Question

Glycerin is being heated by flowing between two parallel 1 -m-wide and 10 -m-long plates with \(12.5-\mathrm{mm}\) spacing. The glycerin enters the parallel plates with a temperature of \(25^{\circ} \mathrm{C}\) and a mass flow rate of \(0.7 \mathrm{~kg} / \mathrm{s}\). The plates have a constant surface temperature of \(40^{\circ} \mathrm{C}\). Determine the outlet mean temperature of the glycerin and the total rate of heat transfer. Evaluate the properties for glycerin at \(30^{\circ} \mathrm{C}\). Is this a good assumption?

Short answer

Answer: The outlet mean temperature of glycerin is 25.1 °C, and the total rate of heat transfer is 174.6 W.

Step-by-step solution

Calculate the volume and area of glycerin flow

We'll first determine the volume flow rate and the flow area between the plates. Thickness of the plates = 12.5 mm Width of the plates = 1 m Length of the plates = 10 m Mass flow rate (m$'\dot')==0.7 \ kg/s The flow volume can be determined using the given plate dimensions: Flow volume = Thickness × Width × Length Flow volume = (0.0125 m) (1 m) (10 m) Flow volume = 0.125 m³ Now, we will find the flow area: Flow area = Thickness × Width Flow area = (0.0125 m) (1 m) Flow area = 0.0125 m²

Find the glycerin properties at 30\(^{\circ} \mathrm{C}\)

To evaluate the properties of glycerin, we need to search in a glycerin properties table or use an online properties calculator. At 30\(^{\circ} \mathrm{C}\), glycerin properties are as follows: Density (ρ) = 1257 kg/m³ Specific heat capacity (c_p) = 2410 J/(kg·K) Thermal conductivity (k) = 0.292 W/(m·K) Dynamic viscosity (μ) = 0.479 Pa·s

Calculate the velocity and Reynolds number of flow

We first need to determine the velocity of glycerin flow using mass flow rate and density: Velocity (V) = Mass flow rate / (Density × Flow area) V = (0.7 kg/s) / (1257 kg/m³ × 0.0125 m²) V = 0.055 m/s Now, we will calculate the Reynolds number. To do so, we need the hydraulic diameter 𝐷\(_h\), which is 2 times the plate spacing for parallel plates: Hydraulic diameter = 2 × Thickness D\(_h\) = 2 × 0.0125 m D\(_h\) = 0.025 m Now, we will find the Reynolds number (Re): Re = (Density × Velocity × Hydraulic diameter) / Dynamic viscosity Re = (1257 kg/m³ × 0.055 m/s × 0.025 m) / (0.479 Pa·s) Re = 3.62 This small Reynolds number indicates that we have laminar flow.

Calculate the Nusselt number and heat transfer coefficient

For laminar flow between parallel plates, we can use the Nusselt number relation for constant surface temperature: Nu = 8 Now, find the heat transfer coefficient (h) using the Nusselt number relation: h = (Nu × Thermal conductivity) / Hydraulic diameter h = (8 × 0.292 W/(m·K)) / 0.025 m h = 93.28 W/(m²·K)

Calculate the heat transfer rate

We can now find the heat transfer rate using the heat transfer coefficient and the temperature difference between the plates and glycerin: q = h × Flow area × Length × (T_plate - T_inlet) q = 93.28 W/(m²·K) × 0.0125 m² × 10 m × (40 - 25) °C q = 174.6 W

Determine the outlet mean temperature of glycerin

We can find the outlet mean temperature of glycerin using the heat transfer rate, mass flow rate, and specific heat capacity: ΔT = q / (m$'\dot × c_p) ΔT = 174.6 W / (0.7 kg/s × 2410 J/(kg·K)) ΔT = 0.107 °C T_outlet = T_inlet + ΔT T_outlet = 25 °C + 0.107 °C T_outlet ≈ 25.1 °C

Validity of the glycerin properties assumption

Since the outlet mean temperature T_outlet is close to the 30\(^{\circ} \mathrm{C}\) where we calculated the glycerin properties, this assumption is valid. #Conclusion# The outlet mean temperature of glycerin is 25.1\(^{\circ} \mathrm{C}\), and the total rate of heat transfer is 174.6 W. The assumption of evaluating glycerin properties at 30\(^{\circ} \mathrm{C}\) is valid.