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4 years ago

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A teenager was pulling a prank and placed a large stuffed penguin in the middle of a roadway. A driver is traveling on this leve

l roadway and sees the penguin 525 feet ahead.

1 answer:

Anvisha [2.4K]4 years ago

3 0

Whats the question????

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An overhead 25m long, uninsulated industrial steam pipe of 100mm diameter is routed through a building whose walls and air are a

DedPeter [7]

Answer:

a) he rate of heat loss from the steam line is 18.413588 kW

b) the annual cost of heat loss from line is $12904.25

Explanation:

a)

first we find the area

A = πdL

d is the diameter (0.1m) and L is the length (25m)

so

A = π × 0.1 × 25

A = 7.85 m²

Now rate of heat loss through convection

qconv = hA(Ts -Ta)

h is the convective heat transfer coefficient (10 W/m²K), Ts is surface temperature (150°), Ta is temperature of air (25°)

so we substitute

qconv = 10 W/m²K × 7.85 m² × ( 150° - 25°)

qconv = 9817.477 J/s

Now heat lost through radiation

qrad = ∈Aα ( Ts⁴ - Ta⁴)

∈ is the emissivity (0.8), α is the boltzmann constant ( 5.67×10⁻⁸m⁻²K⁻⁴ ),

first we shall covert our temperatures from Celsius to kelvin scale

Ts is surface temperature (150 + 273K ), Ta is temperature of air (25 + 273K)

so we substitute

qrad = 0.8 × 7.854 × 5.67×10⁻⁸ × ( (423)⁴ - (298)⁴ )

qrad = 3.5625×10⁻⁷ × 2.413×10¹⁰

qrad = 8596.112 J/s

Now to get the total rate of heat loss through convection and radiation, we say

q = qconv + qrad

q = 9817.477 + 8596.112

q = 18413.588 J/s ≈ 18.413588 kW

Therefore the rate of heat loss from the steam line is 18.413588 kW

b)

annual cost of heat lost rate

A = C × q/n × ( 3600 × 24 × 365 )

C is the cost of heat per MJ( $0.02/10⁶) n is broiler efficiency ( 0.9)

so we substitute

A = 0.02/10⁶ × 18413.588/0.9 × ( 3600 × 24 × 365 )

A = $12904.25

Therefore the annual cost of heat loss from line is $12904.25

4 0

3 years ago

The components of an electronic system dissipating 180 W are located in a 1-m-long horizontal duct whose cross section is 16 cm

oee [108]

Answer:

a) The exit temperature is 39.25°C

b) The highest component surface is 132.22°C

c) The average temperature for air equal to 35°C is a good assumption because the air temperature at the inlet will increase due to the result in the heat gain produced by the duct and whose surface is exposed to a flow of hot.

Explanation:

a) The properties of the air at 35°C:

p = density = 1.145 kg/m³

v = 1.655x10⁻⁵m²/s

k = 0.02625 W/m°C

Pr = 0.7268

cp = 1007 J/kg°C

a) The mass flow rate of air is equal to:

$m=\rho *V = 1.145*0.65=0.7443kg/min=0.0124kg/s$

The exit temperature is:

$T=T_{i} +\frac{Q}{m*c_{p} } =27+\frac{0.85*180}{0.0124*1007} =39.25$ °C

b) The mean fluid velocity is:

$V_{m} =\frac{V}{A} =\frac{0.65}{0.16*0.16} =25.4m/min=0.4232m/s$

The hydraulic diameter is:

$D_{h} =\frac{4A}{p} =\frac{4*0.16*0.16}{4*0.16} =0.16m$

The Reynold´s number is:

$Re=\frac{VD_{h} }{v} =\frac{0.4232*0.16}{1.655x10^{-5} } =4091.36$

Assuming fully developed turbulent flow, the Nusselt number is:

$Nu=0.023Re^{0.8} *Pr^{0.4} =0.023*4091.36^{0.8} *0.7268^{0.4} =15.69$

$h=\frac{k*Nu}{D_{h} } =\frac{0.02625*15.69}{0.16} =2.57W/m^{2} C$

The highest component surface temperature is:

$T=T_{e} +\frac{\frac{Q}{A} }{h} =39.2+\frac{0.85*\frac{180}{4*0.16*1} }{2.57} =132.22$ °C

6 0

4 years ago

inysia [295]

Answer:

NOT OSHA

Explanation:

4 0

4 years ago

The collar A, having a mass of 0.75 kg is attached to a spring having a stiffness of k = 200 N/m . When rod BC rotates about the

gladu [14]

Answer:

Speed=1.633 m/s

Force= 20 N

Explanation:

Ideally, $v^{2}=\frac {ks^{2}}{m}$ hence $v=s\sqrt {\frac {k}{m}}$ where v is the speed of collar, m is the mass of collar, k is spring constant and s is the displacement.

In this case, s=100-0=100mm=0.1m since 1 m is equivalent to 1000mm

k is given as 200 N/m and mass is 0.75 Kg

Substituting the given values

$v=0.1 m\sqrt \frac {200 N/m}{0.75 Kg}=1.632993162 m/s\approx 1.633 m/s$

Therefore, <u>the speed is 1.633 m/s</u>

The sum of vertical forces is given by mg where g is acceleration due to gravity and it's value taken as $9.81 m/s^{2}$

Therefore, $F_y=0.75\times 9.81=7.3575 N\approx 7.36 N$

The sum of forces in normal direction is given by $Ma_n=Ks$ therefore

$Ma_n=200*0.1=20 N$

Therefore, <u>normal force on the rod is 20 N</u>

5 0

3 years ago

Technician A says that the starter solenoid switches the high current on and off. Technician B says that the solenoid on the sta

jek_recluse [69]

Technician A is incorrect so that means Technician B is correct.

3 0

3 years ago