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3 years ago

Does a thicker core make an electromagnet stronger?

Engineering

2 answers:

Pie3 years ago

7 0

Yes it does (: hope i helped

iren [92.7K]3 years ago

5 0

Answer:

Yes

Explanation:

The core of an electromagnet serves to stabilize the magnetic field created by the wire. The thicker the core, the more metal there is to amplify the current. Therefore, a thicker core does make an electromagnet stronger. Hope this helps!

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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

Python:

stira [4]

Answer:

# Python Program to Print

# all subsets of given size of a set

import itertools

def findsubsets(s, n):

return list(itertools.combinations(s, n))

# Driver Code

s = {1, 2, 3}

n = 2

print(findsubsets(s, n))

-----------------------------------------------

# Python Program to Print

# all subsets of given size of a set

import itertools

# def findsubsets(s, n):

def findsubsets(s, n):

return [set(i) for i in itertools.combinations(s, n)]

# Driver Code

s = {1, 2, 3, 4}

n = 3

print(findsubsets(s, n))

-------------------------------------------------------------

# Python Program to Print

# all subsets of given size of a set

import itertools

from itertools import combinations, chain

def findsubsets(s, n):

return list(map(set, itertools.combinations(s, n)))

# Driver Code

s = {1, 2, 3}

n = 2

print(findsubsets(s, n))

4 0

4 years ago

__ signals always take priority over signs and traffic signals

Mars2501 [29]

Answer:

Stop singnals

Explanation:

4 0

4 years ago

How much extra water does a 21.5 ft, 175-lb concrete canoe displace compared to an ultra-lightweight 38-lb Kevlar canoe of the s

pantera1 [17]

Answer:

The volume of the extra water is $2.195 ft^{3}$

Solution:

As per the question:

Mass of the canoe, $m_{c} = 175 lb + w$

Height of the canoe, h = 21.5 ft

Mass of the kevlar canoe, $m_{Kc} = 38 lb + w$

Now, we know that, bouyant force equals the weight of the fluid displaced:

Now,

$V\rho g = mg$

$V = \frac{m}{\rho}$ (1)

where

V = volume

$\rho = 62.41 lb/ft^{3}$ = density

m = mass

Now, for the canoe,

Using eqn (1):

$V_{c} = \frac{m_{c} + w}{\rho}$

$V_{c} = \frac{175 + w}{62.41}$

Similarly, for Kevlar canoe:

$V_{Kc} = \frac{38 + w}{62.41}$

Now, for the excess volume:

V = $V_{c} - V_{Kc}$

V = $\frac{175 + w}{62.41} - \frac{38 + w}{62.41} = 2.195 ft^{3}$

8 0

3 years ago

What is a plaster ratio?

Makovka662 [10]

Plaster Mix Ratio
Mix cement and sand in the ratio of 1:6 (1 cement:6 sand) for inner plastering of bricks. And for outer plastering mix it in the ratio of 1:4. On a brick wall never do plastering of thickness more than 12 or 15mm. At one go, avoid plastering of more than 12mm thickness.

6 0

3 years ago