Answer:
a) 3581.15067 kw
b) 95.4%
Explanation:
<u>Given data:</u>
compressor efficiency = 85%
compressor pressure ratio = 10
Air enters at: flow rate of 5m^3/s , pressure = 100kPa, temperature = 300 K
At turbine inlet : pressure = 950 kPa, temperature = 1400k
Turbine efficiency = 88% , exit pressure of turbine = 100 kPa
A) Develop a full accounting of the exergy increase of the air passing through the gas turbine combustor in kW
attached below is a detailed solution to the given question
Answer:
the highest rate of heat transfer allowed is 0.9306 kW
Explanation:
Given the data in the question;
Volume = 4L = 0.004 m³
V = V
= 0.002 m³
Using Table ( saturated water - pressure table);
at pressure p = 175 kPa;
v = 0.001057 m³/kg
v = 1.0037 m³/kg
u = 486.82 kJ/kg
u 2524.5 kJ/kg
h = 2700.2 kJ/kg
So the initial mass of the water;
m₁ = V/v
+ V
/v
we substitute
m₁ = 0.002/0.001057 + 0.002/1.0037
m₁ = 1.89414 kg
Now, the final mass will be;
m₂ = V/v
m₂ = 0.004 / 1.0037
m₂ = 0.003985 kg
Now, mass leaving the pressure cooker is;
m = m₁ - m₂
m = 1.89414 - 0.003985
m = 1.890155 kg
so, Initial internal energy will be;
U₁ = mu
+ m
u
U₁ = (V/v
)u
+ (V
/v
)u
we substitute
U₁ = (0.002/0.001057)(486.82) + (0.002/1.0037)(2524.5)
U₁ = 921.135288 + 5.030387
U₁ = 926.165675 kJ
Now, using Energy balance;
E - E
= ΔE
QΔt - mh
= m₂u₂ - U₁
QΔt - mh
= m₂u
- U₁
given that time = 75 min = 75 × 60s = 4500 sec
so we substitute
Q(4500) - ( 1.890155 × 2700.2 ) = ( 0.003985 × 2524.5 ) - 926.165675
Q(4500) - 5103.7965 = 10.06013 - 926.165675
Q(4500) = 10.06013 - 926.165675 + 5103.7965
Q(4500) = 4187.690955
Q = 4187.690955 / 4500
Q = 0.9306 kW
Therefore, the highest rate of heat transfer allowed is 0.9306 kW
Answer:
26.7 min
Explanation:
First, we will find the <u>time required to drill each hole</u>:
Formula for <u>distance per hole</u>: 0.5 + A + 1.75
Now, we will calculate the <u>time required to draw back the drill form hole</u>:
= 0.112 / 2 = 0.056 min
Time to move between holes = 1.5 / 15 = 0.1 min
For 100 holes, the number of moves between holes = 99
Total time required to drill 100 holes (t):
t = 100 (0.112 + 0.056) + 99 (0.1) = 26.7 min
