1. Starting with Equations, obtain Equation 7.111 for the linearized state derivatives. , 2. The nonlinear tank shown in Figure E7.21a has an adjustable valve in the discharge line. The valve...

1. Starting with Equations, obtain Equation 7.111 for the linearized state derivatives.

,

2. The nonlinear tank shown in Figure E7.21a has an adjustable valve in the discharge line. The valve opening is given by the normalized variable θ (0 ≤ θ ≤ 1) where θ = 0 is a closed valve and θ = 1 represents a fully open valve. The outflow is obtained from F₀=F₀(θ, H)=c(θ)H^1/2.

An expression for ∆F₀(t) in the linearized differential equation model of the tank A(d/dt) ∆H(t) + ∆F₀(t)=∆F₁(t) is obtained as follows:

Data points along the valve-operating characteristic c(θ) are shown in Figure E7.21b:

a. Find the linearized differential equation about the steady-state operating point where θ =0 5. , H =9 ft .

b. Simulate the tank-level response when the inflow increases by 10% and valve opening decreases by 15% with respect to their operating point values. The initial conditions are H(0)=H, θ(0)=θ. Assume both changes are step inputs. The cross-sectional area of the tank is 50 ft².