Water flowing at the rate of 13.85 kg/s is to be heated from 54.5 to 87.8°C in a heat exchanger by 54 to 430 kg/h of hot gas flowing counterflow and entering at 427°C (cpm = 1.005 kJ/kg · K). The overall Uo = 69.1 W/m^2.K. Calculate the exit-gas temperature and the heat-transfer area.

=> 572.83 K (299.83°C).

=> 95.86 m^2.

Explanation:

Parameters given are; Water flowing= 13.85 kg/s, temperature of water entering = 54.5°C and the temperature of water going out = 87.8°C, gas flow rate 54,430 kg/h(15.11 kg/s). Temperature of gas coming in = 427°C = 700K, specific heat capacity of hot gas and water = 1.005 kJ/ kg.K and 4.187 KJ/kg. K, overall heat transfer coefficient = Uo = 69.1 W/m^2.K.

Hence;

Mass of hot gas × specific heat capacity of hot gas × change in temperature = mass of water × specific heat capacity of water × change in temperature.

15.11 × 1.005(700K - x ) = 13.85 × 4.187(33.3).

If we solve for x, we will get the value of x to be;

x = 572.83 K (2.99.83°C).

x is the temperature of the exit gas that is 572.83 K(299.83°C).

(b). ∆T = 339.2 - 245.33/ln (339.2/245.33).

∆T = 93.87/ln 1.38.

∆T = 291.521K.

Heat transfer rate= 15.11 × 1.005 × 10^3 (700 - 572.83) = 1931146.394.

heat-transfer area = 1931146.394/69.1 × 291.521.

heat-transfer area= 95.86 m^2.

the answer is:  

line graphs show patterns in data with variables that change over time; whereas, circle graphs show data as parts out of a whole (such as out of 100%).