Tᵢ) / (xᵢ₊₁ − xᵢ)
"Forward" because it looks ahead to point i+1.
Backward finite difference
The counterpart that looks behind:
dT/dx |ᵢ ≈ (Tᵢ − Tᵢ₋₁) / (xᵢ − xᵢ₋₁)
Central finite difference
Uses points on both sides for better accuracy:
dT/dx |ᵢ ≈ (Tᵢ₊₁ − Tᵢ₋₁) / (xᵢ₊₁ − xᵢ₋₁)
Fourier's Law of thermal conduction
q = k · A · (dT/dx)
where q is the heat transfer rate, k is thermal conductivity, A is cross-sectional area, and dT/dx is the temperature gradient.
Mass balance (steady-state mixing)
For a continuously stirred tank at steady state, the concentration of the outflow equals the ratio of total solute mass inflow rate to total mass inflow rate.
Continuously stirred tank reactor (CSTR) / infinitely large mixer
A model where the contents are perfectly mixed at all times. The outlet concentration equals the concentration inside the tank. "Infinitely large" means the tank never fills up, so you can treat it as reaching a steady-state concentration.
When you have a table of position vs. temperature data and need to compute heat transfer at a specific point, the procedure is:
Identify which point the question asks about.
Apply the correct finite difference formula (forward, backward, or central) using the neighbouring data points.
Plug the resulting derivative into Fourier's Law.
Given: k = 0.053 kW/mC, A = 0.73 m², and the data table:
Point | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
|---|---|---|---|---|---|---|---|
x (m) | 0.01 | 0.02 | 0.03 | 0.04 | 0.05 | 0.06 | 0.07 |
T (C) | 49.2 | 60.3 | 71.9 | 82.6 | 94.5 | 104.5 | 115.6 |
Task: calculate q at point 4 using the forward finite difference.
Step 1: Forward difference at point 4 uses points 4 and 5.
dT/dx = (T₅ − T₄) / (x₅ − x₄) = (94.5 − 82.6) / (0.05 − 0.04) = 11.9 / 0.01 = 1190 C/m
Step 2: Apply Fourier's Law.
q = 0.053 × 0.73 × 1190 = 46.04 kW
Forward: uses point i and i+1. Asked for explicitly or used at the left boundary where no i−1 exists.
Backward: uses point i and i−1. Used at the right boundary where no i+1 exists.
Central: uses i−1 and i+1. More accurate (second-order) but needs data on both sides.
Read the question carefully. If it says "forward finite difference," use forward, even if central would be more accurate.
When two (or more) streams flow into a perfectly mixed tank and the mixture drains out, the steady-state concentration depends only on what flows in, not on the drain rate.
The key insight: set dC/dt = 0 in the mass balance equation. The algebra simplifies to:
C_steady = (total solute mass inflow rate) / (total mass inflow rate)
The drain rate affects how quickly the tank reaches steady state, but not what the steady-state concentration is.
Given:
Apple juice: 10 kg/min at 29% sugar
Orange juice: 18 kg/min at 15% sugar
Drain rate: 20 kg/min
Infinitely large, continuously stirred mixer
Step 1: Compute sugar inflow rate.
Sugar in = (10 × 0.29) + (18 × 0.15) = 2.9 + 2.7 = 5.6 kg/min
Step 2: Compute total mass inflow rate.
Total in = 10 + 18 = 28 kg/min
Step 3: Steady-state concentration.
C = 5.6 / 28 = 0.20 = 20.00%
The formal derivation via differential equations:
Let M = total mass in tank, S = sugar mass, C = S/M.
dS/dt = sugar_in − C × rate_out
At steady state (dC/dt = 0), solving gives C = sugar_in_rate / total_in_rate. The drain rate cancels because the concentration inside the tank stabilises regardless of how fast mass leaves.
Forward finite difference:
dT/dx |ᵢ ≈ (Tᵢ₊₁ − Tᵢ) / (xᵢ₊₁ − xᵢ)
Backward finite difference:
dT/dx |ᵢ ≈ (Tᵢ − Tᵢ₋₁) / (xᵢ − xᵢ₋₁)
Central finite difference:
dT/dx |ᵢ ≈ (Tᵢ₊₁ − Tᵢ₋₁) / (xᵢ₊₁ − xᵢ₋₁)
Fourier's Law:
q = k · A · (dT/dx)
Steady-state mixing concentration:
C = Σ(rate_i × concentration_i) / Σ(rate_i)
⚠️ "Forward finite difference" means use point i and point i+1. If you accidentally use i−1, you have applied the backward scheme and the answer will be wrong.
⚠️ Make sure units are consistent. k in kW/mC and temperatures in C and positions in m gives q in kW directly. If k were in W/mC, the answer would be in W.
⚠️ For mixing problems, "infinitely large mixer" is code for "use the steady-state formula." The drain rate is a distractor in the steady-state case.
⚠️ The concentration formula uses total inflow rates, not the drain rate. This trips people up because the drain rate is given prominently.
Q: Using the data table above, what would the backward finite difference give for dT/dx at point 4?
A: (T₄ − T₃) / (x₄ − x₃) = (82.6 − 71.9) / (0.04 − 0.03) = 10.7 / 0.01 = 1070 C/m. Then q = 0.053 × 0.73 × 1070 = 41.41 kW.
Q: What would the central finite difference give at point 4?
A: (T₅ − T₃) / (x₅ − x₃) = (94.5 − 71.9) / (0.05 − 0.03) = 22.6 / 0.02 = 1130 C/m. Then q = 0.053 × 0.73 × 1130 = 43.72 kW.
Q: In a mixing problem, stream A flows at 5 kg/min with 40% solute, stream B at 15 kg/min with 10% solute. What is the steady-state concentration?
A: (5 × 0.40 + 15 × 0.10) / (5 + 15) = (2.0 + 1.5) / 20 = 3.5 / 20 = 17.50%.
Q: Why is the central finite difference generally more accurate than forward or backward?
A: It uses data on both sides of the point, which cancels first-order truncation error. Forward and backward are first-order accurate; central is second-order accurate.
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