TOPO NÀO CÓ CẤU HÌNH ĐA ĐIỂM
2 answers:
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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:
a) 358.8K
b) 181.1 kJ/kg.K
c) 0.0068 kJ/kg.K
Explanation:
Given:
P1 = 100kPa
P2= 800kPa
T1 = 22°C = 22+273 = 295K
q_out = 120 kJ/kg
∆S_air = 0.40 kJ/kg.k
T2 =??
a) Using the formula for change in entropy of air, we have:
∆S_air =
Let's take gas constant, Cp= 1.005 kJ/kg.K and R = 0.287 kJ/kg.K
Solving, we have:
[/tex] -0.40= (1.005)ln\frac{T_2}{295} ln\frac{800}{100}[/tex]
Solving for T2 we have:
Taking the exponential on the equation (both sides), we have:
[/tex] T_2 = e^5^.^8^8^2^8 = 358.8K[/tex]
b) Work input to compressor:
= 184.1 kJ/kg
c) Entropy genered during this process, we use the expression;
Egen = ∆Eair + ∆Es
Where; Egen = generated entropy
∆Eair = Entropy change of air in compressor
∆Es = Entropy change in surrounding.
We need to first find ∆Es, since it is unknown.
Therefore ∆Es =
∆Es = 0.4068kJ/kg.k
Hence, entropy generated, Egen will be calculated as:
= -0.40 kJ/kg.K + 0.40608kJ/kg.K
= 0.0068kJ/kg.k