Explanation

• Equal percentage valve characteristic is most suitable for precise flow control at low flow rates.
• In this type of valve, equal increments of valve opening produce equal percentage changes in flow rate.
• It provides better control over a wide range of flow conditions.
• It is commonly used in process industries where accurate regulation is required.

Last verified on 26-06-2026

Explanation

BOD₅ = Initial DO - Final DO  

Given:  

Initial DO = 9 mg/L  

DO after 5 days = 3 mg/L  

BOD₅ = 9 - 3 = 6 mg/L

Last verified on 26-06-2026

Explanation

• Cavitation in a centrifugal pump causes reduced flow and possible damage to pump components.
• Cavitation occurs when vapor bubbles form and collapse inside the pump due to low pressure.
• The collapse of bubbles can cause vibration, noise, and erosion of the impeller.
• It reduces pump efficiency and performance.

Last verified on 26-06-2026

Explanation

• A Reynolds number of 1500 indicates laminar flow.
• In pipe flow, Re < 2000 is generally considered laminar.
• In laminar flow, fluid particles move in smooth, parallel layers with little mixing.
• Oil commonly shows laminar behavior because of its higher viscosity.

Last verified on 26-06-2026

Explanation

• Radiative heat emission increases sixteen times when the absolute temperature doubles.
• According to Stefan-Boltzmann law, heat radiation is proportional to the fourth power of absolute temperature (T⁴).
• If temperature becomes 2T, emission becomes (2T)⁴ = 16T⁴.
• This applies to heat emitted by hot surfaces.

Last verified on 26-06-2026

Explanation

• Conduction is the main mode of heat transfer that makes a metal spoon hot in hot tea.
• Heat moves from the hot tea to the spoon through direct contact.
• Metals are good conductors of heat because they have free-moving electrons.
• Heat flows from a region of higher temperature to lower temperature.

Last verified on 26-06-2026

Explanation

Reaction: NH₃ + HNO₃ → NH₄NO₃  

Mole ratio is 1:1  

Molar masses: NH₃ = 17 g/mol, NH₄NO₃ = 80 g/mol  

80 kg NH₄NO₃ = 80/80 = 1 kmol  

So you need 1 kmol NH₃  

Mass NH₃ = 1 kmol × 17 kg/kmol = 17 kg

Last verified on 26-06-2026

Explanation

• In the Haber process, 4 moles of gas (N₂ + 3H₂) produce 2 moles of NH₃, so high pressure shifts equilibrium toward ammonia.
• High pressure also increases the rate of reaction, making industrial production more efficient.
• Although compression requires more energy, the higher yield of ammonia justifies the cost.

Last verified on 26-06-2026

Explanation

• In the Contact process, vanadium pentoxide (V₂O₅) is used as a catalyst for converting SO₂ into SO₃.
• It increases the reaction rate without being consumed.
• It is preferred due to its efficiency and lower cost compared to platinum.

Last verified on 26-06-2026

Explanation

• Steam reforming is an endothermic reaction, so it requires heat.
• When temperature decreases, equilibrium shifts in the backward direction to produce more heat (toward reactants).
• Therefore, hydrogen production decreases.

Last verified on 26-06-2026