Answer:
See explanation
Explanation:
Solution:-
- The shell and tube heat exchanger are designated by the order of tube and shell passes.
- A single tube pass: The fluid enters from inlet, exchange of heat, the fluid exits.
- A multiple tube pass: The fluid enters from inlet, exchange of heat, U bend of the fluid, exchange of heat, .... ( nth order of pass ), and then exits.
- By increasing the number of passes we have increased the "retention time" of a specific volume of tube fluid; hence, providing sufficient time for the fluid to exchange heat with the shell fluid.
- By making more U-turns we are allowing greater length for the fluid flow to develop with " constriction and turns " into turbulence. This turbulence usually at the final passes allows mixing of fluid and increases the heat transfer coefficient by:
U ∝ v^( 0.8 ) .... ( turbulence )
- The higher the velocity of the fluids the greater the heat transfer coefficient. The increase in the heat transfer coefficient will allow less heat energy carried by either of the fluids to be wasted ; hence, reduced losses.
Thereby, increases the thermal efficiency of the heat exchanger ( higher NTU units ).
Answer:
The amount of energy transferred to the water is 4.214 J
Explanation:
The given parameters are;
The mass of the object that drops = 5 kg
The height from which it drops = 86 mm (0.086 m)
The potential energy P.E. is given by the following formula
P.E = m·g·h
Where;
m = The mass of the object = 5 kg
g = The acceleration de to gravity = 9.8 m/s²
h = The height from which the object is dropped = 0.086 m
Therefore;
P.E. = 5 kg × 9.8 m/s² × 0.086 m = 4.214 J
Given that the potential energy is converted into heat energy, that raises the 1 g of water by 1°C, we have;
The amount of energy transferred to the water = The potential energy, P.E. = 4.214 J.
Answer:
vB = - 0.176 m/s (↓-)
Explanation:
Given
(AB) = 0.75 m
(AB)' = 0.2 m/s
vA = 0.6 m/s
θ = 35°
vB = ?
We use the formulas
Sin θ = Sin 35° = (OA)/(AB) ⇒ (OA) = Sin 35°*(AB)
⇒ (OA) = Sin 35°*(0.75 m) = 0.43 m
Cos θ = Cos 35° = (OB)/(AB) ⇒ (OB) = Cos 35°*(AB)
⇒ (OB) = Cos 35°*(0.75 m) = 0.614 m
We apply Pythagoras' theorem as follows
(AB)² = (OA)² + (OB)²
We derive the equation
2*(AB)*(AB)' = 2*(OA)*vA + 2*(OB)*vB
⇒ (AB)*(AB)' = (OA)*vA + (OB)*vB
⇒ vB = ((AB)*(AB)' - (OA)*vA) / (OB)
then we have
⇒ vB = ((0.75 m)*(0.2 m/s) - (0.43 m)*(0.6 m/s) / (0.614 m)
⇒ vB = - 0.176 m/s (↓-)
The pic can show the question.
Answer:
If Reynolds number increases the extent of the region around the object that is affected by viscosity decreases.
Explanation:
Reynolds number is an important dimensionless parameter in fluid mechanics.
It is calculated as;
where;
ρ is density
v is velocity
d is diameter
μ is viscosity
All these parameters are important in calculating Reynolds number and understanding of fluid flow over an object.
In aerodynamics, the higher the Reynolds number, the lesser the viscosity plays a role in the flow around the airfoil. As Reynolds number increases, the boundary layer gets thinner, which results in a lower drag. Or simply put, if Reynolds number increases the extent of the region around the object that is affected by viscosity decreases.
Answer:
Angle grinders are used mostly for copper, iron, steel, lead, and other metals.
Explanation:
I hope it helps! Have a great day!
Lilac~
