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A quarter-scale turbine model is tested under a head of \(10.8 \mathrm{~m}\). The full-scale turbine is required to work under a head of \(30 \mathrm{~m}\) and to run at \(44.86 \mathrm{rad} \cdot \mathrm{s}^{-1}(7.14 \mathrm{rev} / \mathrm{s}) .\) At what speed must the model be run? If it develops \(100 \mathrm{~kW}\) and uses water at \(1.085 \mathrm{~m}^{3} \cdot \mathrm{s}^{-1}\) at this speed, what power will be obtained from the full-scale turbine, its efficiency being \(3 \%\) better than that of the model? What is the power specific speed of the full-scale, turbine?

Short Answer

Expert verified
The model must be run at approximately 26.94 rad/s. The full-scale turbine will develop approximately 18347.875 kW, and its power specific speed is approximately 0.602 rad路s鈦宦孤(m鲁/s)鹿/虏路(m)鈦烩伒/鈦.

Step by step solution

01

- Determine the model speed

Using the similarity laws for turbines, we can find the speed of the model. The relation is: \[ \frac{N_m}{N_f} = \frac{\text{(Head of Model)}^{1/2}}{\text{(Head of Full Scale)}^{1/2}} \] where \( N_m \) is the model speed, and \( N_f \) is the full-scale speed. Thus, \[ \frac{N_m}{44.86} = \frac{(10.8)^{1/2}}{(30)^{1/2}} \] Solving for \( N_m \) gives us: \[ N_m = 44.86 \times \frac{(10.8)^{1/2}}{(30)^{1/2}} \] \[ N_m = 44.86 \times \frac{3.29}{5.48} \] \[ N_m \approx 26.94 \text{ rad} \cdot \text{s}^{-1} \]
02

- Calculate the power scale factor

The power developed by the turbine scales with the ratio of the heads and the cube of the linear scale. Since the model is quarter-scale, the scale factor for linear dimensions is \( \frac{1}{4} \). The power scale factor is given by: \[ \frac{P_m}{P_f} = \frac{(Head_{model}) \times (Linear Scale)^2}{(Head_{full}) \times (Linear Scale)^5} \] Simplifying the ratio \[ \frac{P_m}{P_f} = \frac{(10.8) \times (1/4)^2}{(30) \times (1/4)^5} \] \[ \frac{P_m}{P_f} = \frac{10.8 \times 1/16}{30 \times 1/1024} \] \[ P_f = P_m \times \frac{30 \times 1/1024}{10.8 \times 1/16} \] \[ P_f = 100 \text{kW} \times \frac{30 \times 1024}{10.8 \times 16} \] \[ P_f \approx 17812.5 \text{kW} \]
03

- Adjust for efficiency

The efficiency of the full-scale turbine is 3% better than the model. If we denote the model's efficiency as \( \text{Eff}_{m} \), then the full-scale efficiency is \( \text{Eff}_{f} = \text{Eff}_{m} + 0.03 \). If the developed power without efficiency adjustments is \( 17812.5 \text{ kW} \), then the final power output of the full-scale turbine is: \[ \text{Power}_{final} = 17812.5 \times (1 + 0.03) \] \[ \text{Power}_{final} = 17812.5 \times 1.03 \] \[ \text{Power}_{final} \approx 18347.875 \text{kW} \]
04

- Determine the power specific speed

The power specific speed (n_s) of the turbine is given by: \[ n_s = N \times \frac{Q^{1/2}}{H^{5/4}} \] where \( N \) is the rotational speed of the turbine, \( Q \) is the volume flow rate, and \( H \) is the head. We need to find this for the full-scale turbine: \[ n_s = 44.86 \times \frac{(1.085)^{1/2}}{(30)^{5/4}} \] \[ n_s = 44.86 \times \frac{(1.041)}{(77.46)} \] \[ n_s \approx 0.602 \text{ rad} \cdot \text{s}^{-1} \cdot \text{(m}^3\text{/s)}^{1/2} \cdot \text{(m)}^{-5/4} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

similarity laws for turbines
The similarity laws, or affinity laws, are essential for understanding how models of turbines relate to their full-scale counterparts. These laws help us predict the performance of a full-scale turbine based on tests conducted on a scaled-down model. These laws assume geometric and dynamic similitude between the model and the prototype.

The key relationship used here is: \ \ \(\frac{N_m}{N_f} = \frac{\text{(Head of Model)}^{1/2}}{\text{(Head of Full Scale)}^{1/2}}\) \ \ This formula allows us to determine the rotational speed of the model (\(N_m\)) if we know the rotational speed of the full-scale turbine (\(N_f\)) and the heads under which both are operating.
model speed calculation
To calculate the speed at which a scaled-down model turbine must be run, we use the similarity laws. Given the full-scale turbine operates at a specific head and speed, we can calculate the appropriate model speed. Here's how we do it:

Switching from heads to speeds: \ \ \(N_m = N_f \times \left( \frac{\text{(Head of Model)}^{1/2}}{\text{(Head of Full Scale)}^{1/2}} \right)\) \ \ With a full-scale speed of 44.86 rad/s and heads of 10.8 m (model) and 30 m (full-scale), we substitute and solve: \ \ \(N_m = 44.86 \times \left( \frac{10.8^{1/2}}{30^{1/2}} \right) \) \ \ Simplifying leads to: \ \ \(N_m = 44.86 \times \frac{3.29}{5.48} \approx 26.94 \text{ rad} \cdot \text{s}^{-1}\) \ \ Thus, the model should run at 26.94 rad/s.
power scale factor
The power scale factor helps us to translate the power output from a model turbine to what a full-scale turbine would produce. This scaling takes into account the difference in both the operating head and the size scale. The power scales with the product of the heads and the fifth power of the geometric scale ratio:

Using the formula: \ \ \(\frac{P_m}{P_f} = \frac{\text{(Head of Model)} \times \text{(Linear Scale)}^2}{\text{(Head of Full Scale)} \times \text{(Linear Scale)}^5}\) \ \ Given a quarter-scale model (linear scale = 1/4) and heads: \ \ \(\frac{P_m}{P_f} = \frac{10.8 \times (1/16)}{30 \times (1/1024)}\) \ \ This simplifies further to: \ \ \(P_f = 100 \times \frac{30 \times 1024}{10.8 \times 16} \approx 17812.5 \text{kW}\)
efficiency adjustment
When transferring results from the model to the full-scale turbine, we need to account for any efficiency differences. In our case, the full-scale turbine's efficiency is 3% better than the model鈥檚. Assuming the model develops a power of 17812.5 kW without efficiency adjustments, the formula for the final power is:

\(\text{Power} = P_f \times (1 + 0.03)\) \ \ Substituting gives: \ \ \(\text{Power} = 17812.5 \times 1.03 \) \ \ Which simplifies to roughly: \ \ \(\text{Power } \approx 18347.875 \text{kW}\)
power specific speed
The power specific speed of a turbine (\(n_s\)) is a crucial non-dimensional number that helps in comparing different designs. It combines turbine speed, flow rate, and head. The equation is:

\(n_s = N \times \frac{Q^{1/2}}{H^{5/4}}\) \ \ For the full-scale turbine, using the given full-scale values: \ \ \(n_s = 44.86 \times \frac{(1.085)^{1/2}}{(30)^{5/4}}\) \ \ Upon simplifying, we have: \ \ \(n_s \approx 0.602 \text{ rad} \cdot \text{s}^{-1} \cdot (\text{m}^3/\text{s})^{1/2} \cdot (\text{m})^{-5/4}\)

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Most popular questions from this chapter

The following duties are to be performed by rotodynamic pumps driven by electric synchronous motors, speed \(100 \pi / n \mathrm{rad} \cdot \mathrm{s}^{-1}(=50 / n \mathrm{rev} / \mathrm{s})\), where \(n\) is an integer: (a) \(14 \mathrm{~m}^{3} \cdot \mathrm{s}^{-1}\) of water against \(1.5 \mathrm{~m}\) head; \((b)\) oil (relative density \(0.80\) ) at \(11.3 \mathrm{~L} \cdot \mathrm{s}^{-1}\) against \(70 \mathrm{kPa}\) pressure; \((c)\) water at \(5.25 \mathrm{~L} \cdot \mathrm{s}^{-1}\) against \(5.5 \mathrm{MPa}\). Designs of pumps are available with specific speeds of \(0.20,0.60,1.20,2.83,4.0\) rad. Which design and speed should be used for each duty?

A single-acting reciprocating water pump, with a bore and stroke of \(150 \mathrm{~mm}\) and \(300 \mathrm{~mm}\) respectively, runs at \(2.51 \mathrm{rad} \cdot \mathrm{s}^{-1}(0.4 \mathrm{rev} / \mathrm{s}) .\) Suction and delivery pipes are each \(75 \mathrm{~mm}\) diameter. The former is \(7.5 \mathrm{~m}\) long and the suction lift is \(3 \mathrm{~m}\). There is no air vessel on the suction side. The delivery, pipe is \(300 \mathrm{~m}\) long, the outlet (at atmospheric pressure) being \(13.5 \mathrm{~m}\) above the level of the pump, and a large air vessel is connected to the delivery pipe at a point \(15 \mathrm{~m}\) from the pump. Calculate the absolute pressure head in the cylinder at beginning, middle and end of each stroke. Assume that the motion of the piston is simple harmonic, that losses at inlet and outlet of each pipe are negligible, that the slip is \(2 \%\), and that \(f\). for both pipes is constant at 0.01. (Atmospheric pressure \(10.33 \mathrm{~m}\) water head.)

\(13.23\) A fluid coupling is to be used to transmit \(150 \mathrm{~kW}\) between an engine and a gear-box when the engine speed is \(251.3 \mathrm{rad} \cdot \mathrm{s}^{-1}\) (40 rev/s). The mean diameter at the outlet of the primary member is \(380 \mathrm{~mm}\) and the cross-sectional area of the flow passage is constant at \(0.026 \mathrm{~m}^{2}\). The relative density of the oil is \(0.85\) and the efficiency of the coupling \(96.5 \%\). Assuming that the shock losses under steady conditions are negligible and that the friction loss round the fluid circuit is four times the mean velocity head, calculate the mean diameter at inlet to the primary member.

In a hydro-electric scheme a number of Pelton wheels are to be used under the following conditions: total output required \(30 \mathrm{MW} ;\) gross head \(245 \mathrm{~m} ;\) speed \(39.27 \mathrm{rad} \cdot \mathrm{s}^{-1}\) \(\left(6.25\right.\) rev/s); 2 jets per wheel; \(C_{\mathrm{Y}}\) of nozzles \(0.97 ;\) maximum overall efficiency (based on conditions immediately before the nozzles) \(81.5 \% ;\) power specific speed for one jet not to exceed \(0.138\) rad \((0.022\) rev); head lost to friction in pipe-line not to exceed \(12 \mathrm{~m} .\) Calculate \((a)\) the number of wheels required, (b) the diameters of the jets and wheels, (c) the hydraulic efficiency, if the blades deflect the water through \(165^{\circ}\) and reduce its relative velocity by \(15 \%,(d)\) the percentage of the input power that remains as kinetic energy of the water at discharge.

A centrifugal fan, for which a number of interchangeable impellers are available, is to supply air at \(4.5 \mathrm{~m}^{3} \cdot \mathrm{s}^{-1}\) to a ventilating duct at a head of \(100 \mathrm{~mm}\) water gauge. For all the impellers the outer diameter is \(500 \mathrm{~mm}\), the breadth \(180 \mathrm{~mm}\) and the blade thickness negligible. The fan runs at \(188.5 \mathrm{rad} \cdot \mathrm{s}^{-1}\) ( \(\left.30 \mathrm{rev} / \mathrm{s}\right)\). Assuming that the conversion of velocity head to pressure head in the volute is counterbalanced by the friction losses there and in the impeller, that there is no whirl at inlet and that the air density is constant at \(1.23 \mathrm{~kg} \cdot \mathrm{m}^{-3}\), determine the most suitable outlet angle of the blades. (Neglect whirl slip.)

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