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BaLLatris [955]
3 years ago
14

One-dimensional, steady-state conduction with uniform internal energy generation occurs in a plane wall with a thickness of 50 m

m and a constant thermal conductivity of 5 W/m*K. For these condtions, the temperature distribution has the form T(x)=a+bx+cx^2. The surface at x=0 has a temperature of T(0)=To=120 deg. (C) and h=500 W/m^2*K. The surface at x=L is well insulated.(a) Applying an overall energy balance to the wall, calculate the volumetric energy generation rate q.(b) Determine the coefficients a,b,c by applying the boundary conditions to the prescribed temperature distribution. Use the results to calculate and plot the temp. distribution.(c) Consider conditions for whic hthe convection coefficient is halved, but q remains unchanged. Determin the new values for a,b,c and plot the temperature dist. (Hint: T(0) is no longer 120)(d) Under conditions for which q is doubled and the convection coeff. remains unchanged (h=500 W/m^2*K) determine new a,b,c and plot temp distribution. Compare (a),(b),and (c) and discuss the effects of h and q on the distributions.
Engineering
1 answer:
uysha [10]3 years ago
4 0

Answer: please help me tooo

Explanation:

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The grinding machine is used for roughing and finishing flat, cylindrical, and conical surfaces; finishing internal cylinders or bores; forming and sharpening cutting tools; snagging or removing rough projections from castings and stampings; and cleaning, polishing, and buffing surfaces.

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3 years ago
The solid cylinders AB and BC are bonded together at B and are attached to fixed supports at A and C. The modulus of rigidity is
romanna [79]

Answer:

a) 0.697*10³ lb.in

b) 6.352 ksi

Explanation:

a)

For cylinder AB:

Let Length of AB = 12 in

c=\frac{1}{2}d=\frac{1}{2} *1.1=0.55in\\ J=\frac{\pi c^4}{2}=\frac{\pi}{2}0.55^4=0.1437\ in^4\\

\phi_B=\frac{T_{AB}L}{GJ}=\frac{T_{AB}*12}{3.3*10^6*0.1437}  =2.53*10^{-5}T_{AB}

For cylinder BC:

Let Length of BC = 18 in

c=\frac{1}{2}d=\frac{1}{2} *2.2=1.1in\\ J=\frac{\pi c^4}{2}=\frac{\pi}{2}1.1^4=2.2998\ in^4\\

\phi_B=\frac{T_{BC}L}{GJ}=\frac{T_{BC}*18}{5.9*10^6*2.2998}  =1.3266*10^{-6}T_{BC}

2.53*10^{-5}T_{AB}=1.3266*10^{-6}T_{BC}\\T_{BC}=19.0717T_{AB}

T_{AB}+T_{BC}-T=0\\T_{AB}+T_{BC}=T\\T_{AB}+T_{BC}=14*10^3\ lb.in\\but\ T_{BC}=19.0717T_{AB}\\T_{AB}+19.0717T_{AB}=14*10^3\\20.0717T_{AB}=14*10^3\\T_{AB}=0.697*10^3\ lb.in\\T_{BC}=13.302*10^3\ lb.in

b) Maximum shear stress in BC

\tau_{BC}=\frac{T_{BC}}{J}c=13.302*10^3*1.1/2.2998=6.352\ ksi

Maximum shear stress in AB

\tau_{AB}=\frac{T_{AB}}{J}c=0.697*10^3*0.55/0.1437=2.667\ ksi

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What are the functions of each computer program
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2 years ago
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