Answer:
a )
b)
Explanation:
given data:
pressure ration rp = 12
inlet temperature = 300 K
TURBINE inlet temperature = 1000 K
AT the end of isentropic process (compression) temperature is
AT the end of isentropic process (expansion) temperature is
isentropic work is given as
w = 1.005(610.18 - 300)
w = 311.73 kJ/kg
w(turbine) = 1.005( 1000 - 491.66)
w(turbine) = 510.88 kJ/kg
a) mass flow rate for isentropic process is given as
b) actual mass flow rate uis given as
Answer:
L = 41.09 Kg m2 / s The angular momentum does not depend on the time
Explanation:
The definition of angular momentum is
L = r x p
Where blacks indicate vectors
Let's apply this definition our case. Linear momentum
p = m v
Let's replace
L = m r x v
The given function is
x = 6.00 i ^ + 4.15 t j ^
We look for speed
v = dx / dt
v = 0 + 4.15 j ^
To evaluate the angular momentum one of the best ways is to use determinants
L = m 6 4.15 k ^
The other products give zero
Let's calculate
L = 1.65 6 4.15 k ^
L = 41.09 Kg m2 / s
The angular momentum does not depend on the time