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Diano4ka-milaya [45]
3 years ago
6

Turbine blades mounted to a rotating disc in a gas turbine engine are exposed to a gas stream that is at [infinity] = 1100°C and

maintains a convection coefficient of 250 W/m2·K over the blade. The blades, which are fabricated from Inconel, 20 W/m·K, have a length of 50 mm. The blade profile has a uniform cross-sectional area of 6 × 10-4 m2 and a perimeter of 110 mm. A proposed blade-cooling scheme, which involves routing air through the supporting disc, is able to maintain the base of each blade at a temperature of 300°C.
a. If the maximum allowable blade temperature is 1050 deg. C and the blade tip may be assumed to be adiabatic, is the proposed cooling scheme satisfactory?
b. For the proposed heating scheme, what is the rate at which heat is transferred from each blade to the coolant?
Engineering
1 answer:
Nataly_w [17]3 years ago
3 0

Answer:

<u><em>note:</em></u>

<u><em>solution is attached in word form due to error in mathematical equation. furthermore i also attach Screenshot of solution in word due to different version of MS Office please find the attachmen</em></u>t

∴ Calculation maybe wrong but method is correct

Download docx
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. Air at 200 C blows over a 50 cm x 75 cm plain carbon steel (AISI 1010) hot plate with a constant surface temperature of 2500 C
MrRissso [65]

Answer:

The inside temperature, T_{in} is approximately 248 °C.

Explanation:

The parameters given are;

Temperature of the air = 20°C

Carbon steel surface temperature 250°C

Area of surface = 50 cm × 75 cm = 0.5 × 0.75 = 0.375 m²

Convection heat transfer coefficient = 25 W/(m²·K)

Heat lost by radiation = 300 W

Assumption,

Air temperature = 20 °C

Hot plate temperature = 250 °C

Thermal conductivity K = 65.2 W/(m·K)

Steady state heat transfer process

One dimensional heat conduction

We have;

Newton's law of cooling;

q = h×A×(T_s - T_{\infty) + Heat loss by radiation

= 25×0.325×(250 - 20) + 300

= 2456.25 W

The rate of energy transfer per second is given by the following relation;

P = \dfrac{K \times A \times \Delta T}{L}

Thermal conductivity K = 65.2 W/(m·K)

Therefore;

2456.25  = \dfrac{65.2 \times 0.375 \times (250 - T_{in})}{0.02}

T_{in} = 250 - \dfrac{2456.25  \times 0.02}{65.2 \times 0.375} = 247.99 ^{\circ}C

The inside temperature, T_{in} = 247.99 °C  ≈ 248 °C.

3 0
4 years ago
Write a SELECT statement that returns the same result set as this SELECT statement. Substitute a subquery in a WHERE clause for
sergiy2304 [10]

Answer:

SELECT distinct VendorName FROM Vendors

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Explanation:

6 0
3 years ago
An Ideal gas is being heated in a circular duct as while flowing over an electric heater of 130 kW. The diameter of duct is 500
Assoli18 [71]

Answer: The exit temperature of the gas in deg C is 32^{o}C.

Explanation:

The given data is as follows.

C_{p} = 1000 J/kg K,   R = 500 J/kg K = 0.5 kJ/kg K (as 1 kJ = 1000 J)

P_{1} = 100 kPa,     V_{1} = 15 m^{3}/s

T_{1} = 27^{o}C = (27 + 273) K = 300 K

We know that for an ideal gas the mass flow rate will be calculated as follows.

     P_{1}V_{1} = mRT_{1}

or,         m = \frac{P_{1}V_{1}}{RT_{1}}

                = \frac{100 \times 15}{0.5 \times 300}  

                = 10 kg/s

Now, according to the steady flow energy equation:

mh_{1} + Q = mh_{2} + W

h_{1} + \frac{Q}{m} = h_{2} + \frac{W}{m}

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(T_{2} - T_{1})C_{p} = \frac{130 - 80}{10}

(T_{2} - T_{1}) = 5 K

T_{2} = 5 K + 300 K

T_{2} = 305 K

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Therefore, we can conclude that the exit temperature of the gas in deg C is 32^{o}C.

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4 years ago
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Answer:

unusual and strange

Explanation:

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katovenus [111]

Answer: Doctrine of ratification

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Doctrine of ratification can either be implied or expressed

Implied ratification is the type of ratification where a persons actions or body language can be seen that he has accepted.

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