(2) the pressure difference is double

K??p???2 K?pso

K′=500

?q??10???q?10???2 ??2q??2(q?10)?10?1.41?250?10?342.5l/m2

V=34.25 l

3.9 Filtration is carried out in a plate and frame filter press, with 20 frames 0.3 m square and 50mm thick. At a constant pressure difference of 248.7kN/m2, one-quarter of the total filtrate per cycle is obtained for the first 300s. Filtration is continued at a constant pressure for a further 1800s, after which the frames are full. The total volume of filtrate per cycle is 0.7 m3 and dismantling and refitting of the press takes 500s

It is decided to use a rotary drum filter, 1.5m long and 2.2m in diameter, in place of the filter press. Assuming that the resistance of the cloth is the same and that the filter cake is incompressible, calculate the speed of rotation of the drum which will result in the same overall rate of filtration as was obtained with the filter press. The filtration in the rotary filter is carried out at a constant pressure difference of 70kN/m2 and with 25% of the drum submerged Solution:

Area of filtration: A=2×0.32×20=3.6m2 Δp=248.7kN/m2

t1=300s, V1=1/4×0.7=0.175m3

t2-t1=1800s, ΔV=0.7-0.175=0.525m3 or t2 =2100s, V2=0.7m3

0.1752?2?0.175Vm?KA2t?3.62?300Kand0.72?2?0.7Vm?3.62?2100KVm=0.2627m3 And

0.72?1.4Vm0.49?1.4?0.2627K???3.1517?10?5 212.96?21003.6?2100capacity

Q?0.7?2.692?10?4m3/s

2100?500

for the rotary drum filter

Area of filtration: A=πdL=3.14×2.2×1.5=10.362m2 Operating pressure:Δp=70kN/m2 And

the drum submerged : φ=25%

For rotary drum filter the filtration coefficient: keep Vm unchanged, K changes with changing in pressure difference

K??K?p?70?3.1517?10?5?8.871?10?6 ?p248.7the capacity of rotary drum filter is equal to that of press filter

2.692?10?4????8.871?10?6?10.3622?0.252?22????nV?n?K?A?Vm?Vm??n?0.2627?0.2627???nn????Solving for n from the equation above by trial and error

???0.0002381??2??? ?A2?Vm2.692?10?4?nV?n?K?V?n?0.069?0.2627m????nn????n=0.0088=0.048rpm(转/min)

3.10 A press filter of 0.093m2 filtering surface was used to separate the suspension containing calcium carbonate, operating in constant pressure. The volume of filtrate collected was 2.27×10-3 m3 during the 50 s, volume of filtrate collected was 3.35×10-3 m3 during the 100 s. How much will the filtrate be obtained as filtering for 200 s? solution：

filtering area A=0.093 m2; filtrate V=2.27×10-3 m3 for t=50s and V=3.35×10-3 m3 for t=100s for constant pressure filtration

V2?2VVm?KA2t

substituting variables given above into the equation gives

2.272?10?6?2?2.27?10?3Vm?0.0932K?50 1 and 3.352?10?6?2?3.35?10?3Vm?0.0932K?100 2

combining equations 1 and 2 Vm=3.85×10-4 m3 and K=1.567×10-5 m2/s

Substitution of the filtering-constants Vm and K into the equation for constant-pressure equation gives

V2?7.70?10?4V?1.567?10?5A2t

solving for volume of filtrate gained for 200s

V=4.835×10-3 m3

3.11 A slurry with incompressible cake is filtered by a 1m2 filter press at constant pressure. If the filtering constant K is 10 m2/min, operating pressure difference ?p is 2 atm and the septum resistance can be ignored. Find

(1) the filtrate volume, in m3 when the filtering time is 10 min.

(2) if the septum resistance must be considered and Vm is 1 m3, what is the filtrate volume V, in

m3 when the filtering time is 10 min ?

(3) what is the filtration rate dV/dt when the filtering time is 10 min for case (1)? Solution:

Equation for the constant-pressure filtration

V2?2VVm?KA2t

filter medium resistance set to be negligible

V2?KA2t

V?AKt?1?10?10?10m3

②The filtering medium resistance is taken into account

V2?2VVm?KA2t

V2?2V?10?12?10 V??2?4?4?100?9.05m3

2③ the equation for constant-pressure filtering is taken derivative as filtering medium resistance doesn’t be taken into account

dVKA210?1???0.5 dt2V2?10

4.2. A flat furnace wall is constructed of a 114-mm layer of Sil-o-cel brick, with a thermal conductivity of 0.138 W/(m·°C) backed by a 229-mm layer of common brick, of conductivity 1.38 W/(m·°C). The temperature of the inner face of the wall is 760°C, and that of the outer face is 76.6°C. (a) What is the heat loss through the wall? (b) What is the temperature of the interface between the refractory brick and the common brick? (c) Supposing that the contact between the two brick layers is poor and that a ―contact resistance‖ of 0.088°C·m2/W is present, what would be the heat loss? Solution:

From equation4.2-11

q(a) heat loss:?A??T?Rii??TB1B2?k1k2?760?76.6?688.89W/m2

0.1140.229?0.1381.38(b) the temperature at interface:

q?T760?t???688.89W/m2

0.114AB10.138k1t=191℃

(c) contact resistance is 0.088°C·m2/W, heat loss:

q?A

??T?Rii??T760?76.6??683.4W/m2

B1B20.1140.229??0.088??0.0880.1381.38k1k24.4. The vapor pipe (do=426 mm) is covered by a 426-mm insulating layer (k=0.615 W/m?C). If the temperature of outer surface of pipe is 177 C and the temperature outside the insulating layer

o

is 38 C, what are the heat loss per meter pipe and the temperature profile within the insulating layer?

Solution：

Similar to compound resistances in series through the flat wall, the total resistance across insulating layer is as follows

o

o

R?B0.4260.2357?? kAm0.615??(1.278?0.426)/ln(1.278/0.426)LLHeat loss per meter tube through the wall is

q?T177?38???589.7W/m LR0.2357and temperature distribution is

q?T??LR177?tr0.615?2??r?0.213?ln0.213?589.7W/m

t?177?2277.5?r?0.213?ln

r 0.2134.5. The outer diameter of a steel tube is 150mm. The tube wall is backed by two insulating layers to reduce the heat loss. The ratio of thermal conductivity of two insulating materials is