Water is to be boiled at sea level in a 30-cm-diameter stainless steel pan placed on top of a 3-kW electric burner. If 60 percen
t of the heat generated by the burner is transferred to the water during boiling, determine the rate of evaporation of water.
1 answer:
Answer:
mevaporation=˙Qhfg=1. 8 kJ /s2269. 6 kJ /kg=0 . 793×10−3kg/ s=2. 855 kg /h
Explanation:
The properties of water at 1 atm and thus at the saturation temperature of 100C are hfg =2256.4 kJ/kg (Table A-4). The net rate of heat transfer to the water is ˙Q=0 . 60×3 kW=1 . 8 kWNoting that it takes 2256.4 kJ of energy to vaporize 1 kg of saturated liquid water, therate of evaporation of water is determined to be mevaporation=˙Qhfg=1. 8 kJ /s2269. 6 kJ /kg=0 . 793×10−3kg/ s=2. 855 kg /h
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Answer:
Explanation:
Given that:
<u>At state 1:</u>
Pressure P₁ = 20 bar
Volume V₁ = 0.03
From the tables at saturated vapour;
Temperature T₁ = 212.4⁰ C ; = 0.0996
/ kg
The mass inside the cylinder is m = 0.3 kg, which is constant.
The specific internal energy u₁ = ug₁ = 2599.2 kJ/kg
<u>At state 2:</u>
Temperature T₂ = 200⁰ C
Since the 1 - 2 occurs in an isochoric process v₂ = v₁ = 0.099 / kg
From temperature T₂ = 200⁰ C
Since , the saturated pressure at state 2 i.e. P₂ = 15.5 bar
Mixture quality
At temperature T₂, the specific internal energy , also
Thus,
<u>At state 3:</u>
Temperature
Specific volume
Thus; ,
SInce , therefore, the phase is in a superheated vapour state.
From the tables of superheated vapour tables; at and T₃ = 200⁰ C
The pressure = 10 bar and v =0.206
The specific internal energy at the pressure of 10 bar = 2622.3 kJ/kg
The changes in the specific internal energy is:
= (2210.686 - 2599.2) kJ/kg
= -388.514 kJ/kg
≅ - 389 kJ/kg
= (2622.3 - 2210.686) kJ/kg
= 411.614 kJ/kg
≅ 410 kJ/kg
We can see the correct sketches of the T-v plot showing the diagrammatic expression in the image attached below.
