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shutvik [7]
3 years ago
6

An electric motor under steady load draws 9.7 amperes at 110 volts; it delivers 1.25(hp) of mechanical energy. The temperature o

f the surroundings is 300 K. What is the total rate of entropy generation in W·K−1?
Engineering
1 answer:
pishuonlain [190]3 years ago
7 0

Answer:

The total rate of entropy generation is 6.665 W/K

Explanation:

Power input = current × voltage = 9.7 × 110 = 1067 W

Power output = 1.25 hp = 1.25 × 746 = 932.5 W

Total power = 1067 + 932.5 = 1999.5 W

Total rate of entropy generation = total power ÷ temperature of the surrounding = 1999.5 W ÷ 300 K = 6.665 W/K

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alekssr [168]

Answer:

It looks like... A machine that reads electric pulse and surge... Not sure though.

Explanation:

8 0
3 years ago
Q-) please give me a reference about Tack coat? Pleae i need it please??!!
Arturiano [62]

Answer:

Tack coat is a sprayed application of an asphalt binder upon an existing asphalt or Portland cement concrete pavement prior to an overlay, or between layers of new asphalt concrete.

Explanation:

4 0
3 years ago
A road has a crest curve, where the PVI station is a 71 35. The road transitions from a 2.1% grade to a -3.4% grade. The highest
sveticcg [70]

Answer:

Stat PVC = Stat(82+98.5)

Stat PVT = Stat(59+71.5)

Explanation

PVI = 71 + 35

Let G1 = Grade 1; G2 = Grade 2

G1 = +2.1% ; G2 = -3.4%

Highest point of curve at station = 74 + 10

General equation of a curve:

y = ax^{2} +bx+c\\dy/dx=2ax+b\\

At highest point of the curve dy/dx=o

2ax+b=0\\x=-b/2a\\x=G1L/(G2-G1)\\x=L/2 +(stat 74+10)-(stat 71+35)\\x=L/2 + 275

-G1L/(G2-G1) = (L/2 + 275)/100\\L = -2327 ft\\Station PVC = Stat(71+35)+(-2327/2)\\\\Stat PVC = 7135-1163.5\\Stat PVC = Stat(82+98.5)\\

Station PVT

Station PVT = Stat PVI + (L/2)\\Station PVT = Stat(71+35)+(-2327/2)\\Station PVT = 7135-1163.5\\Stat PVT = Stat(59+71.5)

3 0
3 years ago
Instead of running blood through a single straight vessel for a distance of 2 mm, one mammalian species uses an array of 100 tin
Marina CMI [18]

Solution:

Given that :

Volume flow is, $Q_1 = 1000 \ mm^3/s$

So, $Q_2= \frac{1000}{100}=10 \ mm^3/s$

Therefore, the equation of a single straight vessel is given by

$F_{f_1}=\frac{8flQ_1^2}{\pi^2gd_1^5}$    ......................(i)

So there are 100 similar parallel pipes of the same cross section. Therefore, the equation for the area is

$\frac{\pi d_1^2}{4}=1000 \times\frac{\pi d_2^2}{4} $

or $d_1=10 \ d_2$

Now for parallel pipes

$H_{f_2}= (H_{f_2})_1= (H_{f_2})_2= .... = = (H_{f_2})_{10}=\frac{8flQ_2^2}{\pi^2 gd_2^5}$  ...........(ii)

Solving the equations (i) and (ii),

$\frac{H_{f_1}}{H_{f_2}}=\frac{\frac{8flQ_1^2}{\pi^2 gd_1^5}}{\frac{8flQ_2^2}{\pi^2 gd_2^5}}$

       $=\frac{Q_1^2}{Q_2^2}\times \frac{d_2^5}{d_1^5}$

       $=\frac{(1000)^2}{(10)^2}\times \frac{d_2^5}{(10d_2)^5}$

       $=\frac{10^6}{10^7}$

Therefore,

$\frac{H_{f_1}}{H_{f_2}}=\frac{1}{10}$

or $H_{f_2}=10 \ H_{f_1}$

Thus the answer is option A). 10

7 0
3 years ago
The specific volume of a system consisting of refrigerant-134a at 1.0 Mpa is 0.01 m /kg. The quality of the R-134a is: (a) 12.6%
Flura [38]

Answer:

option c is correct

47.2%

Explanation:

given data

consisting of refrigerant = 134 a

volume V = 0.01 m³/kg

pressure P = 1MPa = 1000 kPa

to find out

quality of the R 134a

solution

we will get here value of volume Vf and Vv from pressure table 60 kpa to 3 Mpa for 1 Mpa of R134 a

that is

Vf = 0.0008701 m³/kg

Vv = 0.0203 m³/kg

so we will apply here formula that is

quality = (V - Vf) / (Vv - Vf)    ............1

put here value

quality = (0.01 - 0.0008701 ) / ( 0.0203 - 0.0008701 )

quality = 0.4698

so quality is 47 %

SO OPTION C IS CORRECT

4 0
3 years ago
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