Answer:
heat transfer for the process is - 643.3 kJ
Explanation:
given data
mass m = 2 kg
pressure p1 = 500 kPa
temperature t1 = 400°C = 673.15 K
temperature t2 = 40°C = 313.15 K
pressure p2 = 300 kPa
to find out
heat transfer for the process
solution
we know here mass is constant so
m1 = m2
so by energy equation
m ( u2 - u1 ) = Q - W
Q is heat transfer
and in process P = A+ N that is linear spring
so
W = ∫PdV
= 0.5 ( P1+P2) ( V1 - V2)
so for case 1
P1V1 = mRT
put here value
500 V1 = 2 (0.18892) (673.15)
V1 = 0.5087 m³
and
for case 2
P2V2 = nRT
300 V2 = 2 (0.18892) (313.15)
V2 = 0.3944 m³
and
here W will be
W = 0.5 ( 500 + 300 ) ( 0.3944 - 0.5087 )
W = -45.72 kJ
and
Q is here for Cv = 0.83 from ideal gas table
Q = mCv ( T2-T1 ) + W
Q = 2 × 0.83 ( 40 - 400 ) - 45.72
Q = - 643.3 kJ
heat transfer for the process is - 643.3 kJ
It is heat because that is what made the tires lose air
Answer:
The modulus of resilience is 166.67 MPa
Explanation:
Modulus of resilience is given by yield strength ÷ strain
Yield strength = 500 MPa
Strain = 0.003
Modulus of resilience = 500 MPa ÷ 0.003 = 166.67 MPa
Answer:
Explanation:
Obtain the following properties at 6MPa and 600°C from the table "Superheated water".
Obtain the following properties at 10kPa from the table "saturated water"
Calculate the enthalpy at exit of the turbine using the energy balance equation.
Since, the process is isentropic process 
Use the isentropic relations:
Calculate the enthalpy at isentropic state 2s.
a.)
Calculate the isentropic turbine efficiency.
b.)
Find the quality of the water at state 2
since 
at 10KPa <
<
at 10KPa
Therefore, state 2 is in two-phase region.
Calculate the entropy at state 2.
Calculate the rate of entropy production.
since, Q = 0
