Answer:
A) m' = 351.49 kg/s
B) m'= 1036.91 kg/s
Explanation:
We are given;
Pressure Ratio;r_p = 12
Inlet temperature of compressor;T1 = 300 K
Inlet temperature of turbine;T3 = 1000 K
cp = 1.005 kJ/kg·K
k = 1.4
Net power output; W' = 70 MW = 70000 KW
A) Now, the formula for the mass flow rate using the total power output of the compressor and turbine is given as;
m' = W'/[cp(T3(1 - r_p^(-(k - 1)/k)) - T1(r_p^((k - 1)/k))
At, 100% efficiency, plugging in the relevant values, we have;
m' = 70000/(1.005(1000(1 - 12^(-(1.4 - 1)/1.4)) - 300(12^((1.4 - 1)/1.4)))
m' = 70000/199.1508
m' = 351.49 kg/s
B) At 85% efficiency, the formula will now be;
m' = W'/[cp(ηT3(1 - r_p^(-(k - 1)/k)) - (T1/η) (r_p^((k - 1)/k))
Where η is efficiency = 0.85
Thus;
m' = 70000/(1.005(0.85*1000(1 - 12^(-(1.4 - 1)/1.4)) - (300/0.85)(12^((1.4 - 1)/1.4)))
m' = 70000/(1.005*(432.09129 - 364.9189)
m'= 1036.91 kg/s
Answer:
Routine
Explanation:
Loop Structures — The Method Of Repeating Routines In Statements. Repetition of code are called loops, and they are defined as statements that execute lines of code (or routines) repeatedly according to conditions or iterations. ... Take for example a routine that must write as output the string “Hello” 40 times
Answer:
COP of heat pump=3.013
COP of cycle=1.124
Explanation
W = Q2 - Q1 ----- equation 1
W = work done
Q2 = final energy
Q1 = initial energy
A) calculate the COP of the heat pump
COP =Q2/W
from equation 1
Q2 = Q1 + W = 15 + 7.45 = 22.45 KW
therefore COP =22.45/7.45 = 3.013
B) COP when cycle is reversed
COP = Q1/W
from equation 1
Q1 + W = Q2 ------ equation 2
Q2 = 15 Btu/s = 15 * 1.055 = 15.825 KW therefore from equation 2
Q1 = 8.375 KW
COP =8.375/7.45 = 1.124
Answer:i think it is 35
Explanation:
i just guessed sorry im only in 5th grade
