Johor Bahru JB Malaysia Technical Data ~ Sejati Teknik Sdn Bhd in Johor Bahru JB Malaysia

Sejati Teknik Sdn Bhd
10, Jalan Setia 4/6,
Taman Setia Indah,
81100 Johor Bahru, Johor.
Malaysia.

+607-3527113

+607-3516024

+6012-3085077

sejati_teknik@yahoo.com
sales@sejatiteknik.com.my

+60123085077

Message Us

11, Jalan TPK 1/9,
Taman Perindustrian Kinrara,
47100 Puchong,
Selangor, Malaysia.

+603-8070 8975

+603-8070 8973

Associated Companies:
KL- QTC Engineering Sdn Bhd
Penang- Dyna Air Sdn Bhd

Technical Data

Installation

Breaker

Motor output	Breaker		Transformer	Secondary cable		Grounding
	No-fuse breaker		Power supply transformer capacity	If 22kW or less, within 10m	If 37kW or more, within 20m	Cable diameter Terminal screw diameter
	200/220V	400/440V	Power supply transformer capacity	200/220V	400/440V	Cable diameter Terminal screw diameter
7.5kW	100AF-60AT (100AF-60AT)	30AF-30AT (30AF-30AT)	10KVA (17KVA)	8mm² M 5 (14mm² M 5)	3.5mm² M 4 (3.5mm² M 4)	5.5mm² M 4
11kW	100AF-75AT (100AF-75AT)	50AF-50AT (50AF-50AT)	15KVA (20KVA)	14mm² M 5 (14mm² M 5)	5.5mm² M 4 (5.5mm² M 6)	14mm² M 5
15kW	100AF-100AT (225AF125AT) (100AF-100AT)*3	100AF-60AT (100AF-60AT)	20KVA (28KVA) (20kVA)*3	22mm² M 8 (38mm² M 6) (22mm2 M6)*3	14mm² M 5 (14mm² M 5)	14mm² M 5 (14mm2 M6)*3
22kW	NF225--175AT (225AF-175AT)1 (225AF-150AT)*2	100AF-100AT (100AF-100AT)*1	30KVA (41KVA) (32kVA)*3	38mm² M10 (38mm² M 10)	22mm² M 8 (22mm² M 8)	22mm² M 8 (22mm2 M6)*3
30kW	(225AF-175AT)*3	-	(42kVA)*3	(60mm2 M8)*3	-	(38mm2 M6)*3
37kW	NF225--225AT (400AF-300AT)1 (225AF-225AT)*2	225AF-150AT (225AF-175AT)*1	50KVA (80KVA) (52kVA)*3	100mm² M10 (100mm² M10) (80mm2 M10)*3	38mm² M 8 (38mm² M 8)	22mm² M 8 (38mm2 M8)*3
45kW	(400AF-300AT)*3	-	(63kVA)*3	(100mm2 M10)*3	-	(60mm2 M8)*3
55kW	NF400-*-400AT (400AF-350AT)	NF225--225AT (225AF-175AT)1	75KVA (74KVA)	150mm² M12 (150mm² M12)	60mm² M10 (60mm² M 8)	38mm² M 8
75kW	NF400-*-400AT (400AF-400AT)	NF225--225AT (225AF-225AT)1	150KVA (110KVA)	200mm² M12 (200mm² M12)	60mm² M10 (60mm² M12)	38mm² M 8
90kW	－	400AF-350AT	150KVA	200mm² M12	150mm² M12	38mm² M 8
100kW	－	400AF-400AT	150KVA	325mm² M12	150mm² M12	38mm² M 8
120kW	－	400AF-400AT	200KVA	325mm² M12	200mm² M12	60mm² M10
150kW	－	NF400-*-400AT	300KVA	325mm² M12	250mm² M12	60mm² M10

Figures in ( ) are that of inverter specifications.
* (SEW or HEW) Please use breakers we recommend (made by Mitsubishi Electric Co., Ltd.) (if changed in the same frame).
*1: Applied to type 1.
*2: Applied to type 2 (IPM motor machine).
*3: Applied to i-14000 IPM motor machine.

Ventilation

Motor output	Ventilation amount (in case of air cooling)			Cooling tower (in case of water cooling)
Motor output	Heat amount generated (MJ/h)	Exhausted air amount (m³/min)	If t = 5 (m³/min)	CT exit temperature, if 32ºC
7.5kW	32	20	85	-
11kW	47	30	125	-
15kW	68 [54]	45 [85] (80)*3	180 [145]	5t over
22kW	97 [79]	55 [105] (80)*3	260 [210]	10t over
30kW	(121)*3	(100)*3	(340)*3	-
37kW	159 [130]	85 [200] (120)*3	430 [400]	10t over
45kW	(184)*3	(150)*3	(510)*3	-
55kW	201 [Please ask us]	140 [Please ask us]	880 [Please ask us]	15t over
75kW	272 [Please ask us]	190 [Please ask us]	1,200 [Please ask us]	20t over
90kW	-	-	-	30t over
100kW	-	-	-	30t over
120kW	-	-	-	30t over
150kW	-	-	-	30t over

The figures in [ ] are those of ZU and ZUV.
(Please ask us or the designated service factory about the details)
*3: Applied to i-14000 IPM motor machine.

Drain

Why is the power weak in summer? (Temperature increase, occurrence of drain)
A compressor generates heat, it also makes water.

Air contains moisture. The maximum content of the moisture in air is determined by temperature and pressure and it is called the saturated moisture amount (or the saturated moisture pressure if expressed by the partial pressure of the moisture). The saturated moisture amount increases as the temperature increases, and decreases as the pressure increases. This is the reason that the moisture condenses when air is compressed (called condensed water), and the condensed water grows as the compressed air is cooled. How much condensed water is generated when operating a compressor? The answer - the amount of condensed water generated is almost proportional to the amount of air taken in.

Examples of the calculation

If 10m³/min air is compressed to 7kgf/cm².

(1) In summer

Temperature of intake air 30ºC
If the humidity of the intake air is 80%,
The water content in the intake air is 243cc/min.

　                        80
30.4(g/m³)　×   -------- ×　10(m³/min) = 243(g/min)
                          100    　

The water content in the intake air which is compressed to 7kgf/cm2 becomes 38cc/min, if the temperature at the compressor outlet is 30ºC.

　                        1
30.4(g/m³)　× -------- (compression ratio)　×　10(m³/min) = 38(g/min)
                           8

The amount of the condensed water is 243 - 38 = 205cc/min

(2) Autumn

If The temperature of intake air is 20ºC
The humidity of intake air is 60%,
The amount of the condensed water in the intake air is 82.2cc/min, if the outlet temperature is 20ºC,Or 92cc/min, if the outlet temperature is 10ºC.

(3) Winter

If the temperature of the intake air is 10ºC and
The humidity of the intake air is 50%,
The amount of the condensed water in the intake air is 35.2cc/min, if the outlet temperature is 10, or42.1cc/min, if the outlet temperature is 0ºC.

The drain water generated throughout a year blocks the lubrication of the machine, prompts the deterioration of the lubricant, causes the corrosion of the air pressure system, and finally disables the compressor from working in a safe way. Therefore, it must be strictly controlled.

Tank

In general, select the tank whose capacity is 10 - 15% of the discharge volume of the compressor.In general, select the tank whose capacity is 10 - 15% of the discharge volume of the compressor.

Piping

Air Discharge Volume

Normal

The air flow which the compressor takes in a certain time unit is called the air volume (or air quantity), and it is expressed by Q (m³/min). It is generally called the gas volume, but it is specifically called the air volume if the air is taken in. If a fan is used, it is also called the air capacity.

In terms of compressors, the air volume is not expressed by the air discharged, but always expressed by the air converted by the temperature, pressure, and humidity at the air inlet of the compressor, even if the air is used under the pressure of the discharge side. In other words, if the air volume is mentioned, it always means the intake air volume even if it is not specifically said so (JIS-B-8341, JIS-B-0142 No.2008).

Since air changes its volume a great deal depending on the pressure and temperature, to say about the air discharged at a certain place, the pressure and temperature of the place must always be stated. To avoid such trouble, the intake air volume is used.

Thus, the air is expressed converted by the intake conditions of the compressor, but in some cases, it is expressed by converting to a reference condition (0ºC, 760mmHg, dry state). In this case, it is expressed as Nm3/min, where N stands for the N in Normal Condition.
Therefore, if the air volume is expressed as Nm3/min, it must be converted to the intake condition m3/min when selecting compressors or calculating power.

For example, take the case where the air volume expressed as 1000Nm3/min is converted to the air volume under the intake conditions whose temperature is 20ºC, atmosphere pressure is 760mmHg and humidity is 50%.
First, calculate the specific gravity of the air of 20ºC, 760mmHg and 50% humidity from the following formula.

γ =

0.465 ×

750-0.378 × 0.5Χ17.5

273 + 20

=1.185kg/m³

Since the normal condition is γ = 1.293, the intake air volume Q is calculated as follows.

Q =

1000 ×

1.293

1.185

1090m³/min

=1.185kg/m³

Since the volume increases by approximately 10%, you must be careful not to overlook N. In this sense, the intake air volume m³/min is sometimes written as Bm³/min to alert you. Note that the unit m³/h is sometimes expressed as m3/s or by weight kg/s.

The intake air volume means the wet air of temperature 20ºC, absolute pressure 760mmHg, and humidity 65% (JIS-BO142), and the normal condition air volume means the dry air of temperature 0ºC, absolute pressure 760mmHg, and humidity 0% (JIS-BO142). The conversion from the intake air volume is calculated with the following formula using the temperature, atmosphere pressure, moisture pressure and humidity under the intake condition.

Q _s =

Q _n

T_s

273

1.033

P_s–Ψ × P_v

=1.185kg/m³

Where:

Q _s :	Intake air volume	(m³/min)
Q _n :	Normal condition air volume	(Nm³/min)
T_s :	Intake temperature	(°K) = (273 + t_sºC)
P_s :	Intake pressure	(kgf/cm²abs)
		1.033kgf/cm²(abs) = mmHg
Ψ :	Relative humidity	(%)
P_v :	Moisture pressure under intake temperature	(kgf/cm²abs)

*The permissible value of the discharge air volume is ±5% of the intake air volume (JIS-B8341).

Explanation

As shown in the calculation above, the conversion from the intake condition to the normal condition is affected by the temperature, atmosphere pressure and humidity.

If the intake air temperature is higher than the atmosphere temperature, the normal condition air volume further decreases.

If the discharge pressure is changed during operation, the air volume does not change but the horse power consumed changes.

Pressure drop in direct piping

Pressure 7kgf/cm²(700kPa) a pressure descent for pipe length 100m (kgf/cm²)

A free air m³/min	A	20	25	32	38	50	65	80	90	100	115	125	150	175	200	250
	B	3/4	1	1:1/4	1:1/2	2	2:1/2	3	3:1/2	4	4:1/2	5	6	7	8	10
	inside diameter mm	21.6	27.6	35.7	41.6	52.9	67.9	80.7	93.2	105.3	118.1	130.8	155.2	180.2	204.7	254.2
1		0.290	0.081	0.018	0.008
2			0.324	0.077	0.032	0.009
3				0.170	0.072	0.020
4				0.300	0.134	0.036	0.013
5					0.204	0.055	0.021
6						0.079	0.030	0.010
7						0.108	0.041	0.013
8						0.142	0.055	0.017
9						0.180	0.069	0.022
10						0.222	0.085	0.027	0.013
12							0.112	0.039	0.018	0.009
14							0.167	0.053	0.025	0.012
16							0.218	0.069	0.032	0.016	0.009
18								0.087	0.041	0.021	0.012
20								0.107	0.051	0.025	0.014	0.008
25								0.168	0.080	0.040	0.022	0.012
30									0.115	0.051	0.032	0.015
35									0.159	0.078	0.044	0.018
40										0.102	0.057	0.031	0.012
45										0.129	0.072	0.039	0.015
50										0.160	0.089	0.048	0.018	0.008
60											0.129	0.079	0.027	0.012
70												0.095	0.036	0.016	0.008
80												0.123	0.047	0.021	0.011
90													0.060	0.026	0.014
100													0.074	0.033	0.017
150															0.038	0.011
200															0.067	0.020

Calculation Example
There are 3 units of 37kW (Z37) and 2 units of 55kW (Z55). How much should the thickness of the main pipe from the compressor room of a new factory to each factory be?
(6.1m³ / min x 3) + (9.1m³ / min x 2) = 36.5m³ / min
∴ Select 3:1/2B (90A) or more in accordance with the pressure loss table.

The selection conditions of the size of air pressure piping

Equivalency direct pipe length of a pipe belonging(m)

			Pipe diameter(B)
			1/4	3/8	1/2	3/4	1	1:1/4	1:1/2	2	2:1/2	3	4	5	6	8	10	12
90°elbow	Screwing	SS	0.7	0.9	1.1	1.3	1.6	2.0	2.3	2.6	2.9	3.4	4.0
	Screwing	FC										2.7	3.4
	flange	SS			0.3	0.4	0.5	0.6	0.7	0.9	1.1	1.3	1.8	2.2	2.7	3.7	4.3	5.2
	flange	FC										1.1	1.5		2.2	3.0	3.7	4.6
90°long elbow	Screwing	SS	0.5	0.6	0.7	0.7	0.8	1.0	1.0	1.1	1.1	1.2	1.4
	Screwing	FC										1.0	1.1
	flange	SS			0.3	0.4	0.5	0.6	0.7	0.8	0.9	1.0	1.3	1.5	1.7	2.1	2.4	2.7
	flange	FC										0.0	1.0		1.4	1.7	2.1	2.4
45°elbow	Screwing	SS	0.1	0.2	0.2	0.3	0.4	0.5	0.7	0.8	1.0	1.2	1.7
	Screwing	FC										1.0	1.4
	flange	SS			0.1	0.2	0.3.	0.4	0.4	0.5	0.6	0.8	1.1	1.4	1.7	2.4	2.7	3.3
	flange	FC										0.6	0.9		1.4	1.9	2.5	3.0
Tee	Screwing	SS	0.2	0.4	0.5	0.7	1.0	1.4	1.7	2.4	2.8	3.7	5.0
	Screwing	FC										3.0	4.2
	flange	SS			0.2	0.3	0.3	0.4	0.5	0.6	0.6	0.7	0.9	1.0	1.2	1.4	1.6	1.8
	flange	FC										0.	0.		0.	1.	1.	1.
Tee	Screwing	SS	0.7	1.1	1.3	1.6	2.0	2.8	3.0	3.7	4.0	5.2	6.4
	Screwing	FC										4.2	5.2
	flange	SS			0.6	0.8	1.0	1.3	1.6	2.0	2.3	2.9	3.7	4.6	5.5	7.3	9.1	10.3
	flange	FC										2.4	3.0		4.6	6.1	7.6	9.1
180°bend	Screwing	SS	0.7	0.9	1.1	1.3	1.6	2.0	2.3	2.2	2.8	3.4	4.0
	Screwing	FC										2.7	3.4
	flange	SS			0.3	0.4	0.5	0.6	0.7	0.9	1.1	1.3	1.8	2.2	2.7	3.8	4.3	5.2
	flange	FC										1.1	1.5		2.2	3.0	3.7	4.6
	long flange	SS			0.3	0.4	0.5	0.6	0.7	0.8	0.9	1.0	1.3	1.5	1.7	2.2	2.5	2.8
	long flange	FC										0.9	.1.0		1.4	1.7	2.1	2.4
Spherical valve	Screwing	SS	6.4	6.7	6.7	7.3	8.8	11.3	12.8	16.5	18.9	24.1	33.5
	Screwing	FC										19.8	26.2
	flange	SS			11.6	12.2	13.7	16.5	18.0	21.3	23.5	28.6	36.5	45.6	57.8	79.1	94.5	11.9
	flange	FC										23.5	30.2		45.7	64.0	82.2	10.0
Gate valve	Screwing	SS	0.1	0.1	0.2	0.2	0.3	0.3	0.4	0.5	0.5	0.6	0.8
	Screwing	FC										0.5	0.6
	flange	SS								0.8	0.8	0.9	0.9	1.0	1.0	1.0	1.0	1.0
	flange	FC										0.7	0.7		0.8	0.8	0.9	0.9
Angle valve	Screwing	SS	3.9	4.6	4.6	4.6	5.2	5.5	5.5	.5.5	5.5	5.5	5.5
	Screwing	FC										4.6	4.6
	flange	SS			4.6	4.6	5.2	5.5	5.5	6.4	6.7	8.5	11.6	15.2	19.2	27.4	36.6	42.6
	flange	FC										7.0	9.4		15.8	22.6	29.9	36.5
Swing stop valve	Screwing	SS	2.2	2.2	2.4	2.7	3.4	4.0	4.6	5.8	6.7	8.2	11.6
	Screwing	FC										6.7	9.5
	flange	SS			1.2	1.6	2.2	3.1	3.7	5.2	6.4	8.2	11.6	15.2	19.2	27.4	36.6	42.7
	flange	FC										6.7	9.5		15.8	22.6	29.9	36.5
Coupling and unit	Screwing	SS	0.04	0.06	0.06	0.07	0.09	0.1	0.1	0.1	0.1	0.1	0.2
Coupling and unit	Screwing	FC										0.1	0.3

Ventilation volume

Calculation example
H
Q = -----------------
?•Cp(t2-t1)•6

Q :	Ventilation volume	[m/min]
H :	Heat generation	[kcal/h]
? :	Air specific gravity	1.2 [kg/m³]
Cp :	Air specific heat	0.24 [kcal/kg• ºC]
t¹ :	Outdoor temperature	[ºC]
t² :	Indoor temperature	[ºC]
(1kW = 860kcal/h)

Example 1

Entire ventilation without dryer Model: Z55A4 type
Heat generation : H	= 55 [kW] × 860 [kcal/h] = 47300 [kcal/h] = t2–t1 = 5 [ ºC]

Ventilation volume

Q =

47300

1.2 × 0.24 × 5 × 60

= 547[m³/min]

Vent window area

A =

60 × v × α

v :	Air speed m/sec
α :	Vent window opening ratio

Let v = 2 [m/sec], a = 60 [%]

A =

547

60 × 2 × 0.6

= 7.

Example 2

Local ventilation (duct ventilation) without dryer Model: Z55A
Heat generation : H	= 55 [kW] × 860 [kcal/h] × 0.05 = 2365 [kcal/h] = t2–t1 = 5 [ºC]

Ventilation volume

	2365
Q =		= 27.4[m³/min]
	1.2 × 0.24 × 5 × 60

Vent window area

Intake air volume Q1

= Ventilation volume + Outgoing air volume
= 27 + 140 = 167[m³]

A =

60 × v × α

v :	Air speedm/sec
α :	Vent window opening ratio

Let v = 2 [m/sec], a = 60 [%]

A =

167

60 × 2 × 0.6

= 2.3[m²]

Example 3

Entire ventilation with dryer Model: Z55A
Heat generation : H	= 55 [kW] × 860 [kcal/h] = 47300 [kcal/h]
Dryer DN55 11024 [kcal/h]	= t2–t1 = 5 [ºC]

Ventilation volume

Q =

473000 + 11024

1.2 × 0.24 × 5 × 60

= 675[m³/min]

Vent window area

A =

60 × v × α

v :	Air speedm/sec
α :	Vent window opening ratio

Let v = 2 [m / sec], a = 60 [%]

A =

675

60 × 2 × 0.6

= 9.3[m²]

Example 4

Local ventilation with dryer Model: Z55A
Heat generation : H	= 55 [kW] × 860 [kcal/h] × 0.05 = 2365 [kcal/h](the natural heat discharge is supposed to be 5%)
Dryer DN55 11024 [kcal/h[	= t2–t1 = 5 [ºC]

Ventilation volume

Q =

2365 + 11024

1.2 × 0.24 × 5 × 60

= 155[m³/min]

Vent window area

Intake air volumeQ1

= Ventilation volume + Outgoing air volume
= 155 + 140 = 295[m³/min]

v :	Air speed	m/sec
α :	Vent window opening ratio

A =

60 × v × α

Let v = 2 [m/sec], a = 60 [%]

A =

295

60 × 2 × 0.6

= 4.1[m²]

Drain volume

The moisture which is taken in with air for compression becomes water droplets by the compression and is discharged as drain. The drain volume generated changes subject to the temperature and humidity of the intake air and the discharge pressure.

The drain volume which is expected to be generated per 1m3 wet air under the intake condition if the standard atmosphere pressure is compressed to 7kgf/cm².G is calculated as follows.

Saturated moisture volume table (relative humidity 100%)

		0	1	2	3	4	5	6	7	8	9
		Temperature ºC by 1ºC unit
Temperature ºC by 10ºC unit	90	420.1	433.6	448.5	464.3	480.8	496.6	514.3	532.0	550.3	569.7
	80	290.8	301.7	313.3	325.3	337.2	349.9	362.5	375.9	389.7	404.9
	70	197.0	204.9	213.4	222.1	231.1	240.2	249.6	259.4	269.7	280.0
	60	129.8	135.6	141.5	147.6	153.9	160.5	167.3	174.2	181.6	189.0
	50	82.9	86.9	90.9	95.2	99.6	104.2	108.9	114.0	119.1	124.4
	40	51.0	53.6	56.4	59.2	62.2	65.3	68.5	71.8	75.3	78.9
	30	30.3	32.0	33.8	35.6	37.5	39.5	41.6	43.8	46.1	48.5
	20	17.3	18.3	19.4	20.6	21.8	23.0	24.3	25.7	27.2	28.7
	10	9.40	10.0	10.6	11.3	12.1	12.8	13.6	14.5	15.4	16.3
	0	4.85	5.19	5.56	5.95	6.35	6.80	7.26	7.75	8.27	8.82
	-0	4.85	4.52	4.22	3.93	3.66	3.40	3.16	2.94	2.73	2.54
	-10	2.25	2.18	2.02	1.87	1.73	1.60	1.48	1.36	1.26	1.16
	-20	1.067	0.982	0.903	0.829	0.761	0.698	0.640	0.586	0.536	0.490
	-30	0.448	0.409	0.373	0.340	0.309	0.281	0.255	0.232	0.210	0.190
	-40	0.172	0.156	0.141	0.127	0.114	0.103	0.093	0.083	0.075	0.067
	-50	0.060	0.054	0.049	0.043	0.038	0.034	0.030	0.027	0.024	0.021
	-60	0.019	0.017	0.015	0.013	0.011	0.0099	0.0087	0.0076	0.0067	0.0058
	-70	0.0051

Calculation

V :	Drain volume	[/h]
G_S1 :	Saturated moisture volume (By inlet air temperature)	[g/m³]
G_S2 :	Saturated moisture volume (By outlet dew point)	[g/m³]
P :	Air pressure	[kg/cm²]
A :	Air flow	[Nm³/min]

Example

Inlet temperature : 45 [ºC]
Pressure : 6 [kg / cm²]
Flow rate : 20.8 [Nm³ / min]
Outlet dew point : 12 [ºC]

V = (65.3–10.6) ×

1.033

P + 1.033

× 20.8

× 60 ×

= 10.03 [

/h]

Formula to calculate the drain

ω = γ_s × (χ'–100) ×

100

γ_s: Saturated moisture volume under the discharge temperature

Example

Saturated moisture volume under 40ºC
γ_s = 51.0 [g/m³]
χ' = 357.44 [%]

ω	= 51.0 ×	(357.44–100) ×	1 100
	= 131.29 [g/m^3]

In case of ZU11

131.29 1.033 + 7	× 1.4 × 60	= 1372.88 [cc/h]
		approximate 1.37 [/h]

In case of ZU22

131.29 1.033 + 7	× 3.5 × 60	= 3432.20 [cc/h]
		approximate 3432 [/h]