ENGINEERING COLLEGE

A composite wall consists of 20 mm thick steel plate backed by insulation brick (k = 0.39 W/mK) of 50 cm thickness and overlaid by mineral wool of 20 cm thickness (k = 0.05 W/mK) and 70 cm layer of brick of (k = 0.39 W/mK). The inside is exposed to convection at 650°C with h = 65 W/ m2K. The outside is exposed to air at 35°C with a convection coefficient of 15 W/m2K. Determine the heat loss per unit area, interface temperatures and temperature gradients in each materials.

Answers

Answer 1

Answer:

Answer:

Heat loss=85.9W/m^2

ΔT1(Steel)=0.04C

ΔT2(Brick1)=110.13C

ΔT3(Mwood)=343.6C

ΔT1(Brick2)=154.18C

Explanation:

raise the heat transfer equation from the air inside the wall to the outside air from the wall, because that is where you have the temperature data, to find the heat.

To find the temperatures you use the heat found in the previous step, and you use the conduction and convection equations in each wall layer.

I attached the procedure

Answer 2

Answer:

Answer:

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Explanation:

please

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Calculate the "exact" alkalinity of the water in Problem 3-2 if the pH is 9.43.Calculate the "approximate" alkalinity (in mg/L as CaCO3 ) of a water containing 120 mg/L of bicarbonate ion and 15 mg/L of carbonate ion.

Answers

Answer:

A) approximate alkalinity = 123.361 mg/l

B) exact alkalinity = 124.708 mg/l

Explanation:

Given data :

A) determine approximate alkalinity first

Bicarbonate ion = 120 mg/l

carbonate ion = 15 mg/l

Approximate alkalinity = ( carbonate ion ) * 50/30 + ( bicarbonate ion ) * 50/61

= 15 * (50/30) + 120*( 50/61 ) = 123.361 mg/l as CaCO3

B) calculate the exact alkalinity of the water if the pH = 9.43

pH + pOH = 14

9.43 + pOH = 14. therefore pOH = 14 - 9.43 = 4.57

[OH^- ] = 10^-4.57 = 2.692*10^-5 moles/l

[ OH^- ] = 2.692*10^-5 * 179/mole * 10^3 mg/g = 0.458 mg/l

[ H^+ ] = 10^-9.43 * 1 * 10^3 = 3.7154 * 10^-7 mg/l

therefore the exact alkalinity can be calculated as

= ( approximate alkalinity ) + ( [ OH^- ] * 50/17 ) - ( [ H^+ ] * 50/1 )

= 123.361 + ( 0.458 * 50/17 ) - ( 3.7154 * 10^-7 * 50/1 )

= 124.708 mg/l

Cody’s car accelerates from 0m/s to 45 m/s northward in 15 seconds. What is the acceleration of the car

Answers

Answer:

3 m/s²

Explanation:

Acceleration is calculated as :

a= Δv/ t

where ;

Δv = change in velocity

Δv = 45 - 0 = 45 m/s

t= 15 s

a= 45 /15

a= 3 m/s²

You are analyzing an open-return wind tunnel that intakes air at 20 m/s and 320K. When the flow exits the wind tunnel it is moving at a speed of 250 m/s. What is the temperature of the flow exiting that wind tunnel

Answers

The solution is in the attachment

Answer:

please find attached.

Explanation:

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

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

Consider the string length equal to 7. This string is distorted by a function f (x) = 2 sin(2x) - 10sin(10x). What is the wave formed in this string? a. In=12cos (nit ) sin(max) b. 2cos(2t)sin (2x) - 10cos(10t ) sin(10x) c. n 2 sin 2x e' – 10sin 10x e

Answers

Answer:

hello your question has a missing part below is the missing part

Consider the string length equal to $\pi$

answer : 2cos(2t) sin(2x) - 10cos(10t)sin(10x)

Explanation:

Given string length = $\pi$

distorted function f(x) = 2sin(2x) - 10sin(10x)

Determine the wave formed in the string

attached below is a detailed solution of the problem

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Answers

Answer:verify proper cable is hooked between laptop and projector. HDMI ports or 15 pin video output to input.

And laptop is selected to output to respective video output.

Explanation:

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