Which of the following is the primary role of proppants in hydraulic fracturing?
Show Hint
- Keep the fractures open during production
- Decrease the viscosity of fracturing fluid
- Decrease the density of fracturing fluid
- Reduce the viscosity of crude oil in reservoir
The Correct Option is A
Solution and Explanation
- Once the fractures are formed, they tend to close due to the overburden pressure from the surrounding rock layers. If these fractures close, the pathway for oil and gas movement is lost.
- Proppants (typically sand, ceramic beads, or other granular materials) are introduced into the fractures along with the fracturing fluid. Their primary purpose is to lodge within the fractures and "prop them open," ensuring that the channels remain conductive to fluid flow even after the pumping pressure is reduced.
- Option (B): Decreasing viscosity of fracturing fluid is achieved by additives, not proppants.
- Option (C): Density control of fracturing fluid is also managed by fluid composition, not by proppants.
- Option (D): Proppants do not alter the viscosity of crude oil in the reservoir; other enhanced oil recovery (EOR) methods may address that.
- Therefore, the correct and primary role of proppants is to keep the fractures open during production, ensuring continuous hydrocarbon flow.
Top Questions on Petroleum
- A shell-and-tube heat exchanger is used for cooling crude oil from $400~\text{K}$ to $360~\text{K}$. Crude oil flows through the tube at $3650~\text{kg/h}$. Water enters the shell side at $310~\text{K}$ and has a flow rate of $1600~\text{kg/h}$. Assume $c_{p,\text{oil}}=2.5~\text{kJ/(kg.K)}$, $c_{p,\text{w}}=4.187~\text{kJ/(kg.K)}$, overall $U=300~\text{W/(m}^2\!\cdot\!\text{K)}$, and countercurrent flow. The required heat–transfer area is ............... m$^2$ (rounded to one decimal place).
An oil droplet is to be mobilized by injecting water through a pore throat. The oil–water interface has the rear radius of curvature $r_A = 25\times10^{-6}\ \text{m}$ and the forward radius of curvature $r_B = 5\times10^{-6}\ \text{m}$. The pore is completely water-wet (contact angle $=0^\circ$) and interfacial tension is $\sigma = 0.025\ \text{N/m}$. The minimum pressure drop required to mobilize the trapped oil droplet is ________ N/m$^2$ (nearest integer).
- Crude oil having density and viscosity of $850\ \text{kg/m}^3$ and $2\times10^{-3}\ \text{Pa}\!\cdot\!\text{s}$, respectively, is flowing at an average velocity of $0.35\ \text{m/s}$ through a horizontal capillary tube. The inside diameter and length of the capillary tube are $2.5\times10^{-3}\ \text{m}$ and $0.30\ \text{m}$, respectively. The Fanning friction factor is $f=\dfrac{16}{\mathrm{Re}}$. The pressure drop across the capillary tube is ________ Pa (rounded to one decimal place).
A non-Newtonian drilling fluid (Bingham plastic) is between two flat parallel rectangular plates of area $10\ \mathrm{cm^2}$ each, separated by $1\ \mathrm{cm}$. A force of $300$ dyne is required to initiate motion of the upper plate. A force of $600$ dyne keeps the plate moving at a constant velocity of $10\ \mathrm{cm/s}$. The constitutive law is \[ \tau_{yx} = \mu_p \dot{\gamma} + \tau^o_{yx}. \] Find the Bingham plastic viscosity $\mu_p$ in dyne$\cdot$s/cm$^2$ (rounded to the nearest integer).
- Consider a micellar displacement process in a homogeneous reservoir with a porosity of 30%. The volume of the microemulsion slug to be injected is 4% of the pore volume. The slug contains 4 vol% surfactant. The density of the rock and the surfactant is $2.7~\text{g/cm}^3$ and $1.1~\text{g/cm}^3$, respectively.
Assuming that the average surfactant adsorption is 0.25 mg/g of the reservoir rock, the fraction of the injected surfactant that will be adsorbed is ............. (rounded to two decimal places).
Questions Asked in GATE PE exam
Identify the option that has the most appropriate sequence such that a coherent paragraph is formed:
Statement:
P. At once, without thinking much, people rushed towards the city in hordes with the sole aim of grabbing as much gold as they could.
Q. However, little did they realize about the impending hardships they would have to face on their way to the city: miles of mud, unfriendly forests, hungry beasts, and inimical local lords—all of which would reduce their chances of getting gold to almost zero.
R. All of them thought that easily they could lay their hands on gold and become wealthy overnight.
S. About a hundred years ago, the news that gold had been discovered in Kolar spread like wildfire and the whole State was in raptures.
For a hydrocarbon reservoir, the following parameters are used in the general material balance equation (MBE):
\[ \begin{aligned} N & = \text{Initial (original) oil in place, stb} \\ G & = \text{Initial volume of gas cap, scf} \\ m & = \text{Ratio of initial volume of gas cap to volume of oil initial in place, rb/rb} \\ S_{wi} & = \text{Initial water saturation} \\ S_{oi} & = \text{Initial oil saturation} \\ B_{oi} & = \text{Initial oil formation volume factor, rb/stb} \\ B_{gi} & = \text{Initial gas formation volume factor, rb/scf} \end{aligned} \]
The total pore volume (in rb) of the reservoir is:
- GATE PE - 2025
- Reservoir and channel routing
- The eigenvalues of the matrix \[ A = \begin{bmatrix} 3 & -1 & 1 \\ -1 & 5 & -1 \\ 1 & -1 & 3 \end{bmatrix} \] are \(\lambda_1, \lambda_2, \lambda_3\). The value of \(\lambda_1 \lambda_2 \lambda_3 (\lambda_1 + \lambda_2 + \lambda_3)\) is:
- GATE PE - 2025
- Eigenvalues
A stationary tank is cylindrical in shape with two hemispherical ends and is horizontal, as shown in the figure. \(R\) is the radius of the cylinder as well as of the hemispherical ends. The tank is half filled with an oil of density \(\rho\) and the rest of the space in the tank is occupied by air. The air pressure, inside the tank as well as outside it, is atmospheric. The acceleration due to gravity (\(g\)) acts vertically downward. The net horizontal force applied by the oil on the right hemispherical end (shown by the bold outline in the figure) is:
- GATE PE - 2025
- Integration
- If a function \(f(x)\) is continuous in the closed interval \([a, b]\) and the first derivative \(f'(x)\) exists in the open interval \((a, b)\), then according to the Lagrange's Mean Value Theorem: \[ \frac{f(b) - f(a)}{b - a} = f'(c) \] If \(a = 0, b = 1.5\), and \(f(x) = x(x - 1)(x - 2)\), then the value(s) of \(c\) in \([a, b]\) is/are:
- GATE PE - 2025
- Mean value theorems