A closed system undergoes a process in which work is done on the system and the heat transfer Q occurs only at temperature Tb. F
1 answer:
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Answer:
35.7 kg lid we put
Explanation:
given data
temperature = 105 celcius
diameter = 15 cm
Patm = 101 kPa
to find out
How heavy a lid should you put
solution
we know Psaturated from table for temperature is 105 celcius is
Psat = 120.8 kPa
so
area will be here
area = ..................1
here d is diameter
put the value in equation 1
area =
area = 0.01767 m²
so net force is
Fnet = ( Psat - Patm ) × area
Fnet = ( 120.8 - 101 ) × 0.01767
Fnet = 0.3498 KN = 350 N
we know
Fnet = mg
mass =
mass =
mass = 35.7 kg
so 35.7 kg lid we put
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:
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: