Shape factor S=M/N, N isShape factor S=M/N, N is
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Q: ᴀ ʙᴇʟᴛᴇᴅ ᴛʀᴀɴꜱᴍɪꜱꜱɪᴏɴ ꜱʏꜱᴛᴇᴍ ɪꜱ ʟᴏᴀᴅᴇᴅ. ᴛʜᴇ ɪɴᴘᴜᴛ ꜱʜᴀꜰᴛ ᴛᴀᴋᴇꜱ ɪᴛꜱ ᴘᴏᴡᴇʀ ꜰʀᴏᴍ ᴀ 20-ᴋᴡ ᴍᴏᴛᴏʀ ᴛʜᴀᴛ ʀᴜɴꜱ…
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A: Answer is given below
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A: T1T2 = eμθ = e0.3 × 2.88 = 2.372 ---- (a)Tc = m v2 = 0.9 × 26.672 = 640 N
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A: For solution refer below images.
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- Use for the ice? List all sumptions. 1.4 Consider axial flow of water in a cold tube. Write the heat conduction equation for the ice forming axisymmetrically on the inside surface of the tube. ice water 1.5 Consider two-dimensional conduction in the semi-circular cylinder shown. The cylinder is heated with uniform pe here to search 05:48 A qu) * ENG r-ny-E/-F hpPlease solve this numerical problem.One of the strengths of numerical methods is their ability to handle complex boundary conditions. In the sketch, the boundary condition changes from specified heat flux ′′ qs (into the domain) to convection, at the location of the node (m, n). Write the steady-state, two- dimensional finite difference equation at this node.
- The Derivative and Differentiationunder steady-state conditions. If you are given T1 = 200 °C and T2 = 164 °C, determine: a) the conduction heat flux, q,.cond, in m2 W from x = 0 to x = L b) if the dimensions of the triangle ares 15 mm and h 13 mm, calculate the heat transfer due to convection, q,y, in W at x = L Finsulation T2 T T = 20°C h = 500 W/m2.K Triangular Prism x L x 0 L= 50 mm k = 100 W/m-KThe types of scaling transformation are stretch and enlarging O uniform and nonuniform O miniaturizing and enlarging O non of the above
- Choose the correct answer and give step by step solution for this" THEOREMS OF PAPUS " Please refer to the answer key (book from Engineering Mechanics by SInger) Please include the analyzation of figure so that i could undestand the steps and solution. Thankyou!Q1 Passage of an electric current through a long conducting rod of radius r; and thermal conductivity k, results in uniform volumetric heating at a rate of ġ. The conduct- ing rod is wrapped in an electrically nonconducting cladding material of outer radius r, and thermal conduc- tivity k, and convection cooling is provided by an adjoining fluid. Conducting rod, ġ, k, 11 To Čladding, ke For steady-state conditions, write appropriate forms of the heat equations for the rod and cladding. Express ap- propriate boundary conditions for the solution of these equations.
- 4. If all variables of choose the a stream are correct answer dependent of time it is said to be in: (a) Steady flow (b) Unsteady flow (c) Uniform flow (d) Closed flow (e) Constant flow 5. In an irreversible process, there is a: (a) loss of heat (b) No O loss of heat (c) Gain of heat (d) No gain of heatConsider the following linear equations,Give True or False for the following: 1.In liquids and gases, heat transmission is caused by conduction and convection 2.The surface geometry is the important factor in convection heat transfer 3. The heat transfer by conduction from heated surface to the adjacent layer of fluid, 4. The heat transfer is increased in the fin when &> 1 5.The unit of the thermal diffusivity is m²/s 6. Temperature change between the materials interfaces is attributed to the thermal contact resistance 7. A material that has a low heat capacity will have a large thermal diffusivity. 8. Heat conduction flowing from one side to other depends directly on thickness 9.Fin efficiency is the ratio of the fin heat dissipation with that of no fin 10.The critical radius is represented the ratio of the convicted heat transfer to the thermal conductivity