Answer:
Technician A
Explanation: Technician A is correct. Technician B is wrong, as an oil leak can trickle down onto other engine components, away from where the leak actually is.
Answer:
a) -8 lb / ft^3
b) -70.4 lb / ft^3
c) 54.4 lb / ft^3
Explanation:
Given:
- Diameter of pipe D = 12 in
- Shear stress t = 2.0 lb/ft^2
- y = 62.4 lb / ft^3
Find pressure gradient dP / dx when:
a) x is in horizontal flow direction
b) Vertical flow up
c) vertical flow down
Solution:
- dP / dx as function of shear stress and radial distance r:
(dP - y*L*sin(Q))/ L = 2*t / r
dP / L - y*sin(Q) = 2*t / r
Where dP / L = - dP/dx,
dP / dx = -2*t / r - y*sin(Q)
Where r = D /2 ,
dP / dx = -4*t / D - y*sin(Q)
a) Horizontal Pipe Q = 0
Hence, dP / dx = -4*2 / 1 - 62.4*sin(0)
dP / dx = -8 + 0
dP/dx = -8 lb / ft^3
b) Vertical pipe flow up Q = pi/2
Hence, dP / dx = -4*2 / 1 - 62.4*sin(pi/2)
dP / dx = 8 - 62.4
dP/dx = -70.4 lb / ft^3
c) Vertical flow down Q = -pi/2
Hence, dP / dx = -4*2 / 1 - 62.4*sin(-pi/2)
dP / dx = -8 + 62.4
dP/dx = 54.4 lb / ft^3
C, I took the test already.
To solve this problem, we must simply use the concept of Total Energy transferred both in terms of work and heat. This is basically conjugated in the first law of thermodynamics.
If we take the heat absorbed as positive and the expelled as negative we have that the total work done in the heat engine is
For the case of the engine pumps the Energy absorbed is
In this way the ratio between the two would be
So it is reversible, because the state of efficiency of the body is totally efficient.
Answer:
The value of exit temperature from the nozzle = 719.02 K
Explanation:
Temperature at inlet 
= 450°c = 723 K
Velocity at inlet 
= 55 
velocity at outlet 
= 390
Specific heat at constant pressure for steam 
Apply steady flow energy equation for the nozzle
Put all the values in the above formula we get,
⇒ 18723 × 723 + 
= 
+ 
⇒ = 719.02 K
This is the value of exit temperature from the nozzle.
