A turboprop engine consists of a diffuser, compressor, combustor, turbine, and nozzle. The turbine drives a propeller as well as
the compressor. Air enters the diffuser with a volumetric flow rate of 83.7 m^3/s at 40 kPa, 240 K, and a velocity of 180 m/s, and decelerates essentially to zero velocity. The compressor pressure ratio is 10 and the compressor has an isentropic efficiency of 85%. The turbine inlet temperature is 11140 K, and its isentropic efficiency is 85%. The turbine exit pressure is 50 kPa. Flow through the diffuser and nozzle is isentropic. Using an air-standard analysis, determine the power delivered to the propeller, in MW. the velocity at the nozzle exit, in m/s.
1 answer:
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Answer:
The percent elongation in the length of the specimen is 42%
Explanation:
Given that:
The gage length of the original test specimen = 50 mm
The final gage length = 71 mm
The area = 206 mm²
maximum load = 162,699 N
To determine the percent elongation in %, we use the formula:
The percent elongation in the length of the specimen is 42%
Answer:
a.
y[n] = x[n] x[n-1] x[n+1]
(i) Memory-less - It is not memory-less because the given system is depend on past or future values.
(ii) Causal - It is non-casual because the present value of output depend on the future value of input.
(iii) Invertible - It is invertible and the inverse of the given system is
(iv) Stable - It is stable because for all the bounded input, output is bounded.
(v) Time invariant - It is not time invariant because the system is multiplying with a time varying function.
b.
y[n] = cos(x[n])
(i) Memory-less - It is memory-less because the given system is not depend on past or future values.
(ii) Causal - It is casual because the present value of output does not depend on the future value of input.
(iii) Invertible - It is not invertible because two or more than two input values can generate same output values .
For example - for x[n] = 0 , y[n] = cos(0) = 1
for x[n] = 2 , y[n] = cos(2
) = 1
(iv) Stable - It is stable because for all the bounded input, output is bounded.
(v) Time invariant - It is time invariant because the system is not multiplying with a time varying function.
Answer:
a) I_LED= 1/6 A b) Vf= 2.5V
Explanation:
Consider circuit in the attachment.
a) We will simplify current source in paraller with resistor to a voltage source in series with a resistor(see attachment 2)
Solving the circuit in attachment 2 using mesh analysis
-9+2I1+4(I1-I2)-4+2I1=0
8I1 - 4I2= 13 ............... eq 1
4+4(I2-I1)+ I2 + 2=0
4I1- 5I2 = 6 ............ eq 2
I1= 41/24 ; I2 = 1/6; I2= I_LED
b) Solving the circuit in attachment 2 again, this time I2=0
8I1 - 4I2= 13
8I1- 4(0)=13
I1= 13/8
Vf= 4(I1- I2) -4
I2=I_LED=0
Vf= 2.5 V
