Answer:
The rate of condensation of the steam in the condenser is 2.14 kg/s
Explanation:
Given;
mass flow rate of water = 80 kg/s
inlet temperature of water, T₁ = 15°C
outlet temperature of water, T₂ = 30°C
temperature of the steam, ΔHvap. = 60°C
heat of vaporization of water at 60°C = 2357.7 kJ/kg
specific heat capacity of water = 4.2 kJ/kg°C
The rate of condensation can be determined by the rate of heat transfer of water as a cooling system.
[tex]m = \frac{Q}{\delta H_{vap}} = \frac{mc(T_2-T_1)}{\delta H_{vap}}[/tex]
substitute the above values in this equation, we will have;
[tex]m = \frac{Q}{\delta H_{vap}} = \frac{80*4.2(30-15)}{2357.7} = 2.14 \frac{kg}{s}[/tex]
Therefore, the rate of condensation of the steam in the condenser is 2.14 kg/s
Answer:
The rate of condensation is found to be 2.127 kg/s
Explanation:
We take water entering and leaving the condenser in the tubes as the system. Applying the law of conservation of energy to this steady state system:
Ein = Eout
Qin + mh1 = mh2 , neglecting the changes in kinetic and potential energie.
Qin = m(h2 - h1)
Qin = mCp(T2 - T1)
where,
Qin = rate of heat transfer to the cold water from steam
m = mass flow rate of cold water = 80 kg/s
Cp = specific heat capacity of cold water = 4.18 KJ/kg.k
T1 = 15°C
T2 = 30°C
Therefore,
Qin = (80 kg/s)(4.18 KJ/kg.k)(15 k)
Qin = 5016 KJ/s
Now, the rate of condensation of steam is given as:
Qin = (m)(hfg)
m = Qin/hfg
where,
m = rate of condensation of steam
hfg = heat of vaporization of water at 60°C = 2357.7 KJ/kg
Therefore,
m = (5016 KJ/s)/(2357.7 KJ/kg)
m = 2.127 kg/s
