ENGINEERING COLLEGE

How much heat is lost through a 3’× 5' single-pane window with a storm that is exposed to a 60°F temperature differential?A. 450 Btu/hB. 900 Btu/hC. 1350 Btu/hD. 1800 Btu/h

Answers

Answer 1

Answer:

A. 450 btu/h

Explanation:

We solve this problem by using this formula:

Q = U x TD x area

U = U value of used material

TD = Temperature difference = 60°

Q = heat loss

Area = 3x5 = 15

We first find U

R = 1/u

2 = 1/U

U = 1/2 = 0.5

Then when we put these values into the formula above, we would have:

Q = 0.5 x 15 x 60

Q = 450Btu/h

Therefore 450btu/h is the answer

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An activated sludge plant is being designed to handle a feed rate of 0.438 m3 /sec. The influent BOD concentration is 150 mg/L and the cell concentration (MLVSS) is 2,200 mg/L. If you wish to operate the plant with a food-to-microorganism ratio of 0.20 day-1 , what volume of aeration tank should you use? Please give your answer in m3 .

Answers

Answer:

Volume of aeration tank = 1.29 x 10^4 m³

Explanation:

Food/Micro- organism Ratio = 0.2/day

Feed Rate (Q) = 0.438 m³/s

Influent BOD = 150 mg/L

MLVSS = 2200 mg/L

The above mentioned parameters are related by the equation

F/M = QS₀/VX

where S₀ is the influent BOD and X is the cell concentration, with V being the Volume of the tank. Re- arranging the above equation for V and putting the values in, we get

V = 0.4380 x 150/0.2 x 2200

V = 0.1493 (m³/s) x day

V = 0.1493 x 24 x 60 x 60

V = 1.29 x 10^4 m³

What are the general principles of DFA? What are the steps to minimize the number of parts for an assembly?

Answers

Answer Explanation : The general principles for design for assembly (DFA) are,

MINIMIZE NUMBER OF COMPONENT
USE STANDARD COMMERCIALLY AVAILABLE COMPONENTS
USE COMMON PARTS ACROSS PRODUCT LINES
DESIGN FOR EASE OF PART FABRICATION
DESIGN PARTS WITH TOLERANCE THAT ARE WITHIN PROCESS CAPABILITY
MINIMIZE USE OF FLEXIBLE COMPONENT
DESIGN FOR EASE OF ASSEMBLY
USE MODULAR DESIGN
REDUCE ADJUSTMENT REQUIRED

STEPS TO MINIMIZE THE NUMBER OF PARTS

USE OF INCORPORATE HINGS
USE OF INTEGRAL SPRINGS
USE OF SNAP FITS
USE OF GUIDES BEARINGS
USE OF COVERS

Calculate how large a mass would be necessary to obtain a mechanical noise limit of [Equation] = 1 nG, 1 µG, and 1 mG if the mechanical resonance frequency is [Equation] = 100 Hz. If the mass is to be made of cube of `silicon, what would its physical dimensions be?

Answers

Answer:

Mechanical resonance frequency is the frequency of a system to react sharply when the frequency of oscillation is equal to its resonant frequency (natural frequency).

The physical dimension of the silicon is 10kg

Explanation:

Using the formular, Force, F = 1/2π√k/m

At resonance, spring constant, k = mw² ( where w = 2πf), when spring constant, k = centripetal force ( F = mw²r).

Hence, F = 1/2π√mw²/m = f ( f = frequency)

∴ f = F = mg, taking g = 9.8 m/s²

100 Hz = 9.8 m/s² X m

m = 100/9.8 = 10.2kg

Assuming that the following three variables have already been declared, which variable will store a Boolean value after these statements are executed? choice = true;
again = "false";
result = 0;

a. choice
b. again
c. result
d. none of these are Boolean variables

Answers

Answer:

Explanation:

Boolean Algebra deals with either a one or a zero and how to manipulate them in computers or elsewhere. The "choice" option may not work, since for text it must be enclosed in quotation marks, usually. For "again," it's text and not a 1 or 0. So, the answer is C, since this is a 0.

A hydraulic jump is induced in an 80 ft wide channel. The water depths on either side of the jump are 1 ft and 10 ft. Please calculate: a) The velocity of the faster moving flow. b) The flow rate (discharge). c) The Froude number of the sub-critical flow. d) The flow energy dissipated in the hydraulic jump (expressed as percentage of the energy prior to the jump). e) The critical depth.

Answers

Answer:

a) The velocity is 42.0833 ft/s

b) The flow rate is 3366.664 ft³/s

c) The Froude number is 0.2345

d) The flow energy dissipated (expressed as percentage of the energy prior to the jump) is 18.225 ft

e) The critical depth is 3.8030 ft

Explanation:

Given data:

80 ft wide channel, L

1 ft and 10 ft water depths, d₁ and d₂

Questions: a) Velocity of the faster moving flow, v = ?

b) The flow rate (discharge), q = ?

c) The Froude number, F = ?

d) The flow energy dissipated, E = ?

e) The critical depth, dc = ?

a) For the velocity:

$(d_(2) )/(d_(1) ) =(1)/(2) (\sqrt{1+8F^(2) } -1)$

$10*2=\sqrt{1+8F^(2) } -1$

Solving for F:

F = 7.4162

$v=F\sqrt{gd_(1) }$

Here, g = gravity = 32.2 ft/s²

$v=7.4162*√(32.2*1) =42.0833ft/s$

b) The flow rate:

$q=v*L*d_(1) =42.0833*80*1=3366.664ft^(3) /s$

c) The Froude number:

$v_(2) =(q)/(L*d_(2) ) =(3366.664)/(80*10) =4.2083ft/s$

$F=\frac{v_(2)}{\sqrt{gd_(2) } } =(4.2083)/(√(32.2*10) ) =0.2345$

d) The flow energy dissipated:

$E=((d_(2)-d_(1))^(3) )/(4d_(1)d_(2)) =((10-1)^(3) )/(4*1*10) =18.225ft$

e) The critical depth:

$d_(c) =((((q)/(L))^(2) )/(g) )^(1/3) =((((3366.664)/(80))^(2) )/(32.2) )^(1/3) =3.8030ft$

What is the difference between absolute and gage pressure?

Answers

Explanation:

Step1

Absolute pressure is the pressure above zero level of the pressure. Absolute pressure is considering atmospheric pressure in it. Absolute pressure is always positive. There is no negative absolute pressure.

The expression for absolute pressure is given as follows:

$P_(ab)=P_(g)+P_(atm)$

Here, $P_(ab)$ is absolute pressure, $P_(g)$ is gauge pressure and $P_(atm)$ is atmospheric pressure.

Step2

Gauge pressure is the pressure that measure above atmospheric pressure. It is not considering atmospheric pressure. It can be negative called vacuum or negative gauge pressure. Gauge pressure used to simplify the pressure equation for fluid analysis.

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