Answer:
Explanation:
Let assume that heating process occurs at constant pressure, the phenomenon is modelled by the use of the First Law of Thermodynamics:
The specific enthalpies are:
Liquid-Vapor Mixture:
Saturated Vapor:
The thermal energy per unit mass required to heat the steam is:
Answer:
a) a = - 0.0524 m/s²
b) t = 6.17 s
Explanation:
Given
τ = 0.18*τ₀ / x
τ₀ = 3.6 m/s
Determine:
a) The acceleration of the air at x= 2m.
Knowing
a = dτ / dt
Multiplying both the numerator and denominator by dx
a = (dτ / dx) (dx / dt)
Substituting τ for dx / dt
a = τ*(dτ / dx)
⇒ a = τ*(-0.18*τ₀ / x²) = (0.18*τ₀ / x)(-0.18*τ₀ / x²)
⇒ a = - 0.18²*τ₀² / x³ = - 0.18²*(3.6)² / x³
⇒ a = - 0.4199 / x³
If x = 2m
⇒ a = - 0.4199 / (2)³
⇒ a = - 0.0524 m/s²
b) The time required for the air to flow from x=1 to x= 3m.
If
τ = dx / dt = 0.18*τ₀ / x
⇒ dt = (x / 0.18*τ₀) dx
⇒ t = (1 / 0.18*τ₀) ∫x dx
⇒ t = (1 / 0.18*τ₀)*(1 / 2)*x²
then
t = (1 / (0.18*3.6))*(1 / 2)*((3)² - (1)²)
⇒ t = 6.17 s
Answer:
-Differential equation: d²T/dx² = 0
-The boundary conditions are;
1) Heat flux at bottom;
-KAdT(0)/dx = ηq_e
2) Heat flux at top surface;
-KdT(L)/dx = h(T(L) - T(water))
Explanation:
To solve this question, let's work with the following assumptions that we are given;
- Heat transfer is steady and one dimensional
- Thermal conductivity is constant.
- No heat generation exists in the medium
- The top surface which is at x = L will be subjected to convection while the bottom surface which is at x = 0 will be subjected to uniform heat flux.
Will all those assumptions given, the differential equation can be expressed as; d²T/dx² = 0
Now the boundary conditions are;
1) Heat flux at bottom;
q(at x = 0) is;
-KAdT(0)/dx = ηq_e
2) Heat flux at top surface;
q(at x = L):
-KdT(L)/dx = h(T(L) - T(water))
Answer:
speed(0)
square_length = int(input("What is the length of the square?"))
def draw_square():
pendown()
for i in range(4):
forward(square_length)
left(90)
penup()
penup()
setposition(-200, -200)
draw_square()
penup()
setposition(-200, 200 - square_length)
draw_square()
penup()
setposition(200 - square_length, 200 - square_length)
draw_square()
penup()
setposition(200 - square_length, -200)
draw_square()Explanation: