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4 years ago

8

Water enters the tubes of a cold plate at 70°F with an average velocity of 40 ft/min and leaves at 105°F. The diameter of the tu

bes is 0.25 in. Assuming 14 percent of the heat generated is dissipated from the components to the surroundings by convection and radiation and the remaining 86 percent is removed by the cooling water, determine the amount of heat generated by the electronic devices mounted on the cold plate. The properties of water at room temperature are rho = 62.1 lbm/ft3 and cp = 1.00 Btu/lbm·°F.

1 answer:

Fudgin [204]4 years ago

8 0

Answer:

The total amount of heat generated;Q' = 2067 Btu/h

Explanation:

We are given;

Water entering temperature;T1 = 70°F

Water leaving temperature;T2 = 105°F

average velocity of water;V = 40 ft/min

Diameter of tube;D = 0.25 in = 0.25/12 ft = 0.02083 ft

Water density;ρ = 62.1 lbm/ft³

cp = 1.00 Btu/lbm·°F.

Now, the mass flow rate of the water is calculated from;

m' = ρAV

Where ρ is density, A is area and V is velocity

Area = πD²/4 = π*0.02083²/4 = 0.00034077555 ft²

m' = 62.1 * 0.00034077555 * 40

m' = 0.8465 lbm/min

Converting to lbm/hr = 0.8465 * 60 = 50.79 lbm/hr

From energy balance equation, we have;

E_in = E_out

So,

Q_in,w + m'h1 = m'h2

Q_in,w = m'h2 - m'h1

Q_in,w = m'(h2 - h1)

Now, m'(h2 - h1) can be written as;

m'cp(T2 - T1).

Thus ;

Q_in,w = m'cp(T2 - T1)

Plugging in the relevant values, we have;

Q_in,w = (50.79*1)(105 - 70)

Q_in,w = 1777.65 Btu/h

We are told that remaining 86 percent of heat generaged is removed by the cooling water. Thus;

The total amount of heat generated could be defined as;

Q' = Q_in,w/0.86

Q' = 1777.65/0.86

Q' = 2067 Btu/h

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Answer:

a) P₂ = 3219.11 lbf / ft² , b) P₂ = 721.91 lbf / ft² , c) P₂ = 5707.31 lbf / ft²

Explanation:

For this exercise we can use the fluid mechanics equations, let's start with the continuity equation, index 1 is for the starting point and index 2 for the end point of the reduction

A₁ v₁ = A₂ v₂

v₂ = v₁ A₁ / A₂

The area of a circle is

A = π r² = π/4 d²

v₂ = v₁ (d₁ / d₂)²

Let's calculate

v₂ = 12 (7/3)²

v₂ = 65 feet / s

Now let's use Bernoulli's equation

P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

P₁ - P₂ = ρ g (y₂ –y₁) + ½ ρ (v₂² - v₁²)

Case 1. The pipe is horizontal, so

y₁ = y₂

P₁ - P₂ = ½ ρ (v₂² –v₁²)

P₂ = P₁ - ½ ρ (v₂² –v₁²)

ρ = 62.43 lbf / ft³

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P₂ = 3219.11 lbf / ft²

Case 2 vertical pipe with water flow up

y₂ –y₁ = 40 ft

P₁ - P₂ = ρ g (y₂ –y₁) + ½ rho (v₂² - v₁²)

7200 - P₂ = 62.43 (40) + ½ 62.43 / 32 (65 2 - 12 2) =

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P₂ = 5707.31 lbf / ft²

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3 years ago

1. You have a Co-Cr alloy with Young's Modulus: 645 MPa, Poisson ratio 0.28, and yield strength 501 MPa for that alloy when used

bazaltina [42]

Answer:

F_x = 100,200 N

x' = 21.321 cm ... Length

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z' = 1.565 cm

A' = ( 0.783 x 1.565 ) cm

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Given:

- The Modulus of Elasticity E = 645 MPa

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- The Yield Strength Y = 501 MPa

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- The length along y-direction y = 1 cm

- The length along z--direction z = 2 cm

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The maximum tensile load that can be applied in the longitudinal direction of the bar without inducing plastic deformation. b. The length and cross-sectional area of the bar at its tensile elastic limit.

Solution:

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F_i = б_i*A_jk

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F_x = 100,200 N

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ξ_x = Y / E

ξ_x = 501 / 645 = 0.7767

ξ_y = ξ_z = -v*Y / E

ξ_y = ξ_z = -0.28*501 / 645 = - 0.2175

- The corresponding change in lengths at tensile elastic stress are:

Δx = x*ξ_x = 12*0.7767 = 9.321 cm

Δy = y*ξ_y = - 1*0.2175 = -0.2175 cm

Δz = z*ξ_z = - 2*0.2175 = -0.435 cm

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x' = x + Δx = 12 + 9.321 = 21.321 cm

y' = y + Δy = 1 - 0.2175 = 0.7825 cm

z' = z + Δz = 2 - 0.435 = 1.565 cm

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scZoUnD [109]

Answer:

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ZanzabumX [31]

Answer:

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$T_{A} = 0$

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$V_{A} = (V_{A})_{g} + (V_{A})_{c}$

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4 years ago