GATE-CH-1989-5-ii-b-mt-1mark

While drying a material from a moisture content above the critical moisture content to a moisture content very close to the equilibrium moisture content, the surface temperature of the solid (ignoring initial adjustment period):

• Remains a constant

• At first remains a constant and then increases

• Increases continuously

• At first remains a constant and then decreases

GATE-CH-1990-5-iv-mt-2mark

A batch of material is dried under constant drying conditions. When drying is taking place from all the surfaces, the rate of drying during the constant rate period is:

• directly proportional to the solid thickness

• independent of the solid thickness

• inversely proportional to the solid thickness

• directly proportional to the square of solid thickness

GATE-CH-2009-38-mt-2mark

The equilibrium moisture curve for a solid is shown below:

The total moisture content of the solid is \(X\) and it is exposed to air of relative humidity \(H\). In the table below, Group I lists the types of moisture, and Group II represents the region in the graph above.

Group I	Group II
P. Equilibrium moisture	1
Q. Bound moisture	2
R. Unbound moisture	3
S. Free moisture	4

Which ONE of the following is the correct match ?

• P-1, Q-2, R-3, S-4

• P-1, Q-3, R-4, S-2

• P-1, Q-4, R-2, S-3

• P-1, Q-2, R-4, S-3

GATE-CH-2013-20-mt-1mark

A wet solid is dried over a long period of time by unsaturated air of nonzero constant relative humidity. The moisture content eventually attained by the solid is termed as the

• unbound moisture content

• bound moisture content

• free moisture content

• equilibrium moisture content

GATE-CH-1988-15-iv-mt-6mark

In a laboratory drying test with a solid material the following relation for the falling rate period was obtained, \[ \frac {dX}{d\theta } = -0.8 (X - 0.05) \] where \(X\) is the moisture content on dry basis and \(\theta \) is the time in hours. The critical moisture content is 1.4 kg moisture per kg of dry material.
Calculate: 

(a) the time required (in hour) for drying the material from \(X_1 = 4.0\) to \(X_2 = 0.1\).
{#1}

(b) the equilibrium moisture content (kg moisture per kg of dry material).
{#2}

GATE-CH-1990-15-ii-mt-6mark

Drying of a food product is carried out in an insulated tray. The drying air has a partial pressure of water equal to 2360 Pa and a wet bulb temperature of 30\(^\circ \)C. The product has a drying surface of 0.05 m\(^2\)/kg dry solid. The material has a critical moisture content of 0.12 (dry basis) and negligible equilibrium moisture content. The drying rate in the falling rate period is proportional to the moisture content and the mass transfer coefficient is \(5.34 \times 10^{-4}\) kg/(m\(^2\).h.Pa).
Vapor pressure of water at 30\(^\circ \)C = 4232 Pa.
Calculate:
(a) the drying rate in the constant rate period in kg/(m\(^2\).h).
{#1}

(b) the time required (in hour) to dry the material from a moisture of 0.22 to 0.06 (both on dry basis).
{#2}

GATE-CH-2000-2-15-mt-2mark

In a laboratory test run, the rate of drying was found to be \(0.5 \times 10^{-3}\) kg/m².s when the moisture content reduced from 0.4 to 0.1 on a dry basis. The critical moisture content of the material is 0.08 on a dry basis. A tray drier is used to dry 100 kg (dry basis) of the same material under identical conditions. The surface area of the material is 0.04 m²/kg of dry solid. The time required (in seconds) to reduce the moisture content of the solids from 0.3 to 0.2 (dry basis) is

• 2000

• 4000

• 5000

• 6000

GATE-CH-2001-2-13-mt-2mark

200 kg of solid (on dry basis) is subjected to a drying process for a period of 5000 s. The drying occurs in the constant rate period with the drying rate as \(N_c=0.5\times 10^{-3}\) kg/m².s. The initial moisture content of the solid is 0.2 kg moisture/ kg dry solid. The interfacial area available for drying is 4 m²/100 kg of dry solid. The moisture content at the end of the drying period is (in kg moisture/ kg dry solid)

• 0.5

• 0.05

• 0.1

• 0.15

GATE-CH-2003-66-mt-2mark

A solid is being dried in the linear drying rate regime from moisture content \(X_0\) to \(X_F\). The drying rate is zero at \(X=0\) and the critical moisture content is the same as the initial moisture, \(X_0\). The drying time for \(M=(L_S/AR_C)\) is ___________

Where \(L_S\): total mass of dry solids, \(A\): total surface area for drying, \(R_C\): constant; maximum drying rate per unit area, and \(X\): moisture content (in mass of water / mass of dry solids).

• \(M(X_0-X_F)\)

• \(M(X_0/X_F)\)

• \(M\ln (X_0/X_F)\)

• \(MX_0\ln (X_0/X_F)\)

GATE-CH-2004-68-mt-2mark

A 25 cm \(\times \) 25 cm \(\times \) 1 cm flat sheet weighing 1.2 kg initially was dried from both sides under constant drying rate conditions. It took 1500 seconds for the weight of the sheet to reduce to 1.05 kg. Another 1 m \(\times \) 1 m \(\times \) 1 cm flat sheet of the same materials is to be dried from one side only. Under the same constant drying rate conditions, the time required for drying (in seconds) from its initial weight of 19.2 kg to 17.6 kg is

• 1000

• 1500

• 2000

• 2500

GATE-CH-2005-66-mt-2mark

A filter cake is dried with air at wet and dry bulb temperatures of 300 and 323 K respectively. The heat transfer coefficient is 11 W/(m².K) and the latent heat of vapourisation of water is 2500 kJ/kg. Mass transfer does not limit the process. Select the drying rate during the constant rate period. Neglect conduction through the solid and radiation effects.

• \(1.32 \times 10^{-2}\) kg.m\(^{-2}\).s\(^{-1}\)

• \(0.71 \times 10^{-2}\) kg.m\(^{-2}\).s\(^{-1}\)

• \(4.53 \times 10^{-2}\) kg.m\(^{-2}\).s\(^{-1}\)

• \(0.10 \times 10^{-3}\) kg.m\(^{-2}\).s\(^{-1}\)

GATE-CH-2011-41-mt-2mark

A batch of 120 kg wet solid has initial moisture content of 0.2 kg water/kg dry solid. The exposed area for drying is 0.05 m²/kg dry solid. The rate of drying follows the curve given below.

The time required (in hours) for drying this batch to a moisture content of 0.1 kg water/kg dry solid is

• 0.033

• 0.43

• 0.6

• 2.31

GATE-CH-1989-15-i-mt-4mark

A material having a critical moisture content of 24% is dried under constant drying conditions in slabs of dimensions 20cm \(\times \) 20cm \(\times \) 2cm from a moisture content of 40% to 25% in 8 hours. Drying takes place from the two larger surfaces only. The same material is dried in slabs of 40cm \(\times \) 40cm \(\times \) 4cm under similar drying conditions from 50% to 25% moisture content. All moisture content are on wet basis. If the critical moisture content and other physical properties remain unchanged, how long (in hours) will the drying take?

• Answer

GATE-CH-1992-16-c-mt-6mark

Sheet material 0.5 cm thick containing 800 kg of dry stock/m\(^3\) of original wet stock is to be dried at constant drying conditions. The initial drying rate is 4 kg/(h.m\(^2\)) at the initial moisture content of 33%. The final drying rate is 1 kg/(h.m\(^2\)) at 6% final moisture content. The equilibrium moisture content is negligible. If drying is from the two large surfaces only, and if the drying rate in the falling rate period is proportional to the free moisture content, calculate the total drying time (in minutes). All moisture contents are on the dry basis.

• Answer

GATE-CH-1994-9-mt-5mark

Air at 30\(^\circ \)C and 150 kPa in a closed container is compressed and cooled. It is found that the first droplet of water condenses at 200 kPa and 15\(^\circ \)C. Calculate the percent relative humidity of the original air. The vapor pressures of water at 15\(^\circ \)C and 30\(^\circ \)C are 1.7051 kPa and 4.246 kPa respectively.

• Answer

GATE-CH-1996-7-mt-5mark

160 kg of wet solid is to be dried from an initial moisture content of 25% to a final value of 6%. Drying test shows that the rate of drying is constant at \(3 \times 10^{-4}\) kg \(\ce {H2O}\) / (m\(^2\).s) in the region 0.2 – 0.44 kg \(\ce {H2O}\) / kg solid. The drying rate falls linearly in the range 0.01 – 0.2 kg \(\ce {H2O}\) / kg solid. If the equilibrium moisture content is 0.01 kg \(\ce {H2O}\) / kg solid, calculate the time of drying (in hour). The drying surface is 1 m\(^2\) / 30 kg dry weight.

• Answer

GATE-CH-1997-19-mt-1mark

Calculate the time required (in hours) to reduce the moisture content of a solid from 0.66 to 0.25 kg moisture/kg dry solid. The rate of drying \(N\) (kg water evaporated/m\(^2\).s) us given as \[ N = \left \{ \begin {array}{ll} 0.0015 X &\text { for } X \le 0.3 \\ 0.00045 & \text { for } X > 0.3 \end {array} \right . \] where \(X\) = moisture/kg dry solid.
The drying surface is 0.025 m\(^2\)/kg dry solid.

• Answer

GATE-CH-2000-11-mt-5mark

Air at 1 atm is blown past the bulb of a mercury thermometer. The bulb is covered with a wick. The wick is immersed in an organic liquid (molecular weight = 58). The reading of the thermometer is 7.6\(^\circ\)C. At this temperature, the vapor pressure of the liquid is 5 kPa. Find the air temperature (in \(^\circ\)C), given that the ratio of heat transfer coefficient to the mass transfer coefficient (psychrometric ratio) is 2 kJ/kg.K and the latent heat of vaporization of the liquid is 360 kJ/kg. Assume that the air, which is blown, is free from the organic vapor.

• Answer

GATE-CH-2014-32-mt-2mark

A wet solid of 100 kg is dried from a moisture content of 40 wt% to 10 wt%. The critical moisture content is 15 wt% and the equilibrium moisture content is negligible. All moisture contents are on dry basis. The falling rate is considered to be linear. It takes 5 hours to dry the material in the constant rate period. The duration (in hours) of the falling rate period is ____________

• Answer