Air initially at 120 psia and 500o F is expanded by an adiabatic turbine to 15 psia and 200o F. Assuming air can be treated as a
n ideal gas and has variable specific heat. a) Determine the specific work output of the actual turbine (Btu/lbm). b) Determine the amount of specific entropy generation during the irreversible process (Btu/lbm R). c) Determine the isentropic efficiency of this turbine (%). d) Suppose the turbine now operates as an ideal compressor (reversible and adiabatic) where the initial pressure is 15 psia, the initial temperature is 200 o F, and th
1 answer:
Answer:
a. Wa = 73.14 Btu/lbm
b. Sgen = 0.05042 Btu/lbm °R
c. Isentropic efficiency is 70.76%
d. Minimum specific work for compressor W = -146.2698 Btu/lbm [It is negative because work is being done on the compressor]
Explanation:
Complete question is as follows;
Air initially at 120 psia and 500oF is expanded by an adiabatic turbine to 15 psia and 200oF. Assuming air can be treated as an ideal gas and has variable specific heat.
a) Determine the specific work output of the actual turbine (Btu/lbm).
b) Determine the amount of specific entropy generation during the irreversible process (Btu/lbm R).
c) Determine the isentropic efficiency of this turbine (%).
d) Suppose the turbine now operates as an ideal compressor (reversible and adiabatic) where the initial pressure is 15 psia, the initial temperature is 200 oF, and the ideal exit state is 120 psia. What is the minimum specific work the compressor will be required to operate (Btu/lbm)?
solution;
Please check attachment for complete solution and step by step explanation
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Answer:
Diameter of riser =6.02 mm
Explanation:
Given that
Dimensions of rectangular plate is 200mm x 100mm x 20mm.
Volume of rectangle V= 200 x 100 x 20
Surface area of rectangle A
A=2(200 x 100+100 x 20 +20 x 200)
So V/A=7.69
We know that
Solidification times given as
-----1
Lets take diameter of riser is d
Given that riser is in spherical shape so V/A=d/6
And
Time for solidification of rectangle is 3.5 min then time for solidificartion of riser is 4.2 min.
Lets take
Now from equation 1
So by solving this d=6.02 mm
So the diameter of riser is 6.02 mm.
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)