problems based on pumps - fluid and particle mechanics in bioprocess
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This document discusses problems involving pumps. It provides data on the pressure, vacuum gauge reading, correction head, discharge, specific weight of water, input power, and output power for two different pumps. For the first pump, the output power is calculated to be 1.3 kW when the discharge is 0.85 x 10-2 m3/sec. The efficiency is calculated to be 5.3351% when the input power is 0.3267 HP. For the second pump, the discharge is calculated to be 0.0722 m3/sec given the output power of 10.13 HP and other data. The efficiency is calculated to be 13.3% when the input power is 0.7591 HP.
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3) Use continuity equation: A1V1 = A2V2
4) Solve for P2
The pressure at point 2 is 180 kPa.
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1) Determine the vertical rise between the liquid's starting and ending points, assuming an empty tank for the worst case.
2) Calculate friction losses of all pipe and components the liquid encounters on discharge.
3) Add the vertical rise and friction losses together to obtain the total dynamic head. An example calculation is provided to demonstrate determining TDH for 25 GPM flowing from a pump to Tank B.
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1. Constant pressure heat addition in a boiler, heating water to steam.
2. Adiabatic expansion of the steam in a turbine, producing work.
3. Constant pressure heat rejection in a condenser, condensing the steam to water.
4. Adiabatic compression of the water in a pump, returning it to the boiler pressure.
The efficiency of the Rankine cycle depends on the temperature difference between the steam entering and leaving the turbine. Examples are provided to illustrate efficiency calculations for Rankine cycles operating between different pressure and temperature conditions.
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- The main parts of a centrifugal pump are the impeller, casing, suction pipe, foot valve, strainer, and delivery pipe. The impeller increases the fluid's velocity and the casing converts the velocity to pressure.
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Problem solving in fluid mechanics
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1) Determining the filter's loss coefficient
2) Calculating the flow rate if the valve is fully open and the filter's coefficient is given
3) Calculating the percentage change in flow rate between the two scenarios
4) Drawing the total energy line and hydraulic grade line for the system.
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problems based on pumps - fluid and particle mechanics in bioprocess
- 1. Problems based on pumps - DN VAISNAVI 17BBT057
- 2. A Centrifugal pump has a pressure of 9 𝑘𝑔𝑓/𝑚2 and shows vacuum gauge reading of 440 mm Hg . The correction head of pump is 0.45m . Take specific weight of water as 9.81 𝐾𝑁/𝑚3 . 1.Find the output power of the pump when the discharge is 0.85 x 10−2 𝑚3 /𝑠𝑒𝑐. 2.Also find the efficiency if the input power is 0.3267 HP.
- 3. Given data : pressure P = 9 𝑘𝑔𝑓/𝑚2 vacuum gauge reading = 440 mm of Hg = 0.44 x 13.6 m of water = 5.98 m of water. Correction head h(c) = 0.45m discharge Q = 0.85 x 10−2 𝑚3/𝑠𝑒𝑐 specific weight of water = 9.81 𝐾𝑁/𝑚3 .
- 4. Solution : output power p(o) = w x Q x H to find H ( total head ) , H = P + vacuum gauge reading + correction head h(o) = 9 + 5.98 + 0.45 H = 15.43 output power p(o) = w x Q x H = 9.81 x 0.85 x 10−2 x 15.43 = 1.3 KW
- 5. To find efficiency Efficiency in % = Output power / Input power = 1.743 / 0.3267 = 5.3351 1 kilowatt = 1.341 HP 1.3 kW = 1.743 HP
- 6. The output power of gear oil pump is 10.13HP . Find the discharge of pump if the pressure of pump is 12 𝑘𝑔𝑓/𝑚2 and vacuum gauge reading of 170 mm of mercury . The correction head of the pump is negligible . Take specific weight of water as 9.81 𝐾𝑁/𝑚3 . 1.To find the discharge of pump 2.Find the efficiency if input power is 0.7591HP.
- 7. Given data : output power p(o) = 10.13HP W = 9.81 𝐾𝑁/𝑚3 p = 12 𝑘𝑔𝑓/𝑚2 vacuum gauge reading = 170 mm of Hg = 0.17 x 13.6 m of water = 2.31 m of water . Input power , P (i) = 0.7591HP
- 8. Solution : Total head = p + vacuum gauge reading = 12 + 2.3 = 14.3 Output power = Q X H X W Q = output power / H X W = 10.13 / 14.3 x 9.81 = 0.0722 𝑚3/𝑠𝑒𝑐 efficiency in % = output power / input power = 10.13/0.7591 = 13.3