Respuesta :
Temperature and inlet liquid related to transfer functions.
[tex]T(s)=\frac{wc T_i+h_ p{A _P T_w}}{h_P A _P+S mc+wc}}[/tex]
The steady-state value is [tex]h_pA_ p\left \bar T_{}(S)\right.[/tex] = [tex]\frac{w \bar{T}(w)}{h _p A_p} \cdot h_{S} A_ S[/tex]
Explanation:
[tex]m c \frac{d T}{d t}=w((T q-T)+h p A p(T w-T)[/tex] → 1
[tex]m_w c_{w} \frac{d T_{w}}{d t}=h_{s} A_{s}\left(T_{s}-T_{w}\right)-h_pA_p\left(T_{w}-T\right)[/tex] → 2
(a)
Both sides are divided by mc in the equation 1.
[tex]\frac{d T}{d t}=\frac{wc}{mc}\left[T_{i}-T\right]+\frac{h p A_ P}{m c} \quad(T_w-T)[/tex]
[tex]\frac{d T}{d t}=\frac{w}{m} \left[T_{i}-T\right]+\frac{h p A_ P}{m c} \quad(T_w-T)[/tex]
Laplace transform is applied on both sides.
[tex]L\left\{\frac{d T}{d t}\right\}=L\{\frac{\omega}{m}\left(T_{i}-T\right)+{L}\{ \frac{h_{P} A P}{m c}\left(T_{w}-T\right)\right.[/tex]
[tex]S \cdot T(S)-T(0)=\frac{w}{m}\left{(T_i-T(s))}+\frac{h p A p}{m c} {(T_w-T(s)})[/tex]
[tex]\frac{h_pA_p}{mc} T(s) + \frac{w}{m} T(s)[/tex] = [tex]T(0)+\frac{w T_i}{m}+\frac{h p A p}{m c} T w[/tex]
[tex]T(s)\left(\frac{h p A p}{m c}+s+\frac{w}{m}\right)[/tex] [tex]=\frac{w}{m}Ti[/tex][tex]+ \frac{h_pA_p}{mc} Tw[/tex]
[tex]T(s)=\frac{\left[\frac{w cT_i+h_{P} A_ P T_{w}}{m c}\right]}{\left(\frac{h_ P A _P}{mc}+S+\frac{w}{m}\right)}[/tex]
The denominators gets cancelled.
[tex]T(s)=\frac{wc T_i+h_ p{A _P T_w}}{h_P A _P+S mc+wc}}[/tex]
Temperature and inlet liquid related to transfer functions.
b)
[tex]\frac{d T}{d t}=0[/tex] is the value of steady state.
[tex]mc\frac{ d T}{d t}=w(T_i-T)+h_ p A_ p \quad\left(T_{w}-T\right)[/tex] → 1
[tex]m_{w}c_w \frac{d T_{w}}{d t}=h_{s}A_s \left(T_i-T} w\right)-h_p A_ p \left(T_{w}-T\right)\right.[/tex] → 2
[tex]mc\frac{dT}{dt}[/tex] = [tex]w\left(T_{i}-T\right)+ h_p A _P \left(T_{w}-T\right)[/tex]
[tex]\frac{d T}{d t}=0[/tex] for steady state.
0 = [tex]w\left(T_{is}-T(s)\right)+ h_p A _P \left(T_{ws}-T(s)\right)[/tex] →3
0 = [tex]w\bar T_is + h_pA_p \bar T_ws[/tex] → 4
From equation 2
[tex]m_{w}c_w \frac{d T_{w}}{d t}\right.[/tex] = [tex]h_{s} A _s\left(T_{s}-T_{w}\right)-h_{P} A_ p\left(T_{w}-T\right)[/tex]
[tex]\frac{d T}{d t}=0[/tex] for steady state.
0 = [tex]h_{s} A_ s\left(T_{s}-T_{w s}\right)-h_{p} A_ p\left(T_{w s}-T_{s}\right)[/tex]
0 = [tex]h_{S} A_{S}(\bar {T_{w}}(s))[/tex] + [tex]h_pA_p ( }\bar{T}(s))[/tex] → 5
[tex]h_pA_p }(\bar{T} w(s))=-w \bar{T}_{i}(s)[/tex]
[tex]\bar{T} w(s)=\frac{-w \bar{T}_i(s)}{h_p A_ P}[/tex] → 6
Equation 6 is substituted in equation 5
0 = [tex]w(T_i(S))[/tex] + [tex]h_pA_ p\left \bar T_{w}(S)\right.[/tex]
0 = [tex]\frac{-w \bar{T}(w)}{h _p A_p} \cdot h_{S} A_ S[/tex] - [tex]h_pA_ p\left \bar T_{}(S)\right.[/tex]
[tex]h_pA_ p\left \bar T_{}(S)\right.[/tex] = [tex]\frac{w \bar{T}(w)}{h _p A_p} \cdot h_{S} A_ S[/tex]
