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3 years ago

1:

Engineering

1 answer:

Nataly [62]3 years ago

7 0

Answer:

20.87 Pa

Explanation:

The formula for dynamic pressure is given as;

q= 1/2*ρ*v²

where ;

q=dynamic pressure

ρ = density of fluid

v = velocity of fluid

First find v by applying the formula for flow rate as;

Q = v*A where ;

Q= fluid flow rate

v = flow velocity

A= cross-sectional area.

A= cross-sectional vector area of the pipe given by the formula;

A= πr² = 3.14 * 4² = 50.27 in² where r=radius of pipe obtained from the diameter given divided by 2.

Q = fluid flow rate = 105 gpm----change to m³/s as

1 gpm = 0.00006309

105 gpm = 105 * 0.00006309 = 0.006624 m³/s

A= cross-sectional vector area = 50.27 in² -------change to m² as:

1 in² = 0.0006452 m²

50.27 in² = 50.27 * 0.0006452 = 0.03243 m²

Now calculate flow velocity as;

Q =v * A

Q/A = v

0.006624 m³/s / 0.03243 m² =v

0.2043 m/s = v

Now find the dynamic pressure q given as;

q= 1/2 * ρ*v²

q= 1/2 * 1000 * 0.2043² = 20.87 Pa

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A spur gearset has a module of 6 mm and a velocity ratio of 4. The pinion has 16 teeth. Find the number of teeth on the driven g

levacccp [35]

Answer:

NG=64 teeth

dG=384mm

dP=96mm

C=240mm

Explanation:

step one:

given data

module m=6mm

velocity ratio VR=4

number of teeth of pinion Np=16

Step two:

Required

1. Number of teeth on the driven gear

$N_G=N_P*V_R\\\\N_G=16*4\\\\N_G=64$

The driven gear has 64 teeth

2. The pitch diameters

The driven gear diameter

$d_G=N_G*m\\\\d_G=64*6\\\\d_G=384$

The driven gear diameter is 384mm

The pinion diameter

$d_P=N_P*m\\\\d_P=16*6\\\\d_P=96$

Pinion diameter is 96mm

3. Theoretical center-to-center distance

$C=\frac{d_G+d_P}{2} \\\\C=\frac{384+96}{2} \\\\C=\frac{480}{2}\\\\C=240$

The theoretical center-to-center distance is 240mm

5 0

3 years ago

To be safe, the engineers making the ride want to be sure the normal force does not exceed 1.8 times each persons weight - and t

Yuri [45]

Answer:

μ = 0.55

Explanation:

Given that

Normal weight = 1.8 x weight of person

N= 1.8 mg

We know that friction force Fr

Fr= μ N

μ=Coefficient of friction

N=Normal force

To find μ We have to equate friction and gravity force

Fr= Wt

μ N = m g

μ x 1.8 m g = m g

μ = 0.55

So the coefficient of friction will be 0.55.

5 0

3 years ago

A counter-flow double-piped heat exchange is to heat water from 20oC to 80oC at a rate of 1.2 kg/s. The heating is to be accompl

lawyer [7]

Answer:

110 m or 11,000 cm

Explanation:

let mass flow rate for cold and hot fluid = Mc and Mh respectively
let specific heat for cold and hot fluid = Cpc and Cph respectively
let heat capacity rate for cold and hot fluid = Cc and Ch respectively

Mc = 1.2 kg/s and Mh = 2 kg/s

Cpc = 4.18 kj/kg °c and Cph = 4.31 kj/kg °c

Using effectiveness-NUT method

First, we need to determine heat capacity rate for cold and hot fluid, and determine the dimensionless heat capacity rate

Cc = Mc × Cpc = 1.2 kg/s × 4.18 kj/kg °c = 5.016 kW/°c

Ch = Mh × Cph = 2 kg/s × 4.31 kj/kg °c = 8.62 kW/°c

From the result above cold fluid heat capacity rate is smaller

Dimensionless heat capacity rate, C = minimum capacity/maximum capacity

C= Cmin/Cmax

C = 5.016/8.62 = 0.582

.2 Second, we determine the maximum heat transfer rate, Qmax

Qmax = Cmin (Inlet Temp. of hot fluid - Inlet Temp. of cold fluid)

Qmax = (5.016 kW/°c)(160 - 20) °c

Qmax = (5.016 kW/°c)(140) °c = 702.24 kW

.3 Third, we determine the actual heat transfer rate, Q

Q = Cmin (outlet Temp. of cold fluid - inlet Temp. of cold fluid)

Q = (5.016 kW/°c)(80 - 20) °c

Qmax = (5.016 kW/°c)(60) °c = 303.66 kW

.4 Fourth, we determine Effectiveness of the heat exchanger, ε

ε = Q/Qmax

ε = 303.66 kW/702.24 kW

ε = 0.432

.5 Fifth, using appropriate effective relation for double pipe counter flow to determine NTU for the heat exchanger

NTU = $\\ \frac{1}{C-1} ln(\frac{ε-1}{εc -1} )$

NTU = $\frac{1}{0.582-1} ln(\frac{0.432 -1}{0.432 X 0.582 -1} )$

NTU = 0.661

.6 sixth, we determine Heat Exchanger surface area, As

From the question, the overall heat transfer coefficient U = 640 W/m²

As = $\frac{NTU C{min} }{U}$

As = $\frac{0.661 x 5016 W. °c }{640 W/m²}$

As = 5.18 m²

.7 Finally, we determine the length of the heat exchanger, L

L = $\frac{As}{\pi D}$

L = $\frac{5.18 m² }{\pi (0.015 m)}$

L= 109.91 m

L ≅ 110 m = 11,000 cm

3 0

3 years ago

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Pepsi [2]

Answer:

Explanation:

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7 0

4 years ago

A rectangular car-top carrier of 1.7-ft height, 5.0-ft length (front to back), and 4.2-ft width is attached to the top of a car.

Nataliya [291]

Answer:

$\Delta P =1.2 \frac{1.3}{2}(26.822m/s)^2 (4.2*1.7*(0.3048)^2)=13.88 hp$

Explanation:

We can assume that the general formula for the drag force is given by:

$D= C_D \frac{\rho}{2}V^2 A$

And we can see that is proportional to the area. On this case we can calculate the area with the product of the width and the height. And we can express the grad force like this:

$D_1 = C_{D1} \frac{\rho}{2}V^2 (wh)$

Where w is the width and h the height.

The last formula is without consider the area of the carrier, but if we use the area for the carrier we got:

$D_2 = C_{D2} \frac{\rho}{2}V^2 (wh+ A_{carrier})$

If we want to find the additional power added with the carrier we just need to take the difference between the multiplication of drag force by the velocity (assuming equal velocities for both cases) of the two cases, and we got:

$\Delta P = C_{D2} \frac{\rho}{2}V^2 (wh+ A_{carrier}) V- C_{D1} \frac{\rho}{2}V^2 (wh) V$