Solution:

Area of filtration: A=2×0.3×20=3.6mΔp=248.7kN/m2

3

t1=300s, V1=1/4×0.7=0.175m t2-t1=1800s, ΔV=0.7-0.175=0.525m3 or t2 =2100s, V2=0.7m3 0.175and0.7?2?0.7Vm?3.6?2100K22222

?2?0.175Vm?KAt?3.6?300K22

Vm=0.2627m And K?0.7?1.4Vm3.6?2100223

?0.49?1.4?0.262712.96?2100?3.1517?10?5

capacity

Q?0.72100?500?2.692?10?4m/s

3

for the rotary drum filter

Area of filtration: A=πdL=3.14×2.2×1.5=10.362m2

Operating pressure:Δp=70kN/m 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??p?3.1517?10?52

70248.7?8.871?10?6

the capacity of rotary drum filter is equal to that of press filter

2.692?10?4??nV?n???K?A2?n?V2m????Vm?n?????8.871?10?6?10.362n2?0.25?0.26272??0.2627???Solving for n from the equation above by trial and error

2.692?10?4??nV?n???2K?A???2?Vm?Vm??n???n??0.0002381n??0.069?0.2627?

??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

m during the 50 s, volume of filtrate collected was 3.35×10 m during the 100 s. How much will the filtrate be obtained as filtering for 200 s? solution：

2-33-33

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

V23-33

?2VVm?KAt

2substituting variables given above into the equation gives

2.272?10?6?2?2.27?10?2?3.35?10?3Vm?0.093K?50 1 and

223.35?102?6?3Vm?0.093K?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?10-3

?4V?1.567?10?5At

2solving for volume of filtrate gained for 200s V=4.835×10 m

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

V223

?2VVm?KAt

filter medium resistance set to be negligible

V2?KAt

32V?AKt?1?10?10?10m

②The filtering medium resistance is taken into account

V2?2VVm?KAt

2V2?2V?10?1?10

2V??2?4?4?10023?9.05m

③ the equation for constant-pressure filtering is taken derivative as filtering medium resistance doesn’t be taken into account dVdt?KA2V2?10?12?10?0.5

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 (a) heat loss:

qA???T?Rii??TB1k1?B2k2?760?76.60.1140.138?0.2291.38?688.89W/m

2(b) the temperature at interface: qA??TB1k1?760?t0.1140.138?688.89W/m

2t=191℃

(c) contact resistance is 0.088°C·m2/W, heat loss: qA???T?Rii??TB1k1?B2k2?0.088?760?76.60.1140.138?0.2291.38?0.088?683.4W/m

2

4.4. 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 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

o

R?BkAm?0.4260.615??(1.278?0.426)/ln(1.278/0.426)L?0.2357L

Heat loss per meter tube through the wall is

qL??TR?177?380.2357?589.7W/m

and temperature distribution is qL??TR?177?t0.615?2??r?0.213?lnr0.213r0.213?589.7W/m

t?177?2277.5?r?0.213?ln

4.5. 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 k2/k1?2 and both insulating materials have the same thickness 30mm. If the temperature

difference between pipe wall and outer surface of insulating material is constant, which insulating material should be packed inside to enable less heat loss? solution：

k1（lower thermal conductivity）

the insulating material with a lower thermal conductivity placed outside, rate of heat loss Q???t?Ri??tb1k2Am1?b2k1Am2 Am2?2?L(r3?r2)lnr3r2

iand Am1?2?L(r2?r1)lnr2r1?, thickness of two layers are equal b1=b2=b, and k2/k1?2

Q??tb2k1Am1?bk1Am2k1b12Am1?t?1Am2

contrarily, place the material with lower thermal conductivity inside, heat loss Q???tb1k1Am1?b2k2Am2??tbk1Am1?b2k1Am2?k1b1Am1?t?12Am2