HEAT+MASS TRANSFER:FUND.+APPL.(LOOSE)
6th Edition
ISBN: 9781260440027
Author: CENGEL
Publisher: MCG
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Chapter 2, Problem 173P
The thermal conductivity of a solid depends upon the solid’s temperature as
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A flanged bolt coupling consists of two concentric rows of bolts. The inner row has 6 nos. of 16mm diameterbolts spaced evenly in a circle of 250mm in diameter. The outer row of has 10 nos. of 25 mm diameter bolts spaced evenly in a circle of 500mm in diameter. If the allowable shear stress on one bolt is 60 MPa, determine the torque capacity of the coupling. The Poisson’s ratio of the inner row of bolts is 0.2 while that of the outer row is 0.25 and the bolts are steel, E =200 GPa.
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10: The constant wall thickness of a steel tube with the cross sectionshown is 2 mm. If a 600-N-m torque is applied to the tube. Use G = 80 GPa forsteel.1. Find the shear stress (MPa) in the wall of the tube.2. Find the angle of twist, in degrees per meter of length.
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A torque wrench is used to tighten the pipe shown.Dimensions: S1 = 400 mm; S2 = 250 mm; S3 = 100 mmModulus of Rigidity G = 78 GPa1. The diameter of the solid pipe is 20 mm. How much is themaximum force P (N) that can be applied based on theallowable shear stress of 60 MPa?2. For a hollow pipe with 50 mm outside diameter and is 6 mmthick, compute for the maximum force P (kN) that can beapplied such that the angle of twist at A does not exceed 5degrees.3. The torque applied to tighten the hollow pipe is 200 N-m.Given: Pipe outside diameter = 50 mm Pipe thickness = 6 mmSolve for the resulting maximum shear stress (MPa) in the pipe.
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- Correct answer and complete fbd only. I will upvote. 6: The shaft carries a total torque T0 that is uniformly distributedover its length L. Determine the angle of twist (degrees) of the shaft in termsif T0 = 1.2 kN-m, L = 2 m, G = 80 GPa, and diameter = 120 mm.arrow_forward2. Calculate the force in all members of the trusses shown using the method of joints. A 5525 lb C 3500 lb BY 14'-0" D 10'- 0" 6250 lb 10'- 0" Earrow_forwardCorrect answer and complete fbd only. I will upvote. 8: The steel rod fits loosely inside the aluminum sleeve. Both components are attached to a rigid wall at A andjoined together by a pin at B. Because of a slight misalignmentof the pre-drilled holes, the torque T0 = 750 N-m was appliedto the steel rod before the pin could be inserted into theholes. Determine the torque (N-m) in each component afterT0 was removed. Use G = 80 GPa for steel and G = 28 GPa foraluminumarrow_forward
- Correct answer and complete fbd only. I will upvote. 9: The two steel shafts, each with one end builtinto a rigid support, have flanges attached to their freeends. The flanges are to be bolted together. However,initially there is a 6⁰ mismatch in the location of the boltholes as shown in the figure. Determine the maximumshear stress(ksi) in each shaft after the flanges have beenbolted together. The shear modulus of elasticity for steelis 12 x 106 psi. Neglect deformations of the bolts and theflanges.arrow_forwardCorrect answer and complete fbd only. I will upvote. The tapered, wrought iron shaft carriesthe torque T = 2000 lb-in at its free end. Determine theangle of twist (degrees) of the shaft. Use G = 10 x 106psi for wrought ironarrow_forwardCorrect answer and complete fbd only. I will upvote. The compound shaft, consisting of steel and aluminumsegments, carries the two torques shown in the figure. Determine themaximum permissible value of T subject to the following designconditions: τst ≤ 83 MPa, τal ≤ 55 MPa, and θ ≤ 6⁰ (θ is the angle ofrotation of the free end). Use G =83 GPa for steel and G = 28 GPa foraluminum.arrow_forward
- The solid compound shaft, made of threedifferent materials, carries the two torques shown. Theshear moduli are 28 GPa for aluminum, 83 GPa for steel,and 35 GPa for bronze.1. Calculate the maximum shear stress (MPa) in eachmaterial.2. Find the angle of rotation (degrees) of the free endof the shaft.arrow_forwardCorrect answer only please. I will upvote. The velocity of a particle moves along the x-axis and is given by the equation ds/dt = 40 - 3t^2 m/s. Calculate the acceleration at time t=2 s and t=4 s. Calculate also the total displacement at the given interval. Assume at t=0 s=5m.Write the solution using pen and draw the graph if needed.arrow_forwardI want the steps of operation of the circuit, clearly in detail. Please. LV1arrow_forward
- Complet the solution: Vavg Ti Te Ts Qhexp Nuexp htheo Re Nutheo Error (m/s) (°C) (°C) (°C) (W) 2.11 18.8 21.3 45.8 2.61 18.5 20.8 46.3 Heat transfer Given data: a= 10 cm. L= 10 cm. b= 20 cm. H=40cm. ⚫ a = 10, cm: This could represent the width of the duct. ⚫b=20, cm: This might be the height of the duct. ⚫L = 10, cm: This usually stands for the length of the duct in the direction of flow. ⚫H=40, cm: This could indicate the height of some component or another duct-related dimension, but the exact meaning depends on the experiment's context.arrow_forwardplease explain each step and include drawings on the phase diagram. thanksarrow_forwardWrite clearly which points correspond to concentration of solute in front of alpha, concentration of solute in front of beta, amount of solid in the liquid in front of alpha/beta, lowest possible energy (tangent), as well as any other important information. Thank youarrow_forward
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