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2 years ago

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Water (cp = 4180 J/kg·K) is to be heated by solar-heated hot air (cp = 1010 J/kg·K) in a double-pipe counter-flow heat exchanger

. Air enters the heat exchanger at 90°C at a rate of 0.3 kg/s, while water enters at 22°C at a rate of 0.1 kg/s. The overall heat transfer coefficient based on the inner side of the tube is given as 80 W/m2·K. The length of the tube is 12 m and the internal diameter of the tube is 1.2 cm.

1 answer:

Inessa [10]2 years ago

3 0

Answer:

452%

Explanation:

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Assuming the starting height is 0.0 m, calculate the potential energy of the cart after it has been elevated to a height of 0.5

Bogdan [553]

The potential energy is most often referred to as the "energy at rest" and is dependent on the elevation of an object. This can be calculated through the equation,

E = mgh

where E is the potential energy, m is the mass, g is the acceleration due to gravity, and h is the height. In this item, we are not given with the mass of the cart so we assume it to be m. The force is therefore,

E = m(9.8 m/s²)(0.5 m) = 4.9m

Hence, the potential energy is equal to 4.9m.

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2 years ago

A particularly scary roller coaster contains a loop-the-loop in which the car and rider are completely upside down. If the radiu

Pepsi [2]

Answer:

$v = 10.89\ m/s$

Explanation:

given,

radius of loop = 12.1 m

to find the minimum speed transverse by the rider to not to fall out upside down

centripetal force = $\dfrac{mv^2}{r}$

gravitational force = m g

computing both the equation]

$mg = \dfrac{mv^2}{r}$

$v = \sqrt{rg}$

$v = \sqrt{12.1 \times 9.8}$

$v = \sqrt{118.58}$

$v = 10.89\ m/s$

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2 years ago

8.4-1 Consider a magnetic field probe consisting of a flat circular loop of wire with radius 10 cm. The probe’s terminals corres

Vlad1618 [11]

Answer:

B_o = 1.013μT

Explanation:

To find B_o you take into account the formula for the emf:

$\epsilon=-\frac{d\Phi_b}{dt}=-\frac{dBAcos\theta}{dt}=-Acos\theta\frac{dB}{dt}$

where you used that A (area of the loop) is constant, an also the angle between the direction of B and the normal to A.

By applying the derivative you obtain:

$\epsilon=-Acos\theta (2\pi f) B_ocos(2\pi f t+ \alpha)$

when the emf is maximum the angle between B and the normal to A is zero, that is, cosθ = 1 or -1. Furthermore the cos function is 1 or -1. Hence:

$\epsilon=2\pi fAB_o=2\pi (100*10^3Hz)(\pi (0.1m)^2)B_o=19739.20Hzm^2B_o\\\\B_o=\frac{20*10^{-3}V}{19739.20Hzm^2}=1.013*10^{-6}T=1.013\mu T$

hence, B_o = 1.013μT

6 0

2 years ago

A variable-length air column is placed just below a vibrating wire that is fixed at both ends. The length of the air column, ope

pogonyaev

Answer:

The speed of transverse waves in the wire is 234.26 m/s.

Explanation:

Given that,

Length of wire = 129 cm

Speed of sound in air = 340 m/s

First position of resonance = 31.2 cm

We need to calculate the wavelength

For pipe open at one end and closed at other, there is node at closed end and an anti node at open end for 1st resonance.

At 1st resonance,

$L=\dfrac{\lambda}{4}$

$\lambda=4\times L$

Put the value into the formula

$\lambda=4\times31.2\times10^{-2}$

$\lambda=1.248\ m$

We need to calculate the frequency of sound in pipe

Using formula of frequency

$f=\dfrac{v}{\lambda}$

Put the value into the formula

$f=\dfrac{340}{1.248}$

$f=272.4\ Hz$

We need to calculate the node distance

For the wire, there are 3 segments, so 4 nodes

Node-node distance ,

$L=\dfrac{l}{3}$

$L = \dfrac{129}{3}$

$L=43\ cm$

We need to calculate the wavelength of the wire

Using formula of length

$L=\dfrac{\lambda}{2}$

$\lambda=L\times 2$

Put the value into the formula

$\lambda=43\times2$

$\lambda=86\ cm$

We need to calculate the speed of the wave

Using formula of frequency

$f=\dfrac{v}{\lambda}$

$v=f\times\lambda$

Put the value into the formula

$v=272.4\times86\times10^{-2}$

$v=234.26\ m/s$

Hence, The speed of transverse waves in the wire is 234.26 m/s.

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2 years ago

Compute the mean and maximum velocities for a liquid with a flow rate of 20 L/min in a 1.5-in nominal diameter sanitary pipeline

vlada-n [284]

Answer:

Mean velocity = 0.292 m/s

Maximum velocity = 0.584 m/s

The flow is laminar as Re = 229.2

Explanation:

D = 1.5 inches = 0.0381 m

Q = volumetric flow rate = 20 L/min = 0.000333 m³/s

Q = A × v

A = Cross sectional Area = πD²/4 = π(0.0381)²/4 = 0.00114 m²

v = average velocity

v = Q/A = 0.000333/0.00114 = 0.292 m/s

For flow in circular pipes, maximum velocity = 2 × average velocity = 2 × 0.292 = 0.584 m/s

To check if the flow is laminar or turbulent, we need its Reynolds number

Re = (ρvD)/μ

ρ = 1030 kg/m

v = 0.292 m/s

D = 1.5 inches = 0.0381 m

μ = 50 cP = 0.5 poise = 0.05 Pa.s

Re = (1030 × 0.292 × 0.0381)/0.05 = 229.2

For laminar flow, Re < 2100

For turbulent flow, Re > 4000

And 229.2 < 2100, hence, this flow is laminar.

7 0

2 years ago