Question

Steam is to be condensed in the condenser of a steam power plant at a temperature of 60°C with cooling water from a nearby lake, which enters the tubes of the condenser at 15°C at a rate of 80 kg/s and leaves at 30°C. Determine the rate of condensation of the steam in the condenser.

Answer 1

The rate of condensation is found to be 2.127 kg/s

Step-by-step 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

Answer 2

The rate of condensation of the steam in the condenser is 2.14 kg/s

Step-by-step 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.

`m = (Q)/(\delta H_(vap)) = (mc(T_2-T_1))/(\delta H_(vap))`

substitute the above values in this equation, we will have;

`m = (Q)/(\delta H_(vap)) = (80*4.2(30-15))/(2357.7) = 2.14 (kg)/(s)`

Therefore, the rate of condensation of the steam in the condenser is 2.14 kg/s