A 0.2-m^3 rigid tank equipped with a pressure regulator contains steam at 2MPa and 320C. The steam in the tank is now heated. Th
1 answer:
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Answer:
a. V = 109.64 × 10⁵ ft/min
b. Mw = 654519.54 kg/hr
Explanation:
Given Parameters
mass flow rate of water, Mw = 90000g/min = 6607.33 kg/s
inlet temperature of water, T1 = 84 F = 28.89 C
outlet temperature of water, T2 = 68 F = 20 C
specific heat capacity of water, c = 4.18kJ/kgK
rate of heat remover from water, Qw is given by
Qw = 6607.33[28.89 - 20] * 4.18
Qw = 245529.545kw
For air, inlet condition
DBT = 70 F hi = 43.43 kJ/kg
WBT = 60 F wi = 0.00874 kJ/kg
u1 = 0.8445 m/kg
oulet condition,
DBT = 70 F RH = 100.1
h1 = 83.504kJ/kg
Wo = 0.222kJ/kg
check the attached file for complete solution
Answer:
The difference of head in the level of reservoir is 0.23 m.
Explanation:
For pipe 1
For pipe 2
Q=2.8 l/s
We know that Q=AV
head loss (h)
Now putting the all values
So h=0.23 m
So the difference of head in the level of reservoir is 0.23 m.