Answer:
The mass flow rate of the mixture in the manifold is 6.654 kg/min
Explanation;
In this question, we are asked to calculate mass flow rate of the mixture in the manifold
Please check attachment for complete solution and step by step explanation.
Answer:
a) q' = 351.22 W/m
b) q'_total = 1845.56 W / m
c) q'_loss = 254.12 W/m
Explanation:
Given:-
- The diameter of the steam line, d = 100 mm
- The surface emissivity of steam line, ε = 0.8
- The temperature of the steam, Th = 150°C
- The ambient air temperature, T∞ = 20°C
Find:-
(a) Calculate the rate of heat loss per unit length for a calm day.
Solution:-
- Assuming a calm day the heat loss per unit length from the steam line ( q ' ) is only due to the net radiation of the heat from the steam line to the surroundings.
- We will assume that the thickness "t" of the pipe is significantly small and temperature gradients in the wall thickness are negligible. Hence, the temperature of the outside surface Ts = Th = 150°C.
- The net heat loss per unit length due to radiation is given by:
q' = ε*σ*( π*d )* [ Ts^4 - T∞^4 ]
Where,
σ: the stefan boltzmann constant = 5.6703 10-8 (W/m2K4)
Ts: The absolute pipe surface temperature = 150 + 273 = 423 K
T∞:The absolute ambient air temperature = 20 + 273 = 293 K
Therefore,
q' = 0.8*(5.6703 10-8)*( π*0.1 )* [ 423^4 - 293^4 ]
q' = (1.4251*10^-8)* [ 24645536240 ]
q' = 351.22 W / m ... Answer
Find:-
(b) Calculate the rate of heat loss on a breezy day when the wind speed is 8 m/s.
Solution:-
- We have an added heat loss due to the convection current of air with free stream velocity of U∞ = 8 m/s.
- We will first evaluate the following properties of air at T∞ = 20°C = 293 K
Kinematic viscosity ( v ) = 1.5111*10^-5 m^2/s
Thermal conductivity ( k ) = 0.025596
Prandtl number ( Pr ) = 0.71559
- Determine the flow conditions by evaluating the Reynold's number:
Re = U∞*d / v
= ( 8 ) * ( 0.1 ) / ( 1.5111*10^-5 )
= 52941.56574 ... ( Turbulent conditions )
- We will use Churchill - Bernstein equation to determine the surface averaged Nusselt number ( Nu_D ):
- The averaged heat transfer coefficient ( h ) for the flow of air would be:
- The heat loss per unit length due to convection heat transfer is given by:
q'_convec = h*( π*d )* [ Ts - T∞ ]
q'_convec = 36.58951*( π*0.1 )* [ 150 - 20 ]
q'_convec = 11.49493* 130
q'_convec = 1494.3409 W / m
- The total heat loss per unit length ( q'_total ) owes to both radiation heat loss calculated in part a and convection heat loss ( q_convec ):
q'_total = q_a + q_convec
q'_total = 351.22 + 1494.34009
q'_total = 1845.56 W / m ... Answer
Find:-
For the conditions of part (a), calculate the rate of heat loss with a 20-mm-thick layer of insulation (k = 0.08 W/m ⋅ K)
Solution:-
- To reduce the heat loss from steam line an insulation is wrapped around the line which contains a proportion of lost heat within.
- A material with thermal conductivity ( km = 0.08 W/m.K of thickness t = 20 mm ) was wrapped along the steam line.
- The heat loss through the lamination would be due to conduction " q'_t " and radiation " q_rad":
q' = ε*σ*( π*( d + 2t) )* [ Ts^4 - T∞^4 ]
Where,
T_o = T∞ = 20°C
T_s = Film temperature = ( Th + T∞ ) / 2 = ( 150 + 20 ) / 2 = 85°C
r_2 = d/2 + t = 0.1 / 2 + 0.02 = 0.07 m
r_1 = d/2 = 0.1 / 2 = 0.05 m
- The heat loss per unit length would be:
q'_loss = q'_rad - q'_cond
- Compute the individual heat losses:
Therefore,
q'_loss = 351.22 - 97.10
q'_loss = 254.12 W / m .... Answer
- If the wind speed is appreciable the heat loss ( q'_loss ) would increase and the insulation would become ineffective.
Answer:
(a). The value of temperature at the end of heat addition process = 2042.56 K
(b). The value of pressure at the end of heat addition process = 1555.46 k pa
(c). The thermal efficiency of an Otto cycle = 0.4478
(d). The value of mean effective pressure of the cycle = 1506.41
Explanation:
Compression ratio = 8
Initial pressure = 95 k pa
Initial temperature = 278 °c = 551 K
Final pressure = 8 ×
= 8 × 95 = 760 k pa
Final temperature =
×
Final temperature = 551 ×
Final temperature = 998 K
Heat transferred at constant volume Q = 750
(a). We know that Heat transferred at constant volume =
⇒ 1 × 0.718 × ( - 998 ) = 750
⇒ = 2042.56 K
This is the value of temperature at the end of heat addition process.
Since heat addition is constant volume process. so for that process pressure is directly proportional to the temperature.
⇒ P ∝ T
⇒ =
⇒ =
× 760
⇒ = 1555.46 k pa
This is the value of pressure at the end of heat addition process.
(b). Heat rejected from the cycle
For the compression and expansion process,
⇒
⇒
⇒ = 1127.7 K
Heat rejected = 1 × 0.718 × ( 1127.7 - 551)
⇒ = 414.07
Net heat interaction from the cycle =
-
Put the values of &
we get,
⇒ = 750 - 414.07
⇒ = 335.93
We know that for a cyclic process net heat interaction is equal to net work transfer.
⇒ =
⇒ = 335.93
This is the net work output from the cycle.
(c). Thermal efficiency of an Otto cycle is given by
Put the values of &
in the above formula we get,
⇒ = 0.4478
This is the thermal efficiency of an Otto cycle.
(d). Mean effective pressure :-
We know that mean effective pressure of the Otto cycle is given by
=
---------- (1)
where is the swept volume.
=
---------- ( 2 )
From ideal gas equation
= m × R ×
Put all the values in above formula we get,
⇒ 95 × = 1 × 0.287 × 551
⇒ = 0.6
From the same ideal gas equation
= m × R ×
⇒ 760 × = 1 × 0.287 × 998
⇒ = 0.377
Thus swept volume = 0.6 - 0.377
⇒ = 0.223
Thus from equation 1 the mean effective pressure
⇒ =
⇒ = 1506.41
This is the value of mean effective pressure of the cycle.