Heat Transfer Operations Practice Test 5

1. When the bed is fluidized with liquids, it is known as ___________ fluidization, and with gas too it is known as ___________ fluidization.

Homogeneous, Homogeneous

Heterogeneous, Homogeneous

Homogeneous, Heterogeneous

Heterogeneous, Heterogeneous

2. Why do we use Single effect evaporators when we already have multiple effect one?

To have large temperature

To create low pressure

To have higher efficiency with cheap material

To have liberty to use fouling liquids

3. What is the correct expression for pressure drop in a fluidised bed at fluidisation?

ΔP=(1-∈)(ρ_P-ρ_g)gH_C

ΔP=(1-∈)(ρ_P-ρ_g)gH_B

ΔP=(1-∈)(ρ_P)gH_B

ΔP=(∈)(ρ_P-ρ_g)gH_B

4. The basic design of the climbing/falling film plate evaporator consists of two phases, which are?

Liquid to vapour phase while falling

First falling phase then climbing phase

Liquid to vapour phase while climbing

First climbing phase then falling phase

5. In dense phase flow, the bed is ______

Totally fluidised

Fixed bed

Moving as a block

Stationary fluidized bed

6. If we have a steam flow rate of 10kg/hr then what should be the inlet feed rate if we wish to evaporate 4kg of water from the dilute solution in an hour?
Latent heat of vaporization of water = 2,260 kJ/kg
Specific heat capacity = 6 KJ/Kg K

31kg

30kg

33kg

32kg

7. Recognize the following evaporator.

Climbing Film Evaporator

Basket-type Evaporator

Falling Film Evaporator

Short-tube Vertical Evaporator

8. How are the tube surfaces in falling film evaporators heated to enhance evaporation?

Steam condensing at the outer wall

Heaters at about 200℃

Heaters at just above 100℃

Heaters at above 200℃

9. Which one of the following description best suits the definition of slip-stick flow bed?

Dilute phase flow

Sticky flow

Laminar flow

Dense phase flow

10. Gas to particle heat transfer coefficients can be calculated by the equation given below, which one is correct for Re/ϵ < 200.

Nu=0.4 \((\frac{Re}{\in})^\frac{2}{3}\) Pr^0.33

Nu=1.6×10^-2 \((\frac{Re}{\in})^{1.3} Pr^\frac{2}{3}\)

Nu=1.6×10^-2 \((\frac{Re}{\in})^\frac{2}{3}\) Pr^0.33

Nu=1.6×10^-2 \((\frac{Re}{\in})^{1.3}\) Pr^0.33