Answer:64.10 Btu/lbm
Explanation:
Work done in an isothermally compressed steady flow device is expressed as
Work done = P₁V₁ In { P₁/ P₂}
Work done=RT In { P₁/ P₂}
where P₁=13 psia
P₂= 80 psia
Temperature =°F Temperature is convert to °R
T(°R) = T(°F) + 459.67
T(°R) = 55°F+ 459.67
=514.67T(°R)
According to the properties of molar gas, gas constant and critical properties table, R which s the gas constant of air is given as 0.06855 Btu/lbm
Work = RT In { P₁/ P₂}
0.06855 x 514.67 In { 13/ 80}
=0.06855 x 514.67 In {0.1625}
= 0.06855 x 514.67 x -1.817
=- 64.10Btu/lbm
The required work therefore for this isothermal compression is 64.10 Btu/lbm
Answer:
Explanation:
From the question we are told that:
Plane strain fracture toughness of
Y value
Stress level of
Generally the equation for length of a surface crack is mathematically given by
Therefore
in mm
Business plan, a good strategy to help pin point the house best fetchers
Answer:
The correct option is;
Neither Technician A nor B
Explanation:
The evaporative emission monitor or Evaporaive Emission Control System EVAP System monitors enables the Power Control Module of the car to check fuel system leak integrity and the vapor consumption efficiency during engine combustion
It is a requirement of EPA on cars to check the emission of smug forming evaporates from cars
Serious monitor faults can cause the turning on of the check engine lights and the vehicle will not pass OBD II test, but it will not lead to engine shutdown
It runs when the engine is 15 to 85% full and the TP sensor is between 9% and 35%.
Therefore, the correct option is that neither Technician A nor B are correct.
Answer:
See explaination
Explanation:
int RED=10; int BLUE=11; int GREEN=12; int BUTTON1=8; int BUTTON2=9; void setup() { pinMode(RED, OUTPUT); pinMode(BLUE, OUTPUT); pinMode(GREEN, OUTPUT); pinMode(BUTTON1, INPUT); pinMode(BUTTON2, OUTPUT); } void loop() { int BTN1_STATE=digitalRead(BUTTON1); int BTN2_STATE=digitalRead(BUTTON2); if(BTN1_STATE==HIGH) { digitalWrite(BLUE, HIGH); delay(1000); // Wait for 1 second digitalWrite(BLUE, LOW); } if(BTN2_STATE==HIGH) { digitalWrite(RED, HIGH); delay(4000); // Wait for 4 seconds digitalWrite(RED, LOW); } if(BTN1_STATE==HIGH && BTN2_STATE==HIGH) { digitalWrite(GREEN, HIGH); delay(2000); // Wait for 2 second digitalWrite(GREEN, LOW); } }