Brine-loop balancing with a singular valve setting
A small desalination pilot plant circulates brine through three tanks. Let x, y and z be the adjustment flows in L/min added to tanks A, B and C. The flow balances are x + 2y = 3, 2x + ky + z = 5 and y + 3z = 7, where k is the recirculation-valve setting. The control screen must decide whether k = 1 is safe. A technician notices that for one valve setting the three balances may fail to determine the flows uniquely, and proposes locking the valve at k = 4 instead. The supervisor wants a written judgement before the next run. Tanks cannot accept negative adjustment flows, and the plant is designed to operate with 0 ≤ k ≤ 4. The engineer also asks what happens if the third balance becomes y + 3z = 7 + δ because of a small calibration drift, where δ is measured in L/min.
1. Compute the determinant of the coefficient matrix. For which valve setting is the system not uniquely solvable, and what happens at that setting?
Read the worked answer
Write the coefficient matrix A = [[1,2,0],[2,k,1],[0,1,3]]. Expanding along the first row, det A = 1(3k - 1) - 2(6 - 0) + 0 = 3k - 13. Hence the system has a unique solution exactly when k ≠ 13/3.
At k = 13/3, use the first equation to write x = 3 - 2y, and the second to write z = (4 - k)y - 1. Substituting into the third gives y + 3((4 - k)y - 1) = 7, i.e. (13 - 3k)y = 10. At k = 13/3 this is 0 = 10, so the system is inconsistent: there is no flow triple, not infinitely many. The valve setting 13/3 must be avoided.
2. For k = 1, solve the system. If the third right-hand side is 7 + δ, find the range of δ that keeps all three adjustment flows nonnegative.
Read the worked answer
For k = 1, x = 3 - 2y and z = 3y - 1. The third equation gives y + 3(3y - 1) = 7, so 10y = 10 and y = 1. Then x = 1, z = 2.
With 7 replaced by 7 + δ, the same substitution gives y + 3(3y - 1) = 7 + δ, so 10y = 10 + δ, y = 1 + δ/10. Then x = 3 - 2(1 + δ/10) = 1 - δ/5 and z = 3(1 + δ/10) - 1 = 2 + 3δ/10. Nonnegativity requires 1 - δ/5 ≥ 0, 1 + δ/10 ≥ 0 and 2 + 3δ/10 ≥ 0. These give δ ≤ 5, δ ≥ -10 and δ ≥ -20/3. The binding lower bound is -20/3, so δ ∈ [-20/3, 5].
3. Evaluate the technician's proposal to use k = 4. Include the determinant and the solution, and state whether the plant should accept it.
Read the worked answer
At k = 4, det A = 3(4) - 13 = -1 ≠ 0, so a unique solution exists. From (13 - 3k)y = 10, y = 10/(13 - 12) = 10. Then x = 3 - 2(10) = -17 and z = (4 - 4)(10) - 1 = -1. The unique solution is (-17, 10, -1). Although mathematically valid, two adjustment flows are negative, which the tanks cannot accept. The proposal k = 4 should be rejected for operation, even though the matrix is nonsingular.