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Leto [7]
3 years ago
9

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

t 25 *C. Pressurized steam maintains a pipe surface temperature of 150 *C, and the coefficient associated with natural convection is h = 10 W/m^2 * K. The surface emissivity is E = 0.81. What is the rate of heat loss from the steam line?
2. If the steam is gernerated in a gas-fired boiler operating at efficenty of 0.9, and natural gas is priced at C=0.02 per MJ, what is the annual cost of heat loss from line?
Engineering
1 answer:
DedPeter [7]3 years ago
4 0

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

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A tool chest has 650 N weight that acts through the midpoint of the chest. The chest is supported by feet at A and rollers at B.
erica [24]

Answer:

the value of horizontal force P is 170.625 N

the value of horizontal force at P = 227.5 N is that the block moves to right and this motion is due to sliding.

Explanation:

The first diagram attached below shows the free body diagram of the tool chest when it is sliding.

Let start out by calculating the friction force

F_f= \mu N_2

where :

F_f = friction force

\mu = coefficient of friction

N_2 = normal friction

Given that:

\mu = 0.3

F_f = 0.3 N_2

Using the equation of equilibrium along horizontal direction.

\sum f_x = 0

P - F_f = 0

P = 0.3 N_2   ----- Equation (1)

To determine the moment about point B ; we have the expression

\sum M_B  = 0

0 = N_2*70-W*35-P*100

where;

P = horizontal force

N_2 = normal force at support A

W = self- weight of tool chest

Replacing W = 650 N

0 = N_2*70-650*35-100*P

P = \frac{70 N_2-22750}{100} ----- equation (2)

Replacing  \frac{70 N_2-22750}{100}  for P in equation (1)

\frac{70N_2 -22750}{100} =0.3 N_2

N_2 = \frac{22750}{40}N_2 = 568.75 \ N

Plugging the value of N_2 = 568.75 \ N in equation (2)

P = \frac{70(568.75)-22750}{100} \\ \\ P = \frac{39812.5-22750}{100}  \\ \\ P = \frac{17062.5}{100}

P =170.625 N

Thus; the value of horizontal force P is 170.625 N

b)  From the second diagram attached the free body diagram; the free body diagram of the tool chest when it is tipping about point A is also shown below:

Taking the moments about point A:

\sum M_A = 0

-(P × 100)+ (W×35) = 0

P = \frac{W*35}{100}

Replacing 650 N  for W

P = \frac{650*35}{100}

P = 227.5 N

Thus; the value of horizontal force P, when the tool chest tipping about point A is 227.5 N

We conclude that the motion will be impending for the lowest value when P = 170.625 N and when P= 227.5 N

However; the value of horizontal force at P = 227.5 N is that the block moves to right and this motion is due to sliding.

5 0
3 years ago
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