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Norma-Jean [14]
3 years ago
14

In an aircraft jet engine at takeoff, the combustion products expand adiabatically in the exhaust nozzle. At entrance to the noz

zle, the pressure is 0.180 MPa and the temperature is 1200 K. The kinetic energy of the gas entering the nozzle is very much smaller than the kinetic energy of the gas leaving the nozzle. The specific heat of the exhaust gas varies with temperature approximately as follows: Cp = 0.959 + 1.16 X 10-4T + 3.65 X 10-8T2 , in which the units of cP are kJ/kg · K and Tis the temperature in K. The molecular weight of the exhaust gas is 30. Supposing the expansion to be frictionless (reversible) as well as adiabatic, show how the exhaust velocity and pressure will depend on the exhaust temperature for a series of values: 1100, 1000, and 900 K. At each temperature determine also the speed of sound VyRT'.
Engineering
1 answer:
tensa zangetsu [6.8K]3 years ago
5 0

At entrance to the nozzle, the pressure is 0.180 MPa and the temperature is 1200 K. The kinetic energy of the gas entering the nozzle is very much smaller than ... The specific heat of the exhaust gas varies with temperature approximately as follows: ... Problem 4P: In an aircraft jet engine at takeoff, the combustion product.

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An 1800-W toaster, a 1400-W electric frying pan, and a 75-W lamp are plugged into the same outlet in a 15-A, 120-V circuit. The
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Answer:

a) Current drawn by the toaster = 15A

Current drawn by the electric frying pan = 11.67A

Current drawn by the lamp = 0.625A

b) This combination will blow the 15A fuse as the total current requirement for this setup exceeds the 15A rating of the fuse.

Explanation:

a) For parallel connection, there exists, the same voltage and different currents across all the devices.

Voltage cross each of the 3 devices = outlet voltage of 120V

From their respective power rating, current drawn by each device will be calculated.

P = IV

For the toaster, P = 1800 W, V = 120V

I = 1800/120 = 15A

For the electric frying pan, P = 1400 W, V = 120 V

I = 1400/120 = 11.67 A

For the lamp, P = 75 W, V = 120V

I = 75/120 = 0.625 A

b) Total current in a parallel connection setup = Sum total of all the currents.

Total current drawn by all 3 devices = 15 + 11.67 + 0.625 = 27.295A = 27.3 A

This total current requirement surpasses the 15A current rating of the fuse, therefore, this combination will blow the fuse.

Hope this Helps!!!

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