Answer:
The differential equation and the boundary conditions are;
A) -kdT(r1)/dr = h[T∞ - T(r1)]
B) -kdT(r2)/dr = q'_s = 734.56 W/m²
Explanation:
We are given;
T∞ = 70°C.
Inner radii pipe; r1 = 6cm = 0.06 m
Outer radii of pipe;r2 = 6.5cm=0.065 m
Electrical heat power; Q'_s = 300 W
Since power is 300 W per metre length, then; L = 1 m
Now, to the heat flux at the surface of the wire is given by the formula;
q'_s = Q'_s/A
Where A is area = 2πrL
We'll use r2 = 0.065 m
A = 2π(0.065) × 1 = 0.13π
Thus;
q'_s = 300/0.13π
q'_s = 734.56 W/m²
The differential equation and the boundary conditions are;
A) -kdT(r1)/dr = h[T∞ - T(r1)]
B) -kdT(r2)/dr = q'_s = 734.56 W/m²
Answer:
41.5° C
Explanation:
Given data :
1025 steel
Temperature = 4°C
allowed joint space = 5.4 mm
length of rails = 11.9 m
<u>Determine the highest possible temperature </u>
coefficient of thermal expansion ( ∝ ) = 12.1 * 10^-6 /°C
Applying thermal strain ( Δl / l ) = ∝ * ΔT
( 5.4 * 10^-3 / 11.9 ) = 12.1 * 10^-6 * ( T2 - 4 )
∴ ( T2 - 4 ) = ( 5.4 * 10^-3 / 11.9 ) / 12.1 * 10^-6
hence : T2 = 41.5°C
Answer:
<u><em>note:</em></u>
<u><em>solution is attached in word form due to error in mathematical equation. furthermore i also attach Screenshot of solution in word due to different version of MS Office please find the attachment</em></u>
Answer:
(a) 11.437 psia
(b) 13.963 psia
Explanation:
The pressure exerted by a fluid can be estimated by multiplying the density of the fluid, acceleration due to gravity and the depth of the fluid. To determine the fluid density, we have:
fluid density = specific gravity * density of water = 1.25 * 62.4 lbm/ft^3 = 78 lbm/ft^3
height = 28 in * (1 ft/12 in) = 2.33 ft
acceleration due to gravity = 32.174 ft/s^2
The change in pressure = fluid density*acceleration due to gravity*height = 78*32.174*(28/12) = 5855.668 lbm*ft/(s^2 * ft^2) = 5855.668 lbf/ft^2
The we convert from lbf/ft^2 to psi:
(5855.668/32.174)*0.00694 psi = 1.263 psi
(a) pressure = atmospheric pressure - change in pressure = 12.7 - 1.263 = 11.437 psia
(b) pressure = atmospheric pressure + change in pressure = 12.7 + 1.263 = 13.963 psia
