Steam flows steadily through a turbine at a rate of 45,000 lbm/h, entering at 1000 psia and 900°F and leaving at 5 psia as saturated vapor. If the power generated by the turbine is 4.1 MW, determine the rate of heat loss from the steam. The enthalpies are h1 = 1448.6 Btu/lbm and h2 = 1130.7 Btu/lbm.

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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