Please show all step, not Ai generated. The diagram shows a conical section fabricated from pure aluminum. It is of circular cross section having diameter D = ax¹/2, where a = 0.5 m¹/2. The small end is located at x₁ = 25 mm and thelarge end at x₂ = 125 mm. The end temperatures are T₁ = 600 K and T₂ = 400 K, while the lateral surface is well insulated.T₂T₁1.x₂(a) Derive an expression for the temperature distribution 7(x) in symbolic form, assuming one-dimensional conditions. Sketch the temperature distribution.(b) Calculate the heat rate qx²

The temperature at is The temperature at is The diameter is , The lateral surface are insulated

The diagram shows a conical section fabricated from pure aluminum. It is of circular cross section having diameter D = ax¹/2, where a = 0.5 m²/2. The small end is located at x₁ = 25 mm and large end at x₂ = 125 mm. The end temperatures are T₁ = 600 K and 7₂ = 400 K, while the lateral surface is well insulated. (a) Derive an expression for the temperature distribution 7(x) in symbolic form, assuming one-dimensional conditions. Sketch the temperature distribution. (b) Calculate the heat rate q

The diagram shows a conical section fabricated from pure aluminum. It is of circular cross section having diameter D = ax¹/2, where a = 0.5 m²/2. The small end is located at x₁ = 25 mm and large end at x₂ = 125 mm. The end temperatures are T₁ = 600 K and 7₂ = 400 K, while the lateral surface is well insulated. (a) Derive an expression for the temperature distribution 7(x) in symbolic form, assuming one-dimensional conditions. Sketch the temperature distribution. (b) Calculate the heat rate q

Principles of Heat Transfer (Activate Learning with these NEW titles from Engineering!)

8th Edition

ISBN:9781305387102

Author:Kreith, Frank; Manglik, Raj M.

Publisher:Kreith, Frank; Manglik, Raj M.

Chapter3: Transient Heat Conduction

Section: Chapter Questions

Problem 3.4P

See similar textbooks

Related questions

Q: (PID solution with the requirements and simple explanation. Thx) INSTRUCTION: Given three bare…

A: Feedback Solution (1): This feedback control starts with a level transmitter that will measure the…

Q: Produce a 3D solid model of the Guard casing and test boss

A: Given Test boss Guard casing To find 3D solid model

Q: Speed reduction, either through gearing or belt-pulley constructs, is an important element in motion…

Q: 3. Convert C/C++ code to MIPS Code #include using namespace std; main () { int x, y, sum; x = 50; y…

A: The converted MIPS cod edit the given C++ code is given below. I have also attached the output of…

Q: [Chapter 4: Creating Orthograph P4-15) Create an orthographic projection of the following object.…

A: Orthographic view for given isometric view will be as per below

Q: sketch the top view ,front view and side view using catia v5, and i need the orthographic view in…

Q: The fundamentals of CNC programming, as well as G and M codes, are explained in depth.

A: Introduction CNC machining is a fundamental manufacturing technique that produces complicated and…

Q: T For the Robot and work frame shown, the following matrices are given: T, T and $T. Compute = [B]…

Q: Sketch an object that contains all of the necessary elements so that you can use the following line…

Q: Situation 9 - A 6-m long ladder weighing 600 N is shown in the Figure. It is required to determine…

Concept explainers

Axial Load

When a bar of the uniform cross-section is acted by a load, such that the load acts along the longitudinal axis of the bar, such kinds of loadings are known as axial loadings. In general terms, a load is an external force acting on a component. An axial load acts perpendicular to the geometric cross-section of the component. Based on the direction of the application of the load, an axial load can be tensile or compressive. The analysis of bodies under the action of axial loads is generally studied under the branch of mechanics, known as statics.

Question

Please show all step, not Ai generated.

The diagram shows a conical section fabricated from pure aluminum. It is of circular cross section having diameter D = ax¹/2, where a = 0.5 m¹/2. The small end is located at x₁ = 25 mm and the
large end at x₂ = 125 mm. The end temperatures are T₁ = 600 K and T₂ = 400 K, while the lateral surface is well insulated.
T₂
T₁
1.x₂
(a) Derive an expression for the temperature distribution 7(x) in symbolic form, assuming one-dimensional conditions. Sketch the temperature distribution.
(b) Calculate the heat rate qx²

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

SEE SOLUTION Check out a sample Q&A here

Step 1: Given

VIEW

Step 2: Calculation of temperature variation

VIEW

Step 3: Calculation of heat rate

VIEW

Solution

VIEW

Step by step

Solved in 4 steps with 16 images

SEE SOLUTION Check out a sample Q&A here

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.

Similar questions

Problem 3: One hollow cylinder is shrink-fitted inside another. Both cylinders have length L and both the flat faces of each cylinder are constrained in the axial direction. They are free to move in radial and tangential directions. Both cylinders are made of the same material steel. New steel material will be created. Cylinder Inner Cylinder FIGURE 1 SHRINK FIT ASSEMBLY Outside Cylinder Steel: E = 30 x 106 psi Inside Radius r;=4" R₁=6" FIGURE 2 EXPLODED VIEW OF SHRINK FIT ASSEMBLY Outside Radius r=6.005" R₂=8" Length L = 5" L = 5" An internal pressure P = 30,000 psi is applied on the inner surface of the inner cylinder. What to Submit: 1. Find the tangential stress distribution at the outside surface of the outer cylinder (r = 8"). 2. Find the tangential stress distribution at the inner surface of the inner cylinder (r = 4") 3. Find the contact pressure developed at the interference location.
A thin-walled cylinder of a high-strength aluminumalloy (KIC = 24 MPa √m ) has the following dimensions:length = 20 cm; outer diameter = 9 cm; inner diameter7 cm. A semicircular crack of depth a = 0.25 cm is discovered on the inner diameter and oriented along a line parallelto the cylinder axis. If the cylinder is repeatedly pressurized,how many pressure cycles could the cylinder withstandbefore failure? The pressure within the cylinder reaches75 MPa, and the material obeys a fatigue crack propagationrelation of the formda/dN = 5 x 10-39(ΔK)4where da/dN and ΔK have the units of m/cycle and Pa √m ,respectively.
I need a clear answer by hand, not by keyboard and fast answer within 20 minutes. Thank you | dybala
Question 5. A component consists of two horizontal concentric solid cylinders, A and B, fixed together as shown in the diagram below. X mm x Cylinder A 800 mm x = OG Cylinder B mm 500 mm Cylinder A has a diameter of X=50 mm, a length of 800 mm and is made of steel having a density of 7800 kg/m³. Ymm Cylinder B has a diameter of Y=66 mm, a length of 500 mm and is made of aluminium having a density of 2600 kg/m³. The dimension is the distance from the left-hand end of the component to its centre of gravity at point G. Calculate and enter your answer, in mm, correct to three significant figures in the box below.
Please solve and answer the question correctly please. Thank you!!
Problem #5 Consider the aluminum alloy bar with Kic=30 MPa vm, ay=500 MPa. Can you use the LEFM? Explain. 0.375 0.50 P 2.0 1. Units in cm Thickness = 0.75 cm
Hello good evening Sir,Permission, i have a question in my homework related mechanics of material strength or mechanical engineering lesson. The following below is the questions. Please advice. Thank you so much Regards,Irfan
Problem 2. A block of cast iron was wrapped in the thin aluminum foil as shown on the figure. At room temperature there is no stress on the foil. The whole system is then immersed in liquid nitrogen at -196°C. Find the stress on the aluminum foil. Is the foil going to deform permanently? Is it going to break? Stress [MPa] 350 300 250 200 150 100 50 0 0 alu alloy 5083-H34 0.005 0.01 0.015 0.02 Strain Aluminum foil 5 µm thick, this is a continuous strip with the ends welded together Cast iron 20 mm thick a aluminum = 2.4 105 K-¹, Stress MPa 400 300 200 100 a iron 1.2 105 K-1 ↑ 11 0 0 0.05 0.1 0.15 0.2 0.25 Strain Fig. 2. Stress-strain diagrams for aluminum. Left: Low strain region. Right: Full strain range. Hint: you can get the Young's modulus, proportions limit (elastic limit), and break point from the graphs.
Figure 1 shows a system of three bars each made of different materials, and connected together and the temperature is initially kept at 12°C. Thereafter, the temperature of the system was raised to 50°C. Steel Brass Сopper Es = 200 GPa as = 12(10-0)/°C apr = = Epr = 100 GPa 21(10-6)/°C acu = 120 GPa Ecu Ebr %3D %3D 17(10-6)/°C %3D Ast abr Acu = 515 mm² = 450 mm² %3D Ag 200 mm² Apr st 300mm 200 mm 100mm Figure 1: An assembly of three bars 1.1. Determine the thermal stress raising in each bar if the assembly is placed between two rigid walls. 1.2. Determine the force exerted on the right rigid support if an allowance of 0.05 mm is provided between the left end of the steel bar and the left wall.
The figure below shows a cutaway view of an assembly inwhich a solid bronze [G = 6,500 ksi] rod (1) is fitted inside of analuminum alloy [G = 4,000 ksi] tube (2). The tube is attached to afixed plate at C, and both the rod and the tube are welded to a rigidend plate at A. The rod diameter is d1 = 1.75 in. The outside diameterof the tube is D2 = 3.00 in., and its wall thickness is t2 = 0.125 in.Overall dimensions of the assembly are a = 40 in. and b = 8 in. Theallowable shear stresses for the bronze and aluminum materials are10,000 psi and 8,000 psi, respectively. Further, the rotation angle ofend B must be limited to a magnitude of 5°. On the basis of theseconstraints, what is the maximum torque TB that can be applied atend B?
i need the answer quickly
A cylindrical rod of copper (E = 110 GPa, 16 × 106 psi) having a yield strength of 240 MPa (35,000 psi) is to be subjected to a load of 6670 N (1499 lbf). If the length of the rod is 390 mm (15.35 in.), what must be the diameter to allow an elongation of 0.50 mm (0.01969 in.)?