Answer:
The solution to this question is 5.153×10⁻⁴(kmol)/(m²·s)
That is the rate of diffusion of ammonia through the layer is
5.153×10⁻⁴(kmol)/(m²·s)
Explanation:
The diffusion through a stagnant layer is given by
Where
= Diffusion coefficient or diffusivity
z = Thickness in layer of transfer
R = universal gas constant
= Pressure at first boundary
= Pressure at the destination boundary
T = System temperature
= System pressure
Where = 101.3 kPa 
, 
, 
0.5×101.3 = 50.65 kPa
Δz = z₂ - z₁ = 1 mm = 1 × 10⁻³ m
R = 
T = 298 K and = 1.18 
= 1.8×10⁻⁵
= 5.153×10⁻⁴
Hence the rate of diffusion of ammonia through the layer is
5.153×10⁻⁴(kmol)/(m²·s)
Answer:
note:
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Answer:
Explanation:
From the given information:
Water freezing temp. 
Heat rejected temp 
Recall that:
The coefficient of performance is:
Again:
The efficiency given by COP can be represented by:
Finally; the power input in an hour can be determined by using the formula:
In hp; since 1 kW = 1.34102 hp
6.86kW will be = (6.86 × 1.34102) hp
= 9.199 hp
Answer:
Q = -68.859 kJ
Explanation:
given details
mass 
initial pressure P_1 = 104 kPa
Temperature T_1 = 25 Degree C = 25+ 273 K = 298 K
final pressure P_2 = 1068 kPa
Temperature T_2 = 311 Degree C = 311+ 273 K = 584 K
we know that
molecular mass of 
R = 8.314/44 = 0.189 kJ/kg K
c_v = 0.657 kJ/kgK
from ideal gas equation
PV =mRT
WORK DONE
w = 586*(0.1033 -0.514)
W =256.76 kJ
INTERNAL ENERGY IS
HEAT TRANSFER
= 187.902 +(-256.46)
Q = -68.859 kJ
True
the answer to this would be true
