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AleksandrR [38]

3 years ago

6

Oil (SAE 30) at 15.6 oC flows steadily between fixed, horizontal, parallel plates. The pressure drop per unit length along the c

hannel is 35 kPa/m, and the distance between the plates is 9 mm. The flow is laminar. Determine: (a) the volume rate of flow (per meter of width), (b) the magnitude of the shearing stress acting on the bottom plate, and (c) the velocity along the centerline of the channel.

1 answer:

Nitella [24]3 years ago

7 0

Answer:

(a) The volume rate of flow per meter width = 5.6*10⁻³ m²/s

(b) The shear stress acting on the bottom plate = 157.5 N/m²

(c) The velocity along the centerline of the channel = 0.93 m/s

Explanation:

(a)

Calculating the distance of plate from centre line using the formula;

h = d/2

where h = distance of plate

d = diameter of flow = 9 mm

Substituting, we have;

h = 9/2

= 4.5 mm = 4.5*10^-3 m

Calculating the volume flow rate using the formula;

Q = (2h³/3μ)* (Δp/L)

Where;

Q = volume flow rate

h = distance of plate = 4.5*10^-3 m

μ = dynamic viscosity = 0.38 N.s/m²

(Δp/L) = Pressure drop per unit length = 35 kPa/m = 35000 Pa

Substituting into the equation, we have;

Q = (2*0.0045³/3*0.38) *(35000)

= (1.8225*10⁻⁷/1.14) * (35000)

= 1.60*10⁻⁷ * 35000

= 5.6*10⁻³ m²/s

Therefore, the volume flow rate = 5.6*10⁻³ m³/s

(b) Calculating the shear stress acting at the bottom plate using the formula;

τ = h*(Δp/L)

= 0.0045* 35000

= 157.5 N/m²

(c) Calculating the velocity along the centre of the channel using the formula;

u(max) = h²/2μ)* (Δp/L)

= (0.0045²/2*0.38) * 35000

=2.664*10⁻⁵ *35000

= 0.93 m/s

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Answer:

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Explanation:

a) The tank can be modelled by the Principle of Mass Conservation:

$\dot m_{1} + \dot m_{2} - \dot m_{3} = 0$

The mass flow rate exiting the tank is:

$\dot m_{3} = \dot m_{1} + \dot m_{2}$

$\dot m_{3} = 125\,\frac{lbm}{s} + 10\,\frac{lbm}{s}$

$\dot m_{3} = 135\,\frac{lbm}{s}$

b) An expression for the specific enthalpy at outlet is derived from the First Law of Thermodynamics:

$\dot m_{1}\cdot h_{1} + \dot m_{2} \cdot h_{2} - \dot m_{3}\cdot h_{3} = 0$

$h_{3} = \frac{\dot m_{1}\cdot h_{1}+\dot m_{2}\cdot h_{2}}{\dot m_{3}}$

Properties of water are obtained from tables:

$h_{1}=180.16\,\frac{BTU}{lbm}$

$h_{2}=28.08\,\frac{BTU}{lbm} + \left(0.01604\,\frac{ft^{3}}{lbm}\right)\cdot (14.7\,psia-0.25638\,psia)$

$h_{2}=29.032\,\frac{BTU}{lbm}$

The specific enthalpy at outlet is:

$h_{3}=\frac{(125\,\frac{lbm}{s} )\cdot (180.16\,\frac{BTU}{lbm} )+(10\,\frac{lbm}{s} )\cdot (29.032\,\frac{BTU}{lbm} )}{135\,\frac{lbm}{s} }$

$h_{3}=168.965\,\frac{BTU}{lbm}$

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$T = 200.829\,^{\textdegree}F$

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3 years ago

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8 0

3 years ago

P9.28 A large vacuum tank, held at 60 kPa absolute, sucks sea- level standard air through a converging nozzle whose throat diame

eimsori [14]

Answer:

a) $m=0.17kg/s$

b) $Ma=0.89$

Explanation:

From the question we are told that:

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Generally the Power series equation for Mach number is mathematically given by

$\frac{p_0}{p}=(1+\frac{r-1}{2}Ma^2)^{\frac{r}{r-1}}$

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Therefore

Mass flow rate

$\frac{\rho_0}{\rho}=(1+\frac{1.4-1}{2}(0.89)^2)^{\frac{1.4}{1.4-1}}$

$\frac{1.225}{\rho}=(1+\frac{1.4-1}{2}(0.89)^2)^{\frac{1.4}{1.4-1}}$

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$V=Ma(r*T_0\sqrt{T_e}$ )

Where

$T_e=\frac{P_e}{R\rho}\\\\T_e=\frac{60*10^6}{288*0.842\rho}$

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2 years ago

A double-pane insulated window consists of two 1 cm thick pieces of glass separated by a 1.8 cm layer of air. The window measure

Elanso [62]

Answer:

(b). T = 22.55 ⁰C

(c). q = 557.8 W

Explanation:

we take follow a step by step process to solving this problem.

from the question, we have that

The two glass pieces is separated by a 1.8 cm distance layer of air.

the thickness of glass piece is 1 cm

width = 4 m

the height = 3 m

(a). the sketch of the thermal circuit is uploaded in the picture below.

(b). the thermal resistance due to the conduction in the first glass plane is given thus;

R₁ = Lg / Kg A ................(1)

given that Kg rep. the thermal conductivity of the glass plane

A = conduction surface area

Lg = Thickness of glass plane4

taking the thermal conductivity of glass plane as Kg = 0.78 w/mk

inputting values into equation (1) we have,

R₁ = [1 (cm) ˣ 1 (m)/100 (cm)] / [(0.78 w/mk)(4m ˣ 3m)]

R₁ = 1.068 ˣ 10 ⁻³ k/w

Being that we have same thermal resistance in the first and second plane,

therefore R₁ = R₃ = 1.068 ˣ 10 ⁻³ k/w

⇒ Also the thermal resistance between air and glass as a result of the conduction by the layer is given thus

R₂ = La/KaA .....................(2)

given Ka = thermal conductivity of air

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La = thickness of air

substituting values into the equation we have

R₂ = [1.8 (cm) ˣ 1 (m)/100 (cm)] / [(0.0262 w/mk)(4m ˣ 3m)]

R₂ = 5.73 ˣ 10⁻² k/w

Given the thermal resistance on the outer surface due to convection, we have

R₄ = 1/hA

inputting value gives R₄ = 1 / (12 w/m² ˣ 12m) = 6.94 ˣ 10⁻³k/w

R₄ = 6.94 ˣ 10⁻³k/w

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q = 557.8 W

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q = Ti - T₂ / R₃ + R₄

T₂ = Ti - q (R₃ + R₄)

T₂ = 27 - 557.8 (1.068ˣ10⁻³ + 6.94ˣ10⁻³ ) = 22.55°C

T₂ = 22.55°C

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8 0

3 years ago

Which of the following statements about resistance is TRUE?

valentina_108 [34]

Answer:

I think it's the no 3rd

Explanation:

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2 years ago