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3 years ago

How do scientists know our universe is older than 6500 years?

1 answer:

5 0

<span>This is because astronomers have discovered that the Crab Nebula is 6500 light years from earth. This means it takes 6500 earth years for light to get from the Crab Nebula to earth. Therefore, if we are able to observe the Crab Nebula and beyond it then it means that earth has been in existence for 6500 years and over. </span>

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Steam flows steadily through an adiabatic turbine. The inlet conditions of the steam are 4 MPa, 500°C, and 80 m/s and the exit c

FrozenT [24]

Answer:

a) $\Delta \dot K = 24.570\,kW$ , b) $\dot W_{out} = 12729.15\,kW$ , c) $A_{in} = 0.0136\,m^{2}$

Explanation:

A turbine is a device which works usually in steady state and assumption of being adiabatic means no heat interactions between steam through turbine and surroudings and produce mechanical work from fluid energy. Changes in gravitational energy can be neglected. This system can be modelled after the First Law of Thermodynamics:

$-\dot W_{out} + \dot m \cdot (h_{in} - h_{out}) + \frac{1}{2}\cdot \dot m \cdot (v_{in}^{2}-v_{out}^{2}) = 0$

a) Change in kinetic energy

$\Delta \dot K = \frac{1}{2}\cdot \dot m \cdot (v_{in}^{2} - v_{out}^{2})$

$\Delta \dot K = \frac{1}{2} \cdot \left(12.6\,\frac{kg}{s} \right) \cdot \left[\left(80\,\frac{m}{s} \right)^{2}-\left(50\,\frac{m}{s} \right)^{2}\right]$

$\Delta \dot K = 24570\,W$

$\Delta \dot K = 24.570\,kW$

b) Power output

$\dot W_{out} = \dot m \cdot (h_{in} - h_{out}) + \frac{1}{2}\cdot \dot m \cdot (v_{in}^{2}-v_{out}^{2})$

$\dot W_{out} = \left(12.6\,\frac{kg}{s}\right)\cdot \left(3446\,\frac{kJ}{kg} - 2437.7\,\frac{kJ}{kg} \right) + 24.570\,kW$

$\dot W_{out} = 12729.15\,kW$

c) Turbine inlet area

Turbine inlet area can be found by using the following expressions:

$\dot V_{in} = \dot m \cdot \nu_{in}$

$\dot V_{in} = \left(12.6\,\frac{kg}{s}\right) \cdot \left(0.086442\,\frac{m^{3}}{kg} \right)$

$\dot V_{in} = 1.089\,\frac{m^{3}}{s}$

$A_{in} = \frac{\dot V_{in}}{v_{in}}$

$A_{in} = \frac{1.089\,\frac{m^{3}}{s} }{80\,\frac{m}{s} }$

$A_{in} = 0.0136\,m^{2}$

8 0

3 years ago

What is Triangle ill send a picture

ICE Princess25 [194]

A triangle is just a shape...

4 0

3 years ago

Read 2 more answers

A cannon of mass 6.43 x 103 kg is rigidly bolted to the earth so it can recoil only by a negligible amount. The cannon fires a 7

Nata [24]

Answer:

The velocity of the shell when the cannon is unbolted is 500.14 m/s

Explanation:

Given;

mass of cannon, m₁ = 6430 kg

mass of shell, m₂ = 73.8-kg

initial velocity of the shell, u₂ = 503 m/s

Initial kinetic energy of the shell; when the cannon is rigidly bolted to the earth.

K.E = ¹/₂mv²

K.E = ¹/₂ (73.8)(503)²

K.E = 9336032.1 J

When the cannon is unbolted from the earth, we apply the principle of conservation of linear momentum and kinetic energy

change in initial momentum = change in momentum after

0 = m₁u₁ - m₂u₂

m₁v₁ = m₂v₂

where;

v₁ is the final velocity of cannon

v₂ is the final velocity of shell

$v_1 = \frac{m_2v_2}{m_1}$

Apply the principle of conservation kinetic energy

$K = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2\\\\K = \frac{1}{2}m_1(\frac{m_2v_2}{m_1})^2 + \frac{1}{2}m_2v_2^2\\\\K = \frac{1}{2}m_2v_2^2(\frac{m_2}{m_1}) + \frac{1}{2}m_2v_2^2 \\\\K = \frac{1}{2}m_2v_2^2 (\frac{m_2}{m_1} + 1)\\\\2K = m_2v_2^2 (\frac{m_2}{m_1} + 1)\\\\v_2^2 = \frac{2K}{M_2(\frac{m_2}{m_1} + 1)} \\\\v_2^2 = \frac{2*9336032.1}{73.8(\frac{73.8}{6430} + 1)}\\\\$