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NISA [10]
2 years ago
13

What is thermodynamics system? Briefly explain (i) control mass system (ii) control volume system. Comment briefly on control ma

ss system and control volume system at Steady state
Engineering
1 answer:
olga2289 [7]2 years ago
3 0

Explanation:

Thermodynamics system :

 Thermodynamics system is a region or space in which study of matters can be done.The system is separated from surroundings by a boundary this boundary maybe flexible or fixed it depends on situations.The out side the system is called surroundings.

Generally thermodynamics systems are of three types

1.Closed system(control mass system)

  Only energy transfer take place ,no mass transfer take place.

2.Open system(control volume system)

 Both mass as well as energy transfer take place.

3.Isolated system

   Neither mass or nor energy transfer take place.

At steady state ,property is did not changes with respect to time.

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A person walks into a refrigerated warehouse with head uncovered. Model the head as a 25- cm diameter sphere at 35°C with a surf
galina1969 [7]

Answer:

Hello some parts of your question is missing below is the missing part

Convection coefficient = 11 w/m^2. °c

answer : 44.83 watts

Explanation:

Given data :

surface emissivity ( ε )= 0.95

head ( sphere) diameter( D )  = 0.25 m

Temperature of sphere( T )  = 35° C

Temperature of surrounding ( T∞ )  = 25°C

Temperature of surrounding surface ( Ts ) = 15°C

б  = ( 5.67 * 10^-8 )

Determine the total rate of heat loss

First we calculate the surface area of the sphere

As = \pi D^{2}  

= \pi * 0.25^2 =  0.2 m^2

next we calculate heat loss due to radiation

Qrad = ε * б * As( T^{4} - T^{4} _{s} )  ---- ( 1 )

where ;

ε = 0.95

б = ( 5.67 * 10^-8 )

As = 0.2 m^2

T = 35 + 273 = 308 k

Ts = 15 + 273 = 288 k

input values into equation 1

Qrad = 0.95 * ( 5.67 * 10^-8 ) * 0.2 ( (308)^4 - ( 288)^4 )

         = 22.83  watts

Qrad ( heat loss due to radiation ) = 22.83 watts

calculate the heat loss due to convection

Qconv = h* As ( ΔT )

           = 11*0.2 ( 35 -25 )  = 22 watts

Hence total rate of heat loss

=  22 + 22.83

= 44.83 watts

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3 years ago
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3 years ago
A boiler is used to heat steam at a brewery to be used in various applications such as heating water to brew the beer and saniti
Natalija [7]

Answer:

net boiler heat = 301.94 kW

Explanation:

given data

saturated steam = 6.0 bars

temperature = 18°C

flow rate = 115 m³/h = 0.03194 m³/s

heat use by boiler = 90 %

to find out

rate of heat does the boiler output

solution

we can say saturated steam is produce at 6 bar from liquid water 18°C

we know at 6 bar from steam table

hg = 2756 kJ/kg

and

enthalpy of water at 18°C

hf = 75.64 kJ/kg

so heat required for 1 kg is

=hg - hf

= 2680.36 kJ/kg

and

from steam table specific volume of saturated steam at 6 bar is 0.315 m³/kg

so here mass flow rate is

mass flow rate = \frac{0.03194}{0.315}

mass flow rate m = 0.10139 kg/s

so heat required is

H = h × m  

here h is heat required and m is mass flow rate

H = 2680.36  × 0.10139

H =  271.75 kJ/s = 271.75 kW

now 90 % of boiler heat is used for generate saturated stream

so net boiler heat = \frac{H}{0.90}

net boiler heat = \frac{271.75}{0.90}

net boiler heat = 301.94 kW

5 0
2 years ago
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