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Dmitrij [34]
2 years ago
9

Help please i will give brainlist

Engineering
2 answers:
vlada-n [284]2 years ago
5 0

Answer:

Open Circuit

Explanation:

Because the circuit isnt closed, the current wont flow

alekssr [168]2 years ago
3 0
The last one
And nice to meet you I am fatima and I think we are from the same place

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saul85 [17]

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quality ingredients and exceeding industry standards

Explanation:

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2 years ago
A simple ideal Rankine cycle with water as the working fluid operates between the pressure limits of 4 MPa in the boiler and 20
viktelen [127]

Answer:

a) 69,630KW

b) 203 KW

Explanation:

The data obtained from Tables A-4, A-5 and A-6 is as follows:

h_{1} = h_{f,@20KPa} = 251.42 KJ/kg\\v_{1} = v_{f,@20KPa} = 0.001017 KJ/kgK\\\\w_{p,in} = v_{1} * (P_{2} - P_{1})\\w_{p,in} = (0.001017)*(4000-20)\\\\w_{p,in} = 4.05 KJ/kg\\\\h_{2} = h_{1} - w_{p,in} \\h_{2} = 251.42 + 4.05\\\\h_{2}  =  255.47KJ/kg\\\\P_{3} = 4000KPa\\T_{3} = 700 C\\s_{3} = 7.6214 KJ/kgK\\\\h_{3} = 3906.3 KJ/kg\\\\P_{4} = 20 KPa\\s_{3} = s_{4} = 7.6214KJ/kgK\\s_{f} = 0.8320 KJ/kgK\\s_{fg} = 7.0752 KJ/kgK\\\\

x_{4} = \frac{s_{4} - s_{f} }{s_{fg} }  \\\\x_{4} = \frac{7.6214-0.8320}{7.0752} = 0.9596\\\\h_{f} = 251.42KJ/kg \\h_{fg} = 2357.5KJ/kg \\\\h_{4} = h_{f} + x_{4}*h_{fg} = 251.42 + 0.9596*2357.5 = 2513.7KJ/kg\\\\

The power produced and consumed by turbine and pump respectively are:

W_{T,out} = flow(m) *(h_{3} - h_{4}) \\W_{T,out} = 50 *(3906.3-2513.7)\\\\W_{T,out} = 69,630 KW\\\\W_{p,in} = flow(m) *w_{p,in} = 50*4.05 = 203 KW

7 0
3 years ago
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Yuki888 [10]

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C

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Boolean Algebra deals with either a one or a zero and how to manipulate them in computers or elsewhere. The "choice" option may not work, since for text it must be enclosed in quotation marks, usually. For "again," it's text and not a 1 or 0. So, the answer is C, since this is a 0.

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3 years ago
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What is the composition, in atom percent, of an alloy that contains 44.5 lbmof Ag, 83.7 lbmof Au, and 5.3 lbmof Cu? What is the
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