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
You might be interested in
The image is missing, so i have attached it.
Answer:
A) P = 65.11 KN
B) Q = 30 KN
Explanation:
We are given;
The end reaction of the beam; F = 100KN
Coefficient of static friction between two steel surfaces;μ_ss = 0.3
Coefficient of static friction between steel and concrete;μ_sc = 0.6
So, F1 = μ_ss•F =0.3 x 100 = 30 KN
F2 = μ_ss•N_EF = 0.3N_EF
From the screen shot, we see that the angle is 12°
Sum of forces in the Y-direction gives;
F2•sin12 - N_EF•cos12 + 100 = 0
Rearranging gives;
N_EF•cos12 - F2•sin12 = 100
Let's put 0.3N_EF for F2 to give;
N_EF•cos12 - 0.3N_EF•sin12 = 100
Thus;
N_EF(0.9158) - 0.1247 = 100
N_EF(0.9781) = 100 + 0.1247
N_EF = 100.1247/0.9158
N_EF = 109.33 KN
Thus, F2 = 0.3N_EF = 0.3 x 109.33 = 32.8 KN
Wedge will move if;
P = (F1 + F2cos12 + N_EFsin12)
Thus;
P = 10 + (32.8 x 0.9781) + (109.33 x 0.2079)
P ≥ 65.11 KN
B) For static equilibrium, Q = F1
Thus, Q = 30 KN
