Answer:
For detailed answer of "
In subsea oil and natural gas production, hydrocarbon fluids may leave the reservoir with a temperature of 70°C and flow in subsea surrounding of S°C. As a result of the temperature difference between the reservoir and the subsea surrounding, the knowledge of heat transfer is critical to prevent gas hydrate and wax deposition blockages. Consider a subsea pipeline with inner diameter of O.S m and wall thickness of 8 mm is used for transporting liquid hydrocarbon at an average temperature of 70°C, and the average convection heat transfer coefficient on the inner pipeline surface is estimated to be 2SO W/m2.K. The subsea surrounding has a temperature of soc and the average convection heat transfer coefficient on the outer pipeline surface is estimated to be ISO W /m2 .K. If the pipeline is made of material with thermal conductivity of 60 W/m.K, by using the heat conduction equation (a) obtain the temperature variation in the pipeline wall, (b) determine the inner surface temperature of the pipeline wall, (c) obtain the mathematical expression for the rate of heat loss from the liquid hydrocarbon in the pipeline, and (d) determine the heat flux through the outer pipeline surface."
see attachment.
Explanation:
Answer:
Heat required (q) = 471.19kj/kg
Explanation:
Find attached below solution to problem
Answer:
B. V represents volume, v represents velocity
Answer:
Fa = 57.32 N
Explanation:
given data
mass = 5 kg
acceleration = 4 m/s²
angular velocity ω = 2 rad/s
solution
first we take here moment about point A that is
∑Ma = Iα + ∑Mad ...............1
put here value and we get
so here I = ( 
) × m × L² ................2
I = ( ) × 5 × 0.8²
I = 0.267 kg-m²
and
a is = r × α
a = 0.4 α
so now put here value in equation is 1
0 = 0.267 α + m r α (0.4) - m A (0.4)
0 = 0.267 α + 5 (0.4α) (0.4 ) - 5 (4) 0.4
so angular acceleration α = 7.5 rad/s²
so here force acting on x axis will be
∑ F(x) = m a(x) ..............3
a(x) = m a - m rα
put here value
a(x) = 5 × 4 - 5 × 0.4 × 7.5
a(x) = 5 N
and
force acting on y axis will be
∑ F(y) = m a(y) .............. 4
a(y) - mg = mrω²
a(y) - 5 × 9.81 = 5 × 0.4 × 2²
a(y) = 57.1 N
so
total force at A will be
Fa = 
...............5
Fa = 
Fa = 57.32 N
