18
1 Working Principles
The moment M shaft exerted on the shaft by the external motor is then positive. M
and –M d are both positive. The moment M shaft is divided into two parts: the M part
reaches the flow and the –M d part is absorbed by disc friction.
With a driving machine we note
(1.16)
The flow moment M is negative, i.e. the flow exerts a positive moment –M on the
rotor. The shaft moment is negative as well, i.e. the machine exerts a positive moment –M shaft on the coupled load. The moment –M is divided into two parts: the
–M shaft part reaches the shaft. The –M d part is absorbed by disc friction.
We consider a control volume as sketched in Fig. 1.7. As usual, we indicate the
rotor inlet with subscript 1 and the outlet with subscript 2. By taking the moment
around the rotation axis of Newton’s momentum law, we obtain that the change per
time unit of the moment of momentum in the flow equals the moment of the forces
exerted on the control volume:
(1.17)
The moment M encompasses both contributions of pressure forces and of shear
forces on material surfaces. When drawing up of the former expression, it is assumed that shear stresses on the inlet and outlet faces of the control volume do not
form a moment around the rotation axis. This assumption follows from the circumferentially averaged flow representation, where shear stresses have no tangential
component. This assumption is also very good with a real flow.
Multiplication of (1.17) by the speed of rotation Ω results in
P is the power transferred from the rotor to the flow through the flow moment M.
Division by the mass flow rate generates the equation:
(1.18)
with ΔW the work done on the flow by the flow moment M.
Multiplication of equation (1.15) by Ω results in
With a power receiving machine, P shaft is the power supplied to the shaft by the driving motor. The term -P d represents the power associated to the dissipation by disc
friction. After division by the mass flow rate we note
(1.19)
− = −
−
M
M
M
shaft
d .
m r v
rv
M
u
u
(
)
.
2 2
1 1
−
=
m u v
u v
P
u
u
(
)
.
2 2
1 1
−
=
2 2u
1 1u
u v
u v
W ,
D
−
=
shaft
d
P
P P .
= −
o
shaft
irr
W
W q .
D
D
=
+
1 Working Principles
The moment M shaft exerted on the shaft by the external motor is then positive. M
and –M d are both positive. The moment M shaft is divided into two parts: the M part
reaches the flow and the –M d part is absorbed by disc friction.
With a driving machine we note
(1.16)
The flow moment M is negative, i.e. the flow exerts a positive moment –M on the
rotor. The shaft moment is negative as well, i.e. the machine exerts a positive moment –M shaft on the coupled load. The moment –M is divided into two parts: the
–M shaft part reaches the shaft. The –M d part is absorbed by disc friction.
We consider a control volume as sketched in Fig. 1.7. As usual, we indicate the
rotor inlet with subscript 1 and the outlet with subscript 2. By taking the moment
around the rotation axis of Newton’s momentum law, we obtain that the change per
time unit of the moment of momentum in the flow equals the moment of the forces
exerted on the control volume:
(1.17)
The moment M encompasses both contributions of pressure forces and of shear
forces on material surfaces. When drawing up of the former expression, it is assumed that shear stresses on the inlet and outlet faces of the control volume do not
form a moment around the rotation axis. This assumption follows from the circumferentially averaged flow representation, where shear stresses have no tangential
component. This assumption is also very good with a real flow.
Multiplication of (1.17) by the speed of rotation Ω results in
P is the power transferred from the rotor to the flow through the flow moment M.
Division by the mass flow rate generates the equation:
(1.18)
with ΔW the work done on the flow by the flow moment M.
Multiplication of equation (1.15) by Ω results in
With a power receiving machine, P shaft is the power supplied to the shaft by the driving motor. The term -P d represents the power associated to the dissipation by disc
friction. After division by the mass flow rate we note
(1.19)
− = −
−
M
M
M
shaft
d .
m r v
rv
M
u
u
(
)
.
2 2
1 1
−
=
m u v
u v
P
u
u
(
)
.
2 2
1 1
−
=
2 2u
1 1u
u v
u v
W ,
D
−
=
shaft
d
P
P P .
= −
o
shaft
irr
W
W q .
D
D
=
+