Consider the fuel element of Example 5.11, which operates at a uniform volumetric generation rate of q ˙ = 10 7 W / m 3 , until the generation rate suddenly changes to q ˙ = 2 × 10 7 W / m 3 . Use the Finite-Difference Equations, One-Dimensional, Transient conduction model builder of IHT to obtain the implicit form of the finite-difference equations for the 6 nodes, with 2 mm, as shown in the example. (a) Calculate the temperature distribution 1.5 s after the change in operating power, and compare your results with those tabulated in the example. (b) Use the Explore and Graph opt ions of IHT to calculate and plot temperature histories at the midplane (00) and surface (05) nodes for 0 ≤ t ≤ 400 s . What are the steady-state temperatures, and approximately how long does it take to reach the new equilibrium condition after the step change in operating power?
Consider the fuel element of Example 5.11, which operates at a uniform volumetric generation rate of q ˙ = 10 7 W / m 3 , until the generation rate suddenly changes to q ˙ = 2 × 10 7 W / m 3 . Use the Finite-Difference Equations, One-Dimensional, Transient conduction model builder of IHT to obtain the implicit form of the finite-difference equations for the 6 nodes, with 2 mm, as shown in the example. (a) Calculate the temperature distribution 1.5 s after the change in operating power, and compare your results with those tabulated in the example. (b) Use the Explore and Graph opt ions of IHT to calculate and plot temperature histories at the midplane (00) and surface (05) nodes for 0 ≤ t ≤ 400 s . What are the steady-state temperatures, and approximately how long does it take to reach the new equilibrium condition after the step change in operating power?
Solution Summary: The author calculates the energy balance to the center of the rod using one dimensional conduction with uniform heat generation.
Consider the fuel element of Example 5.11, which operates at a uniform volumetric generation rate of
q
˙
=
10
7
W
/
m
3
, until the generation rate suddenly changes to
q
˙
=
2
×
10
7
W
/
m
3
. Use the Finite-Difference Equations, One-Dimensional, Transient conduction model builder of IHT to obtain the implicit form of the finite-difference equations for the 6 nodes, with 2 mm, as shown in the example.
(a) Calculate the temperature distribution 1.5 s after the change in operating power, and compare your results with those tabulated in the example.
(b) Use the Explore and Graph opt ions of IHT to calculate and plot temperature histories at the midplane (00) and surface (05) nodes for
0
≤
t
≤
400
s
. What are the steady-state temperatures, and approximately how long does it take to reach the new equilibrium condition after the step change in operating power?
A truncated solid cone is of circular cross section, and its diameter is related to the axial coordinate by an expression of the form D = ax3/2, where a = 2 m−1/2.
The sides are well insulated, while the top surface of the cone at x1 is maintained at T1 and the bottom surface at x2 is maintained at T2.
Conductivity k = 336 W/m-K
(a) Obtain an expression for the temperature distribution T(x).
(b) What is the rate of heat transfer across the cone if it is constructed of pure aluminum with x1 = 0.086 m, T1 = 113°C, x2 = 0.270 m, and T2 = 25°C?
The walls of a refrigerator are typically constructed with two metal panels enclosing an insulation layer. Consider an insulation layer made of fiberglass (k = 0.046 W/m°C) with a thickness of 50 mm, placed between two metal sheets (k = 60 W/m°C), each 3 mm thick. If the wall separates cold air at 4 °C from ambient air at 25 °C, what is the heat received per unit area, considering internal and external convective coefficients equal to 5 W/m²°C?
Steel pipe 1 cm thick, 1.0 m long and 8 cm deep, quiet with 4 cm thick insulation. The inner wall temperature of the steel pipe is 100 ° C. The ambient temperature around the integrated pipe is 24 ° C. The convection heat transfer coefficient outside the surface is 50 W / (m² K). The thermal conductivity of steel is 54 W / (m K), and the thermal conductivity of the insulation is 0.04 W / (m K). Count; A. Heat loss per meter of pipe = Answer watt. b. Temperature between steel pipe and insulation. = Answer ° C
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