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Diano4ka-milaya [45]
3 years ago
15

Inversion is similar to what other concept discussed in the unit

Engineering
2 answers:
sveticcg [70]3 years ago
8 0

Answer:

reverse engineering

Explanation:

Maurinko [17]3 years ago
6 0

Answer:

reverse engineering

Explanation:

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D scissors or tweezers because in the medical field the EMR will likely need to use the tool for the patients like ripping a clothing
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yuradex [85]

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3 years ago
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

3 0
3 years ago
The hydrofoil boat has an A-36 steel propeller shaft that is 100 ft long. It is connected to an in-line diesel engine that deliv
Nataliya [291]

The question is incomplete. The complete question is :

The hydrofoil boat has an A-36 steel propeller shaft that is 100 ft long. It is connected to an in-line diesel engine that delivers a maximum power of 2590 hp and causes the shaft to rotate at 1700 rpm . If the outer diameter of the shaft is 8 in. and the wall thickness is $\frac{3}{8}$  in.

A) Determine the maximum shear stress developed in the shaft.

$\tau_{max}$ = ?

B) Also, what is the "wind up," or angle of twist in the shaft at full power?

$ \phi $ = ?

Solution :

Given :

Angular speed, ω = 1700 rpm

                              $ = 1700 \frac{\text{rev}}{\text{min}}\left(\frac{2 \pi \text{ rad}}{\text{rev}}\right) \frac{1 \text{ min}}{60 \ \text{s}}$

                              $= 56.67 \pi \text{ rad/s}$

Power $= 2590 \text{ hp} \left( \frac{550 \text{ ft. lb/s}}{1 \text{ hp}}\right)$

          = 1424500 ft. lb/s

Torque, $T = \frac{P}{\omega}$

                 $=\frac{1424500}{56.67 \pi}$

                 = 8001.27 lb.ft

A). Therefore, maximum shear stress is given by :

Applying the torsion formula

$\tau_{max} = \frac{T_c}{J}$

        $=\frac{8001.27 \times 12 \times 4}{\frac{\pi}{2}\left(4^2 - 3.625^4 \right)}$

      = 2.93 ksi

B). Angle of twist :

     $\phi = \frac{TL}{JG}$

         $=\frac{8001.27 \times 12 \times 100 \times 12}{\frac{\pi}{2}\left(4^4 - 3.625^4\right) \times 11 \times 10^3}$

         = 0.08002 rad

         = 4.58°

6 0
3 years ago
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