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Lubov Fominskaja [6]
4 years ago
15

If the operating control on a commercial water heater fails, the gas valve will be de-energized by?

Engineering
1 answer:
kogti [31]4 years ago
5 0

Answer:

Do not operate water heater if exposed to flooding or water damage. ... de-energize the 24-volt gas valve and end the current heating cycle. The control system

Explanation:

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The Keystone Pipeline has an inside diameter of 36 inches andcarries a flow rate of 590,000 barrels of crude oil per day at 40°C
xxTIMURxx [149]

Answer:

The pump horse power required per mile of pipe  45.22\ hp/mile

Explanation:

The explanation is shown on the  second third and fourth image

It make sense that the temperature of the oil is high even though the pipeline is in a cold environment this is so because where the oil is been tapped from is  way hotter than the temperature of the pipeline environment  because it is closer to the earths crust than the environment of the pipeline  

7 0
3 years ago
Two loads connected in parallel draw a total of 2.4 kW at 0.8 pf lagging from a 120-V rms, 60-Hz line. One load absorbs 1.5 kW a
stealth61 [152]

Answer: a) 0.948 b) 117.5µf

Explanation:

Given the load, a total of 2.4kw and 0.8pf

V= 120V, 60 Hz

P= 2.4 kw, cos θ= 80

P= S sin θ - (p/cos θ) sin θ

= P tan θ(cos^-1 (0.8)

=2.4 tan(36.87)= 1.8KVAR

S= 2.4 + j1. 8KVA

1 load absorbs 1.5 kW at 0.707 pf lagging

P= 1.5 kW, cos θ= 0.707 and θ=45 degree

Q= Ptan θ= tan 45°

Q=P=1.5kw

S1= 1.5 +1.5j KVA

S1 + S2= S

2.4+j1.8= 1.5+1.5j + S2

S2= 0.9 + 0.3j KVA

S2= 0.949= 18.43 °

Pf= cos(18.43°) = 0.948

b.) pf to 0.9, a capacitor is needed.

Pf = 0.9

Cos θ= 0.9

θ= 25.84 °

(WC) V^2= P (tan θ1 - tan θ2)

C= 2400 ( tan (36. 87°) - tan (25.84°)) /2 πf × 120^2

f=60, π=22/7

C= 117.5µf

7 0
3 years ago
Which term refers to the impurities found during the welding process ?
Aleks04 [339]

Answer:

idk

Explanation:

idk

6 0
3 years ago
In a semiconductor manufacturing process, three wafers from a lot are tested. Each wafer is classified as pass or fail. Assume t
gladu [14]

Answer:

F(x) = 0           ;  x < 0

         0.064   ;  0 ≤ x < 1

         0.352   ;  1 ≤ x < 2

         0.784   ;  2 ≤ x < 3

            1        ;    x ≥ 3

Explanation:

Each wafer is classified as pass or fail.

The wafers are independent.

Then, we can modelate X : ''Number of wafers that pass the test'' as a Binomial random variable.

X ~ Bi(n,p)

Where n = 3 and p = 0.6 is the success probability

The probatility function is given by :

P(X=x)=f(x)=nCx.p^{x}.(1-p)^{n-x}

Where nCx is the combinatorial number

nCx=\frac{n!}{x!(n-x)!}

Let's calculate f(x) :

f(0)=3C0.(0.6^{0}).(0.4^{3})=0.4^{3}=0.064

f(1)=3C1.(0.6^{1}).(0.4^{2})=0.288

f(2)=3C2.(0.6^{2}).(0.4^{1})=0.432

f(3)=3C3.(0.6^{3}).(0.4^{0})=0.6^{3}=0.216

For the cumulative distribution function that we are looking for :

P(X\leq x)=F(x)

F(0)=f(0)\\F(1)=f(0)+f(1)\\F(2)=f(0)+f(1)+f(2)\\F(3)=f(0)+f(1)+f(2)+f(3)=1

F(0)=0.064\\F(1)=0.064+0.288=0.352\\F(2)=0.064+0.288+0.432=0.784\\F(3)=0.064+0.288+0.432+0.216=1

The cumulative distribution function for X is :

F(x) = 0           ;  x < 0

         0.064   ;  0 ≤ x < 1

         0.352   ;  1 ≤ x < 2

         0.784   ;  2 ≤ x < 3

            1        ;    x ≥ 3

5 0
3 years ago
A hair dryer is basically a duct of constant diameter in which a few layers of electric resistors are placed. A small fan pulls
Inessa05 [86]

Answer:

the percent increase in the velocity of air is 25.65%

Explanation:

Hello!

The first thing we must consider to solve this problem is the continuity equation that states that the amount of mass flow that enters a system is the same as what should come out.

m1=m2

Now remember that mass flow is given by the product of density, cross-sectional area and velocity

(α1)(V1)(A1)=(α2)(V2)(A2)

where

α=density

V=velocity

A=area

Now we can assume that the input and output areas are equal

(α1)(V1)=(α2)(V2)

\frac{V2}{V1} =\frac{\alpha1 }{\alpha 2}

Now we can use the equation that defines the percentage of increase, in this case for speed

i=(\frac{V2}{V1} -1) 100

Now we use the equation obtained in the previous step, and replace values

i=(\frac{\alpha1 }{\alpha 2} -1) 100\\i=(\frac{1.2}{0.955} -1) 100=25.65

the percent increase in the velocity of air is 25.65%

6 0
3 years ago
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