**4.1.1 Instantaneous equation setup of gas solid flow**

According to mass conservation and momentum conservation law, gas and solid phase instantaneous volume averaged conservation equations can be given as following. Gas continuity equation:

$$\frac{\partial \left( \alpha\_{\mathcal{S}} \rho\_{\mathcal{S}} \right)}{\partial t} + \frac{\partial \left( \alpha\_{\mathcal{S}} \rho\_{\mathcal{S}} u\_{\mathcal{S}^j} \right)}{\partial x\_j} = 0 \tag{4}$$

Solid continuity equation:

$$\frac{\partial \left( \alpha\_s \rho\_s \right)}{\partial t} + \frac{\partial \left( \alpha\_s \rho\_s u\_{s\circ} \right)}{\partial x\_j} = 0 \tag{5}$$

Gas momentum equation:

378 Computational Simulations and Applications

As we all know, the physical aspects of fluid flow are governed by three fundamental principles (1) mass is conserved; (2) Newton's second law; (3) energy is conserved. So N-S

To this paper, the suitable mathematical models were selected according to experimental conclusions. As we all know, the Reynolds Number values of gas and granule in dense phase gas solid flow are both high than that of lean phase, which lead to more turbulent motion. Therefore, the addition of solid particles greatly changed gas phase turbulent construction and meanwhile gas fluctuation affect particle motions. So the interaction

When the solid concentration is high enough, interaction among particles affect solid flow characteristics greatly. Consequently, the interaction among particles should be given in simulation process in addition to gas solid interaction and turbulent as the gas solid two

Two-fluid model which based on granule dynamics theory was adapted in the study. Flow parameters such as macroscopic granule transport equations, solid pressure, viscous coefficient, diffusion coefficient, heat conductivity coefficient, granule temperature etc can be obtained through the model. This model was used comprehensively in several fields because it mentioned interation action of gas solid flow, granule turbulent viscosity and

In this study, by using k-ε two phase model, granule dynamics model and gas solid two phase coupled, the gas solid two phase turbulent model were built up. Some reasonable

1. The particles were composed of smooth rigid sphere with the same diameter. And during the flow process, two sphere collisions were mentioned while collision among

2. Gas solid two phase existed in the flow pipe homogeneously with defined physical parameter. Each phase was continuous while the time averaged velocity and volume

3. The acting force of solid phase involved gravity and resistance force. Other kind of force such as buoyancy, false-mass-force, Basset force, thermophoretic force etc was

4. The turbulent impulse of gas solid two phase was isotropy. Diffusion and Brownian movement effect could be neglected. The change of gas phase turbulent energy showed

According to mass conservation and momentum conservation law, gas and solid phase

<sup>0</sup> *g g g g gj*

 

*t x*

*j*

*u*

(4)

instantaneous volume averaged conservation equations can be given as following.

 

assumptions about the flow process must be given as following.

between the two phases leads to the mass, momentum and energy transmission.

**4. Numerical simulation of gas solid flow in pipe** 

**4.1 Mathematical models** 

phase flow turbulent model being set up.

lots of solid particles must be ignored.

ignored according to experimental cases.

**4.1.1 Instantaneous equation setup of gas solid flow** 

the influence of particle to gas phase.

equations are formed.

particle collision roundly.

ratio was different.

Gas continuity equation:

Solid continuity equation:

$$\frac{\partial \left(\alpha\_{\mathcal{S}} \rho\_{\mathcal{S}} \mu\_{\mathcal{S}^i}\right)}{\partial t} + \frac{\partial \left(\alpha\_{\mathcal{S}} \rho\_{\mathcal{S}} \mu\_{\mathcal{S}^i} \mu\_{\mathcal{S}^i}\right)}{\partial \mathbf{x}\_j} = -\alpha\_{\mathcal{S}} \frac{\partial p}{\partial \mathbf{x}\_i} + \alpha\_{\mathcal{S}} \rho\_{\mathcal{S}} g\_i + \frac{\partial \pi\_{ij}}{\partial \mathbf{x}\_j} + F\_i^{cd} \tag{6}$$

Solid momentum equation:

$$\frac{\partial \left(a\_s \rho\_s \mu\_{si}\right)}{\partial t} + \frac{\partial \left(a\_s \rho\_s \mu\_{si} \mu\_{sj}\right)}{\partial \mathbf{x}\_j} = -a\_s \frac{\partial p\_s}{\partial \mathbf{x}\_i} + a\_s \rho\_s g\_i + \frac{\partial \pi\_{sij}}{\partial \mathbf{x}\_j} - F\_i^{cd} \tag{7}$$

Where, *<sup>g</sup>* , *<sup>s</sup>* are gas, solid volume rate, 1 *g s* ; *<sup>g</sup>* , *<sup>s</sup>* are gas density and solid density respectively ; *ugi* , *ugj* , *usi* , *usj* are instantaneous velocity component of gas, solid in *i* , *j* direction. *<sup>g</sup> p* 、 *<sup>s</sup> p* are pressure of gas, solid; *<sup>i</sup> g* is component of gravitation in *i* direction; *cd Fi* is interaction between gas and solid, which includes the inter-phase resistance, false mass force, and pressure gradient force. In horizontal pipe, drag force is the dominant factor. So here

$$F\_i^{cl} = \mathcal{B}\left(\mu\_{gi} - \mu\_{pi}\right) \tag{8}$$

Where is drag force coefficient between the two phase. As *αg*≥0.8, according to experiment results, the expression of can be given as below.

$$\beta = 0.75 C\_D a\_s \frac{\left| u\_g - u\_s \right|}{d\_p} \rho\_g a\_g^{-2.65} \tag{9}$$

CD is single particle drag force coefficient, the calculation equations are given as following.

$$\begin{array}{l} \text{Re}\_{p} > 1000 \\ \text{Re}\_{p} > 1000 \\ 1000 \ge \text{Re}\_{p} > 1 \\ \text{Re}\_{p} \le 1 \end{array} \left| \begin{aligned} \text{C}\_{D} &= 0.44 \\ \text{C}\_{D} &= \frac{24}{\text{Re}\_{p}} (1 + 0.15 \text{Re}\_{p}^{0.68}) \\ \text{C}\_{D} &= \frac{24}{\text{Re}\_{p}} \end{aligned} \right. \tag{10}$$

$$\text{Here,} \quad \mathcal{C}\_{D} = \frac{24}{\text{Re}\_{p}} \Big( 1 + 0.15 \text{Re}\_{p}^{0.68} \Big) \tag{11}$$

Re*p* is defined as particle Reynolds number,

Numerical Simulation of Dense Phase Pneumatic Conveying in Long-Distance Pipe 381

From equations above, time averaged equation setup provided several unknown quantity. Above all factors, turbulent stress played a dominant role to the governing equations. So some assumption or new turbulent model equations must be introduced to accomplish the equations set. Nowadays, Reynolds stress model and vortex model were often employed to agree with different suitable cases. In the study, by contrast to all occasions, Reynolds stress

According to Boussines assumption and some mathematical operation, Reynolds stress

' ' ' ' '' ' '

*uu u uu uu uu t x xx x x x*

*<sup>i</sup> <sup>j</sup> <sup>i</sup> j jj <sup>t</sup> i j kij ik jk*

 

1

3 2

*l*

*C d*

*k kk k k k k*

2

' '

 

*km k m ij ik j k ik i k ij*

33 2 22 3

To this study, the buoyant force and revolution effect were neglected. So the equation above

' ' ' ' '' ''

*uu u uu uu uu t x xx x x x*

 

2. Gas Turbulent Kinetic Energy Equation and Turbulent Dissipation Rate Equation By analogy single flow theory and Boussinesq vortex assumption, the relationship of gas

> <sup>0</sup> ,max 1 /*s s g*

*<sup>i</sup> <sup>j</sup> <sup>i</sup> j jj <sup>t</sup> i j kij ij jk*

*j i i j ij ik jk kk ij ij k k*

*k kk k k k k*

<sup>1</sup> 1/3

2 1 2 3 3 3

*s s si s s si sj s ssi sij dij gi si*

 

*<sup>p</sup> <sup>u</sup> u u <sup>g</sup> u u*

  '

 

(17)

3 2

(18)

(19)

*l*

 

> 

*uu uu u u*

 

' '' ' '

  2 3

*k m k m ij i k i k*

*<sup>k</sup> C u u n n uu nn k C d*

 

*uu uu u u*

(20)

 

 

*u u uu u u u u uu*

*gi si s s si sj s si s sj s sj s si s s si sj*

'' ' ' ' '' '' ' '

*j*

 

*x*

1. Reynolds stress transport equation

transport equation can be given as below.

'' ' '

 

1 2

 

*<sup>u</sup> <sup>u</sup> C uu uu P x x*

 

*tj i*

*j ij*

 

 

*tx x x*

**4.1.3 Turbulent kinetic energy and turbulent dissipation rate equation** 

'' ''

1

1 3

 

Pr 3

*j i ik jk kk ij k k*

2 ,2 ,2 ,2

*<sup>k</sup> C nn nn nn*

'' ''

' ' '' ''

Reynolds stress and averaged velocity gradient was shown as below.

*<sup>u</sup> <sup>u</sup> C uu k C uu uu P k x x*

*<sup>t</sup> i j i j ij*

*g g C uu k xx k*

Solid momentum equation:

model was selected.

2

'

can be rewritten as below.

$$\text{Re}\_p = \frac{\alpha\_{\text{g}} \rho\_{\text{g}} d\_p \left| u\_{\text{g}} - u\_s \right|}{\mu\_{\text{g}}} \tag{12}$$

*τij* is gas viscosity stress.

$$\sigma\_{ij} = \alpha\_g \mu\_g \left\lfloor \left( \frac{\partial u\_{gi}}{\partial \mathbf{x}\_j} + \frac{\partial u\_{gi}}{\partial \mathbf{x}\_i} \right) - \frac{2}{3} \delta\_{ij} \frac{\partial u\_{gk}}{\partial \mathbf{x}\_k} \right\rfloor \tag{13}$$

Where, *<sup>g</sup>* is gas phase kinetic viscosity coefficient.

*ij* is Kronecker delta,

$$
\mathcal{S}\_{ij} = \begin{cases} 1 & \text{i} = \text{j} \\ 0 & \text{i} \neq \text{j} \end{cases}
$$

#### **4.1.2 Time averaged equation setup of gas solid flow**

Gas continuity equation:

$$\frac{\partial \left(a\_{\mathcal{g}} \rho\_{\mathcal{g}}\right)}{\partial t} + \frac{\partial \left(a\_{\mathcal{g}} \rho\_{\mathcal{g}} u\_{\mathcal{g}j}\right)}{\partial \mathbf{x}\_{j}} + \frac{\partial \left(\rho\_{\mathcal{g}} \overline{a\_{\mathcal{g}} u\_{\mathcal{g}j}}\right)}{\partial \mathbf{x}\_{j}} = \mathbf{0} \tag{14}$$

Solid continuity equation:

$$\frac{\partial \left( \alpha\_s \rho\_s \right)}{\partial t} + \frac{\partial \left( \alpha\_s \rho\_s u\_{s\bar{\jmath}} \right)}{\partial \mathbf{x}\_{\bar{\jmath}}} + \frac{\partial \left( \rho\_s \overline{\alpha\_s u\_{s\bar{\jmath}}} \right)}{\partial \mathbf{x}\_{\bar{\jmath}}} = \mathbf{0} \tag{15}$$

Gas momentum equation:

$$\frac{\partial}{\partial t}(\boldsymbol{\alpha}\_{\mathcal{g}}\boldsymbol{\rho}\_{\mathcal{g}}\boldsymbol{u}\_{\mathcal{g}^{\bar{\mathsf{z}}}}) + \frac{\partial}{\partial \boldsymbol{\alpha}\_{\mathcal{j}}}(\boldsymbol{\alpha}\_{\mathcal{g}}\boldsymbol{\rho}\_{\mathcal{g}}\boldsymbol{u}\_{\mathcal{g}^{\bar{\mathsf{z}}}}\boldsymbol{u}\_{\mathcal{g}^{\bar{\mathsf{z}}}}) = -\boldsymbol{\alpha}\_{\mathcal{g}}\frac{\partial p}{\partial \boldsymbol{\alpha}\_{\mathcal{i}}} - \frac{2}{3}\frac{\partial}{\partial \boldsymbol{\alpha}\_{\mathcal{j}}}\left[\boldsymbol{\mu}\_{\mathcal{g}}\boldsymbol{\alpha}\_{\mathcal{g}}\frac{\partial \boldsymbol{u}\_{\mathcal{g}^{\bar{\mathsf{z}}}}}{\partial \boldsymbol{\alpha}\_{\mathcal{k}}} + \boldsymbol{\mu}\_{\mathcal{g}}\boldsymbol{\alpha}\_{\mathcal{g}}\frac{\partial \boldsymbol{u}\_{\mathcal{g}^{\bar{\mathsf{z}}}}}{\partial \boldsymbol{\alpha}\_{\mathcal{k}}}\right] + \tag{16}$$

$$\frac{\partial}{\partial \mathbf{x}\_j} \bigg[ \mu\_g \alpha\_g \Big( \frac{\partial \mathbf{u}\_{g\bar{i}}}{\partial \mathbf{x}\_j} + \frac{\partial \mathbf{u}\_{g\bar{j}}}{\partial \mathbf{x}\_i} \bigg) + \mu\_g \overline{\alpha\_g^{'} \left( \frac{\partial \mathbf{u}\_{g\bar{i}}^{'}}{\partial \mathbf{x}\_j} + \frac{\partial \mathbf{u}\_{g\bar{j}}^{'}}{\partial \mathbf{x}\_i} \right)} \bigg] + \alpha\_g \rho\_g \mathbf{g}\_i + \beta \left( \mathbf{u}\_{si} - \mathbf{u}\_{gi} \right) + \overline{\beta \left( \mathbf{u}\_{si}^{'} - \mathbf{u}\_{gi}^{'} \right)}$$

$$-\frac{\partial}{\partial \boldsymbol{x}\_{j}} \Big( \boldsymbol{\alpha}\_{\boldsymbol{\mathcal{S}}} \boldsymbol{\rho}\_{\boldsymbol{\mathcal{S}}} \overline{\boldsymbol{u}\_{\boldsymbol{\mathcal{S}}^{i}}^{\top} \boldsymbol{u}\_{\boldsymbol{\mathcal{S}}^{j}}^{\top}} + \boldsymbol{\rho}\_{\boldsymbol{\mathcal{S}}} \boldsymbol{u}\_{\boldsymbol{\mathcal{S}}^{i}} \overline{\boldsymbol{a}\_{\boldsymbol{\mathcal{S}}^{i}}^{\top} \boldsymbol{u}\_{\boldsymbol{\mathcal{S}}^{j}}^{\top}} + \boldsymbol{\rho}\_{\boldsymbol{\mathcal{S}}} \boldsymbol{u}\_{\boldsymbol{\mathcal{S}}^{i}} \overline{\boldsymbol{a}\_{\boldsymbol{\mathcal{S}}^{i}}^{\top} \boldsymbol{u}\_{\boldsymbol{\mathcal{S}}^{i}}^{\top}} + \boldsymbol{\rho}\_{\boldsymbol{\mathcal{S}}} \overline{\boldsymbol{a}\_{\boldsymbol{\mathcal{S}}^{i}}^{\top} \boldsymbol{u}\_{\boldsymbol{\mathcal{S}}^{i}}^{\top}} \Big)$$

Solid momentum equation:

380 Computational Simulations and Applications

Re *g gp g <sup>s</sup>*

 

*ij g g ij*

 

<sup>0</sup> *g g g g gj g g gj*

<sup>0</sup> *s s sj s s sj s s*

*tx x*

' <sup>2</sup> '

*g g gi g g gi gj g g g g g*

 

*t x xx x x*

 

 

*tx x*

*j j*

*j j*

*u u*

3

i j ' '' '

*g g g g g g i si gi si gi*

 ' ' ' ' '' ' ' *gggi gj g gi g gj g gj g gi g ggi gj*

*uu u u u u uu*

 

 (16)

*j i j k k*

 

*u u*

 

(14)

(15)

*p u u*

' '

 

 

*gk gk*

 

*g uu uu*

 

*ij* 

> 

*g*

2 3 *gi gi gk*

*ji k*

 ji 0 ji 1

,

*uu u xx x*

 *du u* 

(12)

(13)

*p*

 

*<sup>g</sup>* is gas phase kinetic viscosity coefficient.

**4.1.2 Time averaged equation setup of gas solid flow** 

 

*u u u*

 

j j

*ji i*

*x xx xx*

*j*

*x*

*g g gi gj*

 

*uu uu*

 

 

*τij* is gas viscosity stress.

Where,

*ij* 

is Kronecker delta,

Gas continuity equation:

Solid continuity equation:

Gas momentum equation:

 

$$\begin{split} \frac{\partial}{\partial t} \Big( \alpha\_{s} \rho\_{s} u\_{si} \Big) + \frac{\partial}{\partial x\_{j}} \Big( \alpha\_{s} \rho\_{s} u\_{si} u\_{sj} \Big) &= -\alpha\_{s} \frac{\partial p}{\partial x\_{i}} + \alpha\_{s} \rho\_{s} g\_{i} + \frac{\partial}{\partial x\_{j}} \Big( \overline{\tau\_{sij} + \tau\_{dij}} \Big) + \beta \left( u\_{gi} - u\_{si} \Big) \\ + \overline{\beta \left( \dot{u\_{gj}} - u\_{si} \right)} - \frac{\partial}{\partial x\_{j}} \Big( \alpha\_{s} \rho\_{s} \overline{u\_{si} \dot{u\_{gj}}} + \rho\_{s} u\_{si} \overline{u\_{sj} \dot{u\_{gj}}} + \rho\_{s} u\_{sj} \overline{u\_{sj} \dot{u\_{s}}} + \rho\_{s} \overline{\alpha\_{s} u\_{si} \dot{u\_{gj}}} \Big) \end{split} \tag{17}$$

From equations above, time averaged equation setup provided several unknown quantity. Above all factors, turbulent stress played a dominant role to the governing equations. So some assumption or new turbulent model equations must be introduced to accomplish the equations set. Nowadays, Reynolds stress model and vortex model were often employed to agree with different suitable cases. In the study, by contrast to all occasions, Reynolds stress model was selected.

#### **4.1.3 Turbulent kinetic energy and turbulent dissipation rate equation**

1. Reynolds stress transport equation

According to Boussines assumption and some mathematical operation, Reynolds stress transport equation can be given as below.

 '' '' ' ' ' ' '' ' ' ' ' 1 '' ' ' 2 2 Pr 3 1 3 *<sup>i</sup> <sup>j</sup> <sup>i</sup> j jj <sup>t</sup> i j kij ik jk k kk k k k k <sup>t</sup> i j i j ij tj i j i ik jk kk ij k k uu uu u u uu u uu uu uu t x xx x x x g g C uu k xx k <sup>u</sup> <sup>u</sup> C uu uu P x x* 3 2 ' '' ' ' 1 3 2 ' 2 ,2 ,2 ,2 2 3 33 2 22 3 *k m k m ij i k i k l km k m ij ik j k ik i k ij l <sup>k</sup> C u u n n uu nn k C d <sup>k</sup> C nn nn nn C d* (18)

To this study, the buoyant force and revolution effect were neglected. So the equation above can be rewritten as below.

$$\begin{split} \frac{\partial}{\partial t} \left( \rho \overrightarrow{\dot{u}\_{i} \dot{u}\_{j}^{\prime}} \right) + \frac{\partial}{\partial \mathbf{x}\_{k}} \Big( \rho \mu\_{k} \overrightarrow{\dot{u}\_{i} \dot{u}\_{j}^{\prime}} \Big) &= \frac{\partial}{\partial \mathbf{x}\_{k}} \bigg( \frac{\mu\_{t}}{\sigma\_{k}} \frac{\partial \overrightarrow{\dot{u}\_{i} \dot{u}\_{j}^{\prime}}}{\partial \mathbf{x}\_{k}} + \mu \frac{\partial \overrightarrow{\dot{u}\_{i} \dot{u}\_{j}^{\prime}}}{\partial \mathbf{x}\_{k}} \bigg) - \rho \left( \overrightarrow{\dot{u}\_{i} \dot{u}\_{j}} \frac{\partial \mathbf{u}\_{j}}{\partial \mathbf{x}\_{k}} + \overrightarrow{\dot{u}\_{j} \dot{u}\_{k}^{\prime}} \frac{\partial \mathbf{u}\_{j}}{\partial \mathbf{x}\_{k}} \right) \\ - \mathsf{C}\_{1} \rho \frac{\mathscr{E}}{k} \bigg( \overrightarrow{\dot{u}\_{i} \dot{u}\_{j}^{\prime}} - \frac{2}{3} k \delta\_{ij} \Big) - \mathsf{C}\_{2} \Big[ \rho \left( \overrightarrow{\dot{u}\_{i} \dot{u}\_{k}} \frac{\partial \mathbf{u}\_{j}}{\partial \mathbf{x}\_{k}} + \overrightarrow{\dot{u}\_{j} \dot{u}\_{k}} \frac{\partial \mathbf{u}\_{i}}{\partial \mathbf{x}\_{k}} \right) - \frac{1}{3} P\_{kk} \delta\_{ij} \Big] - \frac{2}{3} \rho \varepsilon \delta\_{ij} \end{split} \tag{19}$$

2. Gas Turbulent Kinetic Energy Equation and Turbulent Dissipation Rate Equation By analogy single flow theory and Boussinesq vortex assumption, the relationship of gas Reynolds stress and averaged velocity gradient was shown as below.

$$\mathcal{g}\_0 = \left[1 - \left(\alpha\_s \mid \alpha\_{s,\text{max}}\right)^{1/3}\right]^{-1} \tag{20}$$

Numerical Simulation of Dense Phase Pneumatic Conveying in Long-Distance Pipe 383

1 2 *s s s s si <sup>p</sup> <sup>s</sup> ks gs s ss*

 *sj sj sj sj s k s k*

 

*ii j i k k*

2

*k*

2

*s k*

 

0 0

Where *g0* stands for granule radial distribution function, which can reflect the effect of solid

5 4 4 <sup>48</sup> <sup>1</sup> 1 1 5 15

0

(36)

*s s*

*d g d e g e e g*

*<sup>t</sup>* . In this paper, by comparing to the pure flow, gas and

*x x x xx x*

*CG G C*

2 3

 

(31)

*st* can be achieved as following equations

(32)

(33)

*gg g <sup>t</sup>* (34)

*s st* (35)

2

  (29)

(30)

2

*i j js*

*ks st st ij p*

*uu u u u u <sup>G</sup>*

 

*G ck k gs s* 2

*T k <sup>L</sup>* 0165 /

When the mathematical model is built up, the key step of the simulation is how to

*gt g g*

 *C*

*<sup>s</sup> st p p*

 *C*

Solid Turbulent Dissipation Rate Equation

Thereinto, *c T* 1 / 1 10 / /

Particle relaxation time,

respectively.

concentration

identify turbulent viscosity

granule phase turbulent viscosity

 

Gks is the generation of solid turbulent kinetic energy.

 

*u*

Ggs is additional source caused by granule adding to gas phase.

 *ss L* ,max Lagrangian time scale of gas turbulent can be written as below.

> *s s* /

> > *gt* ,

*e* 

> *p*

*s ss s s*

The expression of gas effective viscosity coefficient is given as below.

The granule effective viscosity coefficient is the following equation.

The granule phase shear viscosity can be gained by the next equation.

2

 

*t x x xk*

Where *<sup>t</sup> p* is addition pressure which is caused by pulsed velocity.

$$P\_t = \frac{2}{3} \alpha\_{\mathcal{g}} \rho\_{\mathcal{g}} k \tag{21}$$

*k* is turbulent kinetic energy.

$$\hat{\lambda} = \frac{\overline{\dot{u\_i}\dot{u\_j}}}{2} = \frac{1}{2} \left( \overline{\dot{u\_{\mathcal{S}^i}}^2} + \overline{\dot{u\_{\mathcal{S}^j}}^2} + \overline{\dot{u\_{\mathcal{S}^k}}^2} \right) \tag{22}$$

*gt* is turbulent viscosity, which depends on flowage.

Then, turbulent dissipation rate is introduced.

$$
\omega = \frac{\mu\_{\mathcal{S}^t}}{\rho\_{\mathcal{S}}} \left( \frac{\partial u\_i^{\prime}}{\partial \mathbf{x}\_k} \right) \left( \frac{\partial u\_i^{\prime}}{\partial \mathbf{x}\_k} \right) \tag{23}
$$

The transport expression of k and can be given when the addition drag source caused by adding the solid phase. Therefore, the gas turbulent kinetic energy and turbulent dissipation rate in the gas solid flow can be given.

Gas phase turbulent kinetic energy equation:

$$\frac{\partial \left(a\_{\rm g} \rho\_{\rm g} k\right)}{\partial t} + \frac{\partial \left(a\_{\rm g} \rho\_{\rm g} u\_{\rm g} k\right)}{\partial \mathbf{x}\_{j}} = \frac{\partial}{\partial \mathbf{x}\_{j}} \left(\frac{\mu\_{\rm e}}{\sigma\_{k}} \frac{\partial k}{\partial \mathbf{x}\_{j}}\right) + \mathbf{G}\_{\rm kg} + \mathbf{G}\_{\rm sg} - a\_{\rm g} \rho\_{\rm g} x \tag{24}$$

Gas phase turbulent dissipation rate equation:

$$\frac{\partial \left(a\_{\mathcal{S}} \rho\_{\mathcal{S}} x\right)}{\partial t} + \frac{\partial \left(a\_{\mathcal{S}} \rho\_{\mathcal{S}} x a\_{\mathcal{S}j}\right)}{\partial x\_{j}} = \frac{\partial}{\partial x\_{j}} \left(\frac{\mu\_{\varepsilon}}{\sigma\_{x}} \frac{\partial x}{\partial x\_{j}}\right) + \frac{\varepsilon}{k} \left[\mathbf{C}\_{1x} \left(\mathbf{G}\_{\mathcal{S}g} + \mathbf{G}\_{\mathcal{S}g}\right) - \mathbf{C}\_{2x} a\_{\mathcal{S}} \rho\_{\mathcal{S}} x\right] \tag{25}$$

*Gkg* is the generation of gas turbulent kinetic energy.

$$\mathbf{G}\_{\rm kg} = \mu\_{\rm gt} \frac{\partial u\_{\rm gj}}{\partial \mathbf{x}\_i} \left( \frac{\partial u\_{\rm gj}}{\partial \mathbf{x}\_i} + \frac{\partial u\_{\rm gj}}{\partial \mathbf{x}\_j} \right) - \mu\_{\rm gt} \delta\_{\rm ij} \frac{\partial u\_{\rm gi}}{\partial \mathbf{x}\_i} \frac{\partial u\_{\rm gk}}{\partial \mathbf{x}\_k} - \frac{2}{3} \mu\_e \left( \frac{\partial u\_{\rm gk}}{\partial \mathbf{x}\_k} \right)^2 \tag{26}$$

Gsg is additional source caused by granule adding to gas phase turbulent kinetic energy.

$$\mathbf{G}\_{\rm sg} = \mathbf{2}\,\beta \left(k\_s - k\right) \tag{27}$$

Where C1*ε*=1.44,C2*ε*=1.92,*σk*=0.82,*σε*=1.0。

3. Solid turbulent kinetic energy equation and turbulent dissipation rate equation By using the same processing method, Solid turbulent kinetic energy equation can be given as following.

$$\frac{\partial \left(a\_s \rho\_s k\_s\right)}{\partial\_t} + \frac{\partial \left(a\_s \rho\_s u\_{sj} k\_s\right)}{\partial \mathbf{x}\_j} = \frac{\partial}{\partial \mathbf{x}\_j} \left(\frac{\mu\_p}{\sigma\_k} \frac{\partial \mathbf{k}\_s}{\partial \mathbf{x}\_j}\right) + \mathbf{G}\_{ks} + \mathbf{G}\_{\mathcal{S}^s} - a\_s \rho\_s \mathbf{z}\_s \tag{28}$$

Solid Turbulent Dissipation Rate Equation

382 Computational Simulations and Applications

2 3 *P k <sup>t</sup> g g* 

2 2 *i j*

is introduced.

*g g g g gj <sup>e</sup> kg sg gg*

 

*u*

*t x x xk*

*t x xx*

*j jj*

*u u*

' ' <sup>1</sup> '2 '2 '2

' ' *gt i i g k k u u x x*

 

adding the solid phase. Therefore, the gas turbulent kinetic energy and turbulent dissipation

*j jkj k uk <sup>k</sup> G G*

1 2 *g g g g gj <sup>e</sup> kg sg g g*

 

*gj gj gj gi gk gk*

 

*ii j ik k u u u uu u*

*x x x xx x*

*kg gt gt ij e*

Gsg is additional source caused by granule adding to gas phase turbulent kinetic energy.

3. Solid turbulent kinetic energy equation and turbulent dissipation rate equation

*t j jkj*

   

2 *G kk sg s* 

By using the same processing method, Solid turbulent kinetic energy equation can be given

*s ss s s sj s <sup>p</sup> <sup>s</sup> ks gs s ss*

*<sup>k</sup> u k <sup>k</sup> G G x xx*

*gi gj gk*

*k uuu* (22)

can be given when the addition drag source caused by

*CG G C*

2 3

 

(27)

 

 

> 

> > 2

 

(21)

(23)

(24)

(25)

(26)

(28)

Where *<sup>t</sup> p* is addition pressure which is caused by pulsed velocity.

*gt* is turbulent viscosity, which depends on flowage.

*k* is turbulent kinetic energy.

Then, turbulent dissipation rate

The transport expression of k and

*G*

as following.

rate in the gas solid flow can be given. Gas phase turbulent kinetic energy equation:

Gas phase turbulent dissipation rate equation:

 

*Gkg* is the generation of gas turbulent kinetic energy.

Where C1*ε*=1.44,C2*ε*=1.92,*σk*=0.82,*σε*=1.0。

 

$$\frac{\partial \left(a\_s \rho\_s \varepsilon\right)}{\partial t} + \frac{\partial \left(a\_s \rho\_s \varepsilon u\_{si}\right)}{\partial \mathbf{x}\_i} = \frac{\partial}{\partial \mathbf{x}\_j} \left(\frac{\mu\_p}{\sigma\_\varepsilon} \frac{\partial \varepsilon}{\partial \mathbf{x}\_j}\right) + \frac{\varepsilon\_s}{k\_s} \left[\mathbf{C}\_1 \left(\mathbf{G}\_{ks} + \mathbf{G}\_{gs}\right) - \mathbf{C}\_2 a\_s \rho\_s \varepsilon\_s\right] \tag{29}$$

Gks is the generation of solid turbulent kinetic energy.

$$G\_{ks} = \mu\_{st} \frac{\partial \mathbf{u}\_{sj}}{\partial \mathbf{x}\_{i}} \left(\frac{\partial \mathbf{u}\_{sj}}{\partial \mathbf{x}\_{i}} + \frac{\partial \mathbf{u}\_{sj}}{\partial \mathbf{x}\_{j}}\right) - \mu\_{st} \delta\_{ij} \frac{\partial \mathbf{u}\_{sj}}{\partial \mathbf{x}\_{i}} \frac{\partial \mathbf{u}\_{sk}}{\partial \mathbf{x}\_{k}} - \frac{2}{3} \mu\_{\rho} \left(\frac{\partial \mathbf{u}\_{sk}}{\partial \mathbf{x}\_{k}}\right)^{2} \tag{30}$$

Ggs is additional source caused by granule adding to gas phase.

$$\mathbf{G}\_{\mathcal{S}^s} = \mathcal{D}\beta \left(\boldsymbol{c}\boldsymbol{k} - \boldsymbol{k}\_s\right) \tag{31}$$

Thereinto, *c T* 1 / 1 10 / / *ss L* ,max

Lagrangian time scale of gas turbulent can be written as below.

$$T\_L = 0165 \text{k/s}$$

Particle relaxation time, *s s* /

When the mathematical model is built up, the key step of the simulation is how to identify turbulent viscosity *<sup>t</sup>* . In this paper, by comparing to the pure flow, gas and granule phase turbulent viscosity *gt* , *st* can be achieved as following equations respectively.

$$
\mu\_{\rm gt} = \mathbb{C}\_{\mu} \alpha\_{\rm g} \rho\_{\rm g} \frac{k^2}{\varepsilon} \tag{32}
$$

$$
\mu\_{st} = \mathbb{C}\_{\mu} \alpha\_p \rho\_p \frac{k\_s^2}{\varepsilon\_s} \tag{33}
$$

The expression of gas effective viscosity coefficient is given as below.

$$
\mu\_e = \alpha\_\mathcal{g} \mu\_\mathcal{g} + \mu\_\mathcal{g} \tag{34}
$$

The granule effective viscosity coefficient is the following equation.

$$
\mu\_p = \mu\_s + \mu\_{st} \tag{35}
$$

The granule phase shear viscosity can be gained by the next equation.

$$\mu\_s = \frac{4}{5} \alpha\_s^2 \rho\_s g\_0 d\_s (1+e) \sqrt{\frac{\Theta}{\pi}} + \frac{\frac{5\sqrt{\pi}}{48} \rho\_s d\_s \sqrt{\Theta}}{(1+e)g\_0} \left(1 + \frac{4}{5} a\_s g\_0 (1+e)\right)^2 \tag{36}$$

Where *g0* stands for granule radial distribution function, which can reflect the effect of solid concentration

Numerical Simulation of Dense Phase Pneumatic Conveying in Long-Distance Pipe 385

Meanwhile, the grid formation technique has become a critical part in modern computational fluid dynamics. The mesh formation method can be divided two ways. One is algebraic method and the other is differential method. Thereinto, differential method can be used to produce smooth grid to suit complex flow domain. Of course, we can adjust the mesh degree of closeness by changing the control function. And if more accurate solution

In this study, horizontal pipe section with 1 meter length, 0.08 meter diameter and range from 127.7 to 128.7m were chosen as research object. In this pipe section, fly ash was

There are the assumptions that the gas axial velocity cross-section of the entrance with the fully developed turbulent flow of smooth pipe, radial velocity is zero, given the pressure of

<sup>3</sup> <sup>2</sup>

1.5 0.75 *<sup>k</sup> <sup>C</sup> l* 

To the equation, *DH* is hydraulic diameter. From the equation above, it can be concluded, the Reynolds number in the gas turbulent intensity equation is regard hydraulic diameter as

The assumption that the fully developed conditions of the pipe flow, namely the normal

 ( ,,) *g gi u k*

In the study, no-slip condition was adapted. Each parameter near the pipe wall was

0 *i x* 

1/8

<sup>2</sup> *g g k uI* (41)

(42)

0.16 Re *g gDH I* (43)

(44)

conveyed by compressed gas. Fig.5 gave the pipe geometric model with full grid.

Where Ig is gas turbulent intensity, to the fully arisen turbulent flow, then,

needed, mash must be thicker.

**4.2.2 Boundary conditions** 

characteristic length.

*L* is pipe diameter.

*L* is length dimension, to circle pipe,

The gas velocity of inlet is set 9.9m/s. Outlet boundary for gas

Pipe wall surface

0.07 *l L* ,

derivative of the variables solved is zero, given the export pressure.

considered as zero. And wall function method was applied. So

1. Boundary condition for gas phase Inlet boundary for gas

the entrance, turbulent kinetic energy expression is

Turbulent dissipation rate can be expressed by

$$g\_0 = \left[1 - \left(\alpha\_s / \alpha\_{s,\text{max}}\right)^{1/3}\right]^{-1} \tag{37}$$

The expression of granule phase pressure is given as below.

$$p\_s = \alpha\_s \rho\_s \Theta \left[ 1 + 2(1+e)\alpha\_s g\_0 \right] \tag{38}$$

The total viscosity of granule phase is the below equation.

$$
\zeta\_s = \frac{4}{3} \alpha\_s^2 \rho\_s d\_p g\_0 \left( 1 + e \right) \sqrt{\frac{\Theta}{\pi}} \tag{39}
$$

It must be noted that, in the equations above, the symbol of e stands for granule collision recovery coefficient. And it obeys the following rule.

$$\begin{cases} e = 0\\ e = 1\\ 0 < e < 1 \end{cases} \tag{40}$$

When e= 1, it means elastic collision, which is in case of no energy loss. When e=0, it means complete inelastic collision.

When 0 1 *e* , it means energy will diffuse in form of elastic collision.
