【Daily Question: Pump Equipment】Short Answer Question 2019.2.11
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Rules for short-answer questions: 1. If the answer is incorrect, +2 wealth for active participation; if the answer is somewhat reasonable but not specific enough, +5 wealth; if the answer is correct, +8 wealth. 2. Rewards for answers are available once per person only. 3. Each reply allows scoring for only one person. 4. The answer and its explanation will be visible after responding. 5. The reward is valid for 48 hours; no reward will be given after that period. Studying* and winning awards again; come by here every day. Briefly explain why, at the zero-flow rate point (the fully closed condition), the shaft power is not zero according to the formula for calculating shaft power Answer: 1. The shaft power formula is calculated based on the ratio of useful power to efficiency. When the flow rate is zero, the useful power is zero and so is the efficiency; therefore, efficiency cannot be used as a denominator in the formula, and this formula is not applicable when the flow rate is zero. 2. When the flow rate is zero, all of the power is consumed in unnecessary losses (hydraulic losses, volumetric losses, mechanical losses, etc.), so the shaft power is not zero at this time.
At this point, the motor delivers power; the kinetic energy of the liquid is zero. However, the electrical energy of the motor is converted into pressure energy in the liquid, as well as being consumed by the motor itself.
1. When the flow rate is 0, the head is equal to the pressure difference between the inlet and outlet of the pump (expressed in terms of liquid column height). Since the rotating impeller does work on the liquid, increasing its pressure energy, the head is certainly not 0. 2. Since the power outputted as water is 0, the shaft power is equal to the energy consumed by the pump itself. It includes: (1) mechanical losses caused by bearings, seals, etc.; (2) disk losses resulting from friction between the rotating impeller blades and the liquid; (3) volumetric losses caused by the liquid flowing out of the impeller outlet and then returning to that outlet via the mouth ring; (4) hydraulic losses arising from the flow of liquid within the pump, among other types of energy losses. At a flow rate of 0, the third term is the main energy loss. 3. Efficiency is equal to the ratio of the water power output by the pump to its shaft power; since the pump does not generate any water power at all, its efficiency is 0
1. The shaft power formula is calculated based on the ratio of useful power to efficiency. When the flow rate is zero, the useful power is zero and so is the efficiency; therefore, efficiency cannot be used as a denominator in the formula, and this formula is not applicable when the flow rate is zero. 2. When the flow rate is zero, all of the power is consumed in unnecessary losses (hydraulic losses, volumetric losses, mechanical losses, etc.), so the shaft power is not zero at this time.
Since the power output is 0, the shaft power is equal to the energy consumed by the water pump itself. It includes: (1) mechanical losses caused by bearings, seals, etc.; (2) disk losses resulting from friction between the rotating impeller blades and the liquid; (3) volumetric losses caused by the liquid flowing out of the impeller outlet and then returning to that outlet via the mouth ring; (4) hydraulic losses arising from the flow of liquid within the pump, among other types of energy losses. At a flow rate of 0, the third term is the main energy loss.
1. When the flow rate is 0, the head is equal to the pressure difference between the inlet and outlet of the pump (expressed in terms of liquid column height). Since the rotating impeller does work on the liquid, increasing its pressure energy, the head is certainly not 0. 2. Since the power outputted as water is 0, the shaft power is equal to the energy consumed by the pump itself. It includes: (1) mechanical losses caused by bearings, seals, etc.; (2) disk losses resulting from friction between the rotating impeller blades and the liquid; (3) volumetric losses caused by the liquid flowing out of the impeller outlet and then returning to that outlet via the mouth ring; (4) hydraulic losses arising from the flow of liquid within the pump, among other types of energy losses. At a flow rate of 0, the third term is the main energy loss. 3. Efficiency is equal to the ratio of the water power output by the pump to its shaft power; since the pump does not generate any water power at all, its efficiency is 0
Internal losses, such as the work lost due to friction and resistance
As the title suggests, at the zero-flow point (the shut-off point), the output power is 0; therefore, the shaft power equals the energy consumed by the water pump itself. Such losses of capacity include: (1) mechanical losses caused by bearings, seals, etc.; (2) disk losses resulting from friction between the rotating impeller blades and the fluid; (3) volumetric losses caused by the fluid flowing out of the impeller outlet and then returning to that outlet via the mouth ring; (4) hydraulic losses arising from the flow of the fluid within the pump, among other types of energy losses. Therefore, when the flow rate is 0, the shaft power is not zero.
When the shut-off flow rate is zero, the pump’s head is at its maximum; there is a pressure difference between the pump’s inlet and outlet, and the impeller is doing work, so the shaft power is not zero.