In the figure attached for reference, it is given that if $R_1$ is the resistance element of valve 1 and τ ∝ $R_1$, but I am not able to understand it intuitively.
If $R_1$ is high, output flow through that valve must be less, this means it is easier to store the liquid in the tank. Hence, τ ∝ $\frac{1}{R_1}$.
Please tell where I am wrong in my logic. TIA.
I'm an electrical engineer so this problem looks to me like an RC (resistor-capacitor) discharge. In the case of a resistor discharging a capacitor the time constant is given by τ = RC. In your case C represents the tank capacity and R is your R1, the resistance to flow.
Figure 1. The top curve shows the capacitor voltage which is analogous to liquid level height in your tank. The lower trace shows the current which is analogous to the flow out of the tank. (The graph is negative because it's measuring current in.) Image source: Electronics-Circuits.
It should make sense that τ will be proportional to (increase with) both the cross-sectional area and to increasing resistance of the output valve.
Some rules of thumb commonly understood in the electronics world are:
Although Transistor made a perfectly good analogy, I will give my view from an mechanical engineering point of view.
High resistance R means that flow will be restricted. I.e. to empty tank one you would need more time.
The characteristic time $\tau$ in this example is indicative of the time it will take to from empty about 2/3 of the tank.
$\tau$ is related to discharge because $R_1$ in this example controls only the output. This is in contrast to the case electrical case where R controls both charging and discharging.
The actual relationship has the following form: $$h = h_0 (1-e^{-t/\tau})$$ where:
As you can imagine the more restrictive the flow, the more times it takes to empty the tank, the higher the value of $\tau$.
If flow is q, and pressure is p, The linear flow characteristic "R" is defined such that p = (q)(R)
This is directly analogous to V = IR in ohms's law, in the electrical analogy.
If we differentiate p=qR, we get dp = (dq)(R)
In this case, pressure is being reckoned in units of height, for which it should be mentioned that p = (ρ)(g)(h) where the greek letter rho (ρ) is the liquid density, g is the gravity (9.8m/s^2), and h is height.
Thus we write dh = (dq)(R)
We substitute this into equation 2.68:
(A)(dh/dt) = qi - q1
which becomes
(A)(R)(dq/dt) = qi - q1
which is a first order ODE, so the solution is exponential form, and the time constant is AR, which is of course proportional to R.
On an qualitative/intuitive level, bigger R means more constricted flow, which means less flow for any given pressure aka height, which means it takes longer for the tank to drain, which means larger time constant. Similarly, bigger A means there is more liquid to be drained for a given height, which means it takes longer, which means larger time constant.
