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TiliK225 [7]
3 years ago
12

The heat transfer coefficient decreases with increasing x for both the laminar and turbulent regions a. True b. False

Engineering
1 answer:
REY [17]3 years ago
3 0

Answer:

A) True  

Explanation:

Yes this is true when length is creases the heat transfer coefficient decease with length.

The heat transfer(h) coefficient is varying with x by given expression

For Laminar flow

h \alpha \dfrac{1}{x^{\frac{1}{2}}}

For turbulent flow

h \alpha \dfrac{1}{x^{\frac{1}{5}}}

But when flow is in transitional state the heat heat transfer(h) coefficient is increases with x.But for laminar as well as turbulent flow h is decrease when x increases.

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Do the coil resistances have any effect on the plots?
PolarNik [594]
Because of the skin depth effect, the current at high frequency tends to flow at very low depth from radius. Then at high frequency the effective cross section of the wire is narrower than at DC.

Fro example skin depth at 100 kHz is 0.206 mm (0.008”), a wire more thicker than AWG26 could be a waste of copper, better use a bunch of thin wire (Litz wire) to rise the Q factor.
8 0
3 years ago
In a wheatstone bridge three out of four resistors have of 1K ohm each ,and the fourth resistor equals 1010 ohm. If the battery
Dima020 [189]

Answer:

  248.756 mV

  49.7265 µA

Explanation:

The Thevenin equivalent source at one terminal of the bridge is ...

  voltage: (100 V)(1000/(1000 +1000) = 50 V

  impedance: 1000 || 1000 = (1000)(1000)/(1000 +1000) = 500 Ω

The Thevenin equivalent source at the other terminal of the bridge is ...

  voltage = (100 V)(1010/(1000 +1010) = 100(101/201) ≈ 50 50/201 V

  impedance: 1000 || 1010 = (1000)(1010)/(1000 +1010) = 502 98/201 Ω

__

The open-circuit voltage is the difference between these terminal voltages:

  (50 50/201) -(50) = 50/201 V ≈ 0.248756 V . . . . open-circuit voltage

__

The current that would flow is given by the open-circuit voltage divided by the sum of the source resistance and the load resistance:

  (50/201 V)/(500 +502 98/201 +4000) = 1/20110 A ≈ 49.7265 µA

8 0
3 years ago
A three-story school has interior column bays that are spaced 25 ft apart in both directions. If the loading on the flat roof is
Lady_Fox [76]

Answer:

Explanation:

Floor Load:

Lo= 50psf

At= 25x25 = 625 square feet

L= Lo(0.25 +15/\sqrt{KuAt)}

L=50(0.25+15/\sqrt{(4)(625)}= 13.1psf

%reduction= 13.1/50 = 26%

Fr= 3[(13.1psf)(25ft)(25ft)+(20psf)(25ft)(25ft)]= 62k

7 0
3 years ago
It is given that 50 kg/sec of air at 288.2k is iesntropically compressed from 1 to 12 atm. Assuming a calorically perfect gas, d
denis23 [38]

The exit temperature is 586.18K and  compressor input power is 14973.53kW

Data;

  • Mass = 50kg/s
  • T = 288.2K
  • P1 = 1atm
  • P2 = 12 atm

<h3>Exit Temperature </h3>

The exit temperature of the gas can be calculated isentropically as

\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{y-1}{y}\\ y = 1.4\\ C_p= 1.005 Kj/kg.K\\

Let's substitute the values into the formula

\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{y-1}{y} \\\frac{T_2}{288.2} = (\frac{12}{1})^\frac{1.4-1}{1.4} \\ T_2 = 586.18K

The exit temperature is 586.18K

<h3>The Compressor input power</h3>

The compressor input power is calculated as

P= mC_p(T_2-T_1)\\P = 50*1.005*(586.18-288.2)\\P= 14973.53kW

The compressor input power is 14973.53kW

Learn more on exit temperature and compressor input power here;

brainly.com/question/16699941

brainly.com/question/10121263

6 0
3 years ago
You will be observing laminar-turbulent transition for room temperature (about 20°C) water flowing in a 0.602"" ID pipe (Schedul
igomit [66]

Answer:

b) 0.7667gallons/minutes

Explanation:

check the attachment below for other answers

8 0
3 years ago
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