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3 years ago

15

What is 39483048^349374*3948048/3i4u4

1 answer:

Verizon [17]3 years ago

3 0

Answer:

1. $3.81813506×10^2^9$

2. $1.71479428×10^6^5$

3. $9.38483383×10^2^6$

4. $1.150847×10^2^9$

Explanation:

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Steam enters a turbine operating at steady state at 2 MPa, 323 °C with a velocity of 65 m/s. Saturated vapor exits at 0.1 MPa an

Lera25 [3.4K]

Answer:

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Explanation:

The turbine is modelled after the First Law of Thermodynamics:

$-\dot Q_{out} - \dot W_{out} + \dot H_{in} - \dot H_{out} + \dot K_{in} - \dot K_{out} + \dot U_{in} - \dot U_{out} = 0$

The rate of heat transfer between the turbine and its surroundings is:

$\dot Q_{out} = \dot H_{in}-\dot H_{out} + \dot K_{in} - \dot K_{out} - \dot W_{out} + \dot U_{in} - \dot U_{out}$

The specific enthalpies at inlet and outlet are, respectively:

$h_{in} = 3076.41\,\frac{kJ}{kg}$

$h_{out} = 2675.0\,\frac{kJ}{kg}$

The required output is:

$\dot Q_{out} = \left(8\,\frac{kg}{s} \right)\cdot \left\{3076.41\,\frac{kJ}{kg}-2675.0\,\frac{kJ}{kg}+\frac{1}{2}\cdot \left[\left(65\,\frac{m}{s} \right)^{2}-\left(42\,\frac{m}{s} \right)^{2}\right] + \left(9.807\,\frac{m}{s^{2}} \right)\cdot (4\,m) \right\} - 8000\,kW$ $\dot Q_{out} = 13369.104\,kW$

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Answer:

See attachments for step by step procedure into getting answers.

Explanation:

Given that;

R1 = 2.2 kΩ R2 = 4.7 kΩ C = 0.1 F Vdc = +5 V Vac = 5 V peak f = 1 kHz A. Use superposition to calculate the dc voltage at point X: Vdc = ______

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