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konstantin123 [22]

3 years ago

12

An example of the split-off point in oil, gasoline, and kerosene production is that point where crude oil is

1 answer:

eimsori [14]3 years ago

8 0

i believe the correct answer is c but i’m sorry if i’m not correct

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What is true about a point and shoot camera?

mario62 [17]

Answer:

What is true about a point and shoot camera?

They are often smaller than SLR cameras

6 0

2 years ago

A double-threaded Acme stub screw of 2-in. major diameter is used in a jack having a plain thrust collar of 2.5-in. mean diamete

Temka [501]

This is the answer for the question

6 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

64A geothermal pump is used to pump brine whose density is 1050 kg/m3at a rate of 0.3 m3/s from a depth of 200 m. For a pump eff

grin007 [14]

Answer:

835,175.68W

Explanation:

Calculation to determine the required power input to the pump

First step is to calculate the power needed

Using this formula

P=V*p*g*h

Where,

P represent power

V represent Volume flow rate =0.3 m³/s

p represent brine density=1050 kg/m³

g represent gravity=9.81m/s²

h represent height=200m

Let plug in the formula

P=0.3 m³/s *1050 kg/m³*9.81m/s² *200m

P=618,030 W

Now let calculate the required power input to the pump

Using this formula

Required power input=P/μ

Where,

P represent power=618,030 W

μ represent pump efficiency=74%

Let plug in the formula

Required power input=618,030W/0.74

Required power input=835,175.68W

Therefore the required power input to the pump will be 835,175.68W

5 0

2 years ago

Consider a refrigerator that consumes 320 W of electric power when it is running. If the refrigerator runs only one quarterof th

ryzh [129]

Answer:

$5.184

Explanation:

The cost can be calculated using the formula: $Cost = Load \ factor \times Number \ of \ hours \ \\M_{month} = M_{units} \times W\\$

Before using this, we require the following conversions:

<em>320 W → kW:</em>

$\frac {320}{1000} = 0.32$

<em>30 Days → Hours:</em>

$30 \times 24 = 720$

Using the above stated formula:

$M_{month} = 0.09 \times 0.32 \times \frac{1}{4} \times 720 = 5.184$

4 0

2 years ago