Please help me on problem #43, Heat is transferred by radiation and convection b
ID: 1860490 • Letter: P
Question
Please help me on problem #43,
Heat is transferred by radiation and convection between
the inner surface of the nacelle of the wind turbine of
Example 1.3 and the outer surfaces of the gearbox and
generator. The convection heat flux associated with
the gearbox and the generator may be described
by qconv,gb h(Tgb T) and qconv,gen h(Tgen T),
respectively, where the ambient temperature T Ts
(which is the nacelle temperature) and h 40 W/m2 K.
The outer surfaces of both the gearbox and the generator
are characterized by an emissivity of 0.9. If the surface
areas of the gearbox and generator are Agb 6 m2
and Agen 4 m2, respectively, determine their surface
temperatures.
data-mce-bogus="1">
Explanation / Answer
Assumptions:
Steady-state conditions.
Large surroundings.
Surface of the nacelle that is adjacent to the hub is adiabatic.
Analysis: The nacelle temperature represents the minimum possible temperature inside
the nacelle, and the ?rst law of thermodynamics may be used to determine this temperature. The ?rst step is to perform an energy balance on the nacelle to determine the rate of
heat transfer from the nacelle to the air and surroundings under steady-state conditions.
This step can be accomplished using either conservation of WRWDO energy or conservation of
WKHUPDODQGPHFKDQLFDOenergy; we will compare these two approaches.
Conservation of Total Energy The ?rst of the three boxed statements of the ?rst law
in Section 1.3 can be converted to a rate basis and expressed in equation form as follows:
dE/dt = (E)(in to tot) - E'(out to tot)
Under steady-state conditions, this reduces to . The term corresponds to
the mechanical work entering the nacelle , and the term includes the electrical power
output 3 and the rate of heat transfer leaving the nacelle T. Thus
W-P-Q = 0.........(2)
Conservation of Thermal and Mechanical Energy Alternatively, we can express
conservation of thermal and mechanical energy, starting with Equation 1.12c. Under
steady-state conditions, this reduces to
E(in) - E(out) - Eg = 0..........(3)
Here, E(in) once again corresponds to the mechanical work W. However,E(out) now includes
only the rate of heat transfer leaving the nacelle q. It does QRW include the electrical power,
since ( represents only the thermal and mechanical forms of energy. The electrical power
appears in the generation term, because mechanical energy is converted to electrical energy
in the generator, giving rise to a negative source of mechanical energy. That is, .
Thus, Equation (3) becomes
W-q-P = 0............(4)
which is equivalent to Equation (2), as it must be. Regardless of the manner in which the
?rst law of thermodynamics is applied, the following expression for the rate of heat transfer
evolves:
q = W-P ..............(5)
The mechanical work and electrical power are related by the ef?ciencies of the gearbox and
generator,
P = W*T(gb)*T(gen)...................(6)
Equation (5) can therefore be written as
q = P((1/(T(gb)*T(gen)) - 1) = ! 2.5*10^6((1/0.93*0.95) - 1) = 0.33*10^6 W
q = q(rad) + q(conv) = A(q"(rad) + q"(conv))
= PiDL + PiD^2/4[episilon*row*(T(s) - T(sur) +h*(T(s) -T(infinite)))
= 0.33*10^6W
T(s) = 416K = 143degreeC