The water in a large lake is to be used to generate electricity by the installation of a hydraulic turbine-generator. The elevation difference between the free surfaces upstream and downstream of the dam is 165 ft. Water is to be supplied at a rate of 7000 lbm/s. The electrical power generated is measured to be 1546 hp and the generator efficiency of 92%. (1 hp = 550 lbf.ft/s).

Determine:
a. the overall efficiency of the turbine-generator.
b. the mechanical efficiency of the turbine.

Answers

Answer 1

Answer:

a) 75%

b) 82%

Explanation:

Assumptions:

[tex]\text{The mechanical energy for water at turbine exit is negligible.} \\ \\ \text{The elevation of the lake remains constant.}[/tex]

Properties: The density of water [tex]\delta = 1000 kg/m^3[/tex]

Conversions:

[tex]165 \ ft \ to \ meters = 50 m \\ \\7000 \ lbm/s \ to \ kilogram/sec = 3175 kg/s \\ \\1564 \ hp \ to \ kilowatt = 1166 kw \\ \\[/tex]

Analysis:

Note that the bottom of the lake is the reference level. The potential energy of water at the surface becomes gh. Consider that kinetic energy of water at the lake surface & the turbine exit is negligible and the pressure at both locations is the atmospheric pressure and change in the mechanical energy of water between lake surface & turbine exit are:

[tex]e_{mech_{in}} - e_{mech_{out}} = gh - 0[/tex]

Then;

[tex]gh = (9.8 m/s^2) (50 m) \times \dfrac{1 \ kJ/kg}{1000 m^2/s^2}[/tex]

gh = 0.491 kJ/kg

[tex]\Delta E_{mech \ fluid} = m(e_{mech_{in}} - e_{mech_{out}} ) \\ \\ = 3175 kg/s \times 0.491 kJ/kg[/tex]

= 1559 kW

Therefore; the overall efficiency is:

[tex]\eta _{overall} = \eta_{turbine- generator} = \dfrac{W_{elect\ out}}{\Delta E_{mech \fluid}}[/tex]

[tex]= \dfrac{1166 \ kW}{1559 \ kW}[/tex]

= 0.75

= 75%

b) mechanical efficiency of the turbine:

[tex]\eta_{turbine- generator} = \eta_{turbine}\times \eta_{generator}[/tex]

thus;

[tex]\eta_{turbine} = \dfrac{\eta_{[turbine- generator]} }{\eta_{generator}} \\ \\ \eta_{turbine} = \dfrac{0.75}{0.92} \\ \\ \eta_{turbine} = 0.82 \\ \\ \eta_{turbine} = 82\%[/tex]


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Automotive gas turbines have been under development for decades but have not been commonly used in automobiles. Yet helicopters routinely use gas turbines. Explore why different types of engines are used in these respective applications. Compare selection factors such as performance, power-to-weight ratio, space requirements, fuel availability, and environmental impact.

Required:
Summarize your findings in a report with at least three references.

Answers

Answer:

Gas turbines in Helicopters require lesser space.

Explanation:

[1] In terms of Space Requirements:

The gas used in helicopters requires lesser space as compared to Automotive gas turbines. The gas in automobile have higher thermal efficiency.

[2]. In terms of Environmental impact:

The occurrence of environmental solution is very slim when  used in helicopters' engines.

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The vibrations in engines of helicopters make it to have lesser efficiency as compared to automobile.

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Air at 25 m/s and 15°C is used to cool a square hot molded plastic plate 0.5 m to a side having a surface temperature of 140°C. To increase the throughput of the production process, it is proposed to cool the plate using an array of slotted nozzles with width and pitch of 4 mm and 56 mm, respectively, and a nozzle-to-plate separation of 40 mm. The air exits the nozzle at a temperature of 15°C and a velocity of 10 m/s.

Required:
a. Determine the improvement in cooling rate that can be achieved using the slotted nozzle arrangement in lieu of turbulated air at 10 m/s and 15°C in parallel flow over the plate.
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c. What is the air mass rate requirement for the slotted nozzle arrangement?

Answers

Answer:

a. 2.30

b. decreases with increasing velocity.

c. 0.179 kg/s.

Explanation:

Without mincing let's dive straight into the solution to the question above.

                                                         [a].

The improvement in cooling rate that can be achieved using the slotted nozzle arrangement in lieu of turbulated air at 10 m/s and 15°C in parallel flow over the plate can be determined by calculating turbulent flow:

The turbulent flow over the plate= 10 × 0.5/ 20.92 × 10⁻6 = 2.39 × 10⁵.

While the turbulent flow correlation = 0.037( 2.39 × 10⁵)^[tex]\frac{4}{5}[/tex] (0.7)^[tex]\frac{1}{3}[/tex] = 659.6.

Array of slot noozle = [10 × (2  × 0.004)]/ 20.92  × 10^-6] = 3824.

where A = 4/56 =0.714.

And Ar = [ 60 + 4 (40/2  × 4) - 2 ]^2 ]-1/2 = 0.1021.

N = 2/3 (0.1021)^3/4 [ 2  ×  3824/ ( 0.0714 / 0.1021) + (.1021/0.0714)] (0.700)^0.42 =24.3.

h = 24.3  ×  0.030/0.004 = 91.1 W/m^2k.

Therefore; 659.6  × 0.030/0.5 = 39.0 W/m²k.

The turbulent flow = 0.5 × 39.6 × 0.5( 140 -15) = 1237.5 W.

The slot noozle = 91.1  ×  0.5  ×  0.5 [ 140 -15] = 2846.87W.

The improvement in cooling rate = 2846.87/ 1237.5 = 2.30.

                                                     [b].

2.3 [ (2^2/3)/ 2^4/5] = 2.1

Thus, it decreases with increasing velocity

                                                      [c].

The  air mass rate requirement for the slotted nozzle arrangement = 9 × 0.995 (0.5 × 0.004)10 = 0.179 kg/s.

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Answers

Answer:

Attached below is the derived next state equations

Explanation:

Attached below is the derived next state equations

used for the solution of the given problem.

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