Answer:
(a) - A12 = A21 = 2.747
(b) - A12 = 2.148; A21 = 2.781
(c)- A12 = 2.781; A21 = 2.148
Explanation:
(a) - x1(a) = 0.1 | x2(a) = 0.9 | x1(b) = 0.9 | x2(b) = 0.1
LLE equations:
x2(a)*γ2(a) = x2(b)γ2(b)
(b) - x1(a) = 0.2 | x2(a) = 0.8 | x1(b) = 0.9 | x2(b) = 0.1
LLE equations:
x2(a)*γ2(a) = x2(b)γ2(b)
(c) - x1(a) = 0.1 | x2(a) = 0.9 | x1(b) = 0.8 | x2(b) = 0.2
LLE equations:
x2(a)*γ2(a) = x2(b)γ2(b)
Answer:

Explanation:
Soil bearing pressure=
Since we're given pressure of 2500 psf and load of 45000 pounds
The area=
Therefore, the smallest area of safe footings should not be less than 
B) Oxygen combines with nitrogen in the air to form NOx at about 2500 degrees Fahrenheit.
Answer:

Explanation:
We are given:
m = 1.06Kg

T = 22kj
Therefore we need to find coefficient performance or the cycle


= 5
For the amount of heat absorbed:

= 5 × 22 = 110KJ
For the amount of heat rejected:

= 110 + 22 = 132KJ
[tex[ q_H = \frac{Q_L}{m} [/tex];
= 
= 124.5KJ
Using refrigerant table at hfg = 124.5KJ/Kg we have 69.5°c
Convert 69.5°c to K we have 342.5K
To find the minimum temperature:
;

= 285.4K
Convert to °C we have 12.4°C
From the refrigerant R -134a table at
= 12.4°c we have 442KPa