Pump sizing
The pumps used to transfer the fluids and slurries in the mixing vessel of the amoxicillin plant
proposed by Epsilon are all centrifugal pump. Indeed, centrifugal pump is widely used for the
transport of both liquids and slurry. Moreover it operates and monitor in a certain simplicity.
Furthermore it requires few maintenance and valves to monitor the flow.
This section shows the sizing of the P-102 (the pump between the storage vessel (S-102) of
hydrochloric acid and the mixing vessel (M-101)). Pump P-102 is used to transfer the HCl to storage
within half hour. Previously, amoxicillin mixture and water were filled simultaneously into the
reactor in 30 min. thus ,the total filling time of the mixing is therefore 1h.
By using the mass balance and the filling time of each stream, the following equations were used to
find the mass flow rate and volumetric flow rate of each stream:
𝑇𝑜𝑡𝑎𝑙 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑡𝑟𝑒𝑎𝑚 𝑖
𝑚𝑖. =
𝑓𝑖𝑙𝑙𝑖𝑛𝑔 𝑡𝑖𝑚𝑒 𝑜𝑓 𝑠𝑡𝑟𝑒𝑎𝑚 𝑖
𝑚𝑖.
𝑄𝑖 =
𝑇𝑜𝑡𝑎𝑙 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑠𝑢𝑏𝑠𝑡𝑎𝑛𝑐𝑒 𝑖𝑛 𝑠𝑡𝑟𝑒𝑎𝑚 𝑖
Where:
𝑘𝑔
𝑚𝑖. = 𝑚𝑎𝑠𝑠 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑠𝑢𝑏𝑠𝑡𝑎𝑛𝑐𝑒𝑠 𝑖𝑛 𝑠𝑡𝑟𝑒𝑎𝑚 𝑖 ( )
ℎ
𝑚3
𝑄𝑖 = 𝑣𝑜𝑙𝑢𝑚𝑒𝑡𝑟𝑖𝑐 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑠𝑢𝑏𝑠𝑡𝑎𝑛𝑐𝑒𝑠 𝑖𝑛 𝑠𝑡𝑟𝑒𝑎𝑚 𝑖 ( )
ℎ
Stream A (AMOX and Water) was taken to illustrate the formula, the overall calculation of the
streams is tabulated below
𝑚𝐴 10000 + 588.49 1021.27𝐾𝑔
𝜌𝐴 = = =
𝑉𝐴𝑀𝑂𝑋 + 𝑉𝐻2𝑂 0.368 + 10 𝑚3
Table 13: Summary of mass and volumetric flow rates of the different streams.
Stream Total mass Filling time (h) Mass flow rate Volumetric flow
(kg) (kg/h) rate (m3/h)
Stream A (AMOX and 10000+588.49 0.5 21176.98 20.74
water)
Stream B (HCl) and 3.646+1000kg 0.5 2007.29 1.002
water of water
Stream (C) (NaOH) and 3.974+1000kg 0.5 2007.95 0.803
water of water
As mentioned above, the pump transferring hydrochloric acid will be considered, the pump design is
focus on stream B.
With an assumption of 10 m the length of the pipe from the storage tank to the reactor and 50 mm
as the internal diameter of the pipe.
So, the cross-sectional area of the pipe is given by:
𝜋 𝜋
𝐴𝑃 = × 𝐷𝑃2 = × (50 × 10−3 )2 = 1.963 × 10−3
4 4
The minimum velocity of stream B is therefore:
𝑚𝑏. 0.803
𝑢0 = = = 0.114 𝑚/𝑠
𝐴𝑝 1.963 × 10−3
Pressure drop in pipeline (𝑃𝑓)
One of the major concerns of pumps sizing is the relative pressure drop in pipeline, fittings, valves,
heat exchanger and other process equipment.
The pressure loss in pipe can be estimated by using the following equation (Sinnott and Towler,
2009):
𝐿 𝜌 × 𝑢2
∆𝑃𝑓 = 8𝑓 ( )×
𝐷𝑝𝑖 2
Where:
𝑓 = 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟; 𝐿 = 𝑙𝑒𝑛𝑚𝑔𝑡ℎ 𝑜𝑓 𝑝𝑖𝑝𝑒, (𝑚); 𝐷𝑃𝑖 = 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝑝𝑖𝑝𝑒, (𝑚)
𝜌 = 𝑓𝑙𝑢𝑖𝑑 𝑑𝑒𝑛𝑠𝑖𝑡𝑦, (𝑘𝑔/𝑚3);𝑢 = 𝑓𝑙𝑢𝑖𝑑 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦, 𝑚/𝑠
Calculation of friction factor:
The friction factor is function of Reynolds number and pipe relative roughness. The viscosity of HCl at
38% concentration is 2.1 × 10−3 𝑝𝑎. 𝑠.
𝜌𝐻𝐶𝑙 × 𝑢0 × 𝐷𝑝𝑖 1180 × 0.114 × 50 × 10−3
𝑅𝑒 = = = 3202.87 𝑓𝑙𝑜𝑤 𝑖𝑠 𝑡𝑢𝑟𝑏𝑢𝑙𝑒𝑛𝑡
𝜇𝐻𝐶𝑙 2.1 × 10−3
assumption, the line is made of commercial steel pipe, so the absolute roughness is 0.046 mm.
The relative roughness is then:
𝐴𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝑟𝑜𝑢𝑔ℎ𝑛𝑒𝑠𝑠 0.046
𝑒= = = 0.009 𝑚𝑚
𝑝𝑖𝑝𝑒 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 50
From the Moody diagram (Sinnott and Towler, 2009), the friction factor ( f) is found to be 0.0030.
𝐿 𝜌 × 𝑢2 10 1180 × 0.1142
∆𝑃𝑓 = 8𝑓 ( )× = 8 × 0.0030 × × = 36.80 𝑁/𝑚2
𝐷𝑝𝑖 2 50 × 10−3 2
The friction loss in the pipeline is about 0.03680 kPa.
- Miscellaneous pressure losses and total pressure loss
The miscellaneous pressure losses consider the pressure loss due to fittings and valves on the
stream. The pressure drop due to the miscellaneous losses is estimated by the method of number of
velocity heads K (Sinnott and Towler, 2009). It assumed the pipelines between the HCl storage tank
and the mixing vessel contains 2 standard 90° radius elbow, a glove valve 50% open, a plug valve, a
check valve a which pressure drop is assumed to be 0.5 bar(50 Kpa)
Table 14: Miscellaneous losses
Valve and fitting Number Number of velocity heads, K
Entry 1 0.5
Standard 90° radius elbow 2 (2 × 0.8)
Glove valve, ½ open 1 8.5
Plug valve, Open 1 0.4
Check valve 1 1
Exit 1 1.0
Total - 13
As pressure, 𝑀𝑖𝑠𝑐𝑒𝑙𝑙𝑎𝑛𝑒𝑜𝑢𝑠 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑙𝑜𝑠𝑠 = 𝐻𝑒𝑎𝑑 𝑙𝑜𝑠𝑠 × 𝜌 × 𝑔
𝑢𝐵2 0.1142
𝐻𝑒𝑎𝑑 𝑙𝑜𝑠𝑠 = × ∑𝐾 = × 13 = 0.86 𝑚
2×𝑔 2 × 9.81
Therefore, 𝑀𝑖𝑠𝑐𝑒𝑙𝑙𝑎𝑛𝑒𝑜𝑢𝑠 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑙𝑜𝑠𝑠 =0.86 × 1180 × 9.81 = 9.955 𝐾𝑝𝑎
𝑇𝑜𝑡𝑎𝑙 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑙𝑜𝑠𝑠 = 9.955 + 0.03680 + 50 = 60 𝐾𝑝𝑎
