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erastova [34]
3 years ago
14

The velocity profile for a thin film of a Newtonian fluid that is confined between the plate and a fixed surface is defined by u

= (5y-0.5y2) mm/s where y is in mm. Determine the force P that must be applied to the plate to cause this motion. The plate has a surface area of 4000mm2 in contact with the fluid. Take µ= 0.55 N.s/m2
Engineering
1 answer:
zimovet [89]3 years ago
8 0

Answer:

F = 0.0022N

Explanation:

Given:

Surface area (A) = 4,000mm² = 0.004m²

Viscosity = µ = 0.55 N.s/m²

u = (5y-0.5y²) mm/s

Assume y = 4

Computation:

F/A = µ(du/dy)

F = µA(du/dy)

F = µA[(d/dy)(5y-0.5y²)]

F = (0.55)(0.004)[(5-1(4))]

F = 0.0022N

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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
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5 0
2 years ago
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g Four primary sedimentation basins are designed for total average flow of 1.2 m3/s. The TSS concentration in the primary treate
fiasKO [112]

Answer:

Capacity for the sludge pump = 0.217 m³/min

Explanation:

Detailed explanation is given in the attached document.

5 0
3 years ago
Answers:
frutty [35]

Answer:

the chassis ground wire

an inverter

is connected to one side of the power line

failure of a component such as the hash filter

is the hot wire

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3 years ago
When a 60 Hz sinusoidal voltage is applied to the input of a half-wave rectifier, the output frequency is
Ludmilka [50]

Answer:

120 Hz

Explanation:

Given that a 60 Hz sinusoidal voltage is applied to the input of a half-wave rectifier, what will be the output frequency ?

Ideally, the supply frequency will be equal to the ripple frequency.

But in a half - way rectifier, the out put frequency is twice the input sinusoidal voltage frequency.

The output frequency = 2 × 60

Output frequency = 120 Hz

Therefore, When a 60 Hz sinusoidal voltage is applied to the input of a half-wave rectifier, the output frequency is 120 Hz.

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3 years ago
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