Specific Speed
Turbomachinery speed is described by a specific speed 𝑁𝑠 .Dimensioned versions
are used in standard commercial and industrial procedures and are just as
useful. The suction specific speed is often defined by specific speed in pump
and turbine applications.
Derivation of Specific Speed:
Specific speed can be obtained by combination of dimensionless group (𝜋)
Derivation of Pump Specific Speed
𝑄
𝑄
𝜙=
→𝑁=
3
𝑁𝐷
𝜙 𝐷3
𝑔𝐻
𝜓= 2 2
𝑁 𝐷
1
𝑔𝐻 2
→𝑁=(
)
𝜓 𝐷2
𝑁𝜙 = 𝑁𝜓
1
𝑄
𝑔𝐻 2
=
(
)
𝜙 𝐷3
𝜓 𝐷2
1
1 𝑄
𝑔𝐻 2 1
=
(
)
(
)
𝐷3 𝜙
𝜓 𝐷
𝑄
(𝜙 )
𝐷2 =
1
𝑔𝐻 2
(𝜓)
1
𝐷=
1
𝑄2𝜓4
1
1
→ 𝑒𝑞. 1
𝜙 2 (𝑔𝐻)4
Going back to 𝑁𝜙
1
𝑄
𝑄 3
𝑁=
→ 𝐷 = ( ) → 𝑒𝑞. 2
3
𝜙𝐷
𝑁𝜙
𝑒𝑞. 1 𝑎𝑛𝑑 𝑒𝑞. 2
03
1
1
1
𝑄2𝜓4
𝑄3
=
1
1
𝜙 2 (𝑔𝐻)4
1
1
𝑁 3𝜙3
3
1
(𝑔𝐻)4 𝜙 2
𝑁=
1 ∙ 3
𝑄2 𝜓4
1
1
𝑁𝑄 2
=
3
(𝑔𝐻)4
𝜙2
3
𝜓4
1
𝜋5 =
𝜙2
3
𝜓4
1
𝑄 2
(
3)
𝜋5 = 𝑁 𝐷 3
𝑔𝐻 4
( 2 2)
𝑁 𝐷
1
3
3
𝑄2𝑁 2𝐷2
𝜋5 =
1
3
3
𝑁 2 𝐷 2 (𝑔𝐻)4
𝜋5 =
𝑁√𝑄
3
(𝑔𝐻)4
𝑁√𝑄
3
(𝑔𝐻)4
𝑁𝑠 =
𝑁√𝑄
3
∝
𝑁√𝑄
3
𝐻4
→ 𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑆𝑝𝑒𝑒𝑑 𝑜𝑓 𝑃𝑢𝑚𝑝
𝐻4
03
Derivation of Turbine Specific Speed
1
𝑃
𝑃 3
𝜉=
→𝑁=( 5 )
3
5
𝜌𝑁 𝐷
𝜌𝐷 𝜉
𝑔𝐻
𝜓= 2 2
𝑁 𝐷
1
𝑔𝐻 2
→𝑁=(
)
𝜓 𝐷2
𝑁𝜉 = 𝑁𝜓
1
1
𝑃 3
𝑔𝐻 2
( 5 ) = (
)
𝜌𝐷 𝜉
𝜓 𝐷2
1
1
𝐷
1
𝑃 3
𝑔𝐻 2 1
=
(
)
(
)
5 𝜌𝜉
𝜓 𝐷
3
1
2
𝐷3
1
𝑃3 𝜓2
=
1 1 1
(𝑔𝐻)2 𝜌3 𝜉 3
1
3
𝑃2 𝜓4
𝐷=
3
1
→ 𝑒𝑞. 1
1
(𝑔𝐻)4 𝜌2 𝜉 2
Going back to 𝑁𝜙
1
1
𝑔𝐻 2
𝑔𝐻 2
𝑁=(
) → 𝐷=(
) → 𝑒𝑞. 2
2
𝜓𝐷
𝜓 𝑁2
𝑒𝑞. 1 𝑎𝑛𝑑 𝑒𝑞. 2
1
1
𝑔𝐻 2
(
) =
𝜓 𝑁2
1
(𝑔𝐻)2
1
𝑁𝜓 2
1
𝜉2
1
𝜓2
3
𝜓4
3
1
1
(𝑔𝐻)4 𝜌2 𝜉 2
1
=
3
𝑃2 𝜓4
3
𝑃2 𝜓 4
3
1
1
(𝑔𝐻)4 𝜌2 𝜉 2
1
=
𝑁 𝑃2
1
3
1
(𝑔𝐻)2 (𝑔𝐻)4 𝜌2
03
1
𝜉2
5
𝜓4
1
=
𝑁 𝑃2
1
3
(𝑔𝐻)2 (𝑔𝐻)4
→ 𝐷𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛𝑙𝑒𝑠𝑠
1
𝜌2
1
𝜋6 =
𝜉2
5
𝜓4
1
𝜋5 =
2
𝑃
(
)
3
5
𝜌𝑁 𝐷
5
𝑔𝐻 4
( 2 2)
𝑁 𝐷
𝜋5 =
1
5
3
5
5
𝑃2 𝑁 2 𝐷 2
1
5
𝜌2 𝑁 2 𝐷 2 (𝑔𝐻)4
𝜋5 =
𝑁 √𝑃
5
(𝑔𝐻)4
𝑁 √𝑃
5
(𝑔𝐻)4
𝑁𝑠 =
𝑁 √𝑃
5
∝
𝑁 √𝑃
5
𝐻4
→ 𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑆𝑝𝑒𝑒𝑑 𝑜𝑓 𝑇𝑢𝑟𝑏𝑖𝑛𝑒
𝐻4
Were,
𝑄 𝑖𝑠 𝑖𝑛 𝒈𝒑𝒎
𝑃 𝑖𝑠 𝑖𝑛 𝒉𝒑
𝐻 𝑖𝑠 𝑖𝑛 𝒇𝒕
𝑁 𝑖𝑠 𝑖𝑛 𝒓𝒑𝒎
03
Speed Factor or overall head coefficient (Peripheral Coefficient)
𝜙=
√2𝑔𝐻𝑎
𝑣
𝑣 = 𝜋𝐷𝑁
Example:
A pump delivers 0.300
𝑚3
𝑠
against a head of 200 𝑚 with a rotative speed of
2000 𝑟𝑝𝑚. Find the specific speed.
Solution:
𝑁𝑠 =
𝑁√𝑄
3
𝐻4
(2000 𝑟𝑝𝑚)√(0.300
𝑁𝑠 =
𝑚3 1000 𝐿 1 𝑔𝑎𝑙
60𝑠
𝑥
𝑥
𝑥
)
3
𝑠
1𝑚𝑖𝑛
3.785
𝐿
1𝑚
3
3.28 4
(200 𝑚𝑥 1𝑚 )
𝑁𝑠 = 1064.03 𝑟𝑝𝑚
Example:
Select the specific speed of the pump or pumps required to lift 5835 𝑔𝑝𝑚
of water 350 ft through 9400 𝑓𝑡 of 2.5 𝑓𝑡 − 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 pipe (𝑓 = 0.022). The pump
rotative speed is to be 1700 𝑟𝑝𝑚. Consider the following cases:
Single Pump
Two Pumps is Series
Two Pumps in Parallel
03
Given:
𝑄 = 5835 𝑔𝑝𝑚
𝐿 = 9400 𝑓𝑡
𝐷 = 2.5 𝑓𝑡
𝑓 = 0.022
𝑁 = 1700 𝑟𝑝𝑚
Solution
For Single Pump
𝑁𝑠 =
𝑁√𝑄
3
𝐻4
𝐻 𝑖𝑠 𝑡ℎ𝑒 𝑡𝑜𝑡𝑎𝑙 ℎ𝑒𝑎𝑑
𝐻𝑇 = ℎ + ℎ𝑓𝐿
ℎ𝑓𝐿 =
𝑓𝐿𝑣 2
2𝑔𝐷
For velocity 𝑣,
𝑣=
𝑄
𝐴
𝑔𝑎𝑙 3.785 𝐿𝑖𝑡𝑒𝑟𝑠 1 𝑚3 (3.28 𝑓𝑡)3 1𝑚𝑖𝑛
5835 𝑚𝑖𝑛 𝑥
𝑥 1000 𝐿 𝑥
𝑥 60 𝑠𝑒𝑐
1 𝑔𝑎𝑙
1 𝑚3
𝑣=
𝜋
(2.5 𝑓𝑡)2
4
𝑣 = 2.65
ℎ𝑓𝐿 =
𝑓𝑡
𝑠
𝑓𝐿𝑣 2
2𝑔𝐷
03
(0.022)(9400 𝑓𝑡) (2.65
ℎ𝑓𝐿 =
2 (32.2
𝑓𝑡 2
𝑠)
𝑓𝑡
) (2.5𝑓𝑡)
𝑠2
ℎ𝑓𝐿 = 9 𝑓𝑡
𝐻𝑇 = ℎ + ℎ𝑓𝐿
𝐻𝑇 = 350𝑓𝑡 + 9 𝑓𝑡
𝐻𝑇 = 359 𝑓𝑡
𝑁𝑠 =
(1700 𝑟𝑝𝑚)√5835 𝑔𝑝𝑚
3
(359 𝑓𝑡)4
𝑁𝑠 = 1574.5 𝑟𝑝𝑚
For Two Pumps in Series,
𝐻𝑇 = 359 𝑓𝑡 = 𝐻𝑇1 + 𝐻𝑇2
𝐻𝑇 = 𝐻𝑇1 + 𝐻𝑇2 ; 𝐻𝑇1 = 𝐻𝑇2
𝐻𝑇𝑓𝑜𝑟 𝑎 𝑠𝑖𝑛𝑔𝑙𝑒 𝑝𝑢𝑚𝑝 =
𝐻𝑇𝑓𝑜𝑟 𝑎 𝑠𝑖𝑛𝑔𝑙𝑒 𝑝𝑢𝑚𝑝 =
𝐻𝑇
2
359 𝑓𝑡
2
𝐻𝑇𝑓𝑜𝑟 𝑎 𝑠𝑖𝑛𝑔𝑙𝑒 𝑝𝑢𝑚𝑝 = 179.5 𝑓𝑡
𝑁𝑠 =
(1700 𝑟𝑝𝑚)√5835 𝑔𝑝𝑚
3
(179.5 𝑓𝑡)4
𝑁𝑠 = 2648.02 𝑟𝑝𝑚
03
For Parallel Pumps
𝑄𝑇𝑜𝑡𝑎𝑙 = 𝑄1 + 𝑄2 ; 𝑄1 = 𝑄2
𝑄𝑇𝑓𝑜𝑟 𝑎 𝑠𝑖𝑛𝑔𝑙𝑒 𝑝𝑢𝑚𝑝 =
𝑄𝑇𝑓𝑜𝑟 𝑎 𝑠𝑖𝑛𝑔𝑙𝑒 𝑝𝑢𝑚𝑝 =
𝑄𝑇
2
5835 𝑔𝑝𝑚
2
𝑄 = 2917.5 𝑔𝑝𝑚
𝑁𝑠 =
(1700 𝑟𝑝𝑚)√2917.5 𝑔𝑝𝑚
3
(359 𝑓𝑡)4
𝑁𝑠 = 1113.35 𝑟𝑝𝑚
Example:
A double-overhung impulse-turbine installation is to develop 20,000 ℎ𝑝 at
275 𝑟𝑝𝑚 under a net head of 1100 𝑓𝑡. Determine the Specific Speed.
Given:
𝑃 = 20,000 ℎ𝑝
𝑁 = 275 𝑟𝑝𝑚
𝐻 = 1100 𝑓𝑡
Solution:
𝑁𝑠 =
𝑁 √𝑃
5
𝐻4
𝑁𝑠 =
(275 𝑟𝑝𝑚)√20,000 ℎ𝑝
5
(1100 𝑓𝑡)4
03
𝑁𝑠 = 6.14 𝑟𝑝𝑚
Seatwork:
A pump is driven by a two-speed motor having speeds of 1750 𝑟𝑝𝑚 and
1185 𝑟𝑝𝑚. At 1750 𝑟𝑝𝑚, the flow is 𝑄 = 45 𝑔𝑝𝑚, the head is 𝐻 = 90 𝑓𝑡, and the
total efficiency is 𝜂 = 0.60. The pump impeller has a diameter of 10 𝑖𝑛𝑐ℎ𝑒𝑠.
(a) What values of 𝑄, and 𝐻 are obtained if the pump runs at 1185 rpm'?
(b) Find the specific speed of the pump (at 1750 rpm).
A small centrifugal pump is tested at 𝑁 = 2875 𝑟𝑝𝑚 in water. It delivers
0.15 𝑚3/𝑠 at 42 𝑚 of head at its best efficiency point ( 𝜂 = 0.86).
(a) Determine the specific speed of the pump.
(b) Compute the required input power.
(c) Estimate the impeller diameter.
(d) Calculate the Reynolds Number.
03