Question

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Q2. Heat exchanger-NTU method (34 pt). An orange juice is being cooled from 85 °C at a rate of 2000 kg/h in a counter-current heat exchanger by cold water. The juice has a specific heat of 3090 J/(kg.K). The cold water enters the heat exchanger at 8°C at a rate of 2950 kg/h、The overall heat transfer coefficient U-485 w/mK and the area A 4.2 m2. Assuming steady-state conditions, calculate: (a) the heat transfer rate (J/s), (b) the exit juice temperature, and (c) the exit water temperature (and compare to your assumption!).

Answer 1

Ans:

1. Heat transfer rate = hA(T_hot – T_cold)

Here h = Convection coefficient = 485 j/s m² K

            A = Area = 4.2 m².

            T_hot = 85 ⁰C = 358 K

            T_cold = 8 ⁰C = 281 K

So Rate = 1.568 * 10⁵ j/s

2. The exit temperature of juice :

Mcp_juice = 0.555 * 3090 = 1714.95 J/s K

NTU = UA/C = 1.1878

Now efficiency = 1 – e^-NTU = 0.695

So heat transfer rate = C_juice(T_i –T_o) = T_o = 6.43 ⁰C

3. The exit temperature of Water

Mcp_water = 0.8194 * 4186 = 3130.0084 j/s K

Heat transfer rate = C_water(T_i – T_o) = T_o = 37.714 ⁰C