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Minchanka [31]
3 years ago
12

A solid circular cylinder of mass m, radius r, and length l is pivoted about a transverse axis

Engineering
1 answer:
SashulF [63]3 years ago
8 0
To be honest i have no idea
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Why is it important to keep hand tools free of oil and grease?
Bezzdna [24]

Answer:

keeps it from rusting

Explanation:

5 0
3 years ago
A plate and frame heat exchanger has 15 plates made of stainless steel that are 1 m tall. The plates are 1 mm thick and 0.6 m wi
hodyreva [135]

Answer:

14.506°C

Explanation:

Given data :

flow rate of water been cooled = 0.011 m^3/s

inlet temp = 30°C + 273 = 303 k

cooling medium temperature = 6°C  + 273 = 279 k

flow rate of cooling medium = 0.02 m^3/s

Determine the outlet temperature

we can determine the outlet temperature by applying the relation below

Heat gained by cooling medium = Heat lost by water

= ( Mcp ( To - 6 )  =  Mcp ( 30 - To )

since the properties of water and the cooling medium ( water ) is the same

= 0.02 ( To - 6 ) = 0.011 ( 30 - To )

= 1.82 ( To - 6 ) = 30 - To

hence To ( outlet temperature ) = 14.506°C

6 0
3 years ago
A plane wall of thickness 0.1 m and thermal conductivity 25 W/m·K having uniform volumetric heat generation of 0.3 MW/m3 is insu
Contact [7]

Answer:

T = 167 ° C

Explanation:

To solve the question we have the following known variables

Type of surface = plane wall ,

Thermal conductivity k = 25.0 W/m·K,  

Thickness L = 0.1 m,

Heat generation rate q' = 0.300 MW/m³,

Heat transfer coefficient hc = 400 W/m² ·K,

Ambient temperature T∞ = 32.0 °C

We are to determine the maximum temperature in the wall

Assumptions for the calculation are as follows

  • Negligible heat loss through the insulation
  • Steady state system
  • One dimensional conduction across the wall

Therefore by the one dimensional conduction equation we have

k\frac{d^{2}T }{dx^{2} } +q'_{G} = \rho c\frac{dT}{dt}

During steady state

\frac{dT}{dt} = 0 which gives k\frac{d^{2}T }{dx^{2} } +q'_{G} = 0

From which we have \frac{d^{2}T }{dx^{2} }  = -\frac{q'_{G}}{k}

Considering the boundary condition at x =0 where there is no heat loss

 \frac{dT}{dt} = 0 also at the other end of the plane wall we have

-k\frac{dT }{dx } = hc (T - T∞) at point x = L

Integrating the equation we have

\frac{dT }{dx }  = \frac{q'_{G}}{k} x+ C_{1} from which C₁ is evaluated from the first boundary condition thus

0 = \frac{q'_{G}}{k} (0)+ C_{1}  from which C₁ = 0

From the second integration we have

T  = -\frac{q'_{G}}{2k} x^{2} + C_{2}

From which we can solve for C₂ by substituting the T and the first derivative into the second boundary condition s follows

-k\frac{q'_{G}L}{k} = h_{c}( -\frac{q'_{G}L^{2} }{k}  + C_{2}-T∞) → C₂ = q'_{G}L(\frac{1}{h_{c} }+ \frac{L}{2k} } )+T∞

T(x) = \frac{q'_{G}}{2k} x^{2} + q'_{G}L(\frac{1}{h_{c} }+ \frac{L}{2k} } )+T∞ and T(x) = T∞ + \frac{q'_{G}}{2k} (L^{2}+(\frac{2kL}{h_{c} }} )-x^{2} )

∴ Tmax → when x = 0 = T∞ + \frac{q'_{G}}{2k} (L^{2}+(\frac{2kL}{h_{c} }} ))

Substituting the values we get

T = 167 ° C

4 0
3 years ago
A 35-ft³ rigid tank has propane at 25 psia, 540 R and is connected by a valve to another tank of 20 ft³ with propane at 40 psia,
gulaghasi [49]

Answer:

final pressure = 200KPa or 29.138psia

Explanation:

The detailed step by step calculations with appropriate conversion factors applied are as shown in the attachment.

8 0
3 years ago
If a student doesn't major in Engineering as an undergraduate, They could still find a successful
34kurt

Answer:

True

Explanation:

It could either be true or false because you dont really have to be great at something just to do it, you could try new things too.

5 0
2 years ago
Read 2 more answers
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