Answer:
<h2>120°C</h2>
Explanation:
Step one:
given data
T_{wi} = 20^{\circ}C
T_{Ai}=1000K
T_{Ae}= 400kPa
P_{Wi}=200kPa
P_{Ai}=125kPa
P_{We}=200kPa
P_{Ae}=100kPa
m_A=2kg/s
m_W=0.5kg/s
We know that the energy equation is
making the subject of formula we have
from the saturated water table B.1.1 , corresponding to
from the ideal gas properties of air table B.7.1 , corresponding to T=1000K
the enthalpy is:
from the ideal gas properties of air table B.7.1 corresponding to T=400K
Step two:
substituting into the equation we have
from saturated water table B.1.2 at we can obtain the specific enthalpy:
we can see that , hence there are two phases
from saturated water table B.1.2 at
Idk what that means i’m sorry
Should be c.
A detailed product design
And in the question it says it was details design not just a rough sketch.
Answer:
the force conveyed by the fibers is 947.93 lb-f
Explanation:
Given the data in the question;
V_f = 80% = 0.8
V_m = 1 - V_f = 1 - 0.8 = 0.2
Now,
length of fibre L_f = length of Nylon L_n
V_f = A_f × L_f = 0.8
V_m = A_n × L_n = 0.2
so
V_f/V_m = A_f/A_n = 0.8/0.2
A_f/A_n = 4
now, the strains in fibre is equal to strains in nylon
(P/AE)f = (P/AE)n
P_f/A_fE_f = P_n/A_nE_n
P_f = (A_f/A_n)(E_f/E_n)(P_n)
P_f = ( 4 )( 131 / 2.8 )(Pn)
P_f = 187.14Pn
and P_n = Pf / 187.14
Hence
given that P_total = 953 lb-f
P_f + P_n = 953
P_f + ( P_f / 187.14 ) = 953
P_f( 1 + ( 1 / 187.14 ) ) = 953
P_f( 1.00534359 = 953
P_f = 953 / 1.00534359
P_f = 947.93 lb-f
Therefore, the force conveyed by the fibers is 947.93 lb-f