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3 years ago

11

N2 flows through an insulated horizontal tube at steady state. With an inlet temperature of 150°C and a velocity change from 2 m

/s to 50 m/s, what is the exit temperature?

1 answer:

slava [35]3 years ago

4 0

Answer:

The exit temperature is

Explanation:

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A scale model is 4th the size of the pump. Determine the power ratio of the pump and its scale model if the ratio of the heads i

vredina [299]

Given:

size of scale model = 4(size of pump)

power ratio of pump and scale model = 5:1

Solution:

Let the diameter of scale model and pump be $d_{s}$ and $d_{p}$ respectively

and head be $H_{s}$ and $H_{p}$ respectively

Now, power, P is given as a function of head(H) and dischagre(Q)

P = $\rho gQH$ (1)

From eqn (1):

$P \propto QH$

and

$QH \propto \sqrt{H}D^{2}$

So,

$P \propto H^{\frac{3}{2}} D^{2}$

Therefore,

$\frac{P_{s}}{P_{p}}$ = $\frac{D_{s}^{2} H_{s}^{\frac{3}{2}}}{D_{p}^{2} H_{p}^{\frac{3}{2}}}$

$\frac{P_{s}}{P_{p}}$ = $\frac{D_{s}^{2} H_{s}^{\frac{3}{2}}}{D_{p}^{2} H_{p}^{\frac{3}{2}}}$

$\frac{P_{s}}{P_{p}}$ = $\frac{1^{2}\times 5^{\frac{3}{2}}}{4^{2}\times 1^{\frac{3}{2}}}$

$\frac{P_{s}}{P_{p}}$ = $\frac{5\sqrt{5}}{16}$

${P_{s}}:{P_{p}}$ = ${5\sqrt{5}}:{16}$

8 0

3 years ago

9.For a single-frequency sine wave modulating signal of 3 kHz with a carrier frequency of 36 MHz, what is the spacing between si

nikdorinn [45]

The spacing between sidebands is equal to 6 kHz.

<u>Given the following data:</u>

Modulating signal = 3 kHz.

Carrier frequency = 36 MHz.

<h3>What is a sideband?</h3>

A sideband can be defined as a band of frequencies that are lower or higher than the carrier frequency due to the modulation process. Thus, it will either be lower than or higher than the carrier frequency.

Generally, the frequency of the modulating signal is equal to the spacing between the sidebands. Therefore, a modulating signal of 3 kHz simply means that the lower sideband is <u>3 kHz</u> higher while the upper sideband is <u>3 kHz</u> lower.

Spacing = 3 kHz + 3 kHz = 6 kHz.

Read more on frequency here: brainly.com/question/3841958

8 0

2 years ago

Poles are values of Laplace transform variable, s, that make denominator of transfer function zero. Zeros are values of Laplace

Ostrovityanka [42]

Answer:

Zero 1 = -1

Zero 2 = -3

Pole 1 = 0

Pole 2 = -2

Pole 3 = -4

Pole 4 = -6

Gain = 4

Explanation:

For any given transfer function, the general form is given as

T.F = k [N(s)] ÷ [D(s)]

where k = gain of the transfer function

N(s) is the numerator polynomial of the transfer function whose roots are the zeros of the transfer function.

D(s) is the denominator polynomial of the transfer function whose roots are the poles of the transfer function.

k [N(s)] = 4s² + 16s + 12 = 4[s² + 4s + 3]

it is evident that

Gain = k = 4

N(s) = (s² + 4s + 3) = (s² + s + 3s + 3)

= s(s + 1) + 3 (s + 1) = (s + 1)(s + 3)

The zeros are -1 and -3

D(s) = s⁴ + 12s³ + 44s² + 48s

= s(s³ + 12s² + 44s + 48)

= s(s + 2)(s + 4)(s + 6)

The roots are then, 0, -2, -4 and -6.

Hope this Helps!!!

3 0

4 years ago

Given a reservoir watershed of 22 square miles, and assuming a

saveliy_v [14]

Answer:

308 acre-ft of water

Explanation:

Given:

Area of the watershed = 22 square miles

Depth of Rainfall = 0.75 in = $\frac{\textup{0.75}}{\textup{12}}$ =0.0625 ft

Percentage rainfall falling in reservoir as runoff = 35%

Now,

1 square mile = 640 acre

Thus,

22 square miles = 22 × 640 = 14,080 acres

Thus,

The total volume of rainfall = Area of watershed × Depth of the rainfall

or

The total volume of rainfall = 14,080 acres × 0.0625 ft = 880 acre-ft

also,

only 35% of the total rainfall is contributing as runoff

thus,

Runoff = 0.35 × 880 acre-ft = 308 acre-ft of water

8 0

3 years ago

Oil with a density of 850 kg/m3 and kinematic viscosity of 0.00062 m2 /s is being discharged by a 8-mm-diameter, 40-m-long horiz

Naddik [55]

Answer:

Q = 5.06 x 10⁻⁸ m³/s

Explanation:

Given:

v=0.00062 m² /s and ρ= 850 kg/m³

diameter = 8 mm

length of horizontal pipe = 40 m

Dynamic viscosity =

μ = ρv

=850 x 0.00062

= 0.527 kg/m·s

The pressure at the bottom of the tank is:

P₁,gauge = ρ g h = 850 x 9.8 x 4 = 33.32 kN/m²

The laminar flow rate through a horizontal pipe is:

$Q = \dfrac{\Delta P \pi D^4}{128 \mu L}$

$Q= \dfrac{33.32 \times 1000 \pi\times 0.008^4}{128 \times 0.527 \times 40}$

Q = 5.06 x 10⁻⁸ m³/s

4 0

3 years ago