Select the correct answer.

Explain the difference between a rating scale question and a forced choice question.

erik [133]

Answer:

Often times, people use the terms ranking question and rating question interchangeably, when in fact, there is a distinct difference between the two question types. A rating question asks survey respondents to compare different items using a common scale

Explanation:

7 0

3 years ago

A three-phase voltage source with a terminal voltage of 22kV is connected to a three-phase transformer rated 5MVA 22kV/220V. The

amid [387]

Answer:

A) See Attachment B) See Attachment C) 186.153

Explanation:

C) Per phase Voltage= 220/∛3

= 6.037

S=V²/Z

= (6.037)²/(0.01+i0.05)

= 62051.282-i310256.41

Active power losses per phase= 62.051kW

total Active power losses= 62.051×3

=186.153kW

6 0

4 years ago

A boiler is used to heat steam at a brewery to be used in various applications such as heating water to brew the beer and saniti

Natalija [7]

Answer:

net boiler heat = 301.94 kW

Explanation:

given data

saturated steam = 6.0 bars

temperature = 18°C

flow rate = 115 m³/h = 0.03194 m³/s

heat use by boiler = 90 %

to find out

rate of heat does the boiler output

solution

we can say saturated steam is produce at 6 bar from liquid water 18°C

we know at 6 bar from steam table

hg = 2756 kJ/kg

and

enthalpy of water at 18°C

hf = 75.64 kJ/kg

so heat required for 1 kg is

=hg - hf

= 2680.36 kJ/kg

and

from steam table specific volume of saturated steam at 6 bar is 0.315 m³/kg

so here mass flow rate is

mass flow rate = $\frac{0.03194}{0.315}$

mass flow rate m = 0.10139 kg/s

so heat required is

H = h × m

here h is heat required and m is mass flow rate

H = 2680.36 × 0.10139

H = 271.75 kJ/s = 271.75 kW

now 90 % of boiler heat is used for generate saturated stream

so net boiler heat = $\frac{H}{0.90}$

net boiler heat = $\frac{271.75}{0.90}$

net boiler heat = 301.94 kW

5 0

3 years ago

You have a 12-inch PVC water main that is 850 feet long flowing at 5.6 cfs. Point A is at an elevation of 750 ft. Point B is at

alex41 [277]

Known :

D = 12 in = 1 ft

L = 850 ft

Q = 5.6 cfs

hA = 750 ft

hB = 765 ft

PA = 85 psi = 12240 lb/ft²

Solution :

A = πD² / 4 = π(1²) / 4

A = 0.785 ft²

<u>Velocity of water :</u>

U = Q / A = 5.6 / 0.785

U = 7.134 ft/s

<u>Friction loss due to pipe length :</u>

Re = UD / v = (7.134)(1) / (0.511 × 10^(-5))

Re = 1.4 × 10⁶

(From Moody Chart, We Get f = 0.015)

hf = f(L / d)(U² / 2g) = 0.015(850 / 1)((7.134²) / 2(32.2))

hf = 10 ft

PA + γhA = PB + γhB + γhf

PB = PA + γ(hA - hB - hf)

PB = 12240 + (62.4)(750 - 765 - 10)

PB = 10680 lb/ft²

PB = 74.167 psi

8 0

3 years ago

Stainless steel ball bearings (rho = 8085 kg/m3 and cp = 0.480 kJ/kg·°C) having a diameter of 1.2 cm are to be quenched in water

sveticcg [70]

Answer:

$\dot Q = -2.341\,kJ$

Explanation:

The rate of heat transfer from the balls to the air is:

$\dot Q = \left[(800\,\frac{1}{min} )\cdot (\frac{1\,min}{60\,s} )\cdot(8085\,\frac{kg}{m^{3}})\cdot (\frac{4}{3}\pi )\cdot (6\times 10^{-3}\,m)^{3}\right]\cdot (0.480\,\frac{kJ}{kg\cdot ^{\textdegree}C} )\cdot (850\,^{\textdegree}C-900\,^{\textdegree}C)$ $\dot Q = -2.341\,kJ$

3 0

4 years ago

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