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Alchen [17]
3 years ago
15

What does kichfoff's currebt law​

Engineering
1 answer:
QveST [7]3 years ago
4 0

Answer:

Kirchhoff's current law (1st Law) states that the current flowing into a node (or a junction) must be equal to the current flowing out of it

Explanation:

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Consider 4.8 pounds per minute of water vapor at 100 lbf/in2, 500 oF, and a velocity of 100 ft/s entering a nozzle operating at
andriy [413]

Answer:

A) v_2 = 2016.80 ft/s

B) \Delta s = 0.006 Btu/lbm R  

Explanation:

Given data:

P-1 = 100 lbf/in^2

T_1 = 500 degree f

V_1 = 100 ft/s

P_2 = 40 lbf/inc^2

effeciency = 80%

from steady flow enerfy equation

h_1 +\frac{V_1^2}{2} = h_2 + \frac{V_2^2}{2}

where h1 and h2 are inlet and exit enthalpy

for P1 = 100 lbf/in^2 and T1 = 500 degree F

H_1 = 1278.8 Btu/lbm

s_1 = 1.708 Btu/lbm -R

for P1 = 40 lbf/in^2

H_1 = 1193.5 Btu/lbm

s_1 = 1.708 Btu/lbm -R

exit enthalapy h_2

\eta = \frac{h_1 - h'_2}{ h_1 - h_2}

0.80 = \frac{1278.8 - h'_2}{1278.8 -1193.5} = 1197.77 Btu/lbm

from above equation

1278.8 \times 25037 + \frac{100^2}{2} = 1197.77   \times 25037 + \frac{v_2^2}{2}                   [1 Btu/lbm = 25037 ft^2/s^2]

v_2 = 2016.80 ft/s

b) amount of entropy

\Delta s = s_2 - s_1

s_1 = 1.708 Btu/lbm -R

at h_2 = 1197.77 Btu/lbm [\tex]  and [tex]P_2 = 40 lbf/in^2

s_2 is 1.714 Btu/lbm -R

\Delta s = 1.714 - 1.708 = 0.006 Btu/lbm R

6 0
3 years ago
what are 3 things you would like to get out of an ethical engineering class? and why? about a page explanation would do fine.
poizon [28]

Answer:

I don't know

Explanation:

i don't know

8 0
3 years ago
Read 2 more answers
A teenage brain is already fully developed to enable us to manage risks effectively.
Kisachek [45]

Answer:

false

Explanation:

the brain is only really fully devolved by age 26

3 0
2 years ago
Read 2 more answers
A differential amplifier is to have a voltage gain of 100. What will be the feedback resistance required if the input resistance
yuradex [85]

Answer:

required feedback resistance ( R2 ) = 100 k Ω

Explanation:

Given data :

Voltage gain = 100

input resistance ( R1 ) = 1 k ohms

calculate feedback resistance required

voltage gain of differential amplifier

\frac{Vout}{V2 - V1 }  = \frac{R2}{R1}

= Voltage gain =  R2/R1

= 100 = R2/1

hence required feedback resistance ( R2 ) = 100 k Ω

4 0
3 years ago
Help I will brainliest
ValentinkaMS [17]

Answer:

hacking?

Explanation:

jsd

3 0
3 years ago
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