Give an example of a model code.
1 answer:
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Answer:
The conversion factor is 0.00223 ( 1 gallon per minute equals 0.00223 cubic feet per second)
Explanation:
Since the given volume flow rate is gallons per minute.
We know that 1 gallon = 3.785 liters and
1 minute = 60 seconds
Let the flow rate be
Now replacing the gallon and the minute by the above values we get
Thus
Now since we know that 1 liter =
Using this in above relation we get
From the above relation we can see that flow rate of 1 gallons per minute equals flow rate of 0.00223 cubic feet per second. Thus the conversion factor is 0.00223.
Answer:
Answer for the question:
Given a 8-bit ripple carry adder and the following four input scenarios: (i) A4 + 1F, (ii) AB+55, (iii) CA+34, (iv) 6D+29. a) Under which input scenario can adder generate correct output with the minimal delay? b) Under which input scenario can adder generate correct output with the maximum delay?
Is given in the attachment.
Explanation:
Answer:
A) condenser pressure = 9.73 kPa,
B) 10242 kw
C) 36.9%
Explanation:
given data
entrance pressure of steam = 12.5 MPa
temperature of steam = 550⁰c
flow rate of steam = 7.7 kg/s
outer pressure = 2 MPa
reheated steam temperature = 450⁰c
isentropic efficiency of turbine( nt ) = 85% = 0.85
isentropic efficiency of pump = 90% = 0.90
From steam tables
at 12.5 MPa and 550⁰c ; h3 = 3476.5 kJ/kg, S3 = 6.6317 kJ/kgK
also for an Isentropic expansion
S4s = S3 .
therefore when S4s = 6.6317 kJ/kg and P4 = 2 MPa
h4s = 2948.1 kJ/kg
nt = 0.85
nt (0.85) = =
making h4 subject of the equation
h4 = 3476.5 - 0.85 (3476.5 - 2948.1)
h4 = 3027.3 kj/kg
at P5 = 2 MPa and T5 = 450⁰c
h5 = 3358.2 kj/kg, s5 = 7.2815 kj/kgk
at P6 , x6 = 0.95 and s5 = s6
using nt = 0.85 we can calculate for h6 and h6s
from the chart attached below we can see that
p6 = 9.73 kPa, h6 = 2463.3 kj/kg
B) the net power output
solution is attached below
c) thermal efficiency
thermal efficiency = 1 - = 1 - ( 2273.7/ 3603.8) = 36.9% ≈ 37%
