Explanation Given: The bottom of the domain is held by condensing steam at temperature, Te = 100°C Thermal conductivity isk = 15 W/m.K Convective heat transfer coefficient is h = 240 W/m 2 K. Calculation: Take nodes in the heat transfer systems which are shown below: Apply energy equation for node1, k ( Δ r 2 ) T 4 − T 1 r o Δ ρ + k ( r o Δ ρ 2 ) T 2 − T 1 Δ r = 0 Substitute all the given values as follows: 15 ( 10 × 10 − 3 2 ) × T 4 − T 1 40 × 10 − 3 × π 8 + 15 ( 40 × 10 − 3 × π 8 2 ) × T 2 − T 1 10 × 10 − 3 = 0 3.4674 T 1 − 2.4674 T 2 − T 4 = 0 ...... ( 1 ) Energy balance for the node 2, k ( r Δ ϕ 2 ) T 1 − T 2 Δ r + k ( Δ r ) T 5 − T 2 r Δ ϕ + k ( r Δ ϕ 2 ) T 3 − T 2 Δ r = 0 Substitute all the given values in above equation as follows: 15 ( 30 × 10 − 3 × π 8 2 ) T 1 − T 2 10 × 10 − 3 + 15 ( 10 × 10 − 3 ) T 5 − T 2 30 × 10 − 3 × π 8 + 15 ( 30 × 10 − 3 × π 8 2 ) T 3 − T 2 10 × 10 − 3 = 0 T 1 − 6.323 T 2 + T 3 + 4.323 T 5 = 0 ...... ( 2 ) Energy balance for node 3, k ( r i Δ ϕ 2 ) T 2 − T 3 Δ r + k ( Δ r 2 ) T 6 − T 3 r i Δ ϕ + h ( r i Δ ϕ 2 ) ( T ∞ − T 3 ) = 0 Substitute all the given values in above equation as follows: 15 ( 20 × 10 − 3 × π 8 2 ) T 2 − T 3 10 × 10 − 3 + 15 ( 10 × 10 − 3 2 ) T 6 − T 3 20 × 10 − 3 × π 8 + 240 ( 20 × 10 − 3 × π 8 2 ) ( 20 − T 3 ) = 0 T 2 − 2.7811 T 3 + 1.6211 T 6 + 3.2 = 0 ...... ( 3 ) Apply energy equation for node 4, k ( Δ r 2 ) T 1 − T 4 r o Δ ϕ + k ( Δ r 2 ) T 7 − T 4 r o Δ ϕ + k ( r o Δ ϕ ) T 5 − T 4 Δ r = 0 Substitute all the given values in above equation as follows: 15 ( 10 × 10 − 3 2 ) T 1 − T 4 40 × 10 − 3 × π 8 + 15 ( 10 × 10 − 3 2 ) T 7 − T 4 40 × 10 − 3 × π 8 + 15 ( 40 × 10 − 3 × π 8 ) T 5 − T 4 10 × 10 − 3 = 0 T 1 − 6.934 T 1 + 4.934 T 5 + T 7 = 0 ...... ( 4 ) Node 10 and 7 has same pattern in conduction as that of node 4. Thus, the equation must be same. Energy equation for node 7, T 4 − 6.934 T 7 + 4.934 T 8 + T 10 = 0 Energy equation for node 10, T 7 − 6.934 T 10 + 4.934 T 11 + T 13 = 0 Apply energy equation in node 5, k ( Δ r ) ( T 2 − T 5 ) r Δ ϕ + k ( Δ r ) ( T 8 − T 5 ) r Δ ϕ + k ( r Δ ϕ ) ( T 4 − T 5 ) Δ r + k ( r Δ ϕ ) ( T 6 − T 5 ) Δ r = 0 Substitute the given values in above equation as follows: [ 15 ( 10 × 10 − 3 ) ( T 2 − T 5 ) 30 × 10 − 3 × π 8 + 15 ( 10 × 10 − 3 ) ( T 8 − T 5 ) 30 × 10 − 3 × π 8 + 15 ( 30 × 10 − 3 × π 8 ) ( T 4 − T 5 ) 10 × 10 − 3 + 15 ( 30 × 10 − 3 × π 8 ) ( T 6 − T 5 ) 10 × 10 − 3 = 0 T 2 + 1.387 T 4 − 4.774 T 5 + 1.387 T 6 + T 8 = 0 ...... ( 7 ) Apply the energy equation for node 8, T 5 + 1.387 T 7 − 4.774 T 8 + 1.387 T 9 + T 11 = 0 Apply the energy equation for node 11, T 8 + 1.387 T 10 − 4.774 T 11 + 1.387 T 12 + T 14 = 0 Apply the energy equation for node 6, k ( Δ r 2 ) ( T 3 − T 6 ) r i Δ ϕ + k ( Δ r 2 ) ( T 9 − T 6 ) r i Δ ϕ + k ( r i Δ ϕ ) ( T 5 − T 6 ) Δ r + h ( r i Δ ϕ ) ( T ∞ − T 6 ) = 0 Substitute all the given values in above equation as follows: 15 ( 10 × 10 − 3 2 ) ( T 3 − T 6 ) 20 × 10 − 3 × π 8 + 15 ( 10 × 10 − 3 2 ) ( T 9 − T 6 ) 20 × 10 − 3 × π 8 + 15 ( 20 × 10 − 3 × π 8 ) ( T 5 − T 6 ) 10 × 10 − 3 + 240 ( 20 × 10 − 3 × π 8 ) ( 20 − T 6 ) = 0 T 3 + 1.233 T 5 − 3.4303 T 6 + T 9 + 3.947 = 0 ...... ( 10 ) Apply energy equation in node 9 and 12 as follows: T 6 + 1.233 T 8 − 3.4303 T 9 + T 12 + 3.947 = 0 ...... ( 11 ) T 9 + 1.233 T 11 − 3.4303 T 12 + T 15 + 3.947 = 0 ...... ( 12 ) Apply energy equation for node 13, k ( Δ x 2 ) T s − T 13 Δ y + k ( Δ r 2 ) T 10 − T 13 r o Δ ϕ + k ( Δ y 2 + r 0 Δ ρ 2 ) T 14 − T 13 Δ x = 0 Substitute all the values in the above equation as follows: 15 ( 10 × 10 − 3 2 ) T s − T 13 10 × 10 − 3 + 15 ( 10 × 10 − 3 2 ) T 10 − T 13 40 × 10 − 3 × π 8 + 15 ( 10 × 10 − 3 2 + 40 × 10 − 3 × π 8 2 ) T 14 − T 13 10 × 10 − 3 = 0 0

6th Edition

ISBN: 9780470501962

Author: Frank P. Incropera, David P. DeWitt, Theodore L. Bergman, Adrienne S. Lavine

Publisher: Wiley, John & Sons, Incorporated

See similar textbooks

Videos

Question

Chapter 4, Problem 4.60P

To determine

Heat transfer in tube.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 4, Problem 4.58P

Chapter 4, Problem 4.61P

Students have asked these similar questions

Recall that the CWH equation involves two important assumptions. Let us investigate how these assumptions affect the accuracy of state trajectories under the control inputs optimized in (a) and (b). (c.1): Discuss the assumptions about the chief and deputy orbits that are necessary for deriving CWH.

PROBLEM 2.50 1.8 m The concrete post (E-25 GPa and a = 9.9 x 10°/°C) is reinforced with six steel bars, each of 22-mm diameter (E, = 200 GPa and a, = 11.7 x 10°/°C). Determine the normal stresses induced in the steel and in the concrete by a temperature rise of 35°C. 6c " 0.391 MPa 240 mm 240 mm 6₁ = -9.47 MPa

For some viscoelastic polymers that are subjected to stress relaxation tests, the stress decays with time according to a(t) = a(0) exp(-4) (15.10) where σ(t) and o(0) represent the time-dependent and initial (i.e., time = 0) stresses, respectively, and t and T denote elapsed time and the relaxation time, respectively; T is a time-independent constant characteristic of the material. A specimen of a viscoelastic polymer whose stress relaxation obeys Equation 15.10 was suddenly pulled in tension to a measured strain of 0.5; the stress necessary to maintain this constant strain was measured as a function of time. Determine E (10) for this material if the initial stress level was 3.5 MPa (500 psi), which dropped to 0.5 MPa (70 psi) after 30 s.

Chapter 4 Solutions

Introduction to Heat Transfer

Ch. 4 - In the method of separation of variables (Section...Ch. 4 - A two-dimensional rectangular plate is subjected...Ch. 4 - Consider the two-dimensional rectangular plate...Ch. 4 - A two-dimensional rectangular plate is subjected...Ch. 4 - Prob. 4.5P Ch. 4 - Prob. 4.6P Ch. 4 - Free convection heat transfer is sometimes...Ch. 4 - Prob. 4.8P Ch. 4 - Radioactive wastes are temporarily stored in a...Ch. 4 - Based on the dimensionless conduction heat rates...

Ch. 4 - Prob. 4.11P Ch. 4 - A two-dimensional object is subjected to...Ch. 4 - Prob. 4.13P Ch. 4 - Two parallel pipelines spaced 0.5 m apart are...Ch. 4 - A small water droplet of diameter D=100m and...Ch. 4 - Prob. 4.16P Ch. 4 - Pressurized steam at 450 K flows through a long,...Ch. 4 - Prob. 4.19P Ch. 4 - A furnace of cubical shape, with external...Ch. 4 - Prob. 4.21P Ch. 4 - Prob. 4.22P Ch. 4 - A pipeline, used for the transport of crude oil,...Ch. 4 - A long power transmission cable is buried at a...Ch. 4 - Prob. 4.25P Ch. 4 - A cubical glass melting furnace has exterior...Ch. 4 - Prob. 4.27P Ch. 4 - An aluminum heat sink k=240W/mK, used to coolan...Ch. 4 - Hot water is transported from a cogeneration power...Ch. 4 - Prob. 4.30P Ch. 4 - Prob. 4.31P Ch. 4 - Prob. 4.32P Ch. 4 - An igloo is built in the shape of a hemisphere,...Ch. 4 - Consider the thin integrated circuit (chip) of...Ch. 4 - Prob. 4.35P Ch. 4 - The elemental unit of an air heater consists of a...Ch. 4 - Prob. 4.37P Ch. 4 - Prob. 4.38P Ch. 4 - Prob. 4.39P Ch. 4 - Prob. 4.40P Ch. 4 - Prob. 4.41P Ch. 4 - Determine expressions for...Ch. 4 - Prob. 4.43P Ch. 4 - Prob. 4.44P Ch. 4 - Prob. 4.45P Ch. 4 - Derive the nodal finite-difference equations for...Ch. 4 - Prob. 4.47P Ch. 4 - Prob. 4.48P Ch. 4 - Consider a one-dimensional fin of uniform...Ch. 4 - Prob. 4.50P Ch. 4 - Prob. 4.52P Ch. 4 - Prob. 4.53P Ch. 4 - Prob. 4.54P Ch. 4 - Prob. 4.55P Ch. 4 - Prob. 4.56P Ch. 4 - Steady-state temperatures at selected nodal points...Ch. 4 - Prob. 4.58P Ch. 4 - Prob. 4.60P Ch. 4 - The steady-state temperatures C associated with...Ch. 4 - A steady-state, finite-difference analysis has...Ch. 4 - Prob. 4.64P Ch. 4 - Consider a long bar of square cross section (0.8 m...Ch. 4 - Prob. 4.66P Ch. 4 - Prob. 4.67P Ch. 4 - Prob. 4.68P Ch. 4 - Prob. 4.69P Ch. 4 - Consider Problem 4.69. An engineer desires to...Ch. 4 - Consider using the experimental methodology of...Ch. 4 - Prob. 4.72P Ch. 4 - Prob. 4.73P Ch. 4 - Prob. 4.74P Ch. 4 - Prob. 4.75P Ch. 4 - Prob. 4.76P Ch. 4 - Prob. 4.77P Ch. 4 - Prob. 4.78P Ch. 4 - Prob. 4.79P Ch. 4 - Prob. 4.80P Ch. 4 - Spheres A and B arc initially at 800 K, and they...Ch. 4 - Spheres of 40-mm diameter heated to a uniform...Ch. 4 - To determine which parts of a spiders brain are...Ch. 4 - Prob. 4.84P

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.

Heat transfer in tube.

Solution Summary: The author calculates the energy balance for nodes in the heat transfer systems.

Videos

Heat transfer in tube.

Want to see the full answer?

Chapter 4 Solutions