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liraira [26]
3 years ago
10

A rigid tank with a total volume of 0.05 m3 initially contains a two-phase liquid-vapor mixture of water at a pressure of 15 bar

and a quality of 0.2. The tank is heated at a constant rate while a pressure-regulating valve is used to maintain a constant tank pressure by allowing saturated vapor to escape. Kinetic and potential energy effects can be neglected. If the tank is heated until the quality reaches 0.5, determine the final mass (kg) within the tank and the total amount of heat transfer required (kJ).
Engineering
1 answer:
Westkost [7]3 years ago
5 0

Answer:

a) m_{2} = 0.753\,kg, b) Q_{in} = 2122.963\,kJ

Explanation:

A rigid tank means a storage whose volume is constant. Process is entirely isobaric. Initial and final properties of water are included below:

State 1 - Gas-Vapor Mixture

P = 1500\,kPa

T = 198.29^{\textdegree}C

\nu = 0.02726\,\frac{m^{3}}{kg}

u = 1192.94\,\frac{kJ}{kg}

h_{g} = 2791.0\,\frac{kJ}{kg}

x = 0.2

State 2 - Gas-Vapor Mixture

P = 1500\,kPa

T = 198.29^{\textdegree}C

\nu = 0.06643\,\frac{m^{3}}{kg}

u = 1718.12\,\frac{kJ}{kg}

h_{g} = 2791.0\,\frac{kJ}{kg}

x = 0.5

The model for the rigid tank is created by using the First Law of Thermodynamics:

Q_{in} - (m_{1}-m_{2})\cdot h_{g} = m_{2}\cdot u_{2}-m_{1}\cdot u_{1}

Initial and final masses are:

m_{1} = \frac{V_{1}}{\nu_{1}}

m_{1} = \frac{0.05\,m^{3}}{0.02726\,\frac{m^{3}}{kg} }

m_{1} = 1.834\,kg

m_{2} = \frac{V_{2}}{\nu_{2}}

m_{2} = \frac{0.05\,m^{3}}{0.06643\,\frac{m^{3}}{kg} }

m_{2} = 0.753\,kg

a) The final mass within the tank is:

m_{2} = 0.753\,kg

b) The total amount of heat transfer is:

Q_{in} = m_{2}\cdot u_{2}-m_{1}\cdot u_{1}+ (m_{1}-m_{2})\cdot h_{g}

Q_{in} = (0.753\,kg)\cdot (1718.12\,\frac{kJ}{kg} )- (1.834\,kg)\cdot (1192.94\,\frac{kJ}{kg} ) + (1.081\,kg)\cdot (2791.0\,\frac{kJ}{kg} )

Q_{in} = 2122.963\,kJ

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Answer:

TBC thickness of 4 mm is insufficient to prevent fire hazard

Explanation:

Given:

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- The thermal conductivity of engine cover k_1 = 14 W/mK

- The thickness of engine cover L_1 = 0.01 m

- The thermal conductivity of TBC layer k_2 = 1.1 W/mK  ... (Typing error)

- The thickness of TBC layer L_2 = 0.004 m

- Temperature of ambient air outer surface T_o = 69°C

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Would a TBC layer of 4 mm thickness be sufficient to keep the engine cover surface below autoignition temperature of 200°C to prevent fire hazard?

Solution:

- We will use thermal circuit analogy for the 1-D problem and steady state conduction with no heat generation in the cover or TBC layer.

 The temperature at each medium interface and the Thermal resistance for each medium is given in the attachment schematic and circuit analogy.

 - We will calculate the total heat flux for the entire system q:

                       q = ( T_i - T_o ) / R_total

- R_total is the equivalent thermal resistance of the entire circuit. Since all resistances are in series we have:

                       R_total = 1 / h_i + L_1 / k_1 + L_2 / k_2 + 1 / h_o

- Plug in the values and compute:

                       R_total = 1 / 7 + 0.01 / 14 + 0.004 / 1.1 + 1 / 7

                       R_total = 0.2900649351 T-m^2 / W

- Calculate the Total heat flux q:

                       q = ( 333 - 69 ) / 0.2900649351

                       q = 910.141 W / m^2

- Just like the total current in a circuit remains same, the total heat flux remains same. We will use the total heat flux q to calculate the temperature of outer engine surface T_2 as follows:

                      q = ( T_i - T_2 ) / R_i2

Where,

                      R_i2 = 1 / h_i + L_1 / k_1

                      R_i2 = 1 / 7 + 0.01 / 14 = 0.14357 T-m^2 / W

Hence,

                      ( T_i - T_2 ) = q*R_i2

                        T_2 = T_i - q*R_i2

Plug the values in:

                        T_2 = 333 - 910.141*0.14357

                        T_2 = 202.33°C

- The outer surface of the engine cover has a temperature above T_ignition = 200°C. Hence, the TBC thickness of 4 mm is insufficient to prevent fire hazard

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I m from India<br><br><br>5603642259 pd 123456 ​
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Answer:

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Explanation:

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6 0
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