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
Answer:

Explanation:
Given data:
Diffusion constant for nitrogen is 
Diffusion flux 
concentration of nitrogen at high presuure = 2 kg/m^3
location on which nitrogen concentration is 0.5 kg/m^3 ......?
from fick's first law

Take C_A as point on which nitrogen concentration is 2 kg/m^3

Assume X_A is zero at the surface


Answer:
a)We know that acceleration a=dv/dt
So dv/dt=kt^2
dv=kt^2dt
Integrating we get
v(t)=kt^3/3+C
Puttin t=0
-8=C
Putting t=2
8=8k/3-8
k=48/8
k=6