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3 years ago

13

An air-standard Otto cycle has a compression ratio of 6 and the temperature and pressure at the beginning of the compression pro

cess are 520 deg R and 14.2 lbf/in^2, respectively. The heat addition per unit mass of air is 600 Btu/lb. Determine (a) the maximum temperature, in deg R.

1 answer:

tangare [24]3 years ago

3 0

Answer:

The maximum temperature of the cycle is 1065⁰R

Explanation:

The maximum temperature in degree Rankin can be obtained using the formula below;

$\frac{T_2}{T_1} =[\frac{V_1}{V_2}]^{1.4-1}$

Where;

T₂ is the maximum temperature of the cycle

T₁ is the initial temperature of the cycle = 520 deg R = 520 ⁰R

V₁/V₂ is the compression ratio = 6

$T_2 = T_1(\frac{V_1}{V_2})^{0.4}$

$T_2=T_1(6)^{0.4}$

$T_2=520^0R(6)^{0.4}$

T₂ = 1064.96 ⁰R

Therefore, the maximum temperature of the cycle is 1065⁰R

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Answer:

The flux (volume of water per unit time) through the hoop will also double.

Explanation:

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This means that

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This flow rate = volume of water per unit time = Δv/Δt =Q

From all the above statements, we can say

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Steam enters a turbine steadily at 7 MPa and 600°C with a velocity of 60 m/s and leaves at 25 kPa with a quality of 95 percent.

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Answer:

a) $\dot m = 16.168\,\frac{kg}{s}$ , b) $v_{out} = 680.590\,\frac{m}{s}$ , c) $\dot W_{out} = 18276.307\,kW$

Explanation:

A turbine is a steady-state devices which transforms fluid energy into mechanical energy and is modelled after the Principle of Mass Conservation and First Law of Thermodynamics, whose expressions are described hereafter:

Mass Balance

$\frac{v_{in}\cdot A_{in}}{\nu_{in}} - \frac{v_{out}\cdot A_{out}}{\nu_{out}} = 0$

Energy Balance

$-q_{loss} - w_{out} + h_{in} - h_{out} = 0$

Specific volumes and enthalpies are obtained from property tables for steam:

Inlet (Superheated Steam)

$\nu_{in} = 0.055665\,\frac{m^{3}}{kg}$

$h_{in} = 3650.6\,\frac{kJ}{kg}$

Outlet (Liquid-Vapor Mix)

$\nu_{out} = 5.89328\,\frac{m^{3}}{kg}$

$h_{out} = 2500.2\,\frac{kJ}{kg}$

a) The mass flow rate of the steam is:

$\dot m = \frac{v_{in}\cdot A_{in}}{\nu_{in}}$

$\dot m = \frac{\left(60\,\frac{m}{s} \right)\cdot (0.015\,m^{2})}{0.055665\,\frac{m^{3}}{kg} }$

$\dot m = 16.168\,\frac{kg}{s}$

b) The exit velocity of steam is:

$\dot m = \frac{v_{out}\cdot A_{out}}{\nu_{out}}$

$v_{out} = \frac{\dot m \cdot \nu_{out}}{A_{out}}$

$v_{out} = \frac{\left(16.168\,\frac{kg}{s} \right)\cdot \left(5.89328\,\frac{m^{3}}{kg} \right)}{0.14\,m^{2}}$

$v_{out} = 680.590\,\frac{m}{s}$

c) The power output of the steam turbine is:

$\dot W_{out} = \dot m \cdot (-q_{loss} + h_{in}-h_{out})$

$\dot W_{out} = \left(16.168\,\frac{kg}{s} \right)\cdot \left(-20\,\frac{kJ}{kg} + 3650.6\,\frac{kJ}{kg} - 2500.2\,\frac{kJ}{kg}\right)$

$\dot W_{out} = 18276.307\,kW$

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2 years ago

Given a 12-bit A/D converter operating over a voltage range from ????5 V to 5 V, how much does the input voltage have to change,

Doss [256]

Answer:

2.44 mV

Explanation:

This question has to be one of analog quantization size questions and as such, we use the formula

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