Just see my handwriting need text form answer not handwriting retype it. Dont mis all retype A to z Ⓒ Usingconditionsof 6 butthe length of the channel is&m.Acs = width x height6.01) (0.5)0.00532balance quationUsing the energynixep x (rout-fin)= Qcom (AAT)0.05x (005) (Tout-300) = (0.075)(0.005) (600-300)Tout=300.612calculate the hydraulic diameter60 ungthegiven dimensionsD=4. AesPINION MAN41Aes0.005m² (210) (10g)(P= 2xl0.01 +0.5)(00D=Given5.Juol) (2000) x 20=1·0234(0.005)2.0216-002D = 0.04963.h2 85-57 10/2Aes=Is =0.00532600kTin = 300k20.05kg/5. Using the energy.balanceequationUsing theenergybalance equation:mx cpx Tout-300)=hx Asx (600-200)005x005) (Tout-300)IU65-57) (0.005) (300)Tout=302.82kFlow rate=Using the energymx cp xl Tout-300)1.6k93600s= 0·00+kgle.balance equation-=600) x Choos) x (fout-300)Ag x (600-300)=85-87 (0.005) (300)Fout8427-716KⒸ Using the conditions of exceptinelengthis 4.5m.Ausz0.00532P= 2(15+0.01)D==3.02mAcsP=(42(0.005)3.02= 0.0066m.h= NukD=(252.97)(0.0257)0.006621083-55426.

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Engineering

Mechanical Engineering

Just see my handwriting need text form answer not handwriting retype it. Dont mis all retype A to z

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Concept explainers

Layout

The layout is a way of organizing the elements, products, and raw materials in a design. The layout is the representation of information in a structural, logical, descriptive, coherent form.The layout is a mapping of various equipment arrangements, workstations, machinery, plants, and more. It secures the continuous movement of a product quantity along with the quality.

Question

Just see my handwriting need text form answer not handwriting retype it. Dont mis all retype A to z

Ⓒ Using
conditions
of 6 but
the length of the channel is&m.
Acs = width x height
6.01) (0.5)
0.00532
balance quation
Using the energy
nixep x (rout-fin)
= Qcom (AAT)
0.05x (005) (Tout-300) = (0.075)(0.005) (600-300)
Tout=
300.61
2
calculate the hydraulic diameter
60 ung
the
given dimensions
D=
4. Aes
PINION MAN
41
Aes
0.005m² (210) (10g)
(P= 2xl0.01 +0.5)
(00
D=
Given
5.
Juol) (2000) x 20
=
1·023
4(0.005)
2.02
16-002
D = 0.04963.
h2 85-57 10/2
Aes=
Is =
0.00532
600k
Tin = 300k
20.05kg/5.

Using the energy.
balance
equation
Using the
energy
balance equation:
mx cpx Tout-300)
=
hx Asx (600-200)
005x005) (Tout-300)
IU
65-57) (0.005) (300)
Tout=
302.82k
Flow rate
=
Using the energy
mx cp xl Tout-300)
1.6k9
3600s
= 0·00+kgle.
balance equation-
=
600) x Choos) x (fout-300)
Ag x (600-300)
=
85-87 (0.005) (300)
Fout
8427-716K
Ⓒ Using the conditions of exceptine
length
is 4.5m.
Ausz
0.00532
P= 2(15+0.01)
D=
=
3.02m
Acs
P
=
(42(0.005)
3.02
= 0.0066m.
h= Nuk
D
=(252.97)(0.0257)
0.0066
2
1083-5542
6.

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