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3 years ago

If a lever operates at a mechanical disadvantage, it means that the ________.

Engineering

2 answers:

alekssr [168]3 years ago

4 0

Answer:

The correct answer is option 'B': Load is far from fulcrum and the effort is applied near the fulcrum

Explanation:

A lever works on the principle of balancing of torques. The torque about the fulcrum by the load should be equal to the torque by the applied effort. Since we know that the torque is proportional to both the force and the distance it is applied from the distance from the axis of rotation. A lever is used when we need to lift a heavy load by utilizing this effect of the lever arm.

A mechanical disadvantage occurs when we are not able to lift the weight easily due to the fact we apply effort near the fulcrum.

kherson [118]3 years ago

4 0

Answer:

Explanation:

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Consider a steam turbine, with inflow at 500oC and 7.9 MPa. The machine has a total-to-static efficiency ofηts=0.91, and the pre

sergiy2304 [10]

Answer: $\dot m_{in} = 23.942 \frac{kg}{s}$ , $\dot H_{out} = 39632.62 kW$

Explanation:

Since there is no information related to volume flow to and from turbine, let is assume that volume flow at inlet equals to $\dot V = 1 \frac{m^{3}}{s}$ . Turbine is a steady-flow system modelled by using Principle of Mass Conservation and First Law of Thermodynamics:

Principle of Mass Conservation

$\dot m_{in} - \dot m_{out} = 0$

First Law of Thermodynamics

$- \dot W_{out} + \eta\cdot (\dot m_{in} \dot h_{in} - \dot m_{out} \dot h_{out}) = 0$

This 2 x 2 System can be reduced into one equation as follows:

$-\dot W_{out} + \eta \cdot \dot m \cdot ( h_{in}- h_{out})=0$

The water goes to the turbine as Superheated steam and goes out as saturated vapor or a liquid-vapor mix. Specific volume and specific enthalpy at inflow are required to determine specific enthalpy at outflow and mass flow rate, respectively. Property tables are a practical form to get information:

Inflow (Superheated Steam)

$\nu_{in} = 0.041767 \frac{m^{3}}{kg} \\h_{in} = 3399.5 \frac{kJ}{kg}$

The mass flow rate can be calculated by using this expression:

$\dot m_{in} =\frac{\dot V_{in}}{\nu_{in}}$

$\dot m_{in} = 23.942 \frac{kg}{s}$

Afterwards, the specific enthalpy at outflow is determined by isolating it from energy balance:

$h_{out} =h_{in}-\frac{\dot W_{out}}{\eta \cdot \dot m}$

$h_{out} = 1655.36 \frac{kJ}{kg}$

The enthalpy rate at outflow is:

$\dot H_{out} = \dot m \cdot h_{out}$

$\dot H_{out} = 39632.62 kW$

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Yuliya22 [10]

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