In poor weather, you should _______ your following distance.

A well insulated turbine operates at steady state. Steam enters the turbine at 4 MPa with a specific enthalpy of 3015.4 kJ/kg an

Anarel [89]

Answer:

power developed by the turbine = 6927.415 kW

Explanation:

given data

pressure = 4 MPa

specific enthalpy h1 = 3015.4 kJ/kg

velocity v1 = 10 m/s

pressure = 0.07 MPa

specific enthalpy h2 = 2431.7 kJ/kg

velocity v2 = 90 m/s

mass flow rate = 11.95 kg/s

solution

we apply here thermodynamic equation that

energy equation that is

$h1 + \frac{v1}{2} + q = h2 + \frac{v2}{2} + w$

put here value with

turbine is insulated so q = 0

so here

$3015.4 *1000 + \frac{10^2}{2} = 2431.7 * 1000 + \frac{90^2}{2} + w$

solve we get

w = 579700 J/kg = 579.7 kJ/kg

and

W = mass flow rate × w

W = 11.95 × 579.7

W = 6927.415 kW

power developed by the turbine = 6927.415 kW

7 0

2 years ago

-Mn has a cubic structure with a0 = 0.8931 nm and a density of 7.47 g/cm3. -Mn has a different cubic structure, with a0 = 0.63

Fudgin [204]

Answer:

The percentage volume change is -3.0%

Explanation: We are to determine the percentage change that will occurs is alpha-Mn is transformed to beta-Mn

Value are defined as;

Cubic structure (a0) for alpha-Mn = 0.8931nm = 0.8931e-9m = 7.1236e-28cm3

Cubic structure (a0) for beta-Mn = 0.6326nm = 0.6326e-9m = 2.5316e-28cm3

Density of alpha-Mn = 7.47g/cm3

Density of beta-Mn = 7.26g/cm3

Atomic weight of Mn = 54.938g/mol

Atomic radius of Mn = 0.112nm

STEP1: CALCULATE THE ATOM NUMBER PER CELL IN THE ALPHA-Mn;

Atom per cell = (density × cubic structure × Avogadro's constant) ÷ (atomic weight ) × 100000

(7.47× 7.1236e-28 × 6.02e23) ÷ 54.938 = 58.31

Therefore the number of Atom in alpha-Mn is 58.31 atom per cell

STEP2: CALCULATE THE NUMBER OF ATOM PER CELL IN THE BETA-Mn

Atom per cell = (density × cubic structure × Avogadro's constant) ÷ (atomic weight) × 1000000

(7.26 × 2.5316e-28 × 6.02e23) ÷ 54.938 = 20.14

Therefore the number of Atom in beta-Mn is 20.14 atom per cell

STEP3: CALCULATE THE PERCENTAGE VOLUME OF ALPHA-Mn AND BETA-Mn

V% = [(volume of atom × number of atom per cell) ÷ volume of unit cell] × 1000

For Alpha-Mn:

[(1.4049e-30 × 58.31) ÷ 7.1236e-28] × 1000 = 114.998%

For Beta-Mn:

[(1.4049e-30 × 20.14) ÷ 2.5316e-28] × 1000 = 111.766%

STEP4: CALCULATE THE CHANGE IN PERCENTAGE VOLUME FOR ALPHA TO TRANSFORM TO BETA

change = final state - initial state

Therefore;

Change = 111.766 - 114.998 = -3.23%

Therefore for a transformation of Alpha-Mn to Beta-Mn they will be a decrease in volume

3 0

2 years ago

Which of the following explains why material properties present challenges for engineers?

Maurinko [17]

Answer:

Explanation:

They are altered by variables such as temperature hence making materials challenging when dealing with them.

4 0

3 years ago

Free Brainliest Again

svetlana [45]

Answer:

Explanation:

thxs

4 0

2 years ago

Read 2 more answers

You have a Switch LED Lightbulb with the following information: It is equivalent to a 40 Watt incandescent lightbulb. It uses 8

DochEvi [55]

The lightning efficiency based on the scenario depicted will be C. 56 lumens/Watt, more efficient.

<h3>How to calculate the lightning efficiency</h3>

The efficiency of the incandescent bulb will be:

= 450/40 = 11.25 lumens per watt.

The efficiency of the LED bulb will be:

= 450/8 = 56 lumens per watt.

In this case, the LED bulb is more efficient than the incandescent bulb.

Therefore, the lighting efficiency will be 56 lumens/Watt, more efficient

Learn more lightning efficiency on:

brainly.com/question/25927632

4 0

2 years ago