Answer:
18.6h
Explanation:
To solve this Duck's second law in form of Diffusion will be used.
Also note that since the temperature is constant D (change) will also be constant.
Please go through the attached files for further explanation and how the answer Is gotten.
The pressure of water is 7.3851 kPa
<u>Explanation:</u>
Given data,
V = 150×
m = 1 Kg
= 2 MPa
= 40°C
The waters specific volume is calculated:
= V/m
Here, the waters specific volume at initial condition is , the containers volume is V, waters mass is m.
= 150× /1
= 0.15 / Kg
The temperature from super heated water tables used in interpolation method between the lower and upper limit for the specific volume corresponds 0.15 / Kg and 0.13 / Kg.
= 350+(400-350)
= 395.17°C
Hence, the initial temperature is 395.17°C.
The volume is constant in the rigid container.
= = 0.15 / Kg
In saturated water labels for = 40°C.
= 0.001008 / Kg
= 19.515 / Kg
The final state is two phase region < < .
In saturated water labels for = 40°C.
= = 7.3851 kPa
= 7.3851 kPa
Answer:
(a) Rate of heat transfer assuming there is no wind is 152.70J/s
(b) Rate of heat transfer assuming the stack is exposed to 20km/h winds is 151.49J/s
Explanation:
Q/t = KA(T2 - T1)/d
Range of heat transfer coefficient (K) for flue gases is 60-180W/mk
Assuming K=180W/mk, diameter (d) =0.6m, A=3.142×0.6^2/4 = 028278m^2, distance (d) = 10m, T2=40°C, T1=10°C
(a) Q/t = 180×0.28278×(40-10)/10 = 180×0.28278×30/10 = 152.70J/s
(b) V= 20km/h=20×1000/3600m/s= 5.6m/s, t=d/v=10/5.6=1.8s
Q = KA(T2 - T1)/V = 180×0.28278×(40 - 10)/5.6 = 180×0.28278×30/5.6 = 272.68J
Q/t = 272.68J/1.8s = 151.48J/s