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Helen [10]
2 years ago
15

4.68 Steam enters a turbine in a vapor power plant operating at steady state at 560°C, 80 bar, and exits as a saturated vapor at

8 kPa. The turbine operates adiabatically, and the power developed is 9.43 kW. The steam leaving the turbine enters a condenser heat exchanger, where it is condensed to saturated liquid at 8 kPa through heat transfer to cooling water passing through the condenser as a separate stream. The cooling water enters at 18°C and exits at 36°C with negligible change in pressure. Ignoring kinetic and potential energy effects and stray heat transfer at the outer surface of the condenser, determine the mass flow rate of cooling water required, in kg/s.
Engineering
1 answer:
garik1379 [7]2 years ago
3 0

Answer:

please mark me as a brainleast

Explanation:

hahahahhahaahhahahahahahahahahahahahahahahahaahhhahhhahaahahhaahhhahahaah

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Which of the following describe a keyword<br> argument? Check all that apply.
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Calculate the force of attraction between a Ca^2+ and an 0^2- irons whose centers are sep by a distance of 1.25 nm.
Zarrin [17]

Solution:

Given:

Calcium and Oxygen ions each having valence = 2 or 2 valence electrons

seperation distance, r = 1.25nm = 1.25\times 10^{-9} m

Coulombian force for two point charges seperated by a distance 'r' is given by:

F = \frac{1}{4\pi\epsilon _{_{o}} }\times \frac{q_{1}q_{2}}{r^{2}}      (1)

Now, we know that

\frac{1}{4\pi \epsilon _{o}} = 9\times 10^{^{9}}

q = ne

where, n = no. of electrons

e = charge of an electron = 1.6\times 10^{-19} C

q_{1} = n_{1}e = 2e

q_{2} = n_{2}e = 2e

Using Eqn (1)

F = (9\times 10^{^{9}})\times \frac{4\times (1.6\times 10^{-19})^{^{2}}}{(1.25\times 10^{-9})^{2}}

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Since, the force is between two opposite charged ions, it is attractive in nature.

6 0
3 years ago
A chef needs 20 walnuts for a recipe and has to crack each one. the effort force is 4.1inches from the fulcrum and the walnut si
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The calculated answer is 4 kgf. The greatest resisting force that the walnut may exert before the nut cracks

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If load is 40 kgf and effort is?

effort arm=20 cm, load arm=2 cm

By the moments' principle,

effort arm equals load arm.

40kgf×2cm=E×10cm

∴E=40×220kgf=4kgf.

Learn more about Effort here-

brainly.com/question/9457233

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3 0
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