पाठको तेस्रो अनुच्छेद युन्नुहोस् त्यसको मुख्य कुरा बताउनु होस्य : 6

State two reasons why industrial files must be always be used fitted upon the individual

NikAS [45]

Explanation:

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4 0

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

A strip of chicken skin was excised for mechanical testing in tension. The initial dimensions of the rectangular specimen were 3

Stells [14]

Answer:

Table and chart are attached below

Explanation:

4 0

4 years ago

A rod of aluminum has a square 4.5 mm × 4.5 mm cross section and length of 2.0 m. (A) What is the resistance of the rod along it

zimovet [89]

Answer:

The solution is attached below:

Explanation:

8 0

3 years ago

Do the coil resistances have any effect on the plots?

PolarNik [594]

Because of the skin depth effect, the current at high frequency tends to flow at very low depth from radius. Then at high frequency the effective cross section of the wire is narrower than at DC.

Fro example skin depth at 100 kHz is 0.206 mm (0.008”), a wire more thicker than AWG26 could be a waste of copper, better use a bunch of thin wire (Litz wire) to rise the Q factor.

8 0

3 years ago

पाठको तेस्रो अनुच्छेद युन्नुहोस् त्यसको मुख्य कुरा बताउनु होस्य : 6​

पाठको तेस्रो अनुच्छेद युन्नुहोस् त्यसको मुख्य कुरा बताउनु होस्य : 6