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2 years ago

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How do I cancel my subscription

2 answers:

Aleksandr [31]2 years ago

7 0

Hey, it depends what you subscribed to, or what platform you are trying to unsubscribe from?

Ksju [112]2 years ago

5 0

Answer:

it just depends on what you subscribed to and what aplication....

Explanation:

You might be interested in

Basic concepts surrounding electrical circuitry?

Elanso [62]

Hopefully that helps you out and is this for history or science?

3 0

2 years ago

What is the thermal efficiency of this regeneration cycle in terms of enthalpies and fractions of total flow?

irga5000 [103]

Answer:

$\eta =\dfrac{(h_3-h_4)-(h_2-h_1)}{(h_3-h_5)}$

Explanation:

generally regeneration of cycle is used in the case of gas turbine. due to regeneration efficiency of turbine is increased but there is no effect on the on the net work out put of turbine.Actually in regeneration net heta input is decreases that is why total efficiency increase.

Now from T-S diagram

$W_{net}=W_{out}-W_{in}$

$W_{net}=(h_3-h_4)-(h_2-h_1)$

$Q_{in}=h_3-h_5$

Due to generation $(h_5-h_2)$ amount of energy has been saved.

$Q_{generation}=Q_{saved}$

So efficiency of cycle $\eta =\frac{W_{net}}{Q_{in}}$

$\eta =\dfrac{(h_3-h_4)-(h_2-h_1)}{(h_3-h_5)}$

Effectiveness of re-generator

$\varepsilon =\dfrac{(h_5-h_2)}{(h_4-h_2)}$

So the efficiency of regenerative cycle

$\eta =\dfrac{(h_3-h_4)-(h_2-h_1)}{(h_3-h_5)}$

7 0

2 years ago

To find the quotient of 8 divided by 1/3, multiply 8 by?

iVinArrow [24]

Answer: 8.33333333 or 6.1989778

Explanation:

8 0

1 year ago

1. A soil core sampling tube of 4 cm diameter, 12 cm length and initial mass of 0.525 kg (sample only), was dried at 105o C and

belka [17]

Answer:

porosity = 0.07 or 7%

dry bulk density = 3.25g/cm3]

water content =

Explanation:

bulk density = dry Mass / volume of sample

dry mass = 0.490kg = 490g

volume = πr2h = 3.142 * 2 *2 *12 = 150.8cm3

density = 490/150.8 = 3.25g/cm3

porosity = $\frac{wet mass - dry mass }{wet mass}$ = $\frac{0.525 - 0.49}{0.525}$ = 0.07 or 7%

water content = $\frac{wet mass - dry mass}{wet mass}$ = 7%

8 0

2 years ago

Read 2 more answers

3. (a) (5 points) Suppose N packets arrive simultaneously to a link at which no packets are currently being transmitted or queue

Bezzdna [24]

Answer:

(N-1) × (L/2R) = (N-1)/2

Explanation:

let L is length of packet

R is rate

N is number of packets

then

first packet arrived with 0 delay

Second packet arrived at = L/R

Third packet arrived at = 2L/R

Nth packet arrived at = (n-1)L/R

Total queuing delay = L/R + 2L/R + ... + (n - 1)L/R = L(n - 1)/2R

Now

L / R = (1000) / (10^6 ) s = 1 ms

L/2R = 0.5 ms

average queuing delay for N packets = (N-1) * (L/2R) = (N-1)/2

the average queuing delay of a packet = 0 ( put N=1)

4 0

2 years ago