Answer:
work done = 48.88 ×
J
Explanation:
given data
mass = 100 kN
velocity = 310 m/s
time = 30 min = 1800 s
drag force = 12 kN
descends = 2200 m
to find out
work done by the shuttle engine
solution
we know that work done here is
work done = accelerating work - drag work - descending work
put here all value
work done = ( mass ×velocity ×time - force ×velocity ×time - mass ×descends ) 10³ J
work done = ( 100 × 310 × 1800 - 12×310 ×1800 - 100 × 2200 ) 10³ J
work done = 48.88 ×
J
Answer:
The total tube surface area in m² required to achieve an air outlet temperature of 850 K is 192.3 m²
Explanation:
Here we have the heat Q given as follows;
Q = 15 × 1075 × (1100 -
) = 10 × 1075 × (850 - 300) = 5912500 J
∴ 1100 -
= 1100/3
= 733.33 K

Where
= Arithmetic mean temperature difference
= Inlet temperature of the gas = 1100 K
= Outlet temperature of the gas = 733.33 K
= Inlet temperature of the air = 300 K
= Outlet temperature of the air = 850 K
Hence, plugging in the values, we have;

Hence, from;
, we have
5912500 = 90 × A × 341.67

Hence, the total tube surface area in m² required to achieve an air outlet temperature of 850 K = 192.3 m².
Answer:
a) 280MPa
b) -100MPa
c) -0.35
d) 380 MPa
Explanation:
GIVEN DATA:
mean stress 
stress amplitude 
a) 
--------------1

-----------2
solving 1 and 2 equation we get

b) 
c)
stress ratio

d)magnitude of stress range

= 280 -(-100) = 380 MPa
Answer:
(a) The Final Temperature is 315.25 K.
(b) The amount of mass that has entered 0.5742 Kg.
(c) The work done is 56.52 kJ.
(d) The entrophy generation is 0.0398 kJ/kgK.
Explanation:
Explanation is in the following attachments.
Answer:
I am attaching a file with the solution and explanation as the number character limit is exceeding.
Explanation: