Answer:
a)temperature=69.1C
b)3054Kw
Explanation:
Hello!
To solve this problem follow the steps below, the complete procedure is in the attached image
1. draw a complete outline of the problem
2. to find the temperature at the turbine exit use termodinamic tables to find the saturation temperature at 30kPa
note=Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)
through prior knowledge of two other properties such as pressure and temperature.
3. Using thermodynamic tables find the enthalpy and entropy at the turbine inlet, then find the ideal enthalpy using the entropy of state 1 and the outlet pressure = 30kPa
4. The efficiency of the turbine is defined as the ratio between the real power and the ideal power, with this we find the real enthalpy.
Note: Remember that for a turbine with a single input and output, the power is calculated as the product of the mass flow and the difference in enthalpies.
5. Find the real power of the turbine
Answer:
a cable -stayed bridge has, one or more towers,from which cable support the bridge deck.
Answer:
for 5.6V 9 turns, for 12.0V 19 turns, for 480V 755 turns
Explanation:
Vp/Vs= Np/Ns
Vp: Primary voltage
Vs: Secondary Voltage
Np: number of turns on primary side
Ns: number of turns on secondary side
for output 5.6V
140/5.6= 220/Ns
Ns= 8.8 or 9 Turns
for output 12.0V
140/12= 220/Ns
Ns= 18.9 or 19 turns
for output 480V
140/480= 220/Ns
Ns= 754.3 or 755 turns
I’m not that shareholding about it but I think the answer is C mark it green I hope I don’t get it rong
Answer:
a) it is periodic
N = (20/3)k = 20 { for K =3}
b) it is Non-Periodic.
N = ∞
c) x(n) is periodic
N = LCM ( 5, 20 )
Explanation:
We know that In Discrete time system, complex exponentials and sinusoidal signals are periodic only when ( 2π/w₀) ratio is a rational number.
then the period of the signal is given as
N = ( 2π/w₀)K
k is least integer for which N is also integer
Now, if x(n) = x1(n) + x2(n) and if x1(n) and x2(n) are periodic then x(n) will also be periodic; given N = LCM of N1 and N2
now
a) cos(2π(0.15)n)
w₀ = 2π(0.15)
Now, 2π/w₀ = 2π/2π(0.15) = 1/(0.15) = 1×20 / ( 0.15×20) = 20/3
so, it is periodic
N = (20/3)k = 20 { for K =3}
b) cos(2n);
w₀ = 2
Now, 2π/w₀ = 2π/2) = π
so, it is Non-Periodic.
N = ∞
c) cos(π0.3n) + cos(π0.4n)
x(n) = x1(n) + x2(n)
x1(n) = cos(π0.3n)
x2(n) = cos(π0.4n)
so
w₀ = π0.3
2π/w₀ = 2π/π0.3 = 2/0.3 = ( 2×10)/(0.3×10) = 20/3
∴ N1 = 20
AND
w₀ = π0.4
2π/w₀ = 2π/π0. = 2/0.4 = ( 2×10)/(0.4×10) = 20/4 = 5
∴ N² = 5
so, x(n) is periodic
N = LCM ( 5, 20 )
