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aivan3 [116]
2 years ago
5

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

at25C. pressurized steam matains a pipe surface temperature of 150C and the coeff of nature convection h = 10W/m^2K. the surface emissivity e= 0.8
1. 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:
wariber [46]2 years ago
8 0

Answer:

1) q=18414.93 W

2) C=12920$

Explanation:

Given data:

pipe length L=25m

pipe diameter D=100mm =0.1 m

air temperature T_{s1}=T_{\infty1} }=25 °C.....= 298.15k

pipe surface temp T_{s2}=150 °C.....=423.15k

surface emissivity e= 0.8

boiler efficiency η=0.90

natural gas price Cg=$0.02 per MJ

1) Total heat loss and radiation heat loss combined

          q=q_{conv} +q_{rad}

          q=A[h(T_{s2}-T_{s1})+eб(T_{s2}^4-T_{s1}^4)]....... (1)

б=5.67×10^-8 W/m^2K^4 (boltzmann constant)

area A =L.Dπ=25×0.1π=7.85 m^2

putting all these values in eq (1)

q=18414.93 W

2) suppose boiler is operating non stop annual energy loss will be

               E=q.t

                  =18414.93.3600.24.365

                  =5.81×10^11 J

   to find furnace energy consumption

               Ef =E/η

                  =6.46×10^5 MJ

   annual cost

                  C=Cg. Ef

                    =12920$

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thermal energy is being added to steam at 475.8 kPa and 75% quality. determine the amount of thermal energy to be added to creat
eduard

Answer:

q_{in} = 528.6\,\frac{kJ}{kg}

Explanation:

Let assume that heating process occurs at constant pressure, the phenomenon is modelled by the use of the First Law of Thermodynamics:

q_{in} = h_{g} - h_{mix}

The specific enthalpies are:

Liquid-Vapor Mixture:

h_{mix} = 2217.2\,\frac{kJ}{kg}

Saturated Vapor:

h_{g} = 2745.8\,\frac{kJ}{kg}

The thermal energy per unit mass required to heat the steam is:

q_{in} = 2745.8\,\frac{kJ}{kg} - 2217.2\,\frac{kJ}{kg}

q_{in} = 528.6\,\frac{kJ}{kg}

7 0
2 years ago
Which one of the following activities is not an example of incident coordination
Lady bird [3.3K]
Directing, ordering, or controlling
7 0
2 years ago
On the reality television show "Survivor," two tribes compete for luxuries such as food and shelter. During such challenges, one
Ivan

Answer:

Realistic Group Conflict Theory

Explanation:

Realistic Group Conflict theory is a socio-psychological model of internal conflicts among groups.

These groups may compete for a scarce resources based on reality or perception.

These conflicts may be over political power, capital, or social status, etc.

According to this theory, in the competition, only one group wins over the scarce resources while the other has to lose.

Thus The reality show on the television showing two tribes competing for food and shelter where the success of one is the failure for the other is based on Realistic Group Conflict Theory.

5 0
2 years ago
The 10-kg block slides down 2 m on the rough surface with kinetic friction coefficient μk = 0.2. What is the work done by the fr
Rashid [163]

Answer:

153.2 J

Explanation:

Let's first list our given parameters;

mass (m) of the block = 10 kg

which slides down ( i.e displacement) = 2 m

kinetic coefficient of friction (μk) = 0.2

In the diagram shown below;  if we take an integral look at the component of force in the direction of the displacement; we have

F_x= Fcos 40°

F_x= 100 (cos 40°)

F_x= 76.60 N

Workdone by the friction force can now be determined as:

W = F_x × displacement

W = 76.60 × 2

W = 153.2 J

∴  the work done by the friction force = 153.2 J

7 0
2 years ago
Radioactive wastes are temporarily stored in a spherical container, the center of which is buried a distance of 10 m below the e
puteri [66]

Given:

outer radius, R' = 10 m

inner diameter, d = 2 m

inner radius, R = \frac{d}{2} = 1 m

surface temperature, T' = 20^{\circ}C

Thermal conductivity of soil, K = 0.52 W/mK

Solution:

To calculate the thermal temperature of conductor, T, we know amount of heat, Q is given by :

Q =  \frac{T - T'}{\frac{R' - R}{4\pi KRR'}}

500 =  (T - 20)\frac{4\pi \times0.52\times 1\times 10}{10 - 1}

T = 68.86 +20 = 88.865^{\circ}C  

Therefore, outside surface temperature is 88.865^{\circ}C  

4 0
2 years ago
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