A plant located 900 meters above sea level has a closed gasoline tank having a pressure of

69kpag and temperature of 40°C. the gasoline level in the tank is 2.5m above the pump centerline. The
suction line friction and turbulence losses amount to 0.6 meters. With the vapor pressure equal to
48Kpaa and a relative density of 0.72. Determine: (a) the available suction head of the system, (B)with a
flow rate of 400 L/s and a Thoma Coefficient of 0.10, What driver motor is needed, use motor-pump
efficiency equal to 75%.

Given:

Elevation = 900 m Required: NPSH

P1 = 69kPaa PowerB if Q= 400 L/s or 0.4 m3/s, 𝜎 = 0.10

T= 40°C 𝜂𝑝 𝜂𝑚 = 75%

ZL– Zs = 2.5 m
HL 1-s = 0.6m
Pv=48kPaa
SG = 0.72
Solution:

Note: Barometric reading decreases by 83.312 mmHg per 1000m (1 inHg/1000ft) rise in elevation.
83.312𝑚𝑚𝐻𝑔 101.325𝑘𝑝𝑎
𝑃𝑎𝑡𝑚 = {760 𝑚𝑚𝐻𝑔 − ( ) 900𝑚} ( )
1000𝑚 760 𝑚𝑚𝐻𝑔

𝑃𝑎𝑡𝑚 = 91.33𝐾𝑝𝑎
𝑃1 = 𝑃𝑔𝑎𝑔𝑒 + 𝑃𝑎𝑡𝑚 ; 𝑃1 = 69𝑘𝑝𝑎𝑔 + 91.33𝐾𝑝𝑎 = 160.33𝐾𝑝𝑎𝑎
𝜌
𝑆𝐺 = 𝜌_𝐻𝑠 𝑂 ; 𝜌𝑠 = 𝑆𝐺𝑥𝜌𝐻2 𝑂
2

𝐾𝑔 𝑘𝑔
𝜌𝑠 = 0.72 𝑥 1000 𝑚3 = 720 𝑚3
𝑁
𝑘𝑔 𝑚 7061.04 3 𝐾𝑁
𝑚
𝛾𝑠 = 𝜌𝑥𝑔 = 720 𝑥 9.807 = 𝑁 = 7.061
𝑚3 𝑠2 1000 𝑚3
𝐾𝑁

a)

Consider 1-S:
𝑃 𝑣2 𝑃𝑣
𝑁𝑃𝑆𝐻 = [ 𝛾𝑠 + 2𝑔𝑠 ] − 𝛾

For the total energy at the inlet of the pump

𝑃1 𝑣2 𝑃𝑠 𝑣2
𝛾
+ 2𝑔1 + 𝑧1 = 𝛾
+ 2𝑔2 + 𝑍𝑠 + 𝐻𝐿1−𝑠 ; 𝑣 = 0
 𝑃 𝑣2 𝑃1
[ 𝛾𝑠 + 2𝑔𝑠 ] = 𝛾
+ (𝑧1 − 𝑧𝑠 ) − 𝐻𝐿1−𝑠

𝐾𝑁
𝑃 𝑣2 160.33𝐾𝑝𝑎𝑎 𝑚2
[ 𝛾𝑠 + 2𝑔𝑠 ] = 𝐾𝑁 + 2.5𝑚 − 0.6𝑚 ; 𝐾𝑁 +𝑚−𝑚 =𝑚
7.061 3
𝑚 𝑚3

𝑃 𝑣2
[ 𝛾𝑠 + 2𝑔𝑠 ] = 24.6𝑚

𝑃𝑣
For the vapor energy of the liquid
𝛾

𝐾𝑁
𝑃𝑣 48𝑘𝑝𝑎𝑎 𝑚2
𝛾
= 𝐾𝑁 = 6.8𝑚 ; 𝐾𝑁 =𝑚
7.061 3
𝑚 𝑚3

𝑃 𝑣2 𝑃𝑣
∴𝑁𝑃𝑆𝐻 = [ 𝛾𝑠 + 2𝑔𝑠 ] − 𝛾
= 24.6 𝑚 − 6.8

𝑵𝑷𝑺𝑯 = 𝟏𝟕. 𝟖 𝒎
b)
𝑃𝑜𝑤𝑒𝑟 = 𝛾𝑄𝐻𝑝
𝑃𝑜𝑤𝑒𝑟 𝑃𝑜𝑤𝑒𝑟
𝜂𝑚 𝜂𝑝 = ; 𝑃𝑜𝑤𝑒𝑟𝐵 =
𝑃𝑜𝑤𝑒𝑟𝐵 𝜂𝑚 𝜂𝑝
𝑁𝑃𝑆𝐻 𝑁𝑃𝑆𝐻
𝜎= ; 𝐻𝑝 =
𝐻𝑝 𝜎
17.8𝑚
𝐻𝑝 = = 178𝑚
0.10
𝐾𝑁 𝑚3 𝑘𝑗 𝐾𝑤. 𝑠
𝑃𝑜𝑤𝑒𝑟 = 𝛾𝑄𝐻𝑝 = 7.061 3 𝑥 0.4 𝑥 178 𝑚 𝑥 𝑥
𝑚 𝑠 𝐾𝑁. 𝑚 𝐾𝐽
𝑃𝑜𝑤𝑒𝑟 = 502.75 𝑘𝑤
𝑁𝑃𝑆𝐻 502.75𝑘𝑤
∴𝑃𝑜𝑤𝑒𝑟𝐵 = 𝜂 =
𝑚 ηp 0.75

𝑷𝒐𝒘𝒆𝒓𝑩 = 𝟔𝟕𝟎. 𝟑𝟐𝒌𝒘

 5. Calculate the NPSHA for a pump that receives 150 °C water at the rate of 85 Lps through a pipe
whose ID is 25 cm. A suction gauge located 6m below the pump centerline indicates a pressure of 530
kPag. The pump is located at an elevation where the barometer indicates a pressure of 716 mm Hg.
GIVEN: REQ’D:

Patm=716 mm Hg NPSHA

Pg= 530 kPag

T= 150 °C

Q= 85 Lps

ID= 25 cm

SOLUTION:
101.325 𝑘𝑝𝑎
Patm= 716 mm Hg ( )
760 𝑚𝑚 𝐻𝑔

Patm = 95.4588 kpaa

For Ps

Pg= Ps - Patm

Ps = (530 + 95.4588) kpaa

Ps = 625.4588 kpaa
Consider 1- S

𝑃1 𝜈12 𝑃𝑆 𝜈𝑆 2
𝑧1 + + = 𝑧𝑆 + + + 𝐻𝐿 1−𝑆
𝛾 2𝑔 𝛾 2𝑔

𝐻𝐿 1−𝑆 = 0

𝑃𝑆 𝜈𝑆 2 𝑃1 𝜈1 2
+ = + (𝑧1 − 𝑧𝑆 )
𝛾 2𝑔 𝛾 2𝑔

The 𝛾 𝑎𝑡 150 deg 𝐶 𝑖𝑠 8.99 𝑓𝑟𝑜𝑚 𝑠𝑡𝑒𝑎𝑚 𝑡𝑎𝑏𝑙𝑒𝑠

85
( 𝜋 1000
25 2
)
𝑃𝑆 𝜈𝑆 2 625.4588 𝑘𝑝𝑎 ( )
+ = 𝐾𝑁 + 4 100 + (− 6 𝑚)
𝛾 2𝑔 8.99 2(9.81)
𝑚3

𝑃𝑆 𝜈𝑆 2
+ = 63.6563 𝑚
𝛾 2𝑔
For NPSHA
𝑃 𝜈𝑆 2 𝑃𝑣
NPSHA =( 𝛾𝑆 + 2𝑔
) -
𝛾

476.16 𝑘𝑝𝑎
= 63.6563 𝑚 − ( 𝐾𝑁 )
8.99
𝑚3

NPSHA = 10.69 m

6. A pump draws 20 L/s of water (37oC) from reservoir A (El. 10m) to reservoir B (El. 60m) the
suction pipe is 200 mm x 500 m and the discharge pipe is 150 mm x 1200 m. Assuming f = 0.02 and N =
3600 rpm, find: (a) NPSH, (b) σ, and (c) NSS.
m3
Given: Q = 20 L/s = 0.02
s

Water (37oC)

ZA = 10 m

ZB = 60 m

DS = D1 = 200 mm = 0.2 m, LS = L1 = 500 m

DD = D2 = 150 mm = 0.15 m, LD = L2 = 1200 m

f = 0.02

N = 3600 rpm

Required: a) NPSH

b) σ

c) NSS

Solution:
 From Table (3-3) at 37oC

PV = 6.3286 kpaa
𝑘𝑁
Ȣ = 9.7418
𝑚3

a) For NPSH:
𝑃 𝑉𝑆 2 𝑃𝑉
𝑁𝑃𝑆𝐻 = [ Ȣ𝑆 + 2𝑔
] − Ȣ
Eqn. (1)

From Bernoulli’s Equation: (consider A-S)

𝑃𝐴 𝑉𝐴 2 𝑃𝑆 𝑉𝑆 2
+ + 𝑍𝐴 = + + 𝑍𝑆 + 𝐻𝐿𝐴−𝑆 , 𝑃𝐴 = 𝑃𝑎𝑡𝑚 = 101.325 𝑘𝑃𝑎𝑎
Ȣ 2𝑔 Ȣ 2𝑔

𝑃𝑆 𝑉𝑆 2 𝑃𝐴
Ȣ
+ 2𝑔
= Ȣ
+ 𝑍𝐴 − 𝑍𝑆 − 𝐻𝐿𝐴−𝑆

𝑠2
0.0826( )𝑓𝐿1 𝑄 2
𝑚
Where, 𝐻𝐿𝐴−𝑆 = 𝐷1 5

𝑠2 𝑚3 2
𝑃𝑆 𝑉𝑆 2 101.325 𝑘𝑝𝑎𝑎 0.0826( )(0.02)(500𝑚)(0.02
𝑚 𝑠
)
Ȣ
+ 2𝑔
= 𝑘𝑁 + 10𝑚 − 0𝑚 − (0.2𝑚)5
9.7418
𝑚3

= 19.3686 𝑚
From Eqn. (1):
6.3286 𝑘𝑝𝑎𝑎
𝑁𝑃𝑆𝐻 = [19.3686 𝑚] − 𝑘𝑁
9.7418
𝑚3

𝑁𝑃𝑆𝐻 = 18.719 𝑚
b) For σ:
𝑁𝑃𝑆𝐻
𝜎= 𝐻𝑃
Eqn. (2)

From Bernoulli’s Equation: (consider A-B)

𝑃𝐴 𝑉𝐴 2 𝑃𝐵 𝑉𝐵 2
Ȣ
+ 2𝑔
+ 𝑍𝐴 + 𝐻𝑃 = Ȣ
+ 2𝑔
+ 𝑍𝐵 + 𝐻𝐿𝐴−𝑆 + 𝐻𝐿𝐷−𝐵 , 𝑃𝐴 = 𝑃𝐵 𝑎𝑛𝑑 𝑉𝐴 = 𝑉𝐵 = 0

𝐻𝑃 = 𝑍𝐵 − 𝑍𝐴 + 𝐻𝐿𝐴−𝑆 + 𝐻𝐿𝐷−𝐵
𝑠2
0.0826( )𝑓𝐿2 𝑄2
𝑚
Where, 𝐻𝐿𝐷−𝐵 =
𝐷2 5

𝑠2 𝑚3 2 𝑠2 𝑚3 2
0.0826( )(0.02)(500𝑚)(0.02 ) 0.0826( )(0.02)(1200𝑚)(0.02 )
𝑚 𝑠 𝑚 𝑠
𝐻𝑃 = 60 𝑚 − 10 𝑚 + (0.2𝑚)5
+ (0.15𝑚)5

𝐻𝑃 = 61.4748 𝑚
From Eqn. (2):
 18.719 𝑚
𝜎 = 61.4748 𝑚

𝜎 = 0.3045
c) For NSS:
1
51.655𝑁𝑄 ⁄2
N𝑠𝑠 = 3
𝑁𝑃𝑆𝐻 ⁄4

m3 1⁄
51.655(3600 𝑟𝑝𝑚)(0.02 ) 2
s
N𝑠𝑠 = 3
(18.719 𝑚) ⁄4

N𝑠𝑠 = 2922.2535 𝑟𝑝𝑚

7. A pump delivers 9900 L/min of water (50 oC) from reservoir A (El. 10m) to reservoir B
(El. 40m). The suction line is 300mmx180m and the discharge line is 250mmx50m.
Assuming the coefficient of friction for the pipe to be 0.025 and the cavitation parameter
as 0.2, find the reading of the pressure gage (mmHg) at the pump suction.

Given:
𝐿 1𝑚3 1𝑚𝑖𝑛 𝑚3
𝑄 = 9900 𝑚𝑖𝑛 ∗ 1000 𝐿 ∗ = 0.165
60 𝑠 𝑠

𝑍𝑎 = 10𝑚
𝑍𝑏 = 40𝑚
ø𝑎 = 300𝑚𝑚 = 0.3𝑚
𝐿𝑎 = 180𝑚
ø𝑑 = 250𝑚𝑚 = 0.25𝑚
𝐿𝑏 = 50𝑚
𝑓 = 0.025
𝜎 = 0.2
Water @ 50 oC
Req’d: Ps in mmHg

Solution:
Water @ 50oC
𝑘𝑁
𝛾 = 9.69 3
𝑚
𝑃𝑣 = 12.344 𝑘𝑃𝑎𝑎
For Hp:
Use bernoulli’s equation
𝑃𝑎 𝜈𝑎2 𝑃𝑏 𝜈𝑏2
𝑧𝑎 + + + 𝐻𝑃 = 𝑧𝑏 + + + 𝐻𝐿
𝛾 2𝑔 𝛾 2𝑔
𝑧𝑎 + 𝐻𝑃 = 𝑧𝑏 + 𝐻𝐿 (eq. 1)

For HL
0.0826𝑓𝐿𝑎 𝑄 2 0.0826𝑓𝐿𝑏 𝑄 2 𝐿𝑎 𝐿𝑏
𝐻𝐿 = 𝐻𝐿1 + 𝐻𝐿2 = 5 + 5 = 0.0826𝑓𝑄 2 ( 5 + 5 )
𝐷𝑎 𝐷𝑏 𝐷𝑎 𝐷𝑏
180 50
𝐻𝐿 = 0.0826(0.025)(0.165)2 ( + ) = 7.043 𝑚
(0.3)5 (0.25)5

From eq. 1
10 + 𝐻𝑃 = 40 + 7.043
𝐻𝑃 = 37.043 𝑚

𝑁𝑃𝑆𝐻
𝜎=
𝐻𝑃
𝑁𝑃𝑆𝐻 = 𝜎𝐻𝑃 = 0.2(37.043) = 7.4086 𝑚

𝑃 𝜈2 𝑃
𝑁𝑃𝑆𝐻 = [ 𝛾𝑠 + 2𝑔𝑠 ] − 𝛾𝑣 (eq. 2)

For Vs
𝑄 𝑄 0.165 𝑚
𝜈𝑠 = = 𝜋 =𝜋 = 2.3343
𝐴𝑠 𝐷2 4 (0.3)2 𝑠
4
From eq. 2
𝑚
𝑃𝑠 (2.3343 𝑠 )2 12.344 𝑘𝑃𝑎𝑎
7.4086𝑚 = [ 𝑘𝑁 + 𝑚 ]− 𝑘𝑁
9.69 𝑚3 2(9.807 𝑠2 ) 9.69 𝑚3
𝑃𝑠 = 81.441 𝑘𝑃𝑎𝑎
760 𝑚𝑚𝐻𝑔
𝑃𝑠 = (81.441 𝑘𝑃𝑎𝑎 − 101.325𝑘𝑃𝑎) ∗
101.325 𝑘𝑃𝑎

𝑷𝒔 = −𝟏𝟒𝟗. 𝟏𝟒 𝒎𝒎𝑯𝒈 = 𝟏𝟒𝟗. 𝟏𝟒 𝒎𝒎𝑯𝒈, 𝒗𝒂𝒄𝒖𝒖𝒎

8. Calculate the Net Positive Suction Head for a pump handling 100000
kg/hr flow of water coming from an atmospheric storage tank. The water
temperature can be taken as 250C.The suction pipe is galvanized iron
(ε=0.15) with diameter of 150 mm and a length of 20m. The pump suction
is 0.4m above ground level. The tank is elevated from a 1m high platform.
The minimum water level in the tank is 300mm
Given
Q=100 000 kg/hr
P1=101.325 kpa
Water temperature as 250C
ε=0.15
D=150mm
L=20m

Required : NPSH
Solution
𝑃1 𝑃𝑣
𝑁𝑃𝑆𝐻 = ( + (𝑧1 − 𝑧𝑠 ) − 𝐻𝐿1−𝑆 ) −
𝛾 𝛾
From tables
ρ= 997.1 kg/m3
Pv= 3.169 kpa
γ=9.778 kN/m3
𝑚2
𝜐 = 0.897𝑥106
𝑠
 𝑘𝑔 1 1ℎ𝑟 𝑚3
𝑄 = (100 000 ) ( )( ) = 0.0279
ℎ𝑟 997.1 𝑘𝑔 3600𝑠 𝑠
𝑚3
𝑚3
𝑄 0.0279 𝑠
𝑉= =𝜋 = 1.58
𝐴 (.15𝑚)2
4
𝑚3
4𝑄 4(0.0279 )
𝑠
𝑅𝑒 = = 𝑚2
= 264 016.23,
𝜋𝐷𝜐 𝜋(.15𝑚)(0.897𝑥106 )
𝑠

Therefore the flow is turbulent, use Colebrook equation

ε
1 𝐷 2.51
= −2log( + )
√𝑓 3.7 𝑅𝑒√𝑓
.15mm
1 2.51
= −2log( 150𝑚𝑚 + )
√𝑓 3.7 264016.23√𝑓
f=0.021
𝑠2 𝑠2 𝑚3 2
0.0826( )𝑓𝐿𝑄2 0.0826( )(0.021)(20𝑚)(0.0279 )
𝑚 𝑚 𝑠
𝐻𝐿1−𝑆 = =
𝐷5 (.15𝑚)5
𝐻𝐿1−𝑆 = 0.356𝑚
𝑃1 𝑃𝑣
𝑁𝑃𝑆𝐻 = ( + (𝑧1 − 𝑧𝑠 ) − 𝐻𝐿1−𝑆 ) −
𝛾 𝛾
101.235𝑘𝑝𝑎 3.169𝑘𝑝𝑎
𝑁𝑃𝑆𝐻 = ( + (1.3𝑚 − .4𝑚) − 0.356𝑚) −
9.778 𝑘𝑁/𝑚3 9.778 𝑘𝑁/𝑚3
𝑵𝑷𝑺𝑯 = 𝟏𝟎. 𝟓𝟖𝒎