A textile dryer is found to consume 4 m3/hr of natural gas with a calorific valu

Question

A textile dryer is found to consume 4 m3/hr of natural gas with a calorific value of 800 kJ/mole. If thethroughput of the dryer is 60 kg of wet cloth per hour, drying it from 55% moisture to 10% moisture,estimate the overall thermal efficiency of the dryer taking into account the evaporation heat of water only.Heat of evaporation for water 2260 kJ/kg

Ok forst the first part in the solutions it says:

60kg of wet cloth contains:

60 x .55kg water = 33 kg moisture

60x (1-.55) = 27 kg of bone dry cloth

As the final product contains 10% moisture, the moisture in the product is 27/9 = 3kg.

And so the moisture removed/hr = 33-3 = 30 kg/hr

This doesn't make sense to me. 27kg is bone dry, and 10% of that has moisture. So that would be 27*.10 = 2.7kg of extra moisture left over. Since that is more moisture that isn't removed, wouldn't that be 27-2.7= 24.3kg of water actually removed?

I don't understand where they got the 9 from 27/9 either.

Explanation / Answer

You are Wrong this time.

"27kg is bone dry, and 10% of that has moisture"---this doese not mean 10 % of 27 kg.

Total weight is x kg which contains 10 % i. e. 0.1 x kg of water.

So, x - 0.1 x = 27

    Or, 0.9 x = 27

    Or, x = 27/0.9 = 30

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