138
M. Muttakin et al.
T out,cool = T bed,ads −
T bed,ads − T in,cool
exp
−
(U A) bed
˙
m cool C p,cool
(8.46)
Since the adsorber bed is connected to the evaporator, the equilibrium uptake
w* is a function of evaporator pressure and saturation pressure of water at the bed
temperature.
Bed Equations During Mass Recovery
In the mass recovery stage, there is no water flow through any of the bed, as can be
seen in Fig. 8.8. Hence the outlet temperatures of hot and cooling water will remain
the same as their inlet temperatures. The beds are also isolated from the evaporator
and condenser. At the end of the ad/de-sorption phase, the pressure at Bed 1 becomes
higher than that at Bed 2. The opening of the by-pass valve in the mass recovery
stage initiates the flow of excess vapor from Bed 1 to Bed 2 due to pressure swing,
and the process is usually continued (accomplished by appropriate determination of
cycle time) until the mechanical equilibrium between the two beds is attained. The
flow rate of vapor, from Bed 1 to Bed 2, during this stage can be determined from
Thu et al. (2017),
dw
dt
= 1.41576Y A bp
Pρ
K
(8.47)
where Y is the expansion factor (Shashi Menon 2015), A bp is the cross-section area
of the bypass valve, ΔP is the pressure difference between the two beds and K is the
total resistance coefficient. It needs mentioning that the reversed vapor flow due to
pressure changes is also accounted for in the above model. K can be calculated from,
K = f
L
D
+
i
ξ i
where ξ is the minor loss coefficient for different components, L and D are the length
and width of the bypass valve, respectively. f is the friction factor given as,
f =
⎧
⎨
⎩
64
Re
f or Re < 2300
1.325
ln
e
3.7D +
5.74
Re 0.9
2 f or 5000 ≤ Re ≤ 10
8 &10
−6
≤
e
D
≤ 10
−2
The energy balance equations for mass recovery can be written as follows;
for Bed 1,
MC p
bed,des
dT bed,des
dt
= −M bed h ads
dw
dt
(8.48)
M. Muttakin et al.
T out,cool = T bed,ads −
T bed,ads − T in,cool
exp
−
(U A) bed
˙
m cool C p,cool
(8.46)
Since the adsorber bed is connected to the evaporator, the equilibrium uptake
w* is a function of evaporator pressure and saturation pressure of water at the bed
temperature.
Bed Equations During Mass Recovery
In the mass recovery stage, there is no water flow through any of the bed, as can be
seen in Fig. 8.8. Hence the outlet temperatures of hot and cooling water will remain
the same as their inlet temperatures. The beds are also isolated from the evaporator
and condenser. At the end of the ad/de-sorption phase, the pressure at Bed 1 becomes
higher than that at Bed 2. The opening of the by-pass valve in the mass recovery
stage initiates the flow of excess vapor from Bed 1 to Bed 2 due to pressure swing,
and the process is usually continued (accomplished by appropriate determination of
cycle time) until the mechanical equilibrium between the two beds is attained. The
flow rate of vapor, from Bed 1 to Bed 2, during this stage can be determined from
Thu et al. (2017),
dw
dt
= 1.41576Y A bp
Pρ
K
(8.47)
where Y is the expansion factor (Shashi Menon 2015), A bp is the cross-section area
of the bypass valve, ΔP is the pressure difference between the two beds and K is the
total resistance coefficient. It needs mentioning that the reversed vapor flow due to
pressure changes is also accounted for in the above model. K can be calculated from,
K = f
L
D
+
i
ξ i
where ξ is the minor loss coefficient for different components, L and D are the length
and width of the bypass valve, respectively. f is the friction factor given as,
f =
⎧
⎨
⎩
64
Re
f or Re < 2300
1.325
ln
e
3.7D +
5.74
Re 0.9
2 f or 5000 ≤ Re ≤ 10
8 &10
−6
≤
e
D
≤ 10
−2
The energy balance equations for mass recovery can be written as follows;
for Bed 1,
MC p
bed,des
dT bed,des
dt
= −M bed h ads
dw
dt
(8.48)