Respuesta :
Answer:
Q_out = 182.0(Btu/s)
Explanation:
Given:
m = 4500 Ibm/h
P1 = 1000 psiu
T1 = 900°
P2 = 5 psiu
Wout = 4.1 MW
h1 = 1448.6
h2 = 1130.7
We need to find the energy balance first.
Energy balance;
[tex] E_i_n-E_o_u_t= ∆Esystem=0 [/tex];
[tex] Therefore E_i_n=E_o_u_t [/tex]
Therefore;
[tex] mh_1 = Q_o_u_t + W_o_u_t +mh_2 [/tex]
Since we are to find Qout, let's make it the subject of the formula
[tex] Q_o_u_t = -m(h_2-h_1) - W_o_u_t [/tex]
Substituting figures into the equation, we have:
Qout= -(45000/3600 ibm/s)(1130.7 - 1448.6)Btu/ibm - 4100kj/s(1btu/1.055kj)
Qout = 182.0(Btu/s)
Therefore, rate of heat loss of system = 182.0(Btu/s)
The rate of heat loss from the steam will be "182.0 Btu/s".
Power and Heat loss
According to the question,
Steam flows, m = 4500 lbm/h
Enthalpies, h₁ = 1448.6
h₂ = 1130.7
Temperature, T₁ = 900°F
Power, [tex]W_{out}[/tex] = 4.1 MW
P₁ = 1000 psiu
P₂ = 5 psiu
By using energy balance equation,
→ [tex]E_{in} - E_{out}[/tex] = 0
or,
[tex]E_{in} = E_{out}[/tex]
Therefore,
mh₁ = [tex]Q_{out} + W_{out}[/tex] + mh₂
hence,
The heat loss be:
→ [tex]Q_{out}[/tex] = -m(h₂ - h₁) - [tex]W_{out}[/tex]
By substituting the values,
= - ([tex]\frac{45000}{3600}[/tex]) (1130 - 1448.6) - 4100
= 182.0 Btu/s
Thus the above answer is correct.
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