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
Input: what is put in, taken in, or operated on by any process or system.
Output: the amount of something produced by a person, machine, or industry.
Answer:
So the rate of heat transfer to an object is equal to the thermal conductivity of the material the object is made from, multiplied by the surface area in contact, multiplied by the difference in temperature between the two objects, divided by the thickness of the material.
Answer:
64640.92 psi
Explanation:
True stress ( psi ) True strain
49300 0.11
61300 0.21
<u>Determine the true stress necessary to produce a true plastic strain of 0.25</u>
бT1 = 49300
бT2 = 61300
бT3 = ?
∈T1 = 0.11
∈T2 = 0.21
∈T3 = 0.25
note : бTi = k ∈Ti^h
∴ 49300 = k ( 0.11 )^h ----- ( 1 )
61300 = k ( 0.21)^h ------ ( 2 )
solving equations 1 and 2 simultaneously
49300/61300 = ( 0.11 / 0.21 )^h
0.804 = (0.52 )^h
next step : apply logarithm
log ( 0.804 ) = log(0.52)^h
h = log 0.804 / log (0.52)
= 0.33
back to equation 1
49300 = k ( 0.11 )^0.33
k = 49300 / (0.11)^0.33
= 102138
therefore бT3 = K (0.25)^h
= 102138 ( 0.25 )^ 0.33
= 64640.92
