Is an example of an electrical device.

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 following figures were obtained in a standard tensile test on a specimen of low carbon steel with a circular sectional area:

liq [111]

Answer:

See Explaination

Explanation:

1)here for given stress strain curve graph is given as follows

where for getting stress,S=F/A=4F/(pi*(50*10^-3)^2)

for strain=e=dl/l=dl*10^-3/100 mm/mm or m/m

2)so graph is as follows

3)for getting youngs modulus of elasticity we must know slope of graph stress verses strain and for straight line in elastic region upto 12 point we have elastic region and from that we get E as

E=slope of graph for first 12 points=S/e=14.5665*10^9/.812=17.9390*10^9 N/m2

4)for getting ultimate tensile stress at which specimen bears maximum load without failure so we get UTS as

UTS=maximum load/area=40*10^6/1.9634=20.3728*10^6 N/m2

5)percentage reduction in area is given by

percentage reduction in area=[original area-final area/original area]*100

Percent reduction=[5062-10^2]*100/50^2=96%

6)percentage elongation is given by

percent elongation=[final length-original length/original length]*100

final length at fractureis=14.56+100=114.56 mm

so we get percent elongation as=[114.56-100/100]*100=14.56%

7)true fracture stress is given by load at fracture devided by true area at fracture

Sf=load/(true area)=4*28*10^3/(pi*(10*10^-3)^2)=356.5070*10^6 N/m2

8 0

4 years ago

I want to solve the question

DedPeter [7]

Answer:

yes.

Explanation:

5 0

3 years ago

Various factors to be considered in deciding the factor of safety?

Eduardwww [97]

Answer with Explanation:

There are various factors that needed to be taken into account while deciding the factor of safety some of which are summarized below as:

1) Importance of the structure: When we design any structure different structures have different importance in our society. Take an example of hospital, in case a natural disaster struck's a place the hospital should be the designed to withstand the disaster as it's role in the crisis management following a disaster is well understood. Thus while designing it we need it to have a higher factor of safety against failure when compared to a local building.

2) Errors involved in estimation of strength of materials: when we design any component of any machine or a structure we need to have an exact idea of the behavior of the material and know the value of the strength of the material. But many materials that we use in structure such as concrete in buildings have a very complex behavior and we cannot estimate the strength of the concrete absolutely, thus we tend to decrease the strength of the concrete more if errors involved in the estimation of strength are more to give much safety to the structure.

3) Variability of the loads that may act on the structure: If the loads that act on the structure are highly variable such as earthquake loads amd dynamic loads then we tend to increase the factor of safety while estimating the loads on the structure while designing it.

4) Economic consideration: If our project has abundant funds then we can choose a higher factor of safety while designing the project.

4 0

3 years ago

A Charpy V Notch test was conducted for an ASTM A572 Grade 50 bridge steel. The average values of the test results at four diffe

goldenfox [79]

Answer:

Explanation: see attachment below

5 0

4 years ago