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svlad2 [7]
3 years ago
14

The roof of a refrigerated truck compartment consists of a layer of foamed urethane insulation (t2 = 21 mm, ki = 0.026 W/m K) be

tween aluminum alloy panels (t1 = 9 mm, kp = 180 W/m K). The length of the roof is L = 13 m. The net solar radiation into the roof is 107 W/m2. The temperature of the inner surface is Ts,i = -4 oC and the air temperature is T[infinity] = 29 oC. The convective heat transfer coefficient for the top surface of the truck is h = 47 W/m2 K. Find Ts,o in oC.
Engineering
1 answer:
lakkis [162]3 years ago
3 0

Answer:

Tso = 28.15°C

Explanation:

given data

t2 = 21 mm

ki = 0.026 W/m K

t1 = 9 mm

kp = 180 W/m K

length of the roof is L = 13 m

net solar radiation into the roof = 107 W/m²

temperature of the inner surface Ts,i = -4°C

air temperature is T[infinity] = 29°C

convective heat transfer coefficient h = 47 W/m² K

solution

As when energy on the outer surface at roof of a refrigerated truck that is balance as

Q = \frac{T \infty - T si }{\frac{1}{hA}+\frac{t1}{AKp}+\frac{t2}{AKi}+\frac{t1}{aKp}}       .....................1

Q = \frac{T \infty - Tso}{\frac{1}{hA}}                         .....................2

now we compare both equation 1 and 2 and put here value

\frac{29-(-4)}{\frac{1}{47}+\frac{2\times0.009}{180}+\frac{0.021}{0.026}} = \frac{29-Tso}{\frac{1}{47}}            

solve it and we get

Tso = 28.153113

so Tso = 28.15°C

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Consider the freeway in Problem 1. At one point along this freeway there is a 4% upgrade with a directional hourly traffic volum
ryzh [129]

Answer:

The Question is incomplete, the complete question is as follows:

<em>Consider the freeway in Problem 1. At one point along this freeway there is a 4% upgrade with a directional hourly traffic volume of 5,435 vehicles. If all other conditions are as described in Problem 1, how long can this grade be without the freeway LOS dropping to F? </em>

A six-lane rural freeway (three lanes in each direction) has regular weekday users and currently operates at maximum LOS C conditions. The base free-flow speed is 65 mi/h, lanes are 11 ft wide, the right-side shoulder is 4 ft wide, and the interchange density is 0.25 per mile. The highway is one rolling terrain with 10% large trucks and buses (no recreational vehicles), and the peak-hour factor is 0.90. Determine the hourly volume for these conditions

Explanation:

<em>Make the assumption Base continuous flow velocity (BFFS)= 65 mph. </em>

Pitch width= 11 ft.

Decrease in lane width pace,fLW= 1.9 mph.

Complete Lateral clearance= 4 ft. Lateral clearance speed reduction, fLC= 0.8 mph.

Complete Width of the Ramp= 0.25 mile.

Velocity reduction proportional to the ramp height, f ID= 0 mph.

Assume lane number to be = 3.

Reduction in speed corresponding to no. of lanes, fN = 3 mph

Free Flow Speed (FFS) = BFFS – fLW – fLC – fN – fID = 65 – 1.9 – 0.8 – 3 – 0 = 59.3 mph

Peak Flow, V veh/hr

Peak-hour factor = 0.90

Trucks = 10%

Rolling Terrain

fHV = 1/ (1 + 0.10 (2.5-1)) = 1/1.15 = 0.8696

fP = 1.0

Peak Flow Rate, Vp = V / (PHV*n*fHV*fP) = V/ (0.90*3*0.8696*1.0) = 0.426V veh/hr/ln

Average speed of vehicles, S = FFS = 59.3 mph

Level of service C

Density of LOS C lies between 18 - 25 veh/mi/ln

Maximum density = 25 veh/mi/ln

Density = V​​​​​​p /S = 25

0.426V = 25 * 59.3

V = 3480 veh/hr

b) V = 5435 veh/hr

LOS dropping to F

Max density = 45 veh/mi/ln

Density = Vp/S = 45

V​​​​​​p = 45 * 59.3 = 2668.5 veh/hr/ln

V/(PHF * n * f​​​​​​HV * f​​​​​​P​​​) = 2668.5

f​​​​​​HV = 5435/(0.9*3*2668.5*1.0) = 0.754

1/(1+0.10 (E​​​​​​T -1)) = 0.754

E​​​​​​T = 4.26 ~ 3.5

<em>For 4% upgrade and 10% trucks with E​​​​​​T = 3.5, length of the grade is Greater than 1.0 miles</em>

6 0
3 years ago
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Explanation:

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3 years ago
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2 years ago
Technician A says that the definition of torque is how far the crankshaft twists in degrees.Technician B says that torque can re
leonid [27]
Technician B is correct because torque is a force of an object.
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1 year ago
The diameter of an extruder barrel = 85 mm and its length = 2.00 m. The screw rotates at 55 rev/min, its channel depth = 8.0 mm,
babunello [35]

Answer:

Qx = 9.109.10^5 \times 10^{-6} m³/s  

Explanation:

given data

diameter = 85 mm

length = 2 m

depth = 9mm

N = 60 rev/min

pressure p = 11 × 10^6 Pa

viscosity n = 100 Pas

angle = 18°

so  Qd will be

Qd = 0.5 × π² ×D²×dc × sinA × cosA   ..............1

put here value and we get

Qd = 0.5 × π² × ( 85 \times 10^{-3} )²× 9  \times 10^{-3}  × sin18 × cos18

Qd = 94.305 × 10^{-6} m³/s

and

Qb = p × π × D × dc³ × sin²A ÷  12  × n × L    ............2

Qb = 11 × 10^{6} × π × 85 \times 10^{-3}  × ( 9  \times 10^{-3} )³ × sin²18 ÷  12  × 100 × 2

Qb = 85.2 × 10^{-6} m³/s

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volume flow rate Qx = Qd - Qb   ..............3

Qx =  94.305 × 10^{-6}  - 85.2 × 10^{-6}  

Qx = 9.109.10^5 \times 10^{-6} m³/s  

8 0
3 years ago
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