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zavuch27 [327]
3 years ago
14

Which basic principle influences how all HVACR systems work?

Engineering
1 answer:
bezimeni [28]3 years ago
6 0

Answer:

B) An increase in pressure can lower the boiling point of a liquid and change the temperature at which it turns to a gas.

Explanation:

B) An increase in pressure can lower the boiling point of a liquid and change the temperature at which it turns to a gas.

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The loneliest people are to kindest
valkas [14]

Answer:

The most damaged people are the wisest is a fact

Explanation:

5 0
3 years ago
Read 2 more answers
The Stefan-Boltzmann law can be employed to estimate the rate of radiation of energy H from a surface, as in
Mazyrski [523]

Explanation:

A.

H = Aeσ^4

Using the stefan Boltzmann law

When we differentiate

dH/dT = 4AeσT³

dH/dT = 4(0.15)(0.9)(5.67)(10^-8)(650)³

= 8.4085

Exact error = 8.4085x20

= 168.17

H(650) = 0.15(0.9)(5.67)(10^-8)(650)⁴

= 1366.376watts

B.

Verifying values

H(T+ΔT) = 0.15(0.9)(5.67)(10)^-8(670)⁴

= 1542.468

H(T+ΔT) = 0.15(0.9)(5.67)(10^-8)(630)⁴

= 1205.8104

Error = 1542.468-1205.8104/2

= 168.329

ΔT = 40

H(T+ΔT) = 0.15(0.9)(5.67)(10)^-8(690)⁴

= 1735.05

H(T-ΔT) = 0.15(0.9)(5.67)(10^-8)(610)⁴

= 1735.05-1059.83/2

= 675.22/2

= 337.61

5 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
Use the concept that y = c, −[infinity] < x < [infinity], is a constant function if and only if y' = 0 to determine whethe
attashe74 [19]

Answer:

attached below

Explanation:

5 0
3 years ago
For a flow rate of 212 cfs find the critical depth in (a) a rectangular channel with ????=6.5 ft, (b) a triangular channel with
Fofino [41]

Answer:

A. 3.21ft

B. 3.51ft

C. 2.95ft

D. 1.5275ft

Explanation:

A) Q =212 cu.f/s

Formula for critical depth of rectangular section is: dc =[(Q^2) /(b^2(g))]^1/3

Where dc =critical depth, ft

Q= quantity of flow or discharge, ft3/s

B= width of channel, ft (m)

g = acceleration due to gravity which is 9.81m/s2 or 32.185ft/s2

Now, from the question,

Q = 212 cu.f/s and b=6.5ft

Therefore, the critical depth is: [(212^2)/(6.5^2 x32. 185)]^(1/3)

To give ; critical depth= (44,944/1359.82)^(1/3) = 3.21ft

B. Formula for critical depth of a triangular section; dc = (2Q^2/gm^2)^(1/5)

From the question, Q =212 cu.f/s and m=1.6ft while g= 32.185ft/s2

Therefore, critical depth = [(212^2) /(1.6^2 x32. 185)] ^(1/5) = (44,944/84.466)^(1/5) = 3.51ft

C. For trapezoidal channel, critical depth(y) is derived from (Q^2 /g) = (A^3/T)

Where A= (B + my)y and T=(B+2my)

Now from the question, B=6.5ft and m=5ft.

Therefore, A= (6.5 + 2y)y and T=(6. 5 + 2(5y))= 6.5 + 10y

Now, let's plug the value of A and T into the initial equation to derive the critical depth ;

(212^2 /32.185) = [((6.5 + 2y)^3)y^3]/ (6.5 + 10y)

Which gives;

1396.43 = [((6.5 + 2y)^3)y^3]/ (6.5 + 10y)

Multiply both sides by 6.5 + 10y to get;

1396.43(6.5 + 10y) = [((6.5 + 2y)^3)y^3]

Factorizing this, we get y = 2. 95ft

D) Formula for critical depth of a circular section; dc =D/2[1 - cos(Ѳ/2)]

Where D is diameter of pipe and Ѳ is angle at critical depth in radians.

Angle not given, so we assume it's perpendicular angle is 90.

Since angle is in radians, therefore Ѳ/2 = 90/2 = 45 radians ; converting to degree, = 2578. 31

Therefore, dc = (6.5/2) (1 - cos (2578.31))

dc = 3.25(1 - 0.53) = 3.25 x 0.47 = 1.5275ft

8 0
3 years ago
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