ENGINEERING COLLEGE

A test bottle containing just seeded dilution water has its DO level drop by 0.8 mg/L in a 5-day test. A 300 mL BOD bottle filled with 30 mL of wastewater and the rest with seeded dilution water experiences a drop of 7.3 mg/L in the same period (5-day). Calculate the BOD5 of the wastewater.

Answers

Answer 1

Answer:

Answer:

BOD_5 = =65.8 mg/l

Explanation:

dilution water DO level = 0.8 m/l

BOD level drop to 7.3 mg/l

we know that BOD at 5th day can be clculated by using following relation $BOD_5 = ((D_1 -D_2) -(B_1-B_2)(1-P))/(P)$

$D_1 -D_2$ - DO drop in BOD bottle

$B_1-B_2$ - dilution water drop

P= 30/300 = 0.1

$BOD_5 = ((7.3) -(0.8)(1-0.1))/(0.1)$

$BOD_5 = =65.8 mg/l$

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At steady state, air at 200 kPa, 325 K, and mass flow rateof 0.5 kg/s enters an insulated duct having differing inletand exit cross-sectional areas. The inlet cross-sectional area is6 cm26cm 2. At the duct exit, the pressure of the air is 100 kPa and the velocity is 250 m/s. Neglecting potential energyeffects and modeling air as an ideal gas with constant cp=1.008 kJ/kg⋅Kc p =1.008kJ/kg⋅K, determine(a) the velocity of the air at the inlet, in m/s. (b) the temperature of the air at the exit, in K. (c) the exit cross-sectional area, in cm2 (a) the velocity of the air at the inlet, in m/s.(b) the temperature of the air at the exit, in K.(c) the exit cross-sectional area, in cm

A wall in a house contains a single window. The window consists of a single pane of glass whose area is 0.15 m2 and whose thickness is 5 mm. Treat the wall as a slab of the insulating material Styrofoam whose area and thickness are 10 m2 and 0.15 m, respectively. Heat is lost via conduction through the wall and the window. The temperature difference between the inside and outside is the same for the wall and the window. Of the total heat lost by the wall and the window, what is the percentage lost by the window

Answers

Answer:

The percentage loss of the window is $E = 97.3$ %

Explanation:

From the question we are told that

The area of pane of glass is $A = 0.15 m^2$

The thickness is $d = 5mm = (5)/(1000) = 0.005m$

The thickness of the wall is $D = 0.15m$

The area of the wall is $a = 10m^2$

Generally the heat lost as a result of conduction of the window is

$Q_(window) = (j_(glass) * A * (\Delta T) )/(d)$

Where $j_(glass)$ is the thermal conductivity of glass which has a constant value of

$j_(glass) = 0.80 J/(s \cdot m \cdot C^o)$

Substituting values

$Q_(window) = ( 0.80 * 0.15 * (\Delta T) )/(0.005)$

$Q_(window) = 24 \Delta T$

Generally the heat lost as a result of conduction of the wall is

$Q_(wall) = (j_(styrofoam) * A * (\Delta T) )/(d)$

$j_(styrofoam)$ s the thermal conductivity of Styrofoam which has a constant value of $j_(styrofoam) = 0.010J / (s \cdot m \cdot C^o)$

Substituting values

$Q_(wall) = ( 0.010 * 10 * (\Delta T) )/(0.15)$

$Q_(wall) = 0.667 \ \Delta T$

Now the net loss of heat is

$Q_(net) = Q_(window) + Q_(wall)$

Substituting values

$Q_(net) = 24 + 0.667$

$Q_(net) = 24.667$

Now the percentage loss by the window is

$E = (Q_(window) )/(Q_(net)) * 100$

Substituting value

$E = (24)/(24 .667) * 100$

$E = 97.3$ %

Calculate the angle of banking on a bend of 100m radius so that vehicles can travel round the bend at 50km/hr without side thrust on the tyres.

Answers

Answer:

11.125°

Explanation:

Given:

Radius of bend, R = 100 m

Speed around the bend = 50 Km/hr = $(5)/(18)*50$ = 13.89 m/s

Now,

We have the relation

$\tan\theta=(v^2)/(gR)$

where,

θ = angle of banking

g is the acceleration due to gravity

on substituting the respective values, we get

$\tan\theta=(13.89^2)/(9.81*100)$

$\tan\theta=0.1966$

θ = 11.125°

The sports car has a weight of 4500-lb and a center of gravity at G. If it starts from rest it causes the rear wheels to slip as it accelerates. Determine how long it takes for it to reach a speed of 10 ft/s. Also, what are the normal reactions at each of the four wheels on the road? The coefficients of static and kinetic friction at the road are us=0.5 and uk=0.3, respectively. Neglect the mass of the wheels.

Answers

Answer:

Time=2.72 seconds

Front wheel reactions= 1393 lb

Rear wheel reactions= 857 lb

Explanation:

The free body diagram is assumed to be the one attached here

The mass, m of the car is

$M=\frac {W}{g}$ where W is weight and g is acceleration due to gravity

Taking g as $32.2 ft/s^(2)$ then

$M=\frac {4500}{32.2}=139.75 lbm$

Considering equilibrium in x-axis

$Ma_G-f=0$

$Ma_G-(\mu_g* 2N_B)=0$

$139.75* a_G-(0.3* 2* N_B)=0$

$0.6N_B=139.75a_g$

$N_B=232.92a_g$

At point A using the law of equilibrium, the sum of moments is 0 hence

$-2N_B(6)+4500(2)=-Ma_G(2.5)$

$-12N_B+9000=-139.75a_G* 2.5$

$-12(232.92a_G)+900=-349.375a_G$

$a_g\approx 3.68 ft/s^(2)$

The normal reaction at B is therefore

$N_B=232.92a_G=232.92* 3.68\approx 857 lb$

Consider equilibrium in y-axis

$4500-2N_A-2N_B=0$

$N_A+N_B=2250$

$N_A+857=2250$

$N_A=1393 lb$

To find time that the car takes to a speed of 10 ft/s

Using kinematic equation

V=u+at

10=0+3.68t

$t=\frac {10}{3.68}\approx 2.72 s$

What does Enter key do? You cannot click Enter key to start a line if your current is blank?

This is spot to do today

Answers

Answer:

See below

Explanation:

Enter-key also called the "Return key," it is the keyboard key that is pressed to signal the computer to input the line of data or the command that has just been typed.It Was the Return KeyThe Enter key was originally the "Return key" on a typewriter, which caused the carriage to return to the beginning of the next line on the paper. In a word processing or text editing application, pressing Enter ends a paragraph. A character code for return/end-of-line, which is different in Windows than it is in the Mac, Linux or Unix, is inserted into the text at that point.

Answer:

True

Explanation:

Once there are two yellow lines having inner broken lines on the two sides of a center traffic lane, what this is trying to tell you is that you can use those lanes to start a left hand turn, or a U-turn from the both directions of traffic. However you cannot use it for passing. This is sometimes misunderstood by road users and drivers.

An automobile engine consumes fuel at a rate of 27.4 L/h and delivers 55 kW of power to the wheels. If the fuel has a heating value of 44,000 kJ/kg and a density of 0.73 g/cm3, deter- mine the efficiency of this engine in percentage(

Answers

Answer:

The efficiency of the engine is 22.5%.

Explanation:

Efficiency = power output ÷ power input

power output = 55 kW

power input = specific energy×volumetric flow rate×density

specific energy = 44,000 kJ/kg

volumetric flow rate = 27.4 L/h = 27.4 L/h × 1000 cm^3/1 L × 1 h/3600 s = 7.61 cm^3/s

density = 0.73 g/cm^3 = 0.73 g/cm^3 × 1 kg/1000g = 7.3×10^-4 kg/cm^3

power input = 44,000 kJ/kg × 7.61 cm^3/s × 7.3×10^-4 kg/cm^3 = 244.4332 kJ/s = 244.4332 kW

Efficiency = 55 ÷ 244.4332 = 0.225 × 100 = 22.5%

Convert the angles of a triangle to radians.Part A31∘43′53′′, 90∘32′11′′, 57∘43′56′′Express your answers, separated by commas, to six significant figures.nothingrad, rad, radRequest AnswerPart B94∘22′19′′, 40∘54′53′′, 44∘42′48′′Express your answers, separated by commas, to six significant figures.

Answers

Answer:

Explanation:

To convert to radians

A31∘43′53′′, 90∘32′11′′, 57∘43′56′′

using DMS approach ; 1degree = 60minutes = 3600 seconds

1° = 60' = 3600"

And degree to radian = multiply by π/180

A) 31∘43′53′′ = 31degree + 43minutes + 53 seconds

= 31 degree + 43minutes + 53/60

= 31 degree + 43.88minutes

= 31 degree + 43.88/60 = 31.73 degree x π/180 = 0.5534radians

FOR 90∘32′11′′ = 90 degree + 32minutes + 11seconds

= 90degree + 32minutes + 11/60

= 90 degree + 32.183minutes

= 90 degree + 32.183/60 = 90.54degree x π/180

= 1.580radians

FOR 57∘43′56′′ = 57degree + 43minutes+ 56seconds

= 57degree + 43minutes + 56/60

= 57 degree + 43.93minutes

= 57degree + 43.93/60 = 57.73degree X π/180

= 1.00radians

PART B

FOR 94∘22′19′′ = 94degree + 22minutes + 19seconds

= 94degree + 22minutes + 19/60

= 94degree + 22.32minutes

= 94degree + 22.32/60

= 94.37degree X π/180 = 1.65radians

FOR 40∘54′53′′ = 40degree + 54minutes + 53seconds

= 40 degree + 54minutes + 53/60

= 40 degree + 54.88minutes = 40 degree + 54.88/60

= 40.91degree X π/180 = 0.714radians

FOR 44∘42′48′′ = 44degree + 42minutes + 48seconds

= 44degree + 42.8minutes

= 44.71degree X π/180 = 0.780radians

Answer:

0.176270π rad, 0.502980π rad, 0.320735π rad

0.524289π rad, 0.227304π rad, 0.248407π rad

Explanation:

We know that,

1° = 60' 180° = π

1 ' = 1°/60 1° = π/180

a. 31°43'53''

Step 1

53'' = 53 * 1/60

= 53'/60

Step 2

43'53''

= 43'+53'/60

= (2580+43)/60

= 2623'/60

-------- Convert to degrees

= 2623/60 * 1/60

= 2623/3600

Step 3

31°43'53''

= 31+ 2623/3600

= (111600 + 2623)/3600

= 114223°/3600

Now, we convert to radians

= 114223/3600 * π/180°

= 0.176270π rad

90°32'11''

Step 1.

11' = 11 * 1/60

= 11/60

Step 2

32'11'

= 32 + 11/60

= 1931/60

-------- Convert to degrees

= 1931/60 * 1/60

= 1931/3600

Step 3

90°31'11''

= 90 + 1931/3600

= 325931°/3600

Now we convert to radians

= 325931°/3600 * π/180°

= 0.502980π rad

57°43'56''

Step 1

56' = 56 * 1/60

= 56/60

= 14/15

Step 2

43'56''

= 43 + 14/15

= 659/15

Now we convert to degrees

= 659/15 * 1/60

= 659°/900

Step 3

57°43'56''

= 57 + 659/900

= 51959/900

Now we convert to radians

= 51959°/900 * π/180°

= 0.320735π rad

94∘22′19′′

Step 1

19'' = 19/60

Step 2

22'19''

= 22 + 19/60

= 1339/60

Now we convert to degrees

= 1339/60 * 1/60

= 1339°/3600

Step 3

94°22'19"

= 94 + 1339/3600

= 339739°/3600

Now we convert to radians

= 339739°/3600 * π/180

= 0.524289π rad

40∘54′53′′

Step 1

53" = 53/60

Step 2

54'53"

= 54'+ 53/60

= 3293/60

Now we convert to degrees

= 3293/60 * 1/60

= 3293/3600

Step 3

40°54'53"

= 40 + 3293/3600

= 147293/3600

Now we convert to radians

= 147293/3600 * π/180

= 0.227304π rad

44∘42′48′

Step 1

48' = 48/69

= 4/5

Step 2

42'48"

= 42 + 4/5

=214/5

Nowz we convert to degrees

= 214/5 * 1/60

= 107/150

Step 3

44°42'48"

= 44 + 107/150

= 6707/150

Now we convert to radians

= 6707/150 * π/180

= 0.248407π rad

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) A flow is divided into two branches, with the pipe diameter and length the same for each branch. A 1/4-open gate valve is installed in line A, and a 1/3-closed ball valve is installed in line B. The head loss due to friction in each branch is negligible compared with the head loss across the valves. Find the ratio of the velocity in line A to that in line B (include elbow losses for threaded pipe fittings).