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3 years ago

11

To properly purge water from the fuel system of an aircraft equipped with fuel tank sumps and a fuel strainer quick drain, it is

necessary to drain fuel from the _________.

1 answer:

Marysya12 [62]3 years ago

3 0

Answer: fuel strainer drain and the fuel tank sumps.

Explanation:

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3 years ago

Water is the working fluid in an ideal Rankine cycle. The condenser pressure is 8 kPa, and saturated vapor enters the turbine at

sergeinik [125]

Explanation:

The obtained data from water properties tables are:

Point 1 (condenser exit) @ 8 KPa, saturated fluid

$h_{f} = 173.358 \\h_{fg} = 2402.522$

Point 2 (Pump exit) @ 18 MPa, saturated fluid & @ 4 MPa, saturated fluid

$h_{2a} = 489.752\\h_{2b} = 313.2$

Point 3 (Boiler exit) @ 18 MPa, saturated steam & @ 4 MPa, saturated steam

$h_{3a} = 2701.26 \\s_{3a} = 7.1656\\h_{3b} = 2634.14\\s_{3b} = 7.6876$

Point 4 (Turbine exit) @ 8 KPa, mixed fluid

$x_{a} = 0.8608\\h_{4a} = 2241.448938\\x_{b} = 0.9291\\h_{4b} = 2405.54119$

Calculate mass flow rates

Part a) @ 18 MPa

mass flow

$\frac{100*10^6 }{w_{T} - w_{P}} = \frac{100*10^3 }{(h_{3a} - h_{4a}) - (h_{2a} - h_{f})}\\\\= \frac{100*10^ 3}{(2701.26 - 2241.448938 ) - (489.752 - 173.358)}\\\\= 697.2671076 \frac{kg}{s} = 2510161.587 \frac{kg}{hr}$

Heat transfer rate through boiler

$Q_{in} = mass flow * (h_{3a} - h_{2a})\\Q_{in} = (697.2671076)*(2701.26-489.752)\\\\Q_{in} = 1542011.787 W$

Heat transfer rate through condenser

$Q_{out} = mass flow * (h_{4a} - h_{f})\\Q_{out} = (697.2671076)*(2241.448938-173.358)\\\\Q_{out} = 1442011.787 W$

Thermal Efficiency

$n = \frac{W_{net} }{Q_{in} } = \frac{100*10^3}{1542011.787} \\\\n = 0.06485$

Part b) @ 4 MPa

mass flow

$\frac{100*10^6 }{w_{T} - w_{P}} = \frac{100*10^3 }{(h_{3b} - h_{4b}) - (h_{2b} - h_{f})}\\\\= \frac{100*10^ 3}{(2634.14 - 2405.54119 ) - (313.12 - 173.358)}\\\\= 1125 \frac{kg}{s} = 4052374.235 \frac{kg}{hr}$

Heat transfer rate through boiler

$Q_{in} = mass flow * (h_{3b} - h_{2b})\\Q_{in} = (1125.65951)*(2634.14-313.12)\\\\Q_{in} = 2612678.236 W$

Heat transfer rate through condenser

$Q_{out} = mass flow * (h_{4b} - h_{f})\\Q_{out} = (1125)*(2405.54119-173.358)\\\\Q_{out} = 2511206.089 W$

Thermal Efficiency

$n = \frac{W_{net} }{Q_{in} } = \frac{100*10^3}{1542011.787} \\\\n = 0.038275$

6 0

3 years ago

A set of civil construction plans shows a distance of 131 meters across the front of a piece of property. You need to convert th

Hoochie [10]

Answer:

429.5 ft

Explanation:

We are told that;

1 ft = 0.305 meters

We want to convert 131 meters to ft, by proportion we have;

Distance = (131 × 1)/0.305

Distance = 429.5082 ft

We want to approximate to the nearest tenth of a foot.

This gives us;

Distance = 429.5 ft

7 0

3 years ago

A crate with dimension 12 x 10 x 4in and a mass of 120 lbs slides on an inclined plane of dimension 80 x 10 x 3in. Answer the fo

crimeas [40]

Answer:

See attachments for answers.

5 0

4 years ago

A two-story hotel has interior columns for the rooms that are spaced 6 m apart in two perpendicular directions. Determine the re

Mama L [17]

Answer:

3021.7 N/m^2 or 3.022 kN/m^2

Explanation:

The area of the interior column is equivalent to 6*6 = 36 m^2. The length $L_{o}$ of the structure is 4790 N/m^2. The live load element factor ( $K_{LL}$ ) is 4. The reduced live load will be:

L = $L_{o}(0.25 + \frac{4.57}{\sqrt{K_{LL}A_{T}}})$ = $= 4790(0.25 + \frac{4.57}{\sqrt{4*36}}) = 4790(0.25 + \frac{4.57}{\sqrt{144}}) = 4790(0.25 + \frac{4.57}{12}) = 4790(0.25 + 0.381) = 3021.7$

Therefore, the value of the reduced live load that will be supported by the column is 3021.7 N/m^2 or 3.022 kN/m^2.

This is less than 0.4* $L_{o}$ = 0.4*4790 = 1916 N/m^2

5 0

3 years ago