Answer:
Q = 5.06 x 10⁻⁸ m³/s
Explanation:
Given:
v=0.00062 m² /s and ρ= 850 kg/m³
diameter = 8 mm
length of horizontal pipe = 40 m
Dynamic viscosity =
μ = ρv
=850 x 0.00062
= 0.527 kg/m·s
The pressure at the bottom of the tank is:
P₁,gauge = ρ g h = 850 x 9.8 x 4 = 33.32 kN/m²
The laminar flow rate through a horizontal pipe is:


Q = 5.06 x 10⁻⁸ m³/s
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Answer:
The solution for the given problem is done below.
Explanation:
M1 = 2.0
= 0.3636
= 0.5289
= 0.7934
Isentropic Flow Chart: M1 = 2.0 ,
= 1.8
T1 =
(1.8)(288K) = 653.4 K.
In order to choke the flow at the exit (M2=1), the above T0* must be stagnation temperature at the exit.
At the inlet,
T02=
= (1.8)(288K) = 518.4 K.
Q= Cp(T02-T01) =
= 135.7*
J/Kg.
Answer:
Explanation:
a) the steady-state, 1-D incompressible and no energy generation equation can be expressed as follows:

b) For a transient, 1-D, constant with energy generation
suppose T = f(x)
Then; the equation can be expressed as:

where;
= heat generated per unit volume
= Thermal diffusivity
c) The heat equation for a cylinder steady-state with 2-D constant and no compressible energy generation is:

where;
The radial directional term =
and the axial directional term is 
d) The heat equation for a wire going through a furnace is:
![\dfrac{\partial ^2 T}{\partial z^2} = \dfrac{1}{\alpha}\Big [\dfrac{\partial ^2 T}{\partial ^2 t}+ V_z \dfrac{\partial ^2T}{\partial ^2z} \Big ]](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cpartial%20%5E2%20T%7D%7B%5Cpartial%20z%5E2%7D%20%3D%20%5Cdfrac%7B1%7D%7B%5Calpha%7D%5CBig%20%5B%5Cdfrac%7B%5Cpartial%20%5E2%20T%7D%7B%5Cpartial%20%5E2%20t%7D%2B%20V_z%20%5Cdfrac%7B%5Cpartial%20%5E2T%7D%7B%5Cpartial%20%5E2z%7D%20%5CBig%20%5D)
since;
the steady-state is zero, Then:
'
e) The heat equation for a sphere that is transient, 1-D, and incompressible with energy generation is:
