散热器恒温阀英文文献和中文翻译(3)
The above model parameters can be identified simply via a step response test as well.
Step response simulations and experiments confirm a first order transfer function between the radiator output heat and input flow rate at a specific operating point as
In the next section, parameters of the above model are formulated based on the closed-form solution of the radiator output heat, Q r (t, q, Ta).
C. Radiator Dynamical Analysis
In this paper, unlike [3], we found the closed-form map between the radiator heat and operating point which is corresponding flow rate q, and room temperature Ta. We, previously, derived this dependency via a simulative study in the form of two profile curves, [3].
To develop Q (t, q, Ta), a step flow is applied to the radiator, i.e. changing the flow rate from q0 to q1, at a constant differential pressure across the valve. Propagating with the speed of sound, the flow shift is seen in a fraction of second all along the radiator. Hence, flow is regarded as a static parameter for t > 0, rather than temperature distribution along radiator.
Consider a small radiator section △x with depth d and height h as shown in Fig. 4. The temperature of incoming flow to this section is T(x), while the outgoing flow is at T(x+△x)℃. Temperature is considered to be constant T(x) in a single partition.
Fig.4.A radiator section area with the heat transfer equation governed by (8)
The corresponding heat balance equation of this section is given as follows.
in which flow rate is q0 at t = 0 and q1 for t > 0. Cr is the heat capacity of water and the radiator material defined as: Cr = c ω ρ ω V ω. Dividing both sides by △x and approaching △x → 0, we have:
with boundary condition T(0， t) = Tin, T(℩， 0一) = Tout，0 and T(℩, ∞) = Tout,1. If there exists a separable solution, it would be like T(x, t) = T(t) × X(x). Substituting it into (9), we achieve:
which implies a contradiction.
Before proceed to solve the full PDE (9), we need to find the two boundary conditions Tout，0 and Tout，1. For this purpose, take the steady state form of (9) as follows.
which can be written as:
with constants β = and γ = . We will be using the two definitions throughout the paper frequently.
Therefore, the steady state temperature, T(x, t) | t→∞ will be achieved as:
at the specific flow rate q. Substituting the above equation in (12) gives c0 = Ta. Knowing T (0) = Tin, c1 is also found. Finally T(x) looks like:
Therefore the two boundary conditions are: Tout，0 = (Tin - Ta) + Ta and Tout，1 = (Tin - Ta) + Ta corresponding to the flow rates q0 and q1.
Generally solving the full PDE (9) in time domain is a difficult task. However we are interested in the radiator transferred heat to the room rather than temperature distribution along the radiator. Instead of T(x, t), therefore, we will find Q (t) which is independent of x. Q (t) can be formulated as:
Taking time derivative of the above equation and using (9):
With β = . The above equation can be rewritten as:
In which Tin is the constant forward temperature. However Tout in the above equation is a function of time. Therefore we need an expression for Tout (t) which is attained in the following. To develop Tout (t), consider (9) at x = ℩:
The first term in the left side of the above equation is an unknown function of time which we call it f (t). Thus the above equation can be rewritten as: