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:
-50.005 KJ
Explanation:
Mass flow rate = 0.147 KJ per kg
mass= 10 kg
Δh= 50 m
Δv= 15 m/s
W= 10×0.147= 1.47 KJ
Δu= -5 kJ/kg
ΔKE + ΔPE+ ΔU= Q-W
0.5×m×(30^2- 15^2)+ mgΔh+mΔu= Q-W
Q= W+ 0.5×m×(30^2- 15^2) +mgΔh+mΔu
= 1.47 +0.5×1/100×(30^2- 15^2)-9.7×50/1000-50
= 1.47 +3.375-4.8450-50
Q=-50.005 KJ
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Explanation:
Answer:
q=2313.04
T=690.86°C
Explanation:
Given that
Thickness t= 20 cm
Thermal conductivity of firebrick= 1.6 W/m.K
Thermal conductivity of structural brick= 0.7 W/m.K
Inner temperature of firebrick=980°C
Outer temperature of structural brick =30°C
We know that thermal resistance

These are connect in series

Heat transfer

So heat flux
q=2313.04
Lets temperature between interface is T
Now by equating heat in both bricks

So T=690.86°C