A 2-m3 rigid tank initially contains air at 100 kPa and 22°C. The tank is connected to a supply line through a valve. Air is flo
wing in the supply line at 600 kPa and 22°C. The valve is opened, and air is allowed to enter the tank until the pressure in the tank reaches the line pressure, at which point the valve is closed. A thermometer placed in the tank indicates that the air temperature at the final state is 77°C. 1) Starting with the most general form of the appropriate mass balance equation, determine the mass of air that has entered the tank (4 points). 2) Starting with the most general form of the appropriate energy balance equation, determine the amount of heat transferred and whether it was heat transferred in or out (8 points). 3) List at least 3 assumptions needed to complete this problem (3 points).
1 answer:
Answer:
Check the explanation
Explanation:
First of all the initial or primary and final masses can be calculated with the use of the ideal gas relations.
The net It transfer is determined from the energy balance. The initial and final internal energies and the enthalpy of the air in the supply line are obtained from A-I] for the given temperatures.
kindly check the attached image below to see working.
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Answer:
We can not improve CPI of FP instructions when we run the program two times faster because it would be negative.
Explanation:
Processor clock rate = 2 GHz
Execution Time = ∑
Clock cycles can be determined using following formula
Clock cycles = ( x No. FP instructions )+ (
x No. INT instructions) + (
x No. L/S instructions ) + (
x No. branch instructions)
Clock cycles = ( 50 x x 1) + ( 110 x
x 1) + ( 80 x
x 4) + ( 16 x
x 2)
Clock cycles = 512 x 10⁶
So,Initial Execution time for FP instructions is,
=
Initial execution Time = 256 x 10⁻³
For 16 processors ,
clock cycle = 512 x 10⁶
Execution Time = 256 x 10⁻³
To run the program two times faster, half the number of clock cycles
( )= (
x No. FP instructions )+ (
x No. INT instructions) + (
x No. L/S instructions ) + (
x No. branch instructions)
x No. FP instructions = (
) -[ (
x No. INT instructions) + (
x No. L/S instructions ) + (
x No. branch instructions)]
x 50 x
= (
) - [ ( 110 x
x 1) + ( 80 x
x 4) + ( 16 x
x 2)]
x 50 x
= - 206 x
= - 206 x
/ 50 x
= - 4.12 < 0