Introduction

A turbomachine is a device in which energy is transferred continuously between a rotating element (rotor, fitted with blades) and a flowing fluid, through the dynamic action of the blades on the fluid. Pumps, fans, compressors, steam turbines, gas turbines and hydraulic turbines are examples.

Main parts

1. Rotor (runner/impeller) with blades - the element that exchanges energy with the fluid.
2. Stator - fixed casing with guide vanes or nozzles that direct the fluid onto the rotor.
3. Shaft and bearings - carry the torque and support the rotor.
4. Casing - encloses the flow path and contains the pressure.
5. Inlet and outlet passages - volute, diffuser, draft tube, exhaust duct.

 inlet --> [ Stator | Rotor ] --> outlet
              nozzle   blades
                |        |
                +--shaft-+--> to load

Classification

The ratio of the pressure (enthalpy) drop in the rotor to the drop in the whole stage is the degree of reaction: zero for a pure impulse stage and 0.5 for a 50% reaction stage.

Both types transfer energy between a mechanical element and a fluid. They differ in the way the energy transfer takes place.

Examples

• Turbomachines: centrifugal pump, axial-flow gas turbine, steam turbine, centrifugal compressor.
• Positive displacement machines: reciprocating compressor, gear pump, IC engine piston, Roots blower.

First law (steady flow energy equation)

For one inlet and one outlet, neglecting potential energy, per unit mass:

$$q - w = (h_2 - h_1) + \frac{C_2^2 - C_1^2}{2}$$

For an adiabatic machine $q = 0$. Using stagnation enthalpy $h_0 = h + C^2/2$:

• Turbine: $w = h_{01} - h_{02}$
• Compressor: $w_{in} = h_{02} - h_{01}$

For a perfect gas, $h_0 = c_p T_0$, so $w = c_p (T_{01} - T_{02})$ for the turbine.

Second law

For an adiabatic machine $s_2 \ge s_1$. The ideal (reversible) machine is isentropic, $s_2 = s_1$. Irreversibility (friction, shock, tip leakage) raises entropy, so the turbine delivers less work and the compressor needs more work than the isentropic value.

 T                      T       2 actual
 |1                     |      /   2s
 | \                    |     /  /
 |  \ 2s                |    /  /
 |   2 actual           |   1
 +------- s             +------- s
   turbine                compressor

Isentropic efficiencies

$$\eta_t = \frac{h_{01} - h_{02}}{h_{01} - h_{02s}} \approx \frac{T_{01} - T_{02}}{T_{01} - T_{02s}}, \qquad \eta_c = \frac{h_{02s} - h_{01}}{h_{02} - h_{01}} \approx \frac{T_{02s} - T_{01}}{T_{02} - T_{01}}$$

Polytropic (small-stage) efficiency

It is the isentropic efficiency of an infinitesimal stage, assumed constant through the machine. For a compressor with pressure ratio $r$ and $\gamma$:

$$\eta_{p} = \frac{\frac{\gamma - 1}{\gamma}\ln r}{\ln (T_{02}/T_{01})}, \qquad \eta_c = \frac{r^{(\gamma-1)/\gamma} - 1}{r^{(\gamma-1)/(\gamma \eta_p)} - 1}$$

For a turbine expanding by ratio $r$:

$$\eta_t = \frac{1 - r^{-\eta_p (\gamma-1)/\gamma}}{1 - r^{-(\gamma-1)/\gamma}}$$

For the same $\eta_p$, the compressor $\eta_c$ is less than $\eta_p$ (reheating effect, work of one stage heats the next), while the turbine $\eta_t$ is greater than $\eta_p$ (reheat factor). The difference grows with pressure ratio.

Data: $T_{01} = 1100$ K, pressure ratio $r = 600/100 = 6$, $\eta_t = 0.88$, $\dot m = 20$ kg/s, $c_p = 1.148$ kJ/kg K, $\gamma = 1.333$. Inlet and exit velocities are taken as negligible (static = stagnation values).

(a) Isentropic exit temperature

$$\frac{\gamma - 1}{\gamma} = \frac{0.333}{1.333} = 0.250$$ $$T_{02s} = T_{01}\left(\frac{1}{r}\right)^{(\gamma-1)/\gamma} = 1100 \times (1/6)^{0.250} = 703.1\ \text{K}$$

Isentropic temperature drop: $T_{01} - T_{02s} = 396.9$ K.

(b) Actual exit temperature

$$T_{01} - T_{02} = \eta_t (T_{01} - T_{02s}) = 0.88 \times 396.9 = 349.3\ \text{K}$$ $$T_{02} = 1100 - 349.3 = 750.7\ \text{K}$$

(c) Work and power

$$w = c_p (T_{01} - T_{02}) = 1.148 \times 349.3 = 401.0\ \text{kJ/kg}$$ $$P = \dot m w = 20 \times 401.0 = 8.02\ \text{MW}$$

Answer: $T_{02s} \approx 703.1$ K, $T_{02} \approx 750.7$ K, $w \approx 401.0$ kJ/kg, power $\approx 8.02$ MW.

Data: $T_{01} = 300$ K, $r = 8$, $\eta_c = 0.82$, $\dot m = 10$ kg/s, $c_p = 1.005$, $\gamma = 1.4$, so $(\gamma-1)/\gamma = 0.2857$.

(a) Exit temperature

Isentropic exit temperature:

$$T_{02s} = T_{01}\, r^{(\gamma-1)/\gamma} = 300 \times 8^{0.2857} = 543.4\ \text{K}$$

Actual temperature rise:

$$T_{02} - T_{01} = \frac{T_{02s} - T_{01}}{\eta_c} = \frac{243.4}{0.82} = 296.9\ \text{K}$$ $$T_{02} = 300 + 296.9 = 596.9\ \text{K}$$

(b) Work and power

$$w = c_p (T_{02} - T_{01}) = 1.005 \times 296.9 = 298.4\ \text{kJ/kg}$$ $$P = \dot m w = 10 \times 298.4 = 2984\ \text{kW}$$

(c) Polytropic efficiency

$$\eta_p = \frac{\frac{\gamma-1}{\gamma}\ln r}{\ln (T_{02}/T_{01})} = \frac{0.2857 \times \ln 8}{\ln (596.9/300)} = \frac{0.5941}{0.6879} = 0.864$$

The polytropic efficiency (86.4%) is higher than the isentropic efficiency (82%) because of the reheat effect in a multistage compressor.

Answer: $T_{02} \approx 596.9$ K; $w \approx 298.4$ kJ/kg; $P \approx 2984$ kW; $\eta_p \approx 86.4\%$.

Dimensional analysis of a turbomachine (variables: $D$, $N$, $\rho$, $\mu$, $Q$, $gH$, $P$) gives dimensionless groups that allow geometrically similar machines to be compared and model test results to be scaled up.

Use in practice

1. Similarity: two machines are dynamically similar if $\phi$, $\psi$ (and $Re$, $Ma$) are equal, so then their efficiencies are equal.
2. Scaling laws: for the same fluid and $\phi$ constant: $Q \propto N D^3$, $H \propto N^2 D^2$, $P \propto N^3 D^5$.
3. Selection of type: a high $N_s$ means large flow and low head, which needs an axial machine; a low $N_s$ means small flow and high head, which needs a radial one.
4. Performance prediction: a single non-dimensional curve of $\psi$ and $\eta$ against $\phi$ represents a whole family of machines.

For compressible flow, the mass flow parameter $\dot m \sqrt{T_{01}}/p_{01}$ and the speed parameter $N/\sqrt{T_{01}}$ are used in place of $\phi$ and $N$.

For geometrically similar machines with the same fluid and equal $\phi$, $\psi$, $\lambda$:

$$\frac{Q_2}{Q_1} = \frac{N_2}{N_1}\left(\frac{D_2}{D_1}\right)^3, \quad \frac{\Delta p_2}{\Delta p_1} = \left(\frac{N_2}{N_1}\right)^2\left(\frac{D_2}{D_1}\right)^2, \quad \frac{P_2}{P_1} = \left(\frac{N_2}{N_1}\right)^3\left(\frac{D_2}{D_1}\right)^5$$

(a) Same fan at 1750 rpm

$N_2/N_1 = 1750/1450 = 1.207$, $D_2/D_1 = 1$.

(b) Larger fan, same speed

$D_2/D_1 = 1.25$, $N_2/N_1 = 1$.

The efficiency is assumed unchanged (equal $\phi$, $\psi$, ignoring Reynolds number effects). Power rises with the cube of speed, so a small speed increase demands a large motor.

Answer: (a) 4.83 m³/s, 437 Pa, 3.16 kW; (b) 7.81 m³/s, 469 Pa, 5.49 kW.